Chapter 27 Current amp Resistance

HW For Chapter 277th6th 56 88 1115 1626 1931

3144 3852 4255 4456 4862 5268 and also do the Multiple Choice and turn it in

Current Dead or AliveDEATH bull NEUROLOGIC CRITERIA An individual with

irreversible cessation of all brain function including the brain stem is dead

bull CARDIOPULMONARY CRITERIA An individual with irreversible cessation of circulatory and respiratory function is dead

Positive Charges move from HI to LOW potential

HI V LOW V

Negative Charges move from LOW to HI potential

HI V LOW V

HOW FAST DO ELECTRONS MOVE IN A CURRENTCARRYING CONDUCTING WIRE

Electron Speed is called the DRIFT Velocity

Drift velocity ~ 001 ms

Electric Fields travel at the speed of light

dqEvm

τ=v = Δpm=Ftm = qEtm

bull Current flows from a higher potential to a lower potential (electrons flow the opposite way) Current carrying wires are neutral

bull DC current flows in one directionbull AC current oscillates back and forthbull Electrons have a drift velocity of 001ms bull Electric Fields travel at speed of light

I = Coulombsecond = Ampere

Current dQIdt

=

Engine Current ProblemThe starter motor of a car engine draws a current of 150 A from the battery The copper wire to the motor is 50 mm in diameter and 12 m long The starter motor runs for 080 s until the car engine starts

a How much charge passes through the starter motor

(a) Current is defined as I = QΔt so the charge delivered in time Δt is

Q = IΔt = (150 A)(080 s) = 120 C

Current is charge in motionbull Charge eg electrons exists in conductors with a

number density ne (ne approx 1029 m-3)bull ldquoSomehowrdquo put that charge in motion

ndash effective picture - all charges move with a velocity vd

ndash real picture - a lot of ldquorandom motionrdquo of charges with a small average equal to vd

bull Current density J is given by J = qenevd = qnvndash unit of J is Cm2sec or Am2 (A equiv Ampere) and 1A equiv 1Csndash current I is J times cross sectional area I = J πr2

ndash for 10 Amp in 1mm x 1mm area J = 10+7 Am2 and ve is about 10-3 ms (Yes the average velocity is only 1mms)

dqEvm

τ=v = Δpm=Ftm = qEtm IJ qnvA

= =

Atomic Vision of Ohmrsquos Lawbull E-field in conductor (resistor) provided by a batterybull Charges are put in motion but scatter in a very short

time from things that get in the wayndash itrsquos crowded inside that metalndash defects lattice vibrations (phonons) etc

bull Typical scattering time τ = 10-14 secbull Charges ballistically accelerated for this time and then

randomly scatteredbull Average velocity attained in this time is v = Ftm = qEτmbull Current density is J = qnv so current is proportional to E

which is proportional to Voltagebull OHMrsquos LAW J = (q2nτm)E or J = σ E σ = conductivity

2q nm

τσ =d

qEvm

τ=IJ E qnvA

σ= = =

Engine Current Problem AgainThe starter motor of a car engine draws a current of 150 A from the battery The copper wire to the motor is 50 mm in diameter and 12 m long The starter motor runs for 080 s until the car engine starts

a How much charge passes through the starter motorb What is the drift speed of the electrons Show how the

units work out

( )28 3electron number density 85 10 mminustimes

( ) ( )( )4

d 2 2 28 3 19

150 A 5617 10 ms00025 m 85 10 m 160 10 C

J I A Ivne ne r neπ π

minus

minus minus= = = = = times

times times

IJ E qnvA

σ= = =

Fuse ProblemYou need to design a 10 A fuse that lsquoblowsrsquo if the current exceeds 10 A The fuse material in your stockroom melts at a current density of 500 Acm2 What diameter wire of

this material will do the job

J I A=

( )( )

2

2

4 10 A4 0050 cm 050 mm4 500 AcmD I IA D

J Jπ

π π= = rArr = = = =

Resistivity

where E = electric field andJ = current density in conductor

LA

E

j

Property of bulk matter related to resistance of a sample is the resistivity (ρ) defined as

Jρ equiv

E2

1 mq n

ρσ τ

= =

eg for a copper wire ρ ~ 10-8 Ω-m 1mm radius 1 m long then R asymp 01Ω

for glass ρ ~ 10+12 Ω-m for semiconductors ρ ~ 1 Ω-m

So in fact we can compute the resistance if we know a bit about the material and YES the property belongs to the material

For uniform case IJA

= ELV =

rArr Jρ ρ ⎛ ⎞= = = = ⎜ ⎟⎝ ⎠

I ρLV EL L L IA A

rArrALR ρ=whereIRV =

2q nm

τσ =J Eσ=

bull The LONGER the wire the GREATER the Rbull The THINNER the wire the GREATER the Rbull The HOTTER the wire the GREATER the R

Resistance ResistivityALR ρ= 2

1 mq n

ρσ τ

= =

Resistivity Values

Engine Current ProblemThe starter motor of a car engine draws a current of 150 A from the battery The copper wire to the motor is 50 mm in diameter and 12 m long The starter motor runs for 080 s until the car engine starts

c What is the resistance in the copper wire

8 32 2

124 17 10 104 10(005 )

4

l l mR x m xDA m

ρ ρππ

minus minus= = = times Ω = Ω

You Try Problem

If the magnitude of the drift velocity of free electrons in a copper wire is 784 times 10^ ndash4 ms what is the electric field in the conductor The number density for copper is 849 times10^28 electronsm3

J Eσ= 2

1 mq n

ρσ τ

= =

0181 V mE =

Resistance Question

bull The resistivity of both resistors is the same (ρ)bull Therefore the resistances are related as

11

1

1

1

2

22 88

)4(2 R

AL

AL

ALR ==== ρρρ

bull The resistors have the same voltage across them therefore

112

2 81

8I

RV

RVI ===

bull Two cylindrical resistors R1 and R2 are made of identical material R2 has twice the length of R1 but half the radius of R1 ndash These resistors are then connected to a battery V as shown

VI1 I2

ndash What is the relation between I1 the current flowing in R1 and I2 the current flowing in R2

(a) I1 lt I2 (b) I1 = I2 (c) I1 gt I2

1 1 2 2 1 1 2 22

1 2

RA A d

ρ ρ ρ ρ+= + =

l l l l

( ) ( ) ( ) ( )

( )3 3

23

400 10 m 0250 m 600 10 m 0400 m378

300 10 mR

minus minus

minus

times Ωsdot + times Ωsdot= = Ω

times

A rod is made of two materials The figure is not drawn to scale Each conductor has a square cross section 300 mm on a side The first material has a resistivity of 400 times 10ndash3 Ω ∙m and is 250 cm long while the second material has a resistivity of 600 times 10ndash3 Ω ∙m and is 400 cm long What is the resistance between the ends of the rod

Radial Resistance of a Coaxial Cable Leakage

bull Assume the silicon between the conductors to be concentric elements of thickness dr

bull The resistance of the hollow cylinder of silicon is

bull The total radial resistance is

2ρdR drπrL

=

ALR ρ= 2

1 mq n

ρσ τ

= =

2ln

b

a

ρ bR dRπL a

⎛ ⎞= = ⎜ ⎟⎝ ⎠int This is fairly high which is desirable since you want the current to

flow along the cable and not radially out of it

Resistors

Current Dead or AliveDEATH bull NEUROLOGIC CRITERIA An individual with

irreversible cessation of all brain function including the brain stem is dead

bull CARDIOPULMONARY CRITERIA An individual with irreversible cessation of circulatory and respiratory function is dead

Positive Charges move from HI to LOW potential

HI V LOW V

Negative Charges move from LOW to HI potential

HI V LOW V

HOW FAST DO ELECTRONS MOVE IN A CURRENTCARRYING CONDUCTING WIRE

Electron Speed is called the DRIFT Velocity

Drift velocity ~ 001 ms

Electric Fields travel at the speed of light

dqEvm

τ=v = Δpm=Ftm = qEtm

bull Current flows from a higher potential to a lower potential (electrons flow the opposite way) Current carrying wires are neutral

bull DC current flows in one directionbull AC current oscillates back and forthbull Electrons have a drift velocity of 001ms bull Electric Fields travel at speed of light

I = Coulombsecond = Ampere

Current dQIdt

=

Engine Current ProblemThe starter motor of a car engine draws a current of 150 A from the battery The copper wire to the motor is 50 mm in diameter and 12 m long The starter motor runs for 080 s until the car engine starts

a How much charge passes through the starter motor

(a) Current is defined as I = QΔt so the charge delivered in time Δt is

Q = IΔt = (150 A)(080 s) = 120 C

Current is charge in motionbull Charge eg electrons exists in conductors with a

number density ne (ne approx 1029 m-3)bull ldquoSomehowrdquo put that charge in motion

ndash effective picture - all charges move with a velocity vd

ndash real picture - a lot of ldquorandom motionrdquo of charges with a small average equal to vd

bull Current density J is given by J = qenevd = qnvndash unit of J is Cm2sec or Am2 (A equiv Ampere) and 1A equiv 1Csndash current I is J times cross sectional area I = J πr2

ndash for 10 Amp in 1mm x 1mm area J = 10+7 Am2 and ve is about 10-3 ms (Yes the average velocity is only 1mms)

dqEvm

τ=v = Δpm=Ftm = qEtm IJ qnvA

= =

Atomic Vision of Ohmrsquos Lawbull E-field in conductor (resistor) provided by a batterybull Charges are put in motion but scatter in a very short

time from things that get in the wayndash itrsquos crowded inside that metalndash defects lattice vibrations (phonons) etc

bull Typical scattering time τ = 10-14 secbull Charges ballistically accelerated for this time and then

randomly scatteredbull Average velocity attained in this time is v = Ftm = qEτmbull Current density is J = qnv so current is proportional to E

which is proportional to Voltagebull OHMrsquos LAW J = (q2nτm)E or J = σ E σ = conductivity

2q nm

τσ =d

qEvm

τ=IJ E qnvA

σ= = =

Engine Current Problem AgainThe starter motor of a car engine draws a current of 150 A from the battery The copper wire to the motor is 50 mm in diameter and 12 m long The starter motor runs for 080 s until the car engine starts

a How much charge passes through the starter motorb What is the drift speed of the electrons Show how the

units work out

( )28 3electron number density 85 10 mminustimes

( ) ( )( )4

d 2 2 28 3 19

150 A 5617 10 ms00025 m 85 10 m 160 10 C

J I A Ivne ne r neπ π

minus

minus minus= = = = = times

times times

IJ E qnvA

σ= = =

Fuse ProblemYou need to design a 10 A fuse that lsquoblowsrsquo if the current exceeds 10 A The fuse material in your stockroom melts at a current density of 500 Acm2 What diameter wire of

this material will do the job

J I A=

( )( )

2

2

4 10 A4 0050 cm 050 mm4 500 AcmD I IA D

J Jπ

π π= = rArr = = = =

Resistivity

where E = electric field andJ = current density in conductor

LA

E

j

Property of bulk matter related to resistance of a sample is the resistivity (ρ) defined as

Jρ equiv

E2

1 mq n

ρσ τ

= =

eg for a copper wire ρ ~ 10-8 Ω-m 1mm radius 1 m long then R asymp 01Ω

for glass ρ ~ 10+12 Ω-m for semiconductors ρ ~ 1 Ω-m

So in fact we can compute the resistance if we know a bit about the material and YES the property belongs to the material

For uniform case IJA

= ELV =

rArr Jρ ρ ⎛ ⎞= = = = ⎜ ⎟⎝ ⎠

I ρLV EL L L IA A

rArrALR ρ=whereIRV =

2q nm

τσ =J Eσ=

bull The LONGER the wire the GREATER the Rbull The THINNER the wire the GREATER the Rbull The HOTTER the wire the GREATER the R

Resistance ResistivityALR ρ= 2

1 mq n

ρσ τ

= =

Resistivity Values

Engine Current ProblemThe starter motor of a car engine draws a current of 150 A from the battery The copper wire to the motor is 50 mm in diameter and 12 m long The starter motor runs for 080 s until the car engine starts

c What is the resistance in the copper wire

8 32 2

124 17 10 104 10(005 )

4

l l mR x m xDA m

ρ ρππ

minus minus= = = times Ω = Ω

You Try Problem

If the magnitude of the drift velocity of free electrons in a copper wire is 784 times 10^ ndash4 ms what is the electric field in the conductor The number density for copper is 849 times10^28 electronsm3

J Eσ= 2

1 mq n

ρσ τ

= =

0181 V mE =

Resistance Question

bull The resistivity of both resistors is the same (ρ)bull Therefore the resistances are related as

11

1

1

1

2

22 88

)4(2 R

AL

AL

ALR ==== ρρρ

bull The resistors have the same voltage across them therefore

112

2 81

8I

RV

RVI ===

bull Two cylindrical resistors R1 and R2 are made of identical material R2 has twice the length of R1 but half the radius of R1 ndash These resistors are then connected to a battery V as shown

VI1 I2

ndash What is the relation between I1 the current flowing in R1 and I2 the current flowing in R2

(a) I1 lt I2 (b) I1 = I2 (c) I1 gt I2

1 1 2 2 1 1 2 22

1 2

RA A d

ρ ρ ρ ρ+= + =

l l l l

( ) ( ) ( ) ( )

( )3 3

23

400 10 m 0250 m 600 10 m 0400 m378

300 10 mR

minus minus

minus

times Ωsdot + times Ωsdot= = Ω

times

A rod is made of two materials The figure is not drawn to scale Each conductor has a square cross section 300 mm on a side The first material has a resistivity of 400 times 10ndash3 Ω ∙m and is 250 cm long while the second material has a resistivity of 600 times 10ndash3 Ω ∙m and is 400 cm long What is the resistance between the ends of the rod

Radial Resistance of a Coaxial Cable Leakage

bull Assume the silicon between the conductors to be concentric elements of thickness dr

bull The resistance of the hollow cylinder of silicon is

bull The total radial resistance is

2ρdR drπrL

=

ALR ρ= 2

1 mq n

ρσ τ

= =

2ln

b

a

ρ bR dRπL a

⎛ ⎞= = ⎜ ⎟⎝ ⎠int This is fairly high which is desirable since you want the current to

flow along the cable and not radially out of it

Resistors

Positive Charges move from HI to LOW potential

HI V LOW V

Negative Charges move from LOW to HI potential

HI V LOW V

HOW FAST DO ELECTRONS MOVE IN A CURRENTCARRYING CONDUCTING WIRE

Electron Speed is called the DRIFT Velocity

Drift velocity ~ 001 ms

Electric Fields travel at the speed of light

dqEvm

τ=v = Δpm=Ftm = qEtm

bull Current flows from a higher potential to a lower potential (electrons flow the opposite way) Current carrying wires are neutral

bull DC current flows in one directionbull AC current oscillates back and forthbull Electrons have a drift velocity of 001ms bull Electric Fields travel at speed of light

I = Coulombsecond = Ampere

Current dQIdt

=

Engine Current ProblemThe starter motor of a car engine draws a current of 150 A from the battery The copper wire to the motor is 50 mm in diameter and 12 m long The starter motor runs for 080 s until the car engine starts

a How much charge passes through the starter motor

(a) Current is defined as I = QΔt so the charge delivered in time Δt is

Q = IΔt = (150 A)(080 s) = 120 C

Current is charge in motionbull Charge eg electrons exists in conductors with a

number density ne (ne approx 1029 m-3)bull ldquoSomehowrdquo put that charge in motion

ndash effective picture - all charges move with a velocity vd

ndash real picture - a lot of ldquorandom motionrdquo of charges with a small average equal to vd

bull Current density J is given by J = qenevd = qnvndash unit of J is Cm2sec or Am2 (A equiv Ampere) and 1A equiv 1Csndash current I is J times cross sectional area I = J πr2

ndash for 10 Amp in 1mm x 1mm area J = 10+7 Am2 and ve is about 10-3 ms (Yes the average velocity is only 1mms)

dqEvm

τ=v = Δpm=Ftm = qEtm IJ qnvA

= =

Atomic Vision of Ohmrsquos Lawbull E-field in conductor (resistor) provided by a batterybull Charges are put in motion but scatter in a very short

time from things that get in the wayndash itrsquos crowded inside that metalndash defects lattice vibrations (phonons) etc

bull Typical scattering time τ = 10-14 secbull Charges ballistically accelerated for this time and then

randomly scatteredbull Average velocity attained in this time is v = Ftm = qEτmbull Current density is J = qnv so current is proportional to E

which is proportional to Voltagebull OHMrsquos LAW J = (q2nτm)E or J = σ E σ = conductivity

2q nm

τσ =d

qEvm

τ=IJ E qnvA

σ= = =

Engine Current Problem AgainThe starter motor of a car engine draws a current of 150 A from the battery The copper wire to the motor is 50 mm in diameter and 12 m long The starter motor runs for 080 s until the car engine starts

a How much charge passes through the starter motorb What is the drift speed of the electrons Show how the

units work out

( )28 3electron number density 85 10 mminustimes

( ) ( )( )4

d 2 2 28 3 19

150 A 5617 10 ms00025 m 85 10 m 160 10 C

J I A Ivne ne r neπ π

minus

minus minus= = = = = times

times times

IJ E qnvA

σ= = =

Fuse ProblemYou need to design a 10 A fuse that lsquoblowsrsquo if the current exceeds 10 A The fuse material in your stockroom melts at a current density of 500 Acm2 What diameter wire of

this material will do the job

J I A=

( )( )

2

2

4 10 A4 0050 cm 050 mm4 500 AcmD I IA D

J Jπ

π π= = rArr = = = =

Resistivity

where E = electric field andJ = current density in conductor

LA

E

j

Property of bulk matter related to resistance of a sample is the resistivity (ρ) defined as

Jρ equiv

E2

1 mq n

ρσ τ

= =

eg for a copper wire ρ ~ 10-8 Ω-m 1mm radius 1 m long then R asymp 01Ω

for glass ρ ~ 10+12 Ω-m for semiconductors ρ ~ 1 Ω-m

So in fact we can compute the resistance if we know a bit about the material and YES the property belongs to the material

For uniform case IJA

= ELV =

rArr Jρ ρ ⎛ ⎞= = = = ⎜ ⎟⎝ ⎠

I ρLV EL L L IA A

rArrALR ρ=whereIRV =

2q nm

τσ =J Eσ=

bull The LONGER the wire the GREATER the Rbull The THINNER the wire the GREATER the Rbull The HOTTER the wire the GREATER the R

Resistance ResistivityALR ρ= 2

1 mq n

ρσ τ

= =

Resistivity Values

Engine Current ProblemThe starter motor of a car engine draws a current of 150 A from the battery The copper wire to the motor is 50 mm in diameter and 12 m long The starter motor runs for 080 s until the car engine starts

c What is the resistance in the copper wire

8 32 2

124 17 10 104 10(005 )

4

l l mR x m xDA m

ρ ρππ

minus minus= = = times Ω = Ω

You Try Problem

If the magnitude of the drift velocity of free electrons in a copper wire is 784 times 10^ ndash4 ms what is the electric field in the conductor The number density for copper is 849 times10^28 electronsm3

J Eσ= 2

1 mq n

ρσ τ

= =

0181 V mE =

Resistance Question

bull The resistivity of both resistors is the same (ρ)bull Therefore the resistances are related as

11

1

1

1

2

22 88

)4(2 R

AL

AL

ALR ==== ρρρ

bull The resistors have the same voltage across them therefore

112

2 81

8I

RV

RVI ===

bull Two cylindrical resistors R1 and R2 are made of identical material R2 has twice the length of R1 but half the radius of R1 ndash These resistors are then connected to a battery V as shown

VI1 I2

ndash What is the relation between I1 the current flowing in R1 and I2 the current flowing in R2

(a) I1 lt I2 (b) I1 = I2 (c) I1 gt I2

1 1 2 2 1 1 2 22

1 2

RA A d

ρ ρ ρ ρ+= + =

l l l l

( ) ( ) ( ) ( )

( )3 3

23

400 10 m 0250 m 600 10 m 0400 m378

300 10 mR

minus minus

minus

times Ωsdot + times Ωsdot= = Ω

times

A rod is made of two materials The figure is not drawn to scale Each conductor has a square cross section 300 mm on a side The first material has a resistivity of 400 times 10ndash3 Ω ∙m and is 250 cm long while the second material has a resistivity of 600 times 10ndash3 Ω ∙m and is 400 cm long What is the resistance between the ends of the rod

Radial Resistance of a Coaxial Cable Leakage

bull Assume the silicon between the conductors to be concentric elements of thickness dr

bull The resistance of the hollow cylinder of silicon is

bull The total radial resistance is

2ρdR drπrL

=

ALR ρ= 2

1 mq n

ρσ τ

= =

2ln

b

a

ρ bR dRπL a

⎛ ⎞= = ⎜ ⎟⎝ ⎠int This is fairly high which is desirable since you want the current to

flow along the cable and not radially out of it

Resistors

Negative Charges move from LOW to HI potential

HI V LOW V

HOW FAST DO ELECTRONS MOVE IN A CURRENTCARRYING CONDUCTING WIRE

Electron Speed is called the DRIFT Velocity

Drift velocity ~ 001 ms

Electric Fields travel at the speed of light

dqEvm

τ=v = Δpm=Ftm = qEtm

bull Current flows from a higher potential to a lower potential (electrons flow the opposite way) Current carrying wires are neutral

bull DC current flows in one directionbull AC current oscillates back and forthbull Electrons have a drift velocity of 001ms bull Electric Fields travel at speed of light

I = Coulombsecond = Ampere

Current dQIdt

=

Engine Current ProblemThe starter motor of a car engine draws a current of 150 A from the battery The copper wire to the motor is 50 mm in diameter and 12 m long The starter motor runs for 080 s until the car engine starts

a How much charge passes through the starter motor

(a) Current is defined as I = QΔt so the charge delivered in time Δt is

Q = IΔt = (150 A)(080 s) = 120 C

Current is charge in motionbull Charge eg electrons exists in conductors with a

number density ne (ne approx 1029 m-3)bull ldquoSomehowrdquo put that charge in motion

ndash effective picture - all charges move with a velocity vd

ndash real picture - a lot of ldquorandom motionrdquo of charges with a small average equal to vd

bull Current density J is given by J = qenevd = qnvndash unit of J is Cm2sec or Am2 (A equiv Ampere) and 1A equiv 1Csndash current I is J times cross sectional area I = J πr2

ndash for 10 Amp in 1mm x 1mm area J = 10+7 Am2 and ve is about 10-3 ms (Yes the average velocity is only 1mms)

dqEvm

τ=v = Δpm=Ftm = qEtm IJ qnvA

= =

Atomic Vision of Ohmrsquos Lawbull E-field in conductor (resistor) provided by a batterybull Charges are put in motion but scatter in a very short

time from things that get in the wayndash itrsquos crowded inside that metalndash defects lattice vibrations (phonons) etc

bull Typical scattering time τ = 10-14 secbull Charges ballistically accelerated for this time and then

randomly scatteredbull Average velocity attained in this time is v = Ftm = qEτmbull Current density is J = qnv so current is proportional to E

which is proportional to Voltagebull OHMrsquos LAW J = (q2nτm)E or J = σ E σ = conductivity

2q nm

τσ =d

qEvm

τ=IJ E qnvA

σ= = =

Engine Current Problem AgainThe starter motor of a car engine draws a current of 150 A from the battery The copper wire to the motor is 50 mm in diameter and 12 m long The starter motor runs for 080 s until the car engine starts

a How much charge passes through the starter motorb What is the drift speed of the electrons Show how the

units work out

( )28 3electron number density 85 10 mminustimes

( ) ( )( )4

d 2 2 28 3 19

150 A 5617 10 ms00025 m 85 10 m 160 10 C

J I A Ivne ne r neπ π

minus

minus minus= = = = = times

times times

IJ E qnvA

σ= = =

Fuse ProblemYou need to design a 10 A fuse that lsquoblowsrsquo if the current exceeds 10 A The fuse material in your stockroom melts at a current density of 500 Acm2 What diameter wire of

this material will do the job

J I A=

( )( )

2

2

4 10 A4 0050 cm 050 mm4 500 AcmD I IA D

J Jπ

π π= = rArr = = = =

Resistivity

where E = electric field andJ = current density in conductor

LA

E

j

Property of bulk matter related to resistance of a sample is the resistivity (ρ) defined as

Jρ equiv

E2

1 mq n

ρσ τ

= =

eg for a copper wire ρ ~ 10-8 Ω-m 1mm radius 1 m long then R asymp 01Ω

for glass ρ ~ 10+12 Ω-m for semiconductors ρ ~ 1 Ω-m

So in fact we can compute the resistance if we know a bit about the material and YES the property belongs to the material

For uniform case IJA

= ELV =

rArr Jρ ρ ⎛ ⎞= = = = ⎜ ⎟⎝ ⎠

I ρLV EL L L IA A

rArrALR ρ=whereIRV =

2q nm

τσ =J Eσ=

bull The LONGER the wire the GREATER the Rbull The THINNER the wire the GREATER the Rbull The HOTTER the wire the GREATER the R

Resistance ResistivityALR ρ= 2

1 mq n

ρσ τ

= =

Resistivity Values

Engine Current ProblemThe starter motor of a car engine draws a current of 150 A from the battery The copper wire to the motor is 50 mm in diameter and 12 m long The starter motor runs for 080 s until the car engine starts

c What is the resistance in the copper wire

8 32 2

124 17 10 104 10(005 )

4

l l mR x m xDA m

ρ ρππ

minus minus= = = times Ω = Ω

You Try Problem

If the magnitude of the drift velocity of free electrons in a copper wire is 784 times 10^ ndash4 ms what is the electric field in the conductor The number density for copper is 849 times10^28 electronsm3

J Eσ= 2

1 mq n

ρσ τ

= =

0181 V mE =

Resistance Question

bull The resistivity of both resistors is the same (ρ)bull Therefore the resistances are related as

11

1

1

1

2

22 88

)4(2 R

AL

AL

ALR ==== ρρρ

bull The resistors have the same voltage across them therefore

112

2 81

8I

RV

RVI ===

bull Two cylindrical resistors R1 and R2 are made of identical material R2 has twice the length of R1 but half the radius of R1 ndash These resistors are then connected to a battery V as shown

VI1 I2

ndash What is the relation between I1 the current flowing in R1 and I2 the current flowing in R2

(a) I1 lt I2 (b) I1 = I2 (c) I1 gt I2

1 1 2 2 1 1 2 22

1 2

RA A d

ρ ρ ρ ρ+= + =

l l l l

( ) ( ) ( ) ( )

( )3 3

23

400 10 m 0250 m 600 10 m 0400 m378

300 10 mR

minus minus

minus

times Ωsdot + times Ωsdot= = Ω

times

A rod is made of two materials The figure is not drawn to scale Each conductor has a square cross section 300 mm on a side The first material has a resistivity of 400 times 10ndash3 Ω ∙m and is 250 cm long while the second material has a resistivity of 600 times 10ndash3 Ω ∙m and is 400 cm long What is the resistance between the ends of the rod

Radial Resistance of a Coaxial Cable Leakage

bull Assume the silicon between the conductors to be concentric elements of thickness dr

bull The resistance of the hollow cylinder of silicon is

bull The total radial resistance is

2ρdR drπrL

=

ALR ρ= 2

1 mq n

ρσ τ

= =

2ln

b

a

ρ bR dRπL a

⎛ ⎞= = ⎜ ⎟⎝ ⎠int This is fairly high which is desirable since you want the current to

flow along the cable and not radially out of it

Resistors

HOW FAST DO ELECTRONS MOVE IN A CURRENTCARRYING CONDUCTING WIRE

Electron Speed is called the DRIFT Velocity

Drift velocity ~ 001 ms

Electric Fields travel at the speed of light

dqEvm

τ=v = Δpm=Ftm = qEtm

bull Current flows from a higher potential to a lower potential (electrons flow the opposite way) Current carrying wires are neutral

bull DC current flows in one directionbull AC current oscillates back and forthbull Electrons have a drift velocity of 001ms bull Electric Fields travel at speed of light

I = Coulombsecond = Ampere

Current dQIdt

=

Engine Current ProblemThe starter motor of a car engine draws a current of 150 A from the battery The copper wire to the motor is 50 mm in diameter and 12 m long The starter motor runs for 080 s until the car engine starts

a How much charge passes through the starter motor

(a) Current is defined as I = QΔt so the charge delivered in time Δt is

Q = IΔt = (150 A)(080 s) = 120 C

Current is charge in motionbull Charge eg electrons exists in conductors with a

number density ne (ne approx 1029 m-3)bull ldquoSomehowrdquo put that charge in motion

ndash effective picture - all charges move with a velocity vd

ndash real picture - a lot of ldquorandom motionrdquo of charges with a small average equal to vd

bull Current density J is given by J = qenevd = qnvndash unit of J is Cm2sec or Am2 (A equiv Ampere) and 1A equiv 1Csndash current I is J times cross sectional area I = J πr2

ndash for 10 Amp in 1mm x 1mm area J = 10+7 Am2 and ve is about 10-3 ms (Yes the average velocity is only 1mms)

dqEvm

τ=v = Δpm=Ftm = qEtm IJ qnvA

= =

Atomic Vision of Ohmrsquos Lawbull E-field in conductor (resistor) provided by a batterybull Charges are put in motion but scatter in a very short

time from things that get in the wayndash itrsquos crowded inside that metalndash defects lattice vibrations (phonons) etc

bull Typical scattering time τ = 10-14 secbull Charges ballistically accelerated for this time and then

randomly scatteredbull Average velocity attained in this time is v = Ftm = qEτmbull Current density is J = qnv so current is proportional to E

which is proportional to Voltagebull OHMrsquos LAW J = (q2nτm)E or J = σ E σ = conductivity

2q nm

τσ =d

qEvm

τ=IJ E qnvA

σ= = =

Engine Current Problem AgainThe starter motor of a car engine draws a current of 150 A from the battery The copper wire to the motor is 50 mm in diameter and 12 m long The starter motor runs for 080 s until the car engine starts

a How much charge passes through the starter motorb What is the drift speed of the electrons Show how the

units work out

( )28 3electron number density 85 10 mminustimes

( ) ( )( )4

d 2 2 28 3 19

150 A 5617 10 ms00025 m 85 10 m 160 10 C

J I A Ivne ne r neπ π

minus

minus minus= = = = = times

times times

IJ E qnvA

σ= = =

Fuse ProblemYou need to design a 10 A fuse that lsquoblowsrsquo if the current exceeds 10 A The fuse material in your stockroom melts at a current density of 500 Acm2 What diameter wire of

this material will do the job

J I A=

( )( )

2

2

4 10 A4 0050 cm 050 mm4 500 AcmD I IA D

J Jπ

π π= = rArr = = = =

Resistivity

where E = electric field andJ = current density in conductor

LA

E

j

Property of bulk matter related to resistance of a sample is the resistivity (ρ) defined as

Jρ equiv

E2

1 mq n

ρσ τ

= =

eg for a copper wire ρ ~ 10-8 Ω-m 1mm radius 1 m long then R asymp 01Ω

for glass ρ ~ 10+12 Ω-m for semiconductors ρ ~ 1 Ω-m

So in fact we can compute the resistance if we know a bit about the material and YES the property belongs to the material

For uniform case IJA

= ELV =

rArr Jρ ρ ⎛ ⎞= = = = ⎜ ⎟⎝ ⎠

I ρLV EL L L IA A

rArrALR ρ=whereIRV =

2q nm

τσ =J Eσ=

bull The LONGER the wire the GREATER the Rbull The THINNER the wire the GREATER the Rbull The HOTTER the wire the GREATER the R

Resistance ResistivityALR ρ= 2

1 mq n

ρσ τ

= =

Resistivity Values

Engine Current ProblemThe starter motor of a car engine draws a current of 150 A from the battery The copper wire to the motor is 50 mm in diameter and 12 m long The starter motor runs for 080 s until the car engine starts

c What is the resistance in the copper wire

8 32 2

124 17 10 104 10(005 )

4

l l mR x m xDA m

ρ ρππ

minus minus= = = times Ω = Ω

You Try Problem

If the magnitude of the drift velocity of free electrons in a copper wire is 784 times 10^ ndash4 ms what is the electric field in the conductor The number density for copper is 849 times10^28 electronsm3

J Eσ= 2

1 mq n

ρσ τ

= =

0181 V mE =

Resistance Question

bull The resistivity of both resistors is the same (ρ)bull Therefore the resistances are related as

11

1

1

1

2

22 88

)4(2 R

AL

AL

ALR ==== ρρρ

bull The resistors have the same voltage across them therefore

112

2 81

8I

RV

RVI ===

bull Two cylindrical resistors R1 and R2 are made of identical material R2 has twice the length of R1 but half the radius of R1 ndash These resistors are then connected to a battery V as shown

VI1 I2

ndash What is the relation between I1 the current flowing in R1 and I2 the current flowing in R2

(a) I1 lt I2 (b) I1 = I2 (c) I1 gt I2

1 1 2 2 1 1 2 22

1 2

RA A d

ρ ρ ρ ρ+= + =

l l l l

( ) ( ) ( ) ( )

( )3 3

23

400 10 m 0250 m 600 10 m 0400 m378

300 10 mR

minus minus

minus

times Ωsdot + times Ωsdot= = Ω

times

A rod is made of two materials The figure is not drawn to scale Each conductor has a square cross section 300 mm on a side The first material has a resistivity of 400 times 10ndash3 Ω ∙m and is 250 cm long while the second material has a resistivity of 600 times 10ndash3 Ω ∙m and is 400 cm long What is the resistance between the ends of the rod

Radial Resistance of a Coaxial Cable Leakage

bull Assume the silicon between the conductors to be concentric elements of thickness dr

bull The resistance of the hollow cylinder of silicon is

bull The total radial resistance is

2ρdR drπrL

=

ALR ρ= 2

1 mq n

ρσ τ

= =

2ln

b

a

ρ bR dRπL a

⎛ ⎞= = ⎜ ⎟⎝ ⎠int This is fairly high which is desirable since you want the current to

flow along the cable and not radially out of it

Resistors

Electron Speed is called the DRIFT Velocity

Drift velocity ~ 001 ms

Electric Fields travel at the speed of light

dqEvm

τ=v = Δpm=Ftm = qEtm

bull Current flows from a higher potential to a lower potential (electrons flow the opposite way) Current carrying wires are neutral

bull DC current flows in one directionbull AC current oscillates back and forthbull Electrons have a drift velocity of 001ms bull Electric Fields travel at speed of light

I = Coulombsecond = Ampere

Current dQIdt

=

Engine Current ProblemThe starter motor of a car engine draws a current of 150 A from the battery The copper wire to the motor is 50 mm in diameter and 12 m long The starter motor runs for 080 s until the car engine starts

a How much charge passes through the starter motor

(a) Current is defined as I = QΔt so the charge delivered in time Δt is

Q = IΔt = (150 A)(080 s) = 120 C

Current is charge in motionbull Charge eg electrons exists in conductors with a

number density ne (ne approx 1029 m-3)bull ldquoSomehowrdquo put that charge in motion

ndash effective picture - all charges move with a velocity vd

ndash real picture - a lot of ldquorandom motionrdquo of charges with a small average equal to vd

bull Current density J is given by J = qenevd = qnvndash unit of J is Cm2sec or Am2 (A equiv Ampere) and 1A equiv 1Csndash current I is J times cross sectional area I = J πr2

ndash for 10 Amp in 1mm x 1mm area J = 10+7 Am2 and ve is about 10-3 ms (Yes the average velocity is only 1mms)

dqEvm

τ=v = Δpm=Ftm = qEtm IJ qnvA

= =

Atomic Vision of Ohmrsquos Lawbull E-field in conductor (resistor) provided by a batterybull Charges are put in motion but scatter in a very short

time from things that get in the wayndash itrsquos crowded inside that metalndash defects lattice vibrations (phonons) etc

bull Typical scattering time τ = 10-14 secbull Charges ballistically accelerated for this time and then

randomly scatteredbull Average velocity attained in this time is v = Ftm = qEτmbull Current density is J = qnv so current is proportional to E

which is proportional to Voltagebull OHMrsquos LAW J = (q2nτm)E or J = σ E σ = conductivity

2q nm

τσ =d

qEvm

τ=IJ E qnvA

σ= = =

Engine Current Problem AgainThe starter motor of a car engine draws a current of 150 A from the battery The copper wire to the motor is 50 mm in diameter and 12 m long The starter motor runs for 080 s until the car engine starts

a How much charge passes through the starter motorb What is the drift speed of the electrons Show how the

units work out

( )28 3electron number density 85 10 mminustimes

( ) ( )( )4

d 2 2 28 3 19

150 A 5617 10 ms00025 m 85 10 m 160 10 C

J I A Ivne ne r neπ π

minus

minus minus= = = = = times

times times

IJ E qnvA

σ= = =

Fuse ProblemYou need to design a 10 A fuse that lsquoblowsrsquo if the current exceeds 10 A The fuse material in your stockroom melts at a current density of 500 Acm2 What diameter wire of

this material will do the job

J I A=

( )( )

2

2

4 10 A4 0050 cm 050 mm4 500 AcmD I IA D

J Jπ

π π= = rArr = = = =

Resistivity

where E = electric field andJ = current density in conductor

LA

E

j

Property of bulk matter related to resistance of a sample is the resistivity (ρ) defined as

Jρ equiv

E2

1 mq n

ρσ τ

= =

eg for a copper wire ρ ~ 10-8 Ω-m 1mm radius 1 m long then R asymp 01Ω

for glass ρ ~ 10+12 Ω-m for semiconductors ρ ~ 1 Ω-m

So in fact we can compute the resistance if we know a bit about the material and YES the property belongs to the material

For uniform case IJA

= ELV =

rArr Jρ ρ ⎛ ⎞= = = = ⎜ ⎟⎝ ⎠

I ρLV EL L L IA A

rArrALR ρ=whereIRV =

2q nm

τσ =J Eσ=

bull The LONGER the wire the GREATER the Rbull The THINNER the wire the GREATER the Rbull The HOTTER the wire the GREATER the R

Resistance ResistivityALR ρ= 2

1 mq n

ρσ τ

= =

Resistivity Values

Engine Current ProblemThe starter motor of a car engine draws a current of 150 A from the battery The copper wire to the motor is 50 mm in diameter and 12 m long The starter motor runs for 080 s until the car engine starts

c What is the resistance in the copper wire

8 32 2

124 17 10 104 10(005 )

4

l l mR x m xDA m

ρ ρππ

minus minus= = = times Ω = Ω

You Try Problem

If the magnitude of the drift velocity of free electrons in a copper wire is 784 times 10^ ndash4 ms what is the electric field in the conductor The number density for copper is 849 times10^28 electronsm3

J Eσ= 2

1 mq n

ρσ τ

= =

0181 V mE =

Resistance Question

bull The resistivity of both resistors is the same (ρ)bull Therefore the resistances are related as

11

1

1

1

2

22 88

)4(2 R

AL

AL

ALR ==== ρρρ

bull The resistors have the same voltage across them therefore

112

2 81

8I

RV

RVI ===

bull Two cylindrical resistors R1 and R2 are made of identical material R2 has twice the length of R1 but half the radius of R1 ndash These resistors are then connected to a battery V as shown

VI1 I2

ndash What is the relation between I1 the current flowing in R1 and I2 the current flowing in R2

(a) I1 lt I2 (b) I1 = I2 (c) I1 gt I2

1 1 2 2 1 1 2 22

1 2

RA A d

ρ ρ ρ ρ+= + =

l l l l

( ) ( ) ( ) ( )

( )3 3

23

400 10 m 0250 m 600 10 m 0400 m378

300 10 mR

minus minus

minus

times Ωsdot + times Ωsdot= = Ω

times

A rod is made of two materials The figure is not drawn to scale Each conductor has a square cross section 300 mm on a side The first material has a resistivity of 400 times 10ndash3 Ω ∙m and is 250 cm long while the second material has a resistivity of 600 times 10ndash3 Ω ∙m and is 400 cm long What is the resistance between the ends of the rod

Radial Resistance of a Coaxial Cable Leakage

bull Assume the silicon between the conductors to be concentric elements of thickness dr

bull The resistance of the hollow cylinder of silicon is

bull The total radial resistance is

2ρdR drπrL

=

ALR ρ= 2

1 mq n

ρσ τ

= =

2ln

b

a

ρ bR dRπL a

⎛ ⎞= = ⎜ ⎟⎝ ⎠int This is fairly high which is desirable since you want the current to

flow along the cable and not radially out of it

Resistors

bull Current flows from a higher potential to a lower potential (electrons flow the opposite way) Current carrying wires are neutral

bull DC current flows in one directionbull AC current oscillates back and forthbull Electrons have a drift velocity of 001ms bull Electric Fields travel at speed of light

I = Coulombsecond = Ampere

Current dQIdt

=

Engine Current ProblemThe starter motor of a car engine draws a current of 150 A from the battery The copper wire to the motor is 50 mm in diameter and 12 m long The starter motor runs for 080 s until the car engine starts

a How much charge passes through the starter motor

(a) Current is defined as I = QΔt so the charge delivered in time Δt is

Q = IΔt = (150 A)(080 s) = 120 C

Current is charge in motionbull Charge eg electrons exists in conductors with a

number density ne (ne approx 1029 m-3)bull ldquoSomehowrdquo put that charge in motion

ndash effective picture - all charges move with a velocity vd

ndash real picture - a lot of ldquorandom motionrdquo of charges with a small average equal to vd

bull Current density J is given by J = qenevd = qnvndash unit of J is Cm2sec or Am2 (A equiv Ampere) and 1A equiv 1Csndash current I is J times cross sectional area I = J πr2

ndash for 10 Amp in 1mm x 1mm area J = 10+7 Am2 and ve is about 10-3 ms (Yes the average velocity is only 1mms)

dqEvm

τ=v = Δpm=Ftm = qEtm IJ qnvA

= =

Atomic Vision of Ohmrsquos Lawbull E-field in conductor (resistor) provided by a batterybull Charges are put in motion but scatter in a very short

time from things that get in the wayndash itrsquos crowded inside that metalndash defects lattice vibrations (phonons) etc

bull Typical scattering time τ = 10-14 secbull Charges ballistically accelerated for this time and then

randomly scatteredbull Average velocity attained in this time is v = Ftm = qEτmbull Current density is J = qnv so current is proportional to E

which is proportional to Voltagebull OHMrsquos LAW J = (q2nτm)E or J = σ E σ = conductivity

2q nm

τσ =d

qEvm

τ=IJ E qnvA

σ= = =

Engine Current Problem AgainThe starter motor of a car engine draws a current of 150 A from the battery The copper wire to the motor is 50 mm in diameter and 12 m long The starter motor runs for 080 s until the car engine starts

a How much charge passes through the starter motorb What is the drift speed of the electrons Show how the

units work out

( )28 3electron number density 85 10 mminustimes

( ) ( )( )4

d 2 2 28 3 19

150 A 5617 10 ms00025 m 85 10 m 160 10 C

J I A Ivne ne r neπ π

minus

minus minus= = = = = times

times times

IJ E qnvA

σ= = =

Fuse ProblemYou need to design a 10 A fuse that lsquoblowsrsquo if the current exceeds 10 A The fuse material in your stockroom melts at a current density of 500 Acm2 What diameter wire of

this material will do the job

J I A=

( )( )

2

2

4 10 A4 0050 cm 050 mm4 500 AcmD I IA D

J Jπ

π π= = rArr = = = =

Resistivity

where E = electric field andJ = current density in conductor

LA

E

j

Property of bulk matter related to resistance of a sample is the resistivity (ρ) defined as

Jρ equiv

E2

1 mq n

ρσ τ

= =

eg for a copper wire ρ ~ 10-8 Ω-m 1mm radius 1 m long then R asymp 01Ω

for glass ρ ~ 10+12 Ω-m for semiconductors ρ ~ 1 Ω-m

So in fact we can compute the resistance if we know a bit about the material and YES the property belongs to the material

For uniform case IJA

= ELV =

rArr Jρ ρ ⎛ ⎞= = = = ⎜ ⎟⎝ ⎠

I ρLV EL L L IA A

rArrALR ρ=whereIRV =

2q nm

τσ =J Eσ=

bull The LONGER the wire the GREATER the Rbull The THINNER the wire the GREATER the Rbull The HOTTER the wire the GREATER the R

Resistance ResistivityALR ρ= 2

1 mq n

ρσ τ

= =

Resistivity Values

Engine Current ProblemThe starter motor of a car engine draws a current of 150 A from the battery The copper wire to the motor is 50 mm in diameter and 12 m long The starter motor runs for 080 s until the car engine starts

c What is the resistance in the copper wire

8 32 2

124 17 10 104 10(005 )

4

l l mR x m xDA m

ρ ρππ

minus minus= = = times Ω = Ω

You Try Problem

If the magnitude of the drift velocity of free electrons in a copper wire is 784 times 10^ ndash4 ms what is the electric field in the conductor The number density for copper is 849 times10^28 electronsm3

J Eσ= 2

1 mq n

ρσ τ

= =

0181 V mE =

Resistance Question

bull The resistivity of both resistors is the same (ρ)bull Therefore the resistances are related as

11

1

1

1

2

22 88

)4(2 R

AL

AL

ALR ==== ρρρ

bull The resistors have the same voltage across them therefore

112

2 81

8I

RV

RVI ===

bull Two cylindrical resistors R1 and R2 are made of identical material R2 has twice the length of R1 but half the radius of R1 ndash These resistors are then connected to a battery V as shown

VI1 I2

ndash What is the relation between I1 the current flowing in R1 and I2 the current flowing in R2

(a) I1 lt I2 (b) I1 = I2 (c) I1 gt I2

1 1 2 2 1 1 2 22

1 2

RA A d

ρ ρ ρ ρ+= + =

l l l l

( ) ( ) ( ) ( )

( )3 3

23

400 10 m 0250 m 600 10 m 0400 m378

300 10 mR

minus minus

minus

times Ωsdot + times Ωsdot= = Ω

times

A rod is made of two materials The figure is not drawn to scale Each conductor has a square cross section 300 mm on a side The first material has a resistivity of 400 times 10ndash3 Ω ∙m and is 250 cm long while the second material has a resistivity of 600 times 10ndash3 Ω ∙m and is 400 cm long What is the resistance between the ends of the rod

Radial Resistance of a Coaxial Cable Leakage

bull Assume the silicon between the conductors to be concentric elements of thickness dr

bull The resistance of the hollow cylinder of silicon is

bull The total radial resistance is

2ρdR drπrL

=

ALR ρ= 2

1 mq n

ρσ τ

= =

2ln

b

a

ρ bR dRπL a

⎛ ⎞= = ⎜ ⎟⎝ ⎠int This is fairly high which is desirable since you want the current to

flow along the cable and not radially out of it

Resistors

Engine Current ProblemThe starter motor of a car engine draws a current of 150 A from the battery The copper wire to the motor is 50 mm in diameter and 12 m long The starter motor runs for 080 s until the car engine starts

a How much charge passes through the starter motor

(a) Current is defined as I = QΔt so the charge delivered in time Δt is

Q = IΔt = (150 A)(080 s) = 120 C

Current is charge in motionbull Charge eg electrons exists in conductors with a

number density ne (ne approx 1029 m-3)bull ldquoSomehowrdquo put that charge in motion

ndash effective picture - all charges move with a velocity vd

ndash real picture - a lot of ldquorandom motionrdquo of charges with a small average equal to vd

bull Current density J is given by J = qenevd = qnvndash unit of J is Cm2sec or Am2 (A equiv Ampere) and 1A equiv 1Csndash current I is J times cross sectional area I = J πr2

ndash for 10 Amp in 1mm x 1mm area J = 10+7 Am2 and ve is about 10-3 ms (Yes the average velocity is only 1mms)

dqEvm

τ=v = Δpm=Ftm = qEtm IJ qnvA

= =

Atomic Vision of Ohmrsquos Lawbull E-field in conductor (resistor) provided by a batterybull Charges are put in motion but scatter in a very short

time from things that get in the wayndash itrsquos crowded inside that metalndash defects lattice vibrations (phonons) etc

bull Typical scattering time τ = 10-14 secbull Charges ballistically accelerated for this time and then

randomly scatteredbull Average velocity attained in this time is v = Ftm = qEτmbull Current density is J = qnv so current is proportional to E

which is proportional to Voltagebull OHMrsquos LAW J = (q2nτm)E or J = σ E σ = conductivity

2q nm

τσ =d

qEvm

τ=IJ E qnvA

σ= = =

Engine Current Problem AgainThe starter motor of a car engine draws a current of 150 A from the battery The copper wire to the motor is 50 mm in diameter and 12 m long The starter motor runs for 080 s until the car engine starts

a How much charge passes through the starter motorb What is the drift speed of the electrons Show how the

units work out

( )28 3electron number density 85 10 mminustimes

( ) ( )( )4

d 2 2 28 3 19

150 A 5617 10 ms00025 m 85 10 m 160 10 C

J I A Ivne ne r neπ π

minus

minus minus= = = = = times

times times

IJ E qnvA

σ= = =

Fuse ProblemYou need to design a 10 A fuse that lsquoblowsrsquo if the current exceeds 10 A The fuse material in your stockroom melts at a current density of 500 Acm2 What diameter wire of

this material will do the job

J I A=

( )( )

2

2

4 10 A4 0050 cm 050 mm4 500 AcmD I IA D

J Jπ

π π= = rArr = = = =

Resistivity

where E = electric field andJ = current density in conductor

LA

E

j

Property of bulk matter related to resistance of a sample is the resistivity (ρ) defined as

Jρ equiv

E2

1 mq n

ρσ τ

= =

eg for a copper wire ρ ~ 10-8 Ω-m 1mm radius 1 m long then R asymp 01Ω

for glass ρ ~ 10+12 Ω-m for semiconductors ρ ~ 1 Ω-m

So in fact we can compute the resistance if we know a bit about the material and YES the property belongs to the material

For uniform case IJA

= ELV =

rArr Jρ ρ ⎛ ⎞= = = = ⎜ ⎟⎝ ⎠

I ρLV EL L L IA A

rArrALR ρ=whereIRV =

2q nm

τσ =J Eσ=

bull The LONGER the wire the GREATER the Rbull The THINNER the wire the GREATER the Rbull The HOTTER the wire the GREATER the R

Resistance ResistivityALR ρ= 2

1 mq n

ρσ τ

= =

Resistivity Values

Engine Current ProblemThe starter motor of a car engine draws a current of 150 A from the battery The copper wire to the motor is 50 mm in diameter and 12 m long The starter motor runs for 080 s until the car engine starts

c What is the resistance in the copper wire

8 32 2

124 17 10 104 10(005 )

4

l l mR x m xDA m

ρ ρππ

minus minus= = = times Ω = Ω

You Try Problem

If the magnitude of the drift velocity of free electrons in a copper wire is 784 times 10^ ndash4 ms what is the electric field in the conductor The number density for copper is 849 times10^28 electronsm3

J Eσ= 2

1 mq n

ρσ τ

= =

0181 V mE =

Resistance Question

bull The resistivity of both resistors is the same (ρ)bull Therefore the resistances are related as

11

1

1

1

2

22 88

)4(2 R

AL

AL

ALR ==== ρρρ

bull The resistors have the same voltage across them therefore

112

2 81

8I

RV

RVI ===

bull Two cylindrical resistors R1 and R2 are made of identical material R2 has twice the length of R1 but half the radius of R1 ndash These resistors are then connected to a battery V as shown

VI1 I2

ndash What is the relation between I1 the current flowing in R1 and I2 the current flowing in R2

(a) I1 lt I2 (b) I1 = I2 (c) I1 gt I2

1 1 2 2 1 1 2 22

1 2

RA A d

ρ ρ ρ ρ+= + =

l l l l

( ) ( ) ( ) ( )

( )3 3

23

400 10 m 0250 m 600 10 m 0400 m378

300 10 mR

minus minus

minus

times Ωsdot + times Ωsdot= = Ω

times

A rod is made of two materials The figure is not drawn to scale Each conductor has a square cross section 300 mm on a side The first material has a resistivity of 400 times 10ndash3 Ω ∙m and is 250 cm long while the second material has a resistivity of 600 times 10ndash3 Ω ∙m and is 400 cm long What is the resistance between the ends of the rod

Radial Resistance of a Coaxial Cable Leakage

bull Assume the silicon between the conductors to be concentric elements of thickness dr

bull The resistance of the hollow cylinder of silicon is

bull The total radial resistance is

2ρdR drπrL

=

ALR ρ= 2

1 mq n

ρσ τ

= =

2ln

b

a

ρ bR dRπL a

⎛ ⎞= = ⎜ ⎟⎝ ⎠int This is fairly high which is desirable since you want the current to

flow along the cable and not radially out of it

Resistors

Current is charge in motionbull Charge eg electrons exists in conductors with a

number density ne (ne approx 1029 m-3)bull ldquoSomehowrdquo put that charge in motion

ndash effective picture - all charges move with a velocity vd

ndash real picture - a lot of ldquorandom motionrdquo of charges with a small average equal to vd

bull Current density J is given by J = qenevd = qnvndash unit of J is Cm2sec or Am2 (A equiv Ampere) and 1A equiv 1Csndash current I is J times cross sectional area I = J πr2

ndash for 10 Amp in 1mm x 1mm area J = 10+7 Am2 and ve is about 10-3 ms (Yes the average velocity is only 1mms)

dqEvm

τ=v = Δpm=Ftm = qEtm IJ qnvA

= =

Atomic Vision of Ohmrsquos Lawbull E-field in conductor (resistor) provided by a batterybull Charges are put in motion but scatter in a very short

time from things that get in the wayndash itrsquos crowded inside that metalndash defects lattice vibrations (phonons) etc

bull Typical scattering time τ = 10-14 secbull Charges ballistically accelerated for this time and then

randomly scatteredbull Average velocity attained in this time is v = Ftm = qEτmbull Current density is J = qnv so current is proportional to E

which is proportional to Voltagebull OHMrsquos LAW J = (q2nτm)E or J = σ E σ = conductivity

2q nm

τσ =d

qEvm

τ=IJ E qnvA

σ= = =

Engine Current Problem AgainThe starter motor of a car engine draws a current of 150 A from the battery The copper wire to the motor is 50 mm in diameter and 12 m long The starter motor runs for 080 s until the car engine starts

a How much charge passes through the starter motorb What is the drift speed of the electrons Show how the

units work out

( )28 3electron number density 85 10 mminustimes

( ) ( )( )4

d 2 2 28 3 19

150 A 5617 10 ms00025 m 85 10 m 160 10 C

J I A Ivne ne r neπ π

minus

minus minus= = = = = times

times times

IJ E qnvA

σ= = =

Fuse ProblemYou need to design a 10 A fuse that lsquoblowsrsquo if the current exceeds 10 A The fuse material in your stockroom melts at a current density of 500 Acm2 What diameter wire of

this material will do the job

J I A=

( )( )

2

2

4 10 A4 0050 cm 050 mm4 500 AcmD I IA D

J Jπ

π π= = rArr = = = =

Resistivity

where E = electric field andJ = current density in conductor

LA

E

j

Property of bulk matter related to resistance of a sample is the resistivity (ρ) defined as

Jρ equiv

E2

1 mq n

ρσ τ

= =

eg for a copper wire ρ ~ 10-8 Ω-m 1mm radius 1 m long then R asymp 01Ω

for glass ρ ~ 10+12 Ω-m for semiconductors ρ ~ 1 Ω-m

So in fact we can compute the resistance if we know a bit about the material and YES the property belongs to the material

For uniform case IJA

= ELV =

rArr Jρ ρ ⎛ ⎞= = = = ⎜ ⎟⎝ ⎠

I ρLV EL L L IA A

rArrALR ρ=whereIRV =

2q nm

τσ =J Eσ=

bull The LONGER the wire the GREATER the Rbull The THINNER the wire the GREATER the Rbull The HOTTER the wire the GREATER the R

Resistance ResistivityALR ρ= 2

1 mq n

ρσ τ

= =

Resistivity Values

Engine Current ProblemThe starter motor of a car engine draws a current of 150 A from the battery The copper wire to the motor is 50 mm in diameter and 12 m long The starter motor runs for 080 s until the car engine starts

c What is the resistance in the copper wire

8 32 2

124 17 10 104 10(005 )

4

l l mR x m xDA m

ρ ρππ

minus minus= = = times Ω = Ω

You Try Problem

If the magnitude of the drift velocity of free electrons in a copper wire is 784 times 10^ ndash4 ms what is the electric field in the conductor The number density for copper is 849 times10^28 electronsm3

J Eσ= 2

1 mq n

ρσ τ

= =

0181 V mE =

Resistance Question

bull The resistivity of both resistors is the same (ρ)bull Therefore the resistances are related as

11

1

1

1

2

22 88

)4(2 R

AL

AL

ALR ==== ρρρ

bull The resistors have the same voltage across them therefore

112

2 81

8I

RV

RVI ===

bull Two cylindrical resistors R1 and R2 are made of identical material R2 has twice the length of R1 but half the radius of R1 ndash These resistors are then connected to a battery V as shown

VI1 I2

ndash What is the relation between I1 the current flowing in R1 and I2 the current flowing in R2

(a) I1 lt I2 (b) I1 = I2 (c) I1 gt I2

1 1 2 2 1 1 2 22

1 2

RA A d

ρ ρ ρ ρ+= + =

l l l l

( ) ( ) ( ) ( )

( )3 3

23

400 10 m 0250 m 600 10 m 0400 m378

300 10 mR

minus minus

minus

times Ωsdot + times Ωsdot= = Ω

times

A rod is made of two materials The figure is not drawn to scale Each conductor has a square cross section 300 mm on a side The first material has a resistivity of 400 times 10ndash3 Ω ∙m and is 250 cm long while the second material has a resistivity of 600 times 10ndash3 Ω ∙m and is 400 cm long What is the resistance between the ends of the rod

Radial Resistance of a Coaxial Cable Leakage

bull Assume the silicon between the conductors to be concentric elements of thickness dr

bull The resistance of the hollow cylinder of silicon is

bull The total radial resistance is

2ρdR drπrL

=

ALR ρ= 2

1 mq n

ρσ τ

= =

2ln

b

a

ρ bR dRπL a

⎛ ⎞= = ⎜ ⎟⎝ ⎠int This is fairly high which is desirable since you want the current to

flow along the cable and not radially out of it

Resistors

Atomic Vision of Ohmrsquos Lawbull E-field in conductor (resistor) provided by a batterybull Charges are put in motion but scatter in a very short

time from things that get in the wayndash itrsquos crowded inside that metalndash defects lattice vibrations (phonons) etc

bull Typical scattering time τ = 10-14 secbull Charges ballistically accelerated for this time and then

randomly scatteredbull Average velocity attained in this time is v = Ftm = qEτmbull Current density is J = qnv so current is proportional to E

which is proportional to Voltagebull OHMrsquos LAW J = (q2nτm)E or J = σ E σ = conductivity

2q nm

τσ =d

qEvm

τ=IJ E qnvA

σ= = =

Engine Current Problem AgainThe starter motor of a car engine draws a current of 150 A from the battery The copper wire to the motor is 50 mm in diameter and 12 m long The starter motor runs for 080 s until the car engine starts

a How much charge passes through the starter motorb What is the drift speed of the electrons Show how the

units work out

( )28 3electron number density 85 10 mminustimes

( ) ( )( )4

d 2 2 28 3 19

150 A 5617 10 ms00025 m 85 10 m 160 10 C

J I A Ivne ne r neπ π

minus

minus minus= = = = = times

times times

IJ E qnvA

σ= = =

Fuse ProblemYou need to design a 10 A fuse that lsquoblowsrsquo if the current exceeds 10 A The fuse material in your stockroom melts at a current density of 500 Acm2 What diameter wire of

this material will do the job

J I A=

( )( )

2

2

4 10 A4 0050 cm 050 mm4 500 AcmD I IA D

J Jπ

π π= = rArr = = = =

Resistivity

where E = electric field andJ = current density in conductor

LA

E

j

Property of bulk matter related to resistance of a sample is the resistivity (ρ) defined as

Jρ equiv

E2

1 mq n

ρσ τ

= =

eg for a copper wire ρ ~ 10-8 Ω-m 1mm radius 1 m long then R asymp 01Ω

for glass ρ ~ 10+12 Ω-m for semiconductors ρ ~ 1 Ω-m

So in fact we can compute the resistance if we know a bit about the material and YES the property belongs to the material

For uniform case IJA

= ELV =

rArr Jρ ρ ⎛ ⎞= = = = ⎜ ⎟⎝ ⎠

I ρLV EL L L IA A

rArrALR ρ=whereIRV =

2q nm

τσ =J Eσ=

bull The LONGER the wire the GREATER the Rbull The THINNER the wire the GREATER the Rbull The HOTTER the wire the GREATER the R

Resistance ResistivityALR ρ= 2

1 mq n

ρσ τ

= =

Resistivity Values

Engine Current ProblemThe starter motor of a car engine draws a current of 150 A from the battery The copper wire to the motor is 50 mm in diameter and 12 m long The starter motor runs for 080 s until the car engine starts

c What is the resistance in the copper wire

8 32 2

124 17 10 104 10(005 )

4

l l mR x m xDA m

ρ ρππ

minus minus= = = times Ω = Ω

You Try Problem

If the magnitude of the drift velocity of free electrons in a copper wire is 784 times 10^ ndash4 ms what is the electric field in the conductor The number density for copper is 849 times10^28 electronsm3

J Eσ= 2

1 mq n

ρσ τ

= =

0181 V mE =

Resistance Question

bull The resistivity of both resistors is the same (ρ)bull Therefore the resistances are related as

11

1

1

1

2

22 88

)4(2 R

AL

AL

ALR ==== ρρρ

bull The resistors have the same voltage across them therefore

112

2 81

8I

RV

RVI ===

bull Two cylindrical resistors R1 and R2 are made of identical material R2 has twice the length of R1 but half the radius of R1 ndash These resistors are then connected to a battery V as shown

VI1 I2

ndash What is the relation between I1 the current flowing in R1 and I2 the current flowing in R2

(a) I1 lt I2 (b) I1 = I2 (c) I1 gt I2

1 1 2 2 1 1 2 22

1 2

RA A d

ρ ρ ρ ρ+= + =

l l l l

( ) ( ) ( ) ( )

( )3 3

23

400 10 m 0250 m 600 10 m 0400 m378

300 10 mR

minus minus

minus

times Ωsdot + times Ωsdot= = Ω

times

A rod is made of two materials The figure is not drawn to scale Each conductor has a square cross section 300 mm on a side The first material has a resistivity of 400 times 10ndash3 Ω ∙m and is 250 cm long while the second material has a resistivity of 600 times 10ndash3 Ω ∙m and is 400 cm long What is the resistance between the ends of the rod

Radial Resistance of a Coaxial Cable Leakage

bull Assume the silicon between the conductors to be concentric elements of thickness dr

bull The resistance of the hollow cylinder of silicon is

bull The total radial resistance is

2ρdR drπrL

=

ALR ρ= 2

1 mq n

ρσ τ

= =

2ln

b

a

ρ bR dRπL a

⎛ ⎞= = ⎜ ⎟⎝ ⎠int This is fairly high which is desirable since you want the current to

flow along the cable and not radially out of it

Resistors

Engine Current Problem AgainThe starter motor of a car engine draws a current of 150 A from the battery The copper wire to the motor is 50 mm in diameter and 12 m long The starter motor runs for 080 s until the car engine starts

a How much charge passes through the starter motorb What is the drift speed of the electrons Show how the

units work out

( )28 3electron number density 85 10 mminustimes

( ) ( )( )4

d 2 2 28 3 19

150 A 5617 10 ms00025 m 85 10 m 160 10 C

J I A Ivne ne r neπ π

minus

minus minus= = = = = times

times times

IJ E qnvA

σ= = =

Fuse ProblemYou need to design a 10 A fuse that lsquoblowsrsquo if the current exceeds 10 A The fuse material in your stockroom melts at a current density of 500 Acm2 What diameter wire of

this material will do the job

J I A=

( )( )

2

2

4 10 A4 0050 cm 050 mm4 500 AcmD I IA D

J Jπ

π π= = rArr = = = =

Resistivity

where E = electric field andJ = current density in conductor

LA

E

j

Property of bulk matter related to resistance of a sample is the resistivity (ρ) defined as

Jρ equiv

E2

1 mq n

ρσ τ

= =

eg for a copper wire ρ ~ 10-8 Ω-m 1mm radius 1 m long then R asymp 01Ω

for glass ρ ~ 10+12 Ω-m for semiconductors ρ ~ 1 Ω-m

So in fact we can compute the resistance if we know a bit about the material and YES the property belongs to the material

For uniform case IJA

= ELV =

rArr Jρ ρ ⎛ ⎞= = = = ⎜ ⎟⎝ ⎠

I ρLV EL L L IA A

rArrALR ρ=whereIRV =

2q nm

τσ =J Eσ=

bull The LONGER the wire the GREATER the Rbull The THINNER the wire the GREATER the Rbull The HOTTER the wire the GREATER the R

Resistance ResistivityALR ρ= 2

1 mq n

ρσ τ

= =

Resistivity Values

Engine Current ProblemThe starter motor of a car engine draws a current of 150 A from the battery The copper wire to the motor is 50 mm in diameter and 12 m long The starter motor runs for 080 s until the car engine starts

c What is the resistance in the copper wire

8 32 2

124 17 10 104 10(005 )

4

l l mR x m xDA m

ρ ρππ

minus minus= = = times Ω = Ω

You Try Problem

If the magnitude of the drift velocity of free electrons in a copper wire is 784 times 10^ ndash4 ms what is the electric field in the conductor The number density for copper is 849 times10^28 electronsm3

J Eσ= 2

1 mq n

ρσ τ

= =

0181 V mE =

Resistance Question

bull The resistivity of both resistors is the same (ρ)bull Therefore the resistances are related as

11

1

1

1

2

22 88

)4(2 R

AL

AL

ALR ==== ρρρ

bull The resistors have the same voltage across them therefore

112

2 81

8I

RV

RVI ===

bull Two cylindrical resistors R1 and R2 are made of identical material R2 has twice the length of R1 but half the radius of R1 ndash These resistors are then connected to a battery V as shown

VI1 I2

ndash What is the relation between I1 the current flowing in R1 and I2 the current flowing in R2

(a) I1 lt I2 (b) I1 = I2 (c) I1 gt I2

1 1 2 2 1 1 2 22

1 2

RA A d

ρ ρ ρ ρ+= + =

l l l l

( ) ( ) ( ) ( )

( )3 3

23

400 10 m 0250 m 600 10 m 0400 m378

300 10 mR

minus minus

minus

times Ωsdot + times Ωsdot= = Ω

times

A rod is made of two materials The figure is not drawn to scale Each conductor has a square cross section 300 mm on a side The first material has a resistivity of 400 times 10ndash3 Ω ∙m and is 250 cm long while the second material has a resistivity of 600 times 10ndash3 Ω ∙m and is 400 cm long What is the resistance between the ends of the rod

Radial Resistance of a Coaxial Cable Leakage

bull Assume the silicon between the conductors to be concentric elements of thickness dr

bull The resistance of the hollow cylinder of silicon is

bull The total radial resistance is

2ρdR drπrL

=

ALR ρ= 2

1 mq n

ρσ τ

= =

2ln

b

a

ρ bR dRπL a

⎛ ⎞= = ⎜ ⎟⎝ ⎠int This is fairly high which is desirable since you want the current to

flow along the cable and not radially out of it

Resistors

Fuse ProblemYou need to design a 10 A fuse that lsquoblowsrsquo if the current exceeds 10 A The fuse material in your stockroom melts at a current density of 500 Acm2 What diameter wire of

this material will do the job

J I A=

( )( )

2

2

4 10 A4 0050 cm 050 mm4 500 AcmD I IA D

J Jπ

π π= = rArr = = = =

Resistivity

where E = electric field andJ = current density in conductor

LA

E

j

Property of bulk matter related to resistance of a sample is the resistivity (ρ) defined as

Jρ equiv

E2

1 mq n

ρσ τ

= =

eg for a copper wire ρ ~ 10-8 Ω-m 1mm radius 1 m long then R asymp 01Ω

for glass ρ ~ 10+12 Ω-m for semiconductors ρ ~ 1 Ω-m

So in fact we can compute the resistance if we know a bit about the material and YES the property belongs to the material

For uniform case IJA

= ELV =

rArr Jρ ρ ⎛ ⎞= = = = ⎜ ⎟⎝ ⎠

I ρLV EL L L IA A

rArrALR ρ=whereIRV =

2q nm

τσ =J Eσ=

bull The LONGER the wire the GREATER the Rbull The THINNER the wire the GREATER the Rbull The HOTTER the wire the GREATER the R

Resistance ResistivityALR ρ= 2

1 mq n

ρσ τ

= =

Resistivity Values

Engine Current ProblemThe starter motor of a car engine draws a current of 150 A from the battery The copper wire to the motor is 50 mm in diameter and 12 m long The starter motor runs for 080 s until the car engine starts

c What is the resistance in the copper wire

8 32 2

124 17 10 104 10(005 )

4

l l mR x m xDA m

ρ ρππ

minus minus= = = times Ω = Ω

You Try Problem

If the magnitude of the drift velocity of free electrons in a copper wire is 784 times 10^ ndash4 ms what is the electric field in the conductor The number density for copper is 849 times10^28 electronsm3

J Eσ= 2

1 mq n

ρσ τ

= =

0181 V mE =

Resistance Question

bull The resistivity of both resistors is the same (ρ)bull Therefore the resistances are related as

11

1

1

1

2

22 88

)4(2 R

AL

AL

ALR ==== ρρρ

bull The resistors have the same voltage across them therefore

112

2 81

8I

RV

RVI ===

bull Two cylindrical resistors R1 and R2 are made of identical material R2 has twice the length of R1 but half the radius of R1 ndash These resistors are then connected to a battery V as shown

VI1 I2

ndash What is the relation between I1 the current flowing in R1 and I2 the current flowing in R2

(a) I1 lt I2 (b) I1 = I2 (c) I1 gt I2

1 1 2 2 1 1 2 22

1 2

RA A d

ρ ρ ρ ρ+= + =

l l l l

( ) ( ) ( ) ( )

( )3 3

23

400 10 m 0250 m 600 10 m 0400 m378

300 10 mR

minus minus

minus

times Ωsdot + times Ωsdot= = Ω

times

A rod is made of two materials The figure is not drawn to scale Each conductor has a square cross section 300 mm on a side The first material has a resistivity of 400 times 10ndash3 Ω ∙m and is 250 cm long while the second material has a resistivity of 600 times 10ndash3 Ω ∙m and is 400 cm long What is the resistance between the ends of the rod

Radial Resistance of a Coaxial Cable Leakage

bull Assume the silicon between the conductors to be concentric elements of thickness dr

bull The resistance of the hollow cylinder of silicon is

bull The total radial resistance is

2ρdR drπrL

=

ALR ρ= 2

1 mq n

ρσ τ

= =

2ln

b

a

ρ bR dRπL a

⎛ ⎞= = ⎜ ⎟⎝ ⎠int This is fairly high which is desirable since you want the current to

flow along the cable and not radially out of it

Resistors

Resistivity

where E = electric field andJ = current density in conductor

LA

E

j

Property of bulk matter related to resistance of a sample is the resistivity (ρ) defined as

Jρ equiv

E2

1 mq n

ρσ τ

= =

eg for a copper wire ρ ~ 10-8 Ω-m 1mm radius 1 m long then R asymp 01Ω

for glass ρ ~ 10+12 Ω-m for semiconductors ρ ~ 1 Ω-m

So in fact we can compute the resistance if we know a bit about the material and YES the property belongs to the material

For uniform case IJA

= ELV =

rArr Jρ ρ ⎛ ⎞= = = = ⎜ ⎟⎝ ⎠

I ρLV EL L L IA A

rArrALR ρ=whereIRV =

2q nm

τσ =J Eσ=

bull The LONGER the wire the GREATER the Rbull The THINNER the wire the GREATER the Rbull The HOTTER the wire the GREATER the R

Resistance ResistivityALR ρ= 2

1 mq n

ρσ τ

= =

Resistivity Values

Engine Current ProblemThe starter motor of a car engine draws a current of 150 A from the battery The copper wire to the motor is 50 mm in diameter and 12 m long The starter motor runs for 080 s until the car engine starts

c What is the resistance in the copper wire

8 32 2

124 17 10 104 10(005 )

4

l l mR x m xDA m

ρ ρππ

minus minus= = = times Ω = Ω

You Try Problem

If the magnitude of the drift velocity of free electrons in a copper wire is 784 times 10^ ndash4 ms what is the electric field in the conductor The number density for copper is 849 times10^28 electronsm3

J Eσ= 2

1 mq n

ρσ τ

= =

0181 V mE =

Resistance Question

bull The resistivity of both resistors is the same (ρ)bull Therefore the resistances are related as

11

1

1

1

2

22 88

)4(2 R

AL

AL

ALR ==== ρρρ

bull The resistors have the same voltage across them therefore

112

2 81

8I

RV

RVI ===

bull Two cylindrical resistors R1 and R2 are made of identical material R2 has twice the length of R1 but half the radius of R1 ndash These resistors are then connected to a battery V as shown

VI1 I2

ndash What is the relation between I1 the current flowing in R1 and I2 the current flowing in R2

(a) I1 lt I2 (b) I1 = I2 (c) I1 gt I2

1 1 2 2 1 1 2 22

1 2

RA A d

ρ ρ ρ ρ+= + =

l l l l

( ) ( ) ( ) ( )

( )3 3

23

400 10 m 0250 m 600 10 m 0400 m378

300 10 mR

minus minus

minus

times Ωsdot + times Ωsdot= = Ω

times

A rod is made of two materials The figure is not drawn to scale Each conductor has a square cross section 300 mm on a side The first material has a resistivity of 400 times 10ndash3 Ω ∙m and is 250 cm long while the second material has a resistivity of 600 times 10ndash3 Ω ∙m and is 400 cm long What is the resistance between the ends of the rod

Radial Resistance of a Coaxial Cable Leakage

bull Assume the silicon between the conductors to be concentric elements of thickness dr

bull The resistance of the hollow cylinder of silicon is

bull The total radial resistance is

2ρdR drπrL

=

ALR ρ= 2

1 mq n

ρσ τ

= =

2ln

b

a

ρ bR dRπL a

⎛ ⎞= = ⎜ ⎟⎝ ⎠int This is fairly high which is desirable since you want the current to

flow along the cable and not radially out of it

Resistors

bull The LONGER the wire the GREATER the Rbull The THINNER the wire the GREATER the Rbull The HOTTER the wire the GREATER the R

Resistance ResistivityALR ρ= 2

1 mq n

ρσ τ

= =

Resistivity Values

Engine Current ProblemThe starter motor of a car engine draws a current of 150 A from the battery The copper wire to the motor is 50 mm in diameter and 12 m long The starter motor runs for 080 s until the car engine starts

c What is the resistance in the copper wire

8 32 2

124 17 10 104 10(005 )

4

l l mR x m xDA m

ρ ρππ

minus minus= = = times Ω = Ω

You Try Problem

If the magnitude of the drift velocity of free electrons in a copper wire is 784 times 10^ ndash4 ms what is the electric field in the conductor The number density for copper is 849 times10^28 electronsm3

J Eσ= 2

1 mq n

ρσ τ

= =

0181 V mE =

Resistance Question

bull The resistivity of both resistors is the same (ρ)bull Therefore the resistances are related as

11

1

1

1

2

22 88

)4(2 R

AL

AL

ALR ==== ρρρ

bull The resistors have the same voltage across them therefore

112

2 81

8I

RV

RVI ===

bull Two cylindrical resistors R1 and R2 are made of identical material R2 has twice the length of R1 but half the radius of R1 ndash These resistors are then connected to a battery V as shown

VI1 I2

ndash What is the relation between I1 the current flowing in R1 and I2 the current flowing in R2

(a) I1 lt I2 (b) I1 = I2 (c) I1 gt I2

1 1 2 2 1 1 2 22

1 2

RA A d

ρ ρ ρ ρ+= + =

l l l l

( ) ( ) ( ) ( )

( )3 3

23

400 10 m 0250 m 600 10 m 0400 m378

300 10 mR

minus minus

minus

times Ωsdot + times Ωsdot= = Ω

times

A rod is made of two materials The figure is not drawn to scale Each conductor has a square cross section 300 mm on a side The first material has a resistivity of 400 times 10ndash3 Ω ∙m and is 250 cm long while the second material has a resistivity of 600 times 10ndash3 Ω ∙m and is 400 cm long What is the resistance between the ends of the rod

Radial Resistance of a Coaxial Cable Leakage

bull Assume the silicon between the conductors to be concentric elements of thickness dr

bull The resistance of the hollow cylinder of silicon is

bull The total radial resistance is

2ρdR drπrL

=

ALR ρ= 2

1 mq n

ρσ τ

= =

2ln

b

a

ρ bR dRπL a

⎛ ⎞= = ⎜ ⎟⎝ ⎠int This is fairly high which is desirable since you want the current to

flow along the cable and not radially out of it

Resistors

Resistivity Values

Engine Current ProblemThe starter motor of a car engine draws a current of 150 A from the battery The copper wire to the motor is 50 mm in diameter and 12 m long The starter motor runs for 080 s until the car engine starts

c What is the resistance in the copper wire

8 32 2

124 17 10 104 10(005 )

4

l l mR x m xDA m

ρ ρππ

minus minus= = = times Ω = Ω

You Try Problem

If the magnitude of the drift velocity of free electrons in a copper wire is 784 times 10^ ndash4 ms what is the electric field in the conductor The number density for copper is 849 times10^28 electronsm3

J Eσ= 2

1 mq n

ρσ τ

= =

0181 V mE =

Resistance Question

bull The resistivity of both resistors is the same (ρ)bull Therefore the resistances are related as

11

1

1

1

2

22 88

)4(2 R

AL

AL

ALR ==== ρρρ

bull The resistors have the same voltage across them therefore

112

2 81

8I

RV

RVI ===

bull Two cylindrical resistors R1 and R2 are made of identical material R2 has twice the length of R1 but half the radius of R1 ndash These resistors are then connected to a battery V as shown

VI1 I2

ndash What is the relation between I1 the current flowing in R1 and I2 the current flowing in R2

(a) I1 lt I2 (b) I1 = I2 (c) I1 gt I2

1 1 2 2 1 1 2 22

1 2

RA A d

ρ ρ ρ ρ+= + =

l l l l

( ) ( ) ( ) ( )

( )3 3

23

400 10 m 0250 m 600 10 m 0400 m378

300 10 mR

minus minus

minus

times Ωsdot + times Ωsdot= = Ω

times

A rod is made of two materials The figure is not drawn to scale Each conductor has a square cross section 300 mm on a side The first material has a resistivity of 400 times 10ndash3 Ω ∙m and is 250 cm long while the second material has a resistivity of 600 times 10ndash3 Ω ∙m and is 400 cm long What is the resistance between the ends of the rod

Radial Resistance of a Coaxial Cable Leakage

bull Assume the silicon between the conductors to be concentric elements of thickness dr

bull The resistance of the hollow cylinder of silicon is

bull The total radial resistance is

2ρdR drπrL

=

ALR ρ= 2

1 mq n

ρσ τ

= =

2ln

b

a

ρ bR dRπL a

⎛ ⎞= = ⎜ ⎟⎝ ⎠int This is fairly high which is desirable since you want the current to

flow along the cable and not radially out of it

Resistors

Engine Current ProblemThe starter motor of a car engine draws a current of 150 A from the battery The copper wire to the motor is 50 mm in diameter and 12 m long The starter motor runs for 080 s until the car engine starts

c What is the resistance in the copper wire

8 32 2

124 17 10 104 10(005 )

4

l l mR x m xDA m

ρ ρππ

minus minus= = = times Ω = Ω

You Try Problem

If the magnitude of the drift velocity of free electrons in a copper wire is 784 times 10^ ndash4 ms what is the electric field in the conductor The number density for copper is 849 times10^28 electronsm3

J Eσ= 2

1 mq n

ρσ τ

= =

0181 V mE =

Resistance Question

bull The resistivity of both resistors is the same (ρ)bull Therefore the resistances are related as

11

1

1

1

2

22 88

)4(2 R

AL

AL

ALR ==== ρρρ

bull The resistors have the same voltage across them therefore

112

2 81

8I

RV

RVI ===

bull Two cylindrical resistors R1 and R2 are made of identical material R2 has twice the length of R1 but half the radius of R1 ndash These resistors are then connected to a battery V as shown

VI1 I2

ndash What is the relation between I1 the current flowing in R1 and I2 the current flowing in R2

(a) I1 lt I2 (b) I1 = I2 (c) I1 gt I2

1 1 2 2 1 1 2 22

1 2

RA A d

ρ ρ ρ ρ+= + =

l l l l

( ) ( ) ( ) ( )

( )3 3

23

400 10 m 0250 m 600 10 m 0400 m378

300 10 mR

minus minus

minus

times Ωsdot + times Ωsdot= = Ω

times

A rod is made of two materials The figure is not drawn to scale Each conductor has a square cross section 300 mm on a side The first material has a resistivity of 400 times 10ndash3 Ω ∙m and is 250 cm long while the second material has a resistivity of 600 times 10ndash3 Ω ∙m and is 400 cm long What is the resistance between the ends of the rod

Radial Resistance of a Coaxial Cable Leakage

bull Assume the silicon between the conductors to be concentric elements of thickness dr

bull The resistance of the hollow cylinder of silicon is

bull The total radial resistance is

2ρdR drπrL

=

ALR ρ= 2

1 mq n

ρσ τ

= =

2ln

b

a

ρ bR dRπL a

⎛ ⎞= = ⎜ ⎟⎝ ⎠int This is fairly high which is desirable since you want the current to

flow along the cable and not radially out of it

Resistors

You Try Problem

If the magnitude of the drift velocity of free electrons in a copper wire is 784 times 10^ ndash4 ms what is the electric field in the conductor The number density for copper is 849 times10^28 electronsm3

J Eσ= 2

1 mq n

ρσ τ

= =

0181 V mE =

Resistance Question

bull The resistivity of both resistors is the same (ρ)bull Therefore the resistances are related as

11

1

1

1

2

22 88

)4(2 R

AL

AL

ALR ==== ρρρ

bull The resistors have the same voltage across them therefore

112

2 81

8I

RV

RVI ===

bull Two cylindrical resistors R1 and R2 are made of identical material R2 has twice the length of R1 but half the radius of R1 ndash These resistors are then connected to a battery V as shown

VI1 I2

ndash What is the relation between I1 the current flowing in R1 and I2 the current flowing in R2

(a) I1 lt I2 (b) I1 = I2 (c) I1 gt I2

1 1 2 2 1 1 2 22

1 2

RA A d

ρ ρ ρ ρ+= + =

l l l l

( ) ( ) ( ) ( )

( )3 3

23

400 10 m 0250 m 600 10 m 0400 m378

300 10 mR

minus minus

minus

times Ωsdot + times Ωsdot= = Ω

times

A rod is made of two materials The figure is not drawn to scale Each conductor has a square cross section 300 mm on a side The first material has a resistivity of 400 times 10ndash3 Ω ∙m and is 250 cm long while the second material has a resistivity of 600 times 10ndash3 Ω ∙m and is 400 cm long What is the resistance between the ends of the rod

Radial Resistance of a Coaxial Cable Leakage

bull Assume the silicon between the conductors to be concentric elements of thickness dr

bull The resistance of the hollow cylinder of silicon is

bull The total radial resistance is

2ρdR drπrL

=

ALR ρ= 2

1 mq n

ρσ τ

= =

2ln

b

a

ρ bR dRπL a

⎛ ⎞= = ⎜ ⎟⎝ ⎠int This is fairly high which is desirable since you want the current to

flow along the cable and not radially out of it

Resistors

Resistance Question

bull The resistivity of both resistors is the same (ρ)bull Therefore the resistances are related as

11

1

1

1

2

22 88

)4(2 R

AL

AL

ALR ==== ρρρ

bull The resistors have the same voltage across them therefore

112

2 81

8I

RV

RVI ===

bull Two cylindrical resistors R1 and R2 are made of identical material R2 has twice the length of R1 but half the radius of R1 ndash These resistors are then connected to a battery V as shown

VI1 I2

ndash What is the relation between I1 the current flowing in R1 and I2 the current flowing in R2

(a) I1 lt I2 (b) I1 = I2 (c) I1 gt I2

1 1 2 2 1 1 2 22

1 2

RA A d

ρ ρ ρ ρ+= + =

l l l l

( ) ( ) ( ) ( )

( )3 3

23

400 10 m 0250 m 600 10 m 0400 m378

300 10 mR

minus minus

minus

times Ωsdot + times Ωsdot= = Ω

times

A rod is made of two materials The figure is not drawn to scale Each conductor has a square cross section 300 mm on a side The first material has a resistivity of 400 times 10ndash3 Ω ∙m and is 250 cm long while the second material has a resistivity of 600 times 10ndash3 Ω ∙m and is 400 cm long What is the resistance between the ends of the rod

Radial Resistance of a Coaxial Cable Leakage

bull Assume the silicon between the conductors to be concentric elements of thickness dr

bull The resistance of the hollow cylinder of silicon is

bull The total radial resistance is

2ρdR drπrL

=

ALR ρ= 2

1 mq n

ρσ τ

= =

2ln

b

a

ρ bR dRπL a

⎛ ⎞= = ⎜ ⎟⎝ ⎠int This is fairly high which is desirable since you want the current to

flow along the cable and not radially out of it

Resistors

1 1 2 2 1 1 2 22

1 2

RA A d

ρ ρ ρ ρ+= + =

l l l l

( ) ( ) ( ) ( )

( )3 3

23

400 10 m 0250 m 600 10 m 0400 m378

300 10 mR

minus minus

minus

times Ωsdot + times Ωsdot= = Ω

times

A rod is made of two materials The figure is not drawn to scale Each conductor has a square cross section 300 mm on a side The first material has a resistivity of 400 times 10ndash3 Ω ∙m and is 250 cm long while the second material has a resistivity of 600 times 10ndash3 Ω ∙m and is 400 cm long What is the resistance between the ends of the rod

Radial Resistance of a Coaxial Cable Leakage

bull Assume the silicon between the conductors to be concentric elements of thickness dr

bull The resistance of the hollow cylinder of silicon is

bull The total radial resistance is

2ρdR drπrL

=

ALR ρ= 2

1 mq n

ρσ τ

= =

2ln

b

a

ρ bR dRπL a

⎛ ⎞= = ⎜ ⎟⎝ ⎠int This is fairly high which is desirable since you want the current to

flow along the cable and not radially out of it

Resistors

Radial Resistance of a Coaxial Cable Leakage

bull Assume the silicon between the conductors to be concentric elements of thickness dr

bull The resistance of the hollow cylinder of silicon is

bull The total radial resistance is

2ρdR drπrL

=

ALR ρ= 2

1 mq n

ρσ τ

= =

2ln

b

a

ρ bR dRπL a

⎛ ⎞= = ⎜ ⎟⎝ ⎠int This is fairly high which is desirable since you want the current to

flow along the cable and not radially out of it

Resistors

Resistors

The HOTTER the wire the GREATER the R

Resistance Dependence on Temperature

0 (1 )R R Tα= + Δ

0 original resistance temperature coefficient of resistivity

temperature change (lt100 C)

R

Tα

==

Δ = o

When are light bulbs more likely to blow

When hot or cold

0 (1 )R R Tα= + Δ

The HOTTER the wire the GREATER the R

At lower Resistance the bulb draws more current and it blows the filament

Ohmic Material

bull An ohmic devicebull The resistance is

constant over a wide range of voltages

bull The relationship between current and voltage is linear

bull The slope is related to the resistance

Nonohmic Material

bull Nonohmic materials are those whose resistance changes with voltage or current

bull The current-voltage relationship is nonlinear

bull A diode is a common example of a nonohmic device

Superconductivitybull 1911 H K Onnes who had figured

out how to make liquid helium used it to cool mercury to 42 K and looked at its resistance

ndashCurrent can flow even if E=0ndashCurrent in superconducting rings can flow for years with no decrease

bull 1957 Bardeen (UIUC) Cooper and Schrieffer (ldquoBCSrdquo) publish theoretical explanation for which they get the Nobel prize in 1972

ndash It was Bardeenrsquos second Nobel prize (1956 ndash transistor)

bull At low temperatures the resistance of some metals 0 measured to be less than 10-16bullρconductor (ie ρlt10-24 Ωm)

Superconductivitybull 1986 ldquoHighrdquo temperature superconductors are

discovered (Tc=77K)ndash Important because liquid nitrogen (77 K) is much cheaper than

liquid heliumndash Highest critical temperature to date 138 K (-135˚ C = -211˚ F)

bull Today Superconducting loops are used to produce ldquolosslessrdquo electromagnets (only need to cool them not fight dissipation of current) for particle physics [Fermilab accelerator IL]

bull The Future Smaller motors ldquolosslessrdquo power transmission lines magnetic levitation trains quantum computers

When are light bulbs more likely to blow

When hot or cold

0 (1 )R R Tα= + Δ

The HOTTER the wire the GREATER the R

At lower Resistance the bulb draws more current and it blows the filament

Ohmic Material

bull An ohmic devicebull The resistance is

constant over a wide range of voltages

bull The relationship between current and voltage is linear

bull The slope is related to the resistance

Nonohmic Material

bull Nonohmic materials are those whose resistance changes with voltage or current

bull The current-voltage relationship is nonlinear

bull A diode is a common example of a nonohmic device

Superconductivitybull 1911 H K Onnes who had figured

out how to make liquid helium used it to cool mercury to 42 K and looked at its resistance

ndashCurrent can flow even if E=0ndashCurrent in superconducting rings can flow for years with no decrease

bull 1957 Bardeen (UIUC) Cooper and Schrieffer (ldquoBCSrdquo) publish theoretical explanation for which they get the Nobel prize in 1972

ndash It was Bardeenrsquos second Nobel prize (1956 ndash transistor)

bull At low temperatures the resistance of some metals 0 measured to be less than 10-16bullρconductor (ie ρlt10-24 Ωm)

Superconductivitybull 1986 ldquoHighrdquo temperature superconductors are

discovered (Tc=77K)ndash Important because liquid nitrogen (77 K) is much cheaper than

liquid heliumndash Highest critical temperature to date 138 K (-135˚ C = -211˚ F)

bull Today Superconducting loops are used to produce ldquolosslessrdquo electromagnets (only need to cool them not fight dissipation of current) for particle physics [Fermilab accelerator IL]

bull The Future Smaller motors ldquolosslessrdquo power transmission lines magnetic levitation trains quantum computers

Ohmic Material

bull An ohmic devicebull The resistance is

constant over a wide range of voltages

bull The relationship between current and voltage is linear

bull The slope is related to the resistance

Nonohmic Material

bull Nonohmic materials are those whose resistance changes with voltage or current

bull The current-voltage relationship is nonlinear

bull A diode is a common example of a nonohmic device

Superconductivitybull 1911 H K Onnes who had figured

out how to make liquid helium used it to cool mercury to 42 K and looked at its resistance

ndashCurrent can flow even if E=0ndashCurrent in superconducting rings can flow for years with no decrease

bull 1957 Bardeen (UIUC) Cooper and Schrieffer (ldquoBCSrdquo) publish theoretical explanation for which they get the Nobel prize in 1972

ndash It was Bardeenrsquos second Nobel prize (1956 ndash transistor)

bull At low temperatures the resistance of some metals 0 measured to be less than 10-16bullρconductor (ie ρlt10-24 Ωm)

Superconductivitybull 1986 ldquoHighrdquo temperature superconductors are

discovered (Tc=77K)ndash Important because liquid nitrogen (77 K) is much cheaper than

liquid heliumndash Highest critical temperature to date 138 K (-135˚ C = -211˚ F)

bull Today Superconducting loops are used to produce ldquolosslessrdquo electromagnets (only need to cool them not fight dissipation of current) for particle physics [Fermilab accelerator IL]

bull The Future Smaller motors ldquolosslessrdquo power transmission lines magnetic levitation trains quantum computers

Nonohmic Material

bull Nonohmic materials are those whose resistance changes with voltage or current

bull The current-voltage relationship is nonlinear

bull A diode is a common example of a nonohmic device

Superconductivitybull 1911 H K Onnes who had figured

out how to make liquid helium used it to cool mercury to 42 K and looked at its resistance

ndashCurrent can flow even if E=0ndashCurrent in superconducting rings can flow for years with no decrease

bull 1957 Bardeen (UIUC) Cooper and Schrieffer (ldquoBCSrdquo) publish theoretical explanation for which they get the Nobel prize in 1972

ndash It was Bardeenrsquos second Nobel prize (1956 ndash transistor)

bull At low temperatures the resistance of some metals 0 measured to be less than 10-16bullρconductor (ie ρlt10-24 Ωm)

Superconductivitybull 1986 ldquoHighrdquo temperature superconductors are

discovered (Tc=77K)ndash Important because liquid nitrogen (77 K) is much cheaper than

liquid heliumndash Highest critical temperature to date 138 K (-135˚ C = -211˚ F)

bull Today Superconducting loops are used to produce ldquolosslessrdquo electromagnets (only need to cool them not fight dissipation of current) for particle physics [Fermilab accelerator IL]

bull The Future Smaller motors ldquolosslessrdquo power transmission lines magnetic levitation trains quantum computers

Superconductivitybull 1911 H K Onnes who had figured

out how to make liquid helium used it to cool mercury to 42 K and looked at its resistance

ndashCurrent can flow even if E=0ndashCurrent in superconducting rings can flow for years with no decrease

bull 1957 Bardeen (UIUC) Cooper and Schrieffer (ldquoBCSrdquo) publish theoretical explanation for which they get the Nobel prize in 1972

ndash It was Bardeenrsquos second Nobel prize (1956 ndash transistor)

bull At low temperatures the resistance of some metals 0 measured to be less than 10-16bullρconductor (ie ρlt10-24 Ωm)

Superconductivitybull 1986 ldquoHighrdquo temperature superconductors are

discovered (Tc=77K)ndash Important because liquid nitrogen (77 K) is much cheaper than

liquid heliumndash Highest critical temperature to date 138 K (-135˚ C = -211˚ F)

bull Today Superconducting loops are used to produce ldquolosslessrdquo electromagnets (only need to cool them not fight dissipation of current) for particle physics [Fermilab accelerator IL]

bull The Future Smaller motors ldquolosslessrdquo power transmission lines magnetic levitation trains quantum computers

Superconductivitybull 1986 ldquoHighrdquo temperature superconductors are

discovered (Tc=77K)ndash Important because liquid nitrogen (77 K) is much cheaper than

liquid heliumndash Highest critical temperature to date 138 K (-135˚ C = -211˚ F)

bull Today Superconducting loops are used to produce ldquolosslessrdquo electromagnets (only need to cool them not fight dissipation of current) for particle physics [Fermilab accelerator IL]

bull The Future Smaller motors ldquolosslessrdquo power transmission lines magnetic levitation trains quantum computers

Ohms Law ΔV = IRALR ρ=

Resistance QUESTIONHow much current will flow

through a lamp that has a resistance of 60 Ohms when

12 Volts are impressed across it

USE OHMS LAW ΔV = IR

12 12 260 60

V V VI AR V A

Δ= = = =

Ω

What makes the Glow

Resistance QUESTIONHow much current will flow

through a lamp that has a resistance of 60 Ohms when

12 Volts are impressed across it

USE OHMS LAW ΔV = IR

12 12 260 60

V V VI AR V A

Δ= = = =

Ω

What makes the Glow

What makes the Glow

Ohmic Heat LossbullThe resistor is normally in contact with the air so its increased temperature will result in a transfer of energy by heat into the airbullThe resistor also emits thermal radiation which can make it glowbullAfter some time interval the resistor reaches a constant temperaturebullThe rate at which the system loses potential energy as the charge passes through the resistor is equal to the rate at which the system gains internal energy in the resistorThe power is the rate at which the energy is delivered to the resistor

Electrical Powerbull As a charge moves from a to b the

electric potential energy of the system increases by QΔV The chemical energy in the battery must decrease by this same amount

bull As the charge moves through the resistor (c to d) the system loses this electric potential energy during collisions of the electrons with the atoms of the resistor

bull This energy is transformed into internal energy in the resistor asincreased vibrational motion of the atoms in the resistor

bull You pay for ENERGY not for ELECTRONSbull Kilowatt-hour is the energy consumed in one

hour [kWh]=J NOT TIME Power x Time

POWER

2V VP I V VR R

Δ Δ⎛ ⎞= Δ = =⎜ ⎟⎝ ⎠

[ ] WattEnergy JPtime s

= = =

2( )P I V I IR I R= Δ = =

P I V= Δ

If V = 120V What is I USE P = IV=gt I = PV

Appliance _ Power Current (A)

Hair Dryer 1600 Watts

Electric Iron 1200 Watts

TV 100 Watts

Computer 45 Watts

If V = 120V What is I USE P = IV=gt I = PV

Appliance _ Power Current (A)

Hair Dryer 1600 Watts 133 A

Electric Iron 1200 Watts 10 A

TV 100 Watts 83 A

Computer 45 Watts 38 A

Electric Shock

bullElectric Shock occurs when current is produced in the body which is caused by an impressed voltage

bullVoltage is the CAUSE bullCurrent does the DAMAGE

What causes electric Shock in the human body Voltage or Current

Current (A) Effect

0001 Can be felt0005 Painful0010 Causes involuntary muscle spasms0015 Causes loss of muscle control0070 If through heart serious

If current lasts for 1 s - FATAL

Dry Skin Body Resistance 500000 ΩWet Skin Body Resistance 1000 Ω

Electric Shock

Electric BillCost to run for 1 hr

$05 per 1 kw-hr Cost = Power x Time x Rate

Appliance _ Power Cost______

Hair Dryer 1600 Watts

Electric Iron 1200 Watts

TV 100 Watts

Computer 45 Watts

Electric BillCost to run for 1 hr

$05 per 1 kw-hr Cost = Power x Time x Rate

Appliance _ Power Cost______

Hair Dryer 1600 Watts $008

Electric Iron 1200 Watts $006

TV 100 Watts $0005

Computer 45 Watts $0003

QUESTION

The voltage and power on a light bulb read ldquo120 V 60 Wrdquo How much current will flow

through the bulb USE P = I ΔV

I = PΔV = 60 W120 V = 12 Amp

QUESTION

The power and voltage on a light bulb read ldquo120 V 60 Wrdquo What is the resistance of the

filament (I = 5 A)Hint USE OHMS LAW V = IR

R = VI = 120 V 5 A = 240 Ω

QUESTION

The power rating for two light bulbs read30W and 60W Which bulb has the greatest

resistance at 120V 2 2 P V R R V P= rarr =

2(120 ) 30 480R V W= = Ω2(120 ) 60 240R V W= = Ω

Which burns brighter and why

Quick Quiz 278

For the two lightbulbs shown in this figure rank the current values at the points from greatest to least

Ia = Ib gt Ic = Id gt Ie = If

The 60 W bulb has the lowest resistance and therefore draws the most current

Which light burns the brightest

bull The voltage of each device is the full voltage of the EMF source (the battery)

bull The total current is divided between each path

Parallel Circuits

1 21 2 1 2

1 1( )P

V V VI I I VR R R R R

= + = + = + =

1 2 3

1 1 1 1 PR R R R

= + + +

Equivalent Resistance

bull The current is the same in each devicebull The equivalent resistance of the circuit is the sum

of the individual resistances R=R1+R2

Series Circuits

1 2 1 2 3+ SV V V R R R R= = + + +

1 2 1 2 1 2( + ) + + I R R IR IR V V= = =

total totalV IR=

bull The current is the same in each devicebull The equivalent resistance of the circuit is the sum

of the individual resistances R=R1+R2

Series Circuits

1 2 1 2 3+ SV V V R R R R= = + + +

To find the current use the total voltage and equivalent resistance

S

VIR

=

Circuits ProblemBulbs in Series vs ParallelA circuit contains a 48-V battery and two 240Ω light bulbs

In which circuit does each bulb burn brighterRULE THE MORE POWER DISSIPATED IN A BULB THE

BRIGHTER IT IS

P IVFind the power in each bulb when in series and in parallel

=

Circuits ProblemBulbs in Series vs ParallelIf a bulb burns out - what happens to the other bulb in

each circuit Does it go out Is it brighter Dimmer Or

In the series circuit the burned out bulb will short the circuit and the other bulb will go out

In the parallel circuit the other bulb will have the same brightness

In SeriesThe Voltage is divided in series each bulb gets half V = 24V

2 2(24 ) 24240

V V VP IV V WR R

= = = = =Ω

Circuits ProblemBulbs in Series vs Parallel

In ParallelVoltage is the same in each bulb 48V

2 2(48 ) 96240

V VP WR

= = =Ω

Parallel Bulbs Burn Brighter

Circuits Problem3 Bulbs in SeriesIf one more bulb is added to each circuit (3 bulbs total) how does the brightness of the bulbs change Or not

In the series circuit the bulbs DIM WHY

( )22 2singlebulb 3 1

9 9PVV VP

R R R= = = =

1 2 3V V V V= + +In series each of the three equal bulbs gets one third of the Voltage

(V3) that a single bulb would get

Note P=VI but I is due to the equivalent Resistance I = VRs =V3RSo the Current through each is 13 the current through a single bulb andP=VI=V3 x I3 = VI9 = P9 The bulbs burn 19 as bright

Circuits Problem3 Bulbs in ParallelIf one more bulb is added to each circuit (3 bulbs total) how does the brightness of the bulbs change Or not

In the parallel circuit the bulbs DO NOT DIM WHY

2

singlebulbVP PR

= =

In parallel each of the three equal bulbs gets the full voltage of the battery source

Is this getting something for nothing

NO Parallel circuits drain the battery faster

Parallel CircuitsbullAs the number of branches is increased the overall resistance of the circuit is DECREASED

bullOverall resistance is lowered with each added path between any two points of the circuit

bullThis means the overall resistance of the circuit is less than the resistance of any one of the branches (Weird)

bullAs overall resistance is lowered more current is drawn This is how you blow fuses

FusesIf the current drawn exceeds safe levels the fuse melts and the

circuit lsquobreaksrdquo ndash most house have switches not fuses

Limitations of DC Transmissionbull Large currents in wires produce heat and energy losses by

P = I2Rbull Large expensive conductors would be needed or else very

high voltage drops (and efficiency losses) would resultbull High loads of direct current could rarely be transmitted for

distances greater than one mile without introducing excessive voltage drops

bull Direct current can not easily be changed to higher or lower voltages Separate electrical lines are needed to distribute power to appliances that used different voltages for example lighting and electric motors

Advantages of AC Transmission

bull Alternating Current can be transformed to lsquosteprsquo the voltage up or down with transformers

bull Power is transmitted at great distances at HIGH voltages and LOW currents and then stepped down to low voltages for use in homes (240V) and industry (440V)

bull Convert AC to DC with a rectifier in appliances

AC is more efficient for Transmission amp Distribution of

electrical power than DC

In The FuturehellipLong Distance AC Power Transmission

may not be needed

War of Currents 1880rsquos

Thomas Edison American inventor and businessman pushed for the development of a DC power network

George WestinghouseAmerican entrepreneur and engineer backed financially the development of a practical AC power network

Nikola Tesla Serbian inventor physicist and electro-mechanical engineer was instrumental in developing AC networks

bull You pay for ENERGY not for ELECTRONSbull Kilowatt-hour is the energy consumed in one

hour [kWh]=J NOT TIME Power x Time

POWER

2V VP I V VR R

Δ Δ⎛ ⎞= Δ = =⎜ ⎟⎝ ⎠

[ ] WattEnergy JPtime s

= = =

2( )P I V I IR I R= Δ = =

P I V= Δ

If V = 120V What is I USE P = IV=gt I = PV

Appliance _ Power Current (A)

Hair Dryer 1600 Watts

Electric Iron 1200 Watts

TV 100 Watts

Computer 45 Watts

If V = 120V What is I USE P = IV=gt I = PV

Appliance _ Power Current (A)

Hair Dryer 1600 Watts 133 A

Electric Iron 1200 Watts 10 A

TV 100 Watts 83 A

Computer 45 Watts 38 A

Electric Shock

bullElectric Shock occurs when current is produced in the body which is caused by an impressed voltage

bullVoltage is the CAUSE bullCurrent does the DAMAGE

What causes electric Shock in the human body Voltage or Current

Current (A) Effect

0001 Can be felt0005 Painful0010 Causes involuntary muscle spasms0015 Causes loss of muscle control0070 If through heart serious

If current lasts for 1 s - FATAL

Dry Skin Body Resistance 500000 ΩWet Skin Body Resistance 1000 Ω

Electric Shock

Electric BillCost to run for 1 hr

$05 per 1 kw-hr Cost = Power x Time x Rate

Appliance _ Power Cost______

Hair Dryer 1600 Watts

Electric Iron 1200 Watts

TV 100 Watts

Computer 45 Watts

Electric BillCost to run for 1 hr

$05 per 1 kw-hr Cost = Power x Time x Rate

Appliance _ Power Cost______

Hair Dryer 1600 Watts $008

Electric Iron 1200 Watts $006

TV 100 Watts $0005

Computer 45 Watts $0003

QUESTION

The voltage and power on a light bulb read ldquo120 V 60 Wrdquo How much current will flow

through the bulb USE P = I ΔV

I = PΔV = 60 W120 V = 12 Amp

QUESTION

The power and voltage on a light bulb read ldquo120 V 60 Wrdquo What is the resistance of the

filament (I = 5 A)Hint USE OHMS LAW V = IR

R = VI = 120 V 5 A = 240 Ω

QUESTION

The power rating for two light bulbs read30W and 60W Which bulb has the greatest

resistance at 120V 2 2 P V R R V P= rarr =

2(120 ) 30 480R V W= = Ω2(120 ) 60 240R V W= = Ω

Which burns brighter and why

Quick Quiz 278

For the two lightbulbs shown in this figure rank the current values at the points from greatest to least

Ia = Ib gt Ic = Id gt Ie = If

The 60 W bulb has the lowest resistance and therefore draws the most current

Which light burns the brightest

bull The voltage of each device is the full voltage of the EMF source (the battery)

bull The total current is divided between each path

Parallel Circuits

1 21 2 1 2

1 1( )P

V V VI I I VR R R R R

= + = + = + =

1 2 3

1 1 1 1 PR R R R

= + + +

Equivalent Resistance

bull The current is the same in each devicebull The equivalent resistance of the circuit is the sum

of the individual resistances R=R1+R2

Series Circuits

1 2 1 2 3+ SV V V R R R R= = + + +

1 2 1 2 1 2( + ) + + I R R IR IR V V= = =

total totalV IR=

bull The current is the same in each devicebull The equivalent resistance of the circuit is the sum

of the individual resistances R=R1+R2

Series Circuits

1 2 1 2 3+ SV V V R R R R= = + + +

To find the current use the total voltage and equivalent resistance

S

VIR

=

Circuits ProblemBulbs in Series vs ParallelA circuit contains a 48-V battery and two 240Ω light bulbs

In which circuit does each bulb burn brighterRULE THE MORE POWER DISSIPATED IN A BULB THE

BRIGHTER IT IS

P IVFind the power in each bulb when in series and in parallel

=

Circuits ProblemBulbs in Series vs ParallelIf a bulb burns out - what happens to the other bulb in

each circuit Does it go out Is it brighter Dimmer Or

In the series circuit the burned out bulb will short the circuit and the other bulb will go out

In the parallel circuit the other bulb will have the same brightness

In SeriesThe Voltage is divided in series each bulb gets half V = 24V

2 2(24 ) 24240

V V VP IV V WR R

= = = = =Ω

Circuits ProblemBulbs in Series vs Parallel

In ParallelVoltage is the same in each bulb 48V

2 2(48 ) 96240

V VP WR

= = =Ω

Parallel Bulbs Burn Brighter

Circuits Problem3 Bulbs in SeriesIf one more bulb is added to each circuit (3 bulbs total) how does the brightness of the bulbs change Or not

In the series circuit the bulbs DIM WHY

( )22 2singlebulb 3 1

9 9PVV VP

R R R= = = =

1 2 3V V V V= + +In series each of the three equal bulbs gets one third of the Voltage

(V3) that a single bulb would get

Note P=VI but I is due to the equivalent Resistance I = VRs =V3RSo the Current through each is 13 the current through a single bulb andP=VI=V3 x I3 = VI9 = P9 The bulbs burn 19 as bright

Circuits Problem3 Bulbs in ParallelIf one more bulb is added to each circuit (3 bulbs total) how does the brightness of the bulbs change Or not

In the parallel circuit the bulbs DO NOT DIM WHY

2

singlebulbVP PR

= =

In parallel each of the three equal bulbs gets the full voltage of the battery source

Is this getting something for nothing

NO Parallel circuits drain the battery faster

Parallel CircuitsbullAs the number of branches is increased the overall resistance of the circuit is DECREASED

bullOverall resistance is lowered with each added path between any two points of the circuit

bullThis means the overall resistance of the circuit is less than the resistance of any one of the branches (Weird)

bullAs overall resistance is lowered more current is drawn This is how you blow fuses

FusesIf the current drawn exceeds safe levels the fuse melts and the

circuit lsquobreaksrdquo ndash most house have switches not fuses

Limitations of DC Transmissionbull Large currents in wires produce heat and energy losses by

P = I2Rbull Large expensive conductors would be needed or else very

high voltage drops (and efficiency losses) would resultbull High loads of direct current could rarely be transmitted for

distances greater than one mile without introducing excessive voltage drops

bull Direct current can not easily be changed to higher or lower voltages Separate electrical lines are needed to distribute power to appliances that used different voltages for example lighting and electric motors

Advantages of AC Transmission

bull Alternating Current can be transformed to lsquosteprsquo the voltage up or down with transformers

bull Power is transmitted at great distances at HIGH voltages and LOW currents and then stepped down to low voltages for use in homes (240V) and industry (440V)

bull Convert AC to DC with a rectifier in appliances

AC is more efficient for Transmission amp Distribution of

electrical power than DC

In The FuturehellipLong Distance AC Power Transmission

may not be needed

War of Currents 1880rsquos

Thomas Edison American inventor and businessman pushed for the development of a DC power network

George WestinghouseAmerican entrepreneur and engineer backed financially the development of a practical AC power network

Nikola Tesla Serbian inventor physicist and electro-mechanical engineer was instrumental in developing AC networks

If V = 120V What is I USE P = IV=gt I = PV

Appliance _ Power Current (A)

Hair Dryer 1600 Watts

Electric Iron 1200 Watts

TV 100 Watts

Computer 45 Watts

If V = 120V What is I USE P = IV=gt I = PV

Appliance _ Power Current (A)

Hair Dryer 1600 Watts 133 A

Electric Iron 1200 Watts 10 A

TV 100 Watts 83 A

Computer 45 Watts 38 A

Electric Shock

bullElectric Shock occurs when current is produced in the body which is caused by an impressed voltage

bullVoltage is the CAUSE bullCurrent does the DAMAGE

What causes electric Shock in the human body Voltage or Current

Current (A) Effect

0001 Can be felt0005 Painful0010 Causes involuntary muscle spasms0015 Causes loss of muscle control0070 If through heart serious

If current lasts for 1 s - FATAL

Dry Skin Body Resistance 500000 ΩWet Skin Body Resistance 1000 Ω

Electric Shock

Electric BillCost to run for 1 hr

$05 per 1 kw-hr Cost = Power x Time x Rate

Appliance _ Power Cost______

Hair Dryer 1600 Watts

Electric Iron 1200 Watts

TV 100 Watts

Computer 45 Watts

Electric BillCost to run for 1 hr

$05 per 1 kw-hr Cost = Power x Time x Rate

Appliance _ Power Cost______

Hair Dryer 1600 Watts $008

Electric Iron 1200 Watts $006

TV 100 Watts $0005

Computer 45 Watts $0003

QUESTION

The voltage and power on a light bulb read ldquo120 V 60 Wrdquo How much current will flow

through the bulb USE P = I ΔV

I = PΔV = 60 W120 V = 12 Amp

QUESTION

The power and voltage on a light bulb read ldquo120 V 60 Wrdquo What is the resistance of the

filament (I = 5 A)Hint USE OHMS LAW V = IR

R = VI = 120 V 5 A = 240 Ω

QUESTION

The power rating for two light bulbs read30W and 60W Which bulb has the greatest

resistance at 120V 2 2 P V R R V P= rarr =

2(120 ) 30 480R V W= = Ω2(120 ) 60 240R V W= = Ω

Which burns brighter and why

Quick Quiz 278

For the two lightbulbs shown in this figure rank the current values at the points from greatest to least

Ia = Ib gt Ic = Id gt Ie = If

The 60 W bulb has the lowest resistance and therefore draws the most current

Which light burns the brightest

bull The voltage of each device is the full voltage of the EMF source (the battery)

bull The total current is divided between each path

Parallel Circuits

1 21 2 1 2

1 1( )P

V V VI I I VR R R R R

= + = + = + =

1 2 3

1 1 1 1 PR R R R

= + + +

Equivalent Resistance

bull The current is the same in each devicebull The equivalent resistance of the circuit is the sum

of the individual resistances R=R1+R2

Series Circuits

1 2 1 2 3+ SV V V R R R R= = + + +

1 2 1 2 1 2( + ) + + I R R IR IR V V= = =

total totalV IR=

bull The current is the same in each devicebull The equivalent resistance of the circuit is the sum

of the individual resistances R=R1+R2

Series Circuits

1 2 1 2 3+ SV V V R R R R= = + + +

To find the current use the total voltage and equivalent resistance

S

VIR

=

Circuits ProblemBulbs in Series vs ParallelA circuit contains a 48-V battery and two 240Ω light bulbs

In which circuit does each bulb burn brighterRULE THE MORE POWER DISSIPATED IN A BULB THE

BRIGHTER IT IS

P IVFind the power in each bulb when in series and in parallel

=

Circuits ProblemBulbs in Series vs ParallelIf a bulb burns out - what happens to the other bulb in

each circuit Does it go out Is it brighter Dimmer Or

In the series circuit the burned out bulb will short the circuit and the other bulb will go out

In the parallel circuit the other bulb will have the same brightness

In SeriesThe Voltage is divided in series each bulb gets half V = 24V

2 2(24 ) 24240

V V VP IV V WR R

= = = = =Ω

Circuits ProblemBulbs in Series vs Parallel

In ParallelVoltage is the same in each bulb 48V

2 2(48 ) 96240

V VP WR

= = =Ω

Parallel Bulbs Burn Brighter

Circuits Problem3 Bulbs in SeriesIf one more bulb is added to each circuit (3 bulbs total) how does the brightness of the bulbs change Or not

In the series circuit the bulbs DIM WHY

( )22 2singlebulb 3 1

9 9PVV VP

R R R= = = =

1 2 3V V V V= + +In series each of the three equal bulbs gets one third of the Voltage

(V3) that a single bulb would get

Note P=VI but I is due to the equivalent Resistance I = VRs =V3RSo the Current through each is 13 the current through a single bulb andP=VI=V3 x I3 = VI9 = P9 The bulbs burn 19 as bright

Circuits Problem3 Bulbs in ParallelIf one more bulb is added to each circuit (3 bulbs total) how does the brightness of the bulbs change Or not

In the parallel circuit the bulbs DO NOT DIM WHY

2

singlebulbVP PR

= =

In parallel each of the three equal bulbs gets the full voltage of the battery source

Is this getting something for nothing

NO Parallel circuits drain the battery faster

Parallel CircuitsbullAs the number of branches is increased the overall resistance of the circuit is DECREASED

bullOverall resistance is lowered with each added path between any two points of the circuit

bullThis means the overall resistance of the circuit is less than the resistance of any one of the branches (Weird)

bullAs overall resistance is lowered more current is drawn This is how you blow fuses

FusesIf the current drawn exceeds safe levels the fuse melts and the

circuit lsquobreaksrdquo ndash most house have switches not fuses

Limitations of DC Transmissionbull Large currents in wires produce heat and energy losses by

P = I2Rbull Large expensive conductors would be needed or else very

high voltage drops (and efficiency losses) would resultbull High loads of direct current could rarely be transmitted for

distances greater than one mile without introducing excessive voltage drops

bull Direct current can not easily be changed to higher or lower voltages Separate electrical lines are needed to distribute power to appliances that used different voltages for example lighting and electric motors

Advantages of AC Transmission

bull Alternating Current can be transformed to lsquosteprsquo the voltage up or down with transformers

bull Power is transmitted at great distances at HIGH voltages and LOW currents and then stepped down to low voltages for use in homes (240V) and industry (440V)

bull Convert AC to DC with a rectifier in appliances

AC is more efficient for Transmission amp Distribution of

electrical power than DC

In The FuturehellipLong Distance AC Power Transmission

may not be needed

War of Currents 1880rsquos

Thomas Edison American inventor and businessman pushed for the development of a DC power network

George WestinghouseAmerican entrepreneur and engineer backed financially the development of a practical AC power network

Nikola Tesla Serbian inventor physicist and electro-mechanical engineer was instrumental in developing AC networks

If V = 120V What is I USE P = IV=gt I = PV

Appliance _ Power Current (A)

Hair Dryer 1600 Watts 133 A

Electric Iron 1200 Watts 10 A

TV 100 Watts 83 A

Computer 45 Watts 38 A

Electric Shock

bullElectric Shock occurs when current is produced in the body which is caused by an impressed voltage

bullVoltage is the CAUSE bullCurrent does the DAMAGE

What causes electric Shock in the human body Voltage or Current

Current (A) Effect

0001 Can be felt0005 Painful0010 Causes involuntary muscle spasms0015 Causes loss of muscle control0070 If through heart serious

If current lasts for 1 s - FATAL

Dry Skin Body Resistance 500000 ΩWet Skin Body Resistance 1000 Ω

Electric Shock

Electric BillCost to run for 1 hr

$05 per 1 kw-hr Cost = Power x Time x Rate

Appliance _ Power Cost______

Hair Dryer 1600 Watts

Electric Iron 1200 Watts

TV 100 Watts

Computer 45 Watts

Electric BillCost to run for 1 hr

$05 per 1 kw-hr Cost = Power x Time x Rate

Appliance _ Power Cost______

Hair Dryer 1600 Watts $008

Electric Iron 1200 Watts $006

TV 100 Watts $0005

Computer 45 Watts $0003

QUESTION

The voltage and power on a light bulb read ldquo120 V 60 Wrdquo How much current will flow

through the bulb USE P = I ΔV

I = PΔV = 60 W120 V = 12 Amp

QUESTION

The power and voltage on a light bulb read ldquo120 V 60 Wrdquo What is the resistance of the

filament (I = 5 A)Hint USE OHMS LAW V = IR

R = VI = 120 V 5 A = 240 Ω

QUESTION

The power rating for two light bulbs read30W and 60W Which bulb has the greatest

resistance at 120V 2 2 P V R R V P= rarr =

2(120 ) 30 480R V W= = Ω2(120 ) 60 240R V W= = Ω

Which burns brighter and why

Quick Quiz 278

For the two lightbulbs shown in this figure rank the current values at the points from greatest to least

Ia = Ib gt Ic = Id gt Ie = If

The 60 W bulb has the lowest resistance and therefore draws the most current

Which light burns the brightest

bull The voltage of each device is the full voltage of the EMF source (the battery)

bull The total current is divided between each path

Parallel Circuits

1 21 2 1 2

1 1( )P

V V VI I I VR R R R R

= + = + = + =

1 2 3

1 1 1 1 PR R R R

= + + +

Equivalent Resistance

bull The current is the same in each devicebull The equivalent resistance of the circuit is the sum

of the individual resistances R=R1+R2

Series Circuits

1 2 1 2 3+ SV V V R R R R= = + + +

1 2 1 2 1 2( + ) + + I R R IR IR V V= = =

total totalV IR=

bull The current is the same in each devicebull The equivalent resistance of the circuit is the sum

of the individual resistances R=R1+R2

Series Circuits

1 2 1 2 3+ SV V V R R R R= = + + +

To find the current use the total voltage and equivalent resistance

S

VIR

=

Circuits ProblemBulbs in Series vs ParallelA circuit contains a 48-V battery and two 240Ω light bulbs

In which circuit does each bulb burn brighterRULE THE MORE POWER DISSIPATED IN A BULB THE

BRIGHTER IT IS

P IVFind the power in each bulb when in series and in parallel

=

Circuits ProblemBulbs in Series vs ParallelIf a bulb burns out - what happens to the other bulb in

each circuit Does it go out Is it brighter Dimmer Or

In the series circuit the burned out bulb will short the circuit and the other bulb will go out

In the parallel circuit the other bulb will have the same brightness

In SeriesThe Voltage is divided in series each bulb gets half V = 24V

2 2(24 ) 24240

V V VP IV V WR R

= = = = =Ω

Circuits ProblemBulbs in Series vs Parallel

In ParallelVoltage is the same in each bulb 48V

2 2(48 ) 96240

V VP WR

= = =Ω

Parallel Bulbs Burn Brighter

Circuits Problem3 Bulbs in SeriesIf one more bulb is added to each circuit (3 bulbs total) how does the brightness of the bulbs change Or not

In the series circuit the bulbs DIM WHY

( )22 2singlebulb 3 1

9 9PVV VP

R R R= = = =

1 2 3V V V V= + +In series each of the three equal bulbs gets one third of the Voltage

(V3) that a single bulb would get

Note P=VI but I is due to the equivalent Resistance I = VRs =V3RSo the Current through each is 13 the current through a single bulb andP=VI=V3 x I3 = VI9 = P9 The bulbs burn 19 as bright

Circuits Problem3 Bulbs in ParallelIf one more bulb is added to each circuit (3 bulbs total) how does the brightness of the bulbs change Or not

In the parallel circuit the bulbs DO NOT DIM WHY

2

singlebulbVP PR

= =

In parallel each of the three equal bulbs gets the full voltage of the battery source

Is this getting something for nothing

NO Parallel circuits drain the battery faster

Parallel CircuitsbullAs the number of branches is increased the overall resistance of the circuit is DECREASED

bullOverall resistance is lowered with each added path between any two points of the circuit

bullThis means the overall resistance of the circuit is less than the resistance of any one of the branches (Weird)

bullAs overall resistance is lowered more current is drawn This is how you blow fuses

FusesIf the current drawn exceeds safe levels the fuse melts and the

circuit lsquobreaksrdquo ndash most house have switches not fuses

Limitations of DC Transmissionbull Large currents in wires produce heat and energy losses by

P = I2Rbull Large expensive conductors would be needed or else very

high voltage drops (and efficiency losses) would resultbull High loads of direct current could rarely be transmitted for

distances greater than one mile without introducing excessive voltage drops

bull Direct current can not easily be changed to higher or lower voltages Separate electrical lines are needed to distribute power to appliances that used different voltages for example lighting and electric motors

Advantages of AC Transmission

bull Alternating Current can be transformed to lsquosteprsquo the voltage up or down with transformers

bull Power is transmitted at great distances at HIGH voltages and LOW currents and then stepped down to low voltages for use in homes (240V) and industry (440V)

bull Convert AC to DC with a rectifier in appliances

AC is more efficient for Transmission amp Distribution of

electrical power than DC

In The FuturehellipLong Distance AC Power Transmission

may not be needed

War of Currents 1880rsquos

Thomas Edison American inventor and businessman pushed for the development of a DC power network

George WestinghouseAmerican entrepreneur and engineer backed financially the development of a practical AC power network

Nikola Tesla Serbian inventor physicist and electro-mechanical engineer was instrumental in developing AC networks

Electric Shock

bullElectric Shock occurs when current is produced in the body which is caused by an impressed voltage

bullVoltage is the CAUSE bullCurrent does the DAMAGE

What causes electric Shock in the human body Voltage or Current

Current (A) Effect

0001 Can be felt0005 Painful0010 Causes involuntary muscle spasms0015 Causes loss of muscle control0070 If through heart serious

If current lasts for 1 s - FATAL

Dry Skin Body Resistance 500000 ΩWet Skin Body Resistance 1000 Ω

Electric Shock

Electric BillCost to run for 1 hr

$05 per 1 kw-hr Cost = Power x Time x Rate

Appliance _ Power Cost______

Hair Dryer 1600 Watts

Electric Iron 1200 Watts

TV 100 Watts

Computer 45 Watts

Electric BillCost to run for 1 hr

$05 per 1 kw-hr Cost = Power x Time x Rate

Appliance _ Power Cost______

Hair Dryer 1600 Watts $008

Electric Iron 1200 Watts $006

TV 100 Watts $0005

Computer 45 Watts $0003

QUESTION

The voltage and power on a light bulb read ldquo120 V 60 Wrdquo How much current will flow

through the bulb USE P = I ΔV

I = PΔV = 60 W120 V = 12 Amp

QUESTION

The power and voltage on a light bulb read ldquo120 V 60 Wrdquo What is the resistance of the

filament (I = 5 A)Hint USE OHMS LAW V = IR

R = VI = 120 V 5 A = 240 Ω

QUESTION

The power rating for two light bulbs read30W and 60W Which bulb has the greatest

resistance at 120V 2 2 P V R R V P= rarr =

2(120 ) 30 480R V W= = Ω2(120 ) 60 240R V W= = Ω

Which burns brighter and why

Quick Quiz 278

For the two lightbulbs shown in this figure rank the current values at the points from greatest to least

Ia = Ib gt Ic = Id gt Ie = If

The 60 W bulb has the lowest resistance and therefore draws the most current

Which light burns the brightest

bull The voltage of each device is the full voltage of the EMF source (the battery)

bull The total current is divided between each path

Parallel Circuits

1 21 2 1 2

1 1( )P

V V VI I I VR R R R R

= + = + = + =

1 2 3

1 1 1 1 PR R R R

= + + +

Equivalent Resistance

bull The current is the same in each devicebull The equivalent resistance of the circuit is the sum

of the individual resistances R=R1+R2

Series Circuits

1 2 1 2 3+ SV V V R R R R= = + + +

1 2 1 2 1 2( + ) + + I R R IR IR V V= = =

total totalV IR=

bull The current is the same in each devicebull The equivalent resistance of the circuit is the sum

of the individual resistances R=R1+R2

Series Circuits

1 2 1 2 3+ SV V V R R R R= = + + +

To find the current use the total voltage and equivalent resistance

S

VIR

=

Circuits ProblemBulbs in Series vs ParallelA circuit contains a 48-V battery and two 240Ω light bulbs

In which circuit does each bulb burn brighterRULE THE MORE POWER DISSIPATED IN A BULB THE

BRIGHTER IT IS

P IVFind the power in each bulb when in series and in parallel

=

Circuits ProblemBulbs in Series vs ParallelIf a bulb burns out - what happens to the other bulb in

each circuit Does it go out Is it brighter Dimmer Or

In the series circuit the burned out bulb will short the circuit and the other bulb will go out

In the parallel circuit the other bulb will have the same brightness

In SeriesThe Voltage is divided in series each bulb gets half V = 24V

2 2(24 ) 24240

V V VP IV V WR R

= = = = =Ω

Circuits ProblemBulbs in Series vs Parallel

In ParallelVoltage is the same in each bulb 48V

2 2(48 ) 96240

V VP WR

= = =Ω

Parallel Bulbs Burn Brighter

Circuits Problem3 Bulbs in SeriesIf one more bulb is added to each circuit (3 bulbs total) how does the brightness of the bulbs change Or not

In the series circuit the bulbs DIM WHY

( )22 2singlebulb 3 1

9 9PVV VP

R R R= = = =

1 2 3V V V V= + +In series each of the three equal bulbs gets one third of the Voltage

(V3) that a single bulb would get

Note P=VI but I is due to the equivalent Resistance I = VRs =V3RSo the Current through each is 13 the current through a single bulb andP=VI=V3 x I3 = VI9 = P9 The bulbs burn 19 as bright

Circuits Problem3 Bulbs in ParallelIf one more bulb is added to each circuit (3 bulbs total) how does the brightness of the bulbs change Or not

In the parallel circuit the bulbs DO NOT DIM WHY

2

singlebulbVP PR

= =

In parallel each of the three equal bulbs gets the full voltage of the battery source

Is this getting something for nothing

NO Parallel circuits drain the battery faster

Parallel CircuitsbullAs the number of branches is increased the overall resistance of the circuit is DECREASED

bullOverall resistance is lowered with each added path between any two points of the circuit

bullThis means the overall resistance of the circuit is less than the resistance of any one of the branches (Weird)

bullAs overall resistance is lowered more current is drawn This is how you blow fuses

FusesIf the current drawn exceeds safe levels the fuse melts and the

circuit lsquobreaksrdquo ndash most house have switches not fuses

Limitations of DC Transmissionbull Large currents in wires produce heat and energy losses by

P = I2Rbull Large expensive conductors would be needed or else very

high voltage drops (and efficiency losses) would resultbull High loads of direct current could rarely be transmitted for

distances greater than one mile without introducing excessive voltage drops

bull Direct current can not easily be changed to higher or lower voltages Separate electrical lines are needed to distribute power to appliances that used different voltages for example lighting and electric motors

Advantages of AC Transmission

bull Alternating Current can be transformed to lsquosteprsquo the voltage up or down with transformers

bull Power is transmitted at great distances at HIGH voltages and LOW currents and then stepped down to low voltages for use in homes (240V) and industry (440V)

bull Convert AC to DC with a rectifier in appliances

AC is more efficient for Transmission amp Distribution of

electrical power than DC

In The FuturehellipLong Distance AC Power Transmission

may not be needed

War of Currents 1880rsquos

Thomas Edison American inventor and businessman pushed for the development of a DC power network

George WestinghouseAmerican entrepreneur and engineer backed financially the development of a practical AC power network

Nikola Tesla Serbian inventor physicist and electro-mechanical engineer was instrumental in developing AC networks

Current (A) Effect

0001 Can be felt0005 Painful0010 Causes involuntary muscle spasms0015 Causes loss of muscle control0070 If through heart serious

If current lasts for 1 s - FATAL

Dry Skin Body Resistance 500000 ΩWet Skin Body Resistance 1000 Ω

Electric Shock

Electric BillCost to run for 1 hr

$05 per 1 kw-hr Cost = Power x Time x Rate

Appliance _ Power Cost______

Hair Dryer 1600 Watts

Electric Iron 1200 Watts

TV 100 Watts

Computer 45 Watts

Electric BillCost to run for 1 hr

$05 per 1 kw-hr Cost = Power x Time x Rate

Appliance _ Power Cost______

Hair Dryer 1600 Watts $008

Electric Iron 1200 Watts $006

TV 100 Watts $0005

Computer 45 Watts $0003

QUESTION

The voltage and power on a light bulb read ldquo120 V 60 Wrdquo How much current will flow

through the bulb USE P = I ΔV

I = PΔV = 60 W120 V = 12 Amp

QUESTION

The power and voltage on a light bulb read ldquo120 V 60 Wrdquo What is the resistance of the

filament (I = 5 A)Hint USE OHMS LAW V = IR

R = VI = 120 V 5 A = 240 Ω

QUESTION

The power rating for two light bulbs read30W and 60W Which bulb has the greatest

resistance at 120V 2 2 P V R R V P= rarr =

2(120 ) 30 480R V W= = Ω2(120 ) 60 240R V W= = Ω

Which burns brighter and why

Quick Quiz 278

For the two lightbulbs shown in this figure rank the current values at the points from greatest to least

Ia = Ib gt Ic = Id gt Ie = If

The 60 W bulb has the lowest resistance and therefore draws the most current

Which light burns the brightest

bull The voltage of each device is the full voltage of the EMF source (the battery)

bull The total current is divided between each path

Parallel Circuits

1 21 2 1 2

1 1( )P

V V VI I I VR R R R R

= + = + = + =

1 2 3

1 1 1 1 PR R R R

= + + +

Equivalent Resistance

bull The current is the same in each devicebull The equivalent resistance of the circuit is the sum

of the individual resistances R=R1+R2

Series Circuits

1 2 1 2 3+ SV V V R R R R= = + + +

1 2 1 2 1 2( + ) + + I R R IR IR V V= = =

total totalV IR=

bull The current is the same in each devicebull The equivalent resistance of the circuit is the sum

of the individual resistances R=R1+R2

Series Circuits

1 2 1 2 3+ SV V V R R R R= = + + +

To find the current use the total voltage and equivalent resistance

S

VIR

=

Circuits ProblemBulbs in Series vs ParallelA circuit contains a 48-V battery and two 240Ω light bulbs

In which circuit does each bulb burn brighterRULE THE MORE POWER DISSIPATED IN A BULB THE

BRIGHTER IT IS

P IVFind the power in each bulb when in series and in parallel

=

Circuits ProblemBulbs in Series vs ParallelIf a bulb burns out - what happens to the other bulb in

each circuit Does it go out Is it brighter Dimmer Or

In the series circuit the burned out bulb will short the circuit and the other bulb will go out

In the parallel circuit the other bulb will have the same brightness

In SeriesThe Voltage is divided in series each bulb gets half V = 24V

2 2(24 ) 24240

V V VP IV V WR R

= = = = =Ω

Circuits ProblemBulbs in Series vs Parallel

In ParallelVoltage is the same in each bulb 48V

2 2(48 ) 96240

V VP WR

= = =Ω

Parallel Bulbs Burn Brighter

Circuits Problem3 Bulbs in SeriesIf one more bulb is added to each circuit (3 bulbs total) how does the brightness of the bulbs change Or not

In the series circuit the bulbs DIM WHY

( )22 2singlebulb 3 1

9 9PVV VP

R R R= = = =

1 2 3V V V V= + +In series each of the three equal bulbs gets one third of the Voltage

(V3) that a single bulb would get

Note P=VI but I is due to the equivalent Resistance I = VRs =V3RSo the Current through each is 13 the current through a single bulb andP=VI=V3 x I3 = VI9 = P9 The bulbs burn 19 as bright

Circuits Problem3 Bulbs in ParallelIf one more bulb is added to each circuit (3 bulbs total) how does the brightness of the bulbs change Or not

In the parallel circuit the bulbs DO NOT DIM WHY

2

singlebulbVP PR

= =

In parallel each of the three equal bulbs gets the full voltage of the battery source

Is this getting something for nothing

NO Parallel circuits drain the battery faster

Parallel CircuitsbullAs the number of branches is increased the overall resistance of the circuit is DECREASED

bullOverall resistance is lowered with each added path between any two points of the circuit

bullThis means the overall resistance of the circuit is less than the resistance of any one of the branches (Weird)

bullAs overall resistance is lowered more current is drawn This is how you blow fuses

FusesIf the current drawn exceeds safe levels the fuse melts and the

circuit lsquobreaksrdquo ndash most house have switches not fuses

Limitations of DC Transmissionbull Large currents in wires produce heat and energy losses by

P = I2Rbull Large expensive conductors would be needed or else very

high voltage drops (and efficiency losses) would resultbull High loads of direct current could rarely be transmitted for

distances greater than one mile without introducing excessive voltage drops

bull Direct current can not easily be changed to higher or lower voltages Separate electrical lines are needed to distribute power to appliances that used different voltages for example lighting and electric motors

Advantages of AC Transmission

bull Alternating Current can be transformed to lsquosteprsquo the voltage up or down with transformers

bull Power is transmitted at great distances at HIGH voltages and LOW currents and then stepped down to low voltages for use in homes (240V) and industry (440V)

bull Convert AC to DC with a rectifier in appliances

AC is more efficient for Transmission amp Distribution of

electrical power than DC

In The FuturehellipLong Distance AC Power Transmission

may not be needed

War of Currents 1880rsquos

Thomas Edison American inventor and businessman pushed for the development of a DC power network

George WestinghouseAmerican entrepreneur and engineer backed financially the development of a practical AC power network

Nikola Tesla Serbian inventor physicist and electro-mechanical engineer was instrumental in developing AC networks

Electric BillCost to run for 1 hr

$05 per 1 kw-hr Cost = Power x Time x Rate

Appliance _ Power Cost______

Hair Dryer 1600 Watts

Electric Iron 1200 Watts

TV 100 Watts

Computer 45 Watts

Electric BillCost to run for 1 hr

$05 per 1 kw-hr Cost = Power x Time x Rate

Appliance _ Power Cost______

Hair Dryer 1600 Watts $008

Electric Iron 1200 Watts $006

TV 100 Watts $0005

Computer 45 Watts $0003

QUESTION

The voltage and power on a light bulb read ldquo120 V 60 Wrdquo How much current will flow

through the bulb USE P = I ΔV

I = PΔV = 60 W120 V = 12 Amp

QUESTION

The power and voltage on a light bulb read ldquo120 V 60 Wrdquo What is the resistance of the

filament (I = 5 A)Hint USE OHMS LAW V = IR

R = VI = 120 V 5 A = 240 Ω

QUESTION

The power rating for two light bulbs read30W and 60W Which bulb has the greatest

resistance at 120V 2 2 P V R R V P= rarr =

2(120 ) 30 480R V W= = Ω2(120 ) 60 240R V W= = Ω

Which burns brighter and why

Quick Quiz 278

For the two lightbulbs shown in this figure rank the current values at the points from greatest to least

Ia = Ib gt Ic = Id gt Ie = If

The 60 W bulb has the lowest resistance and therefore draws the most current

Which light burns the brightest

bull The voltage of each device is the full voltage of the EMF source (the battery)

bull The total current is divided between each path

Parallel Circuits

1 21 2 1 2

1 1( )P

V V VI I I VR R R R R

= + = + = + =

1 2 3

1 1 1 1 PR R R R

= + + +

Equivalent Resistance

bull The current is the same in each devicebull The equivalent resistance of the circuit is the sum

of the individual resistances R=R1+R2

Series Circuits

1 2 1 2 3+ SV V V R R R R= = + + +

1 2 1 2 1 2( + ) + + I R R IR IR V V= = =

total totalV IR=

bull The current is the same in each devicebull The equivalent resistance of the circuit is the sum

of the individual resistances R=R1+R2

Series Circuits

1 2 1 2 3+ SV V V R R R R= = + + +

To find the current use the total voltage and equivalent resistance

S

VIR

=

Circuits ProblemBulbs in Series vs ParallelA circuit contains a 48-V battery and two 240Ω light bulbs

In which circuit does each bulb burn brighterRULE THE MORE POWER DISSIPATED IN A BULB THE

BRIGHTER IT IS

P IVFind the power in each bulb when in series and in parallel

=

Circuits ProblemBulbs in Series vs ParallelIf a bulb burns out - what happens to the other bulb in

each circuit Does it go out Is it brighter Dimmer Or

In the series circuit the burned out bulb will short the circuit and the other bulb will go out

In the parallel circuit the other bulb will have the same brightness

In SeriesThe Voltage is divided in series each bulb gets half V = 24V

2 2(24 ) 24240

V V VP IV V WR R

= = = = =Ω

Circuits ProblemBulbs in Series vs Parallel

In ParallelVoltage is the same in each bulb 48V

2 2(48 ) 96240

V VP WR

= = =Ω

Parallel Bulbs Burn Brighter

Circuits Problem3 Bulbs in SeriesIf one more bulb is added to each circuit (3 bulbs total) how does the brightness of the bulbs change Or not

In the series circuit the bulbs DIM WHY

( )22 2singlebulb 3 1

9 9PVV VP

R R R= = = =

1 2 3V V V V= + +In series each of the three equal bulbs gets one third of the Voltage

(V3) that a single bulb would get

Note P=VI but I is due to the equivalent Resistance I = VRs =V3RSo the Current through each is 13 the current through a single bulb andP=VI=V3 x I3 = VI9 = P9 The bulbs burn 19 as bright

Circuits Problem3 Bulbs in ParallelIf one more bulb is added to each circuit (3 bulbs total) how does the brightness of the bulbs change Or not

In the parallel circuit the bulbs DO NOT DIM WHY

2

singlebulbVP PR

= =

In parallel each of the three equal bulbs gets the full voltage of the battery source

Is this getting something for nothing

NO Parallel circuits drain the battery faster

Parallel CircuitsbullAs the number of branches is increased the overall resistance of the circuit is DECREASED

bullOverall resistance is lowered with each added path between any two points of the circuit

bullThis means the overall resistance of the circuit is less than the resistance of any one of the branches (Weird)

bullAs overall resistance is lowered more current is drawn This is how you blow fuses

FusesIf the current drawn exceeds safe levels the fuse melts and the

circuit lsquobreaksrdquo ndash most house have switches not fuses

Limitations of DC Transmissionbull Large currents in wires produce heat and energy losses by

P = I2Rbull Large expensive conductors would be needed or else very

high voltage drops (and efficiency losses) would resultbull High loads of direct current could rarely be transmitted for

distances greater than one mile without introducing excessive voltage drops

bull Direct current can not easily be changed to higher or lower voltages Separate electrical lines are needed to distribute power to appliances that used different voltages for example lighting and electric motors

Advantages of AC Transmission

bull Alternating Current can be transformed to lsquosteprsquo the voltage up or down with transformers

bull Power is transmitted at great distances at HIGH voltages and LOW currents and then stepped down to low voltages for use in homes (240V) and industry (440V)

bull Convert AC to DC with a rectifier in appliances

AC is more efficient for Transmission amp Distribution of

electrical power than DC

In The FuturehellipLong Distance AC Power Transmission

may not be needed

War of Currents 1880rsquos

Thomas Edison American inventor and businessman pushed for the development of a DC power network

George WestinghouseAmerican entrepreneur and engineer backed financially the development of a practical AC power network

Nikola Tesla Serbian inventor physicist and electro-mechanical engineer was instrumental in developing AC networks

Electric BillCost to run for 1 hr

$05 per 1 kw-hr Cost = Power x Time x Rate

Appliance _ Power Cost______

Hair Dryer 1600 Watts $008

Electric Iron 1200 Watts $006

TV 100 Watts $0005

Computer 45 Watts $0003

QUESTION

The voltage and power on a light bulb read ldquo120 V 60 Wrdquo How much current will flow

through the bulb USE P = I ΔV

I = PΔV = 60 W120 V = 12 Amp

QUESTION

The power and voltage on a light bulb read ldquo120 V 60 Wrdquo What is the resistance of the

filament (I = 5 A)Hint USE OHMS LAW V = IR

R = VI = 120 V 5 A = 240 Ω

QUESTION

The power rating for two light bulbs read30W and 60W Which bulb has the greatest

resistance at 120V 2 2 P V R R V P= rarr =

2(120 ) 30 480R V W= = Ω2(120 ) 60 240R V W= = Ω

Which burns brighter and why

Quick Quiz 278

For the two lightbulbs shown in this figure rank the current values at the points from greatest to least

Ia = Ib gt Ic = Id gt Ie = If

The 60 W bulb has the lowest resistance and therefore draws the most current

Which light burns the brightest

bull The voltage of each device is the full voltage of the EMF source (the battery)

bull The total current is divided between each path

Parallel Circuits

1 21 2 1 2

1 1( )P

V V VI I I VR R R R R

= + = + = + =

1 2 3

1 1 1 1 PR R R R

= + + +

Equivalent Resistance

bull The current is the same in each devicebull The equivalent resistance of the circuit is the sum

of the individual resistances R=R1+R2

Series Circuits

1 2 1 2 3+ SV V V R R R R= = + + +

1 2 1 2 1 2( + ) + + I R R IR IR V V= = =

total totalV IR=

bull The current is the same in each devicebull The equivalent resistance of the circuit is the sum

of the individual resistances R=R1+R2

Series Circuits

1 2 1 2 3+ SV V V R R R R= = + + +

To find the current use the total voltage and equivalent resistance

S

VIR

=

Circuits ProblemBulbs in Series vs ParallelA circuit contains a 48-V battery and two 240Ω light bulbs

In which circuit does each bulb burn brighterRULE THE MORE POWER DISSIPATED IN A BULB THE

BRIGHTER IT IS

P IVFind the power in each bulb when in series and in parallel

=

Circuits ProblemBulbs in Series vs ParallelIf a bulb burns out - what happens to the other bulb in

each circuit Does it go out Is it brighter Dimmer Or

In the series circuit the burned out bulb will short the circuit and the other bulb will go out

In the parallel circuit the other bulb will have the same brightness

In SeriesThe Voltage is divided in series each bulb gets half V = 24V

2 2(24 ) 24240

V V VP IV V WR R

= = = = =Ω

Circuits ProblemBulbs in Series vs Parallel

In ParallelVoltage is the same in each bulb 48V

2 2(48 ) 96240

V VP WR

= = =Ω

Parallel Bulbs Burn Brighter

Circuits Problem3 Bulbs in SeriesIf one more bulb is added to each circuit (3 bulbs total) how does the brightness of the bulbs change Or not

In the series circuit the bulbs DIM WHY

( )22 2singlebulb 3 1

9 9PVV VP

R R R= = = =

1 2 3V V V V= + +In series each of the three equal bulbs gets one third of the Voltage

(V3) that a single bulb would get

Note P=VI but I is due to the equivalent Resistance I = VRs =V3RSo the Current through each is 13 the current through a single bulb andP=VI=V3 x I3 = VI9 = P9 The bulbs burn 19 as bright

Circuits Problem3 Bulbs in ParallelIf one more bulb is added to each circuit (3 bulbs total) how does the brightness of the bulbs change Or not

In the parallel circuit the bulbs DO NOT DIM WHY

2

singlebulbVP PR

= =

In parallel each of the three equal bulbs gets the full voltage of the battery source

Is this getting something for nothing

NO Parallel circuits drain the battery faster

Parallel CircuitsbullAs the number of branches is increased the overall resistance of the circuit is DECREASED

bullOverall resistance is lowered with each added path between any two points of the circuit

bullThis means the overall resistance of the circuit is less than the resistance of any one of the branches (Weird)

bullAs overall resistance is lowered more current is drawn This is how you blow fuses

FusesIf the current drawn exceeds safe levels the fuse melts and the

circuit lsquobreaksrdquo ndash most house have switches not fuses

Limitations of DC Transmissionbull Large currents in wires produce heat and energy losses by

P = I2Rbull Large expensive conductors would be needed or else very

high voltage drops (and efficiency losses) would resultbull High loads of direct current could rarely be transmitted for

distances greater than one mile without introducing excessive voltage drops

bull Direct current can not easily be changed to higher or lower voltages Separate electrical lines are needed to distribute power to appliances that used different voltages for example lighting and electric motors

Advantages of AC Transmission

bull Alternating Current can be transformed to lsquosteprsquo the voltage up or down with transformers

bull Power is transmitted at great distances at HIGH voltages and LOW currents and then stepped down to low voltages for use in homes (240V) and industry (440V)

bull Convert AC to DC with a rectifier in appliances

AC is more efficient for Transmission amp Distribution of

electrical power than DC

In The FuturehellipLong Distance AC Power Transmission

may not be needed

War of Currents 1880rsquos

Thomas Edison American inventor and businessman pushed for the development of a DC power network

George WestinghouseAmerican entrepreneur and engineer backed financially the development of a practical AC power network

Nikola Tesla Serbian inventor physicist and electro-mechanical engineer was instrumental in developing AC networks

QUESTION

The voltage and power on a light bulb read ldquo120 V 60 Wrdquo How much current will flow

through the bulb USE P = I ΔV

I = PΔV = 60 W120 V = 12 Amp

QUESTION

The power and voltage on a light bulb read ldquo120 V 60 Wrdquo What is the resistance of the

filament (I = 5 A)Hint USE OHMS LAW V = IR

R = VI = 120 V 5 A = 240 Ω

QUESTION

The power rating for two light bulbs read30W and 60W Which bulb has the greatest

resistance at 120V 2 2 P V R R V P= rarr =

2(120 ) 30 480R V W= = Ω2(120 ) 60 240R V W= = Ω

Which burns brighter and why

Quick Quiz 278

For the two lightbulbs shown in this figure rank the current values at the points from greatest to least

Ia = Ib gt Ic = Id gt Ie = If

The 60 W bulb has the lowest resistance and therefore draws the most current

Which light burns the brightest

bull The voltage of each device is the full voltage of the EMF source (the battery)

bull The total current is divided between each path

Parallel Circuits

1 21 2 1 2

1 1( )P

V V VI I I VR R R R R

= + = + = + =

1 2 3

1 1 1 1 PR R R R

= + + +

Equivalent Resistance

bull The current is the same in each devicebull The equivalent resistance of the circuit is the sum

of the individual resistances R=R1+R2

Series Circuits

1 2 1 2 3+ SV V V R R R R= = + + +

1 2 1 2 1 2( + ) + + I R R IR IR V V= = =

total totalV IR=

bull The current is the same in each devicebull The equivalent resistance of the circuit is the sum

of the individual resistances R=R1+R2

Series Circuits

1 2 1 2 3+ SV V V R R R R= = + + +

To find the current use the total voltage and equivalent resistance

S

VIR

=

Circuits ProblemBulbs in Series vs ParallelA circuit contains a 48-V battery and two 240Ω light bulbs

In which circuit does each bulb burn brighterRULE THE MORE POWER DISSIPATED IN A BULB THE

BRIGHTER IT IS

P IVFind the power in each bulb when in series and in parallel

=

Circuits ProblemBulbs in Series vs ParallelIf a bulb burns out - what happens to the other bulb in

each circuit Does it go out Is it brighter Dimmer Or

In the series circuit the burned out bulb will short the circuit and the other bulb will go out

In the parallel circuit the other bulb will have the same brightness

In SeriesThe Voltage is divided in series each bulb gets half V = 24V

2 2(24 ) 24240

V V VP IV V WR R

= = = = =Ω

Circuits ProblemBulbs in Series vs Parallel

In ParallelVoltage is the same in each bulb 48V

2 2(48 ) 96240

V VP WR

= = =Ω

Parallel Bulbs Burn Brighter

Circuits Problem3 Bulbs in SeriesIf one more bulb is added to each circuit (3 bulbs total) how does the brightness of the bulbs change Or not

In the series circuit the bulbs DIM WHY

( )22 2singlebulb 3 1

9 9PVV VP

R R R= = = =

1 2 3V V V V= + +In series each of the three equal bulbs gets one third of the Voltage

(V3) that a single bulb would get

Note P=VI but I is due to the equivalent Resistance I = VRs =V3RSo the Current through each is 13 the current through a single bulb andP=VI=V3 x I3 = VI9 = P9 The bulbs burn 19 as bright

Circuits Problem3 Bulbs in ParallelIf one more bulb is added to each circuit (3 bulbs total) how does the brightness of the bulbs change Or not

In the parallel circuit the bulbs DO NOT DIM WHY

2

singlebulbVP PR

= =

In parallel each of the three equal bulbs gets the full voltage of the battery source

Is this getting something for nothing

NO Parallel circuits drain the battery faster

Parallel CircuitsbullAs the number of branches is increased the overall resistance of the circuit is DECREASED

bullOverall resistance is lowered with each added path between any two points of the circuit

bullThis means the overall resistance of the circuit is less than the resistance of any one of the branches (Weird)

bullAs overall resistance is lowered more current is drawn This is how you blow fuses

FusesIf the current drawn exceeds safe levels the fuse melts and the

circuit lsquobreaksrdquo ndash most house have switches not fuses

Limitations of DC Transmissionbull Large currents in wires produce heat and energy losses by

P = I2Rbull Large expensive conductors would be needed or else very

high voltage drops (and efficiency losses) would resultbull High loads of direct current could rarely be transmitted for

distances greater than one mile without introducing excessive voltage drops

bull Direct current can not easily be changed to higher or lower voltages Separate electrical lines are needed to distribute power to appliances that used different voltages for example lighting and electric motors

Advantages of AC Transmission

bull Alternating Current can be transformed to lsquosteprsquo the voltage up or down with transformers

bull Power is transmitted at great distances at HIGH voltages and LOW currents and then stepped down to low voltages for use in homes (240V) and industry (440V)

bull Convert AC to DC with a rectifier in appliances

AC is more efficient for Transmission amp Distribution of

electrical power than DC

In The FuturehellipLong Distance AC Power Transmission

may not be needed

War of Currents 1880rsquos

Thomas Edison American inventor and businessman pushed for the development of a DC power network

George WestinghouseAmerican entrepreneur and engineer backed financially the development of a practical AC power network

Nikola Tesla Serbian inventor physicist and electro-mechanical engineer was instrumental in developing AC networks

QUESTION

The power and voltage on a light bulb read ldquo120 V 60 Wrdquo What is the resistance of the

filament (I = 5 A)Hint USE OHMS LAW V = IR

R = VI = 120 V 5 A = 240 Ω

QUESTION

The power rating for two light bulbs read30W and 60W Which bulb has the greatest

resistance at 120V 2 2 P V R R V P= rarr =

2(120 ) 30 480R V W= = Ω2(120 ) 60 240R V W= = Ω

Which burns brighter and why

Quick Quiz 278

For the two lightbulbs shown in this figure rank the current values at the points from greatest to least

Ia = Ib gt Ic = Id gt Ie = If

The 60 W bulb has the lowest resistance and therefore draws the most current

Which light burns the brightest

bull The voltage of each device is the full voltage of the EMF source (the battery)

bull The total current is divided between each path

Parallel Circuits

1 21 2 1 2

1 1( )P

V V VI I I VR R R R R

= + = + = + =

1 2 3

1 1 1 1 PR R R R

= + + +

Equivalent Resistance

bull The current is the same in each devicebull The equivalent resistance of the circuit is the sum

of the individual resistances R=R1+R2

Series Circuits

1 2 1 2 3+ SV V V R R R R= = + + +

1 2 1 2 1 2( + ) + + I R R IR IR V V= = =

total totalV IR=

bull The current is the same in each devicebull The equivalent resistance of the circuit is the sum

of the individual resistances R=R1+R2

Series Circuits

1 2 1 2 3+ SV V V R R R R= = + + +

To find the current use the total voltage and equivalent resistance

S

VIR

=

Circuits ProblemBulbs in Series vs ParallelA circuit contains a 48-V battery and two 240Ω light bulbs

In which circuit does each bulb burn brighterRULE THE MORE POWER DISSIPATED IN A BULB THE

BRIGHTER IT IS

P IVFind the power in each bulb when in series and in parallel

=

Circuits ProblemBulbs in Series vs ParallelIf a bulb burns out - what happens to the other bulb in

each circuit Does it go out Is it brighter Dimmer Or

In the series circuit the burned out bulb will short the circuit and the other bulb will go out

In the parallel circuit the other bulb will have the same brightness

In SeriesThe Voltage is divided in series each bulb gets half V = 24V

2 2(24 ) 24240

V V VP IV V WR R

= = = = =Ω

Circuits ProblemBulbs in Series vs Parallel

In ParallelVoltage is the same in each bulb 48V

2 2(48 ) 96240

V VP WR

= = =Ω

Parallel Bulbs Burn Brighter

Circuits Problem3 Bulbs in SeriesIf one more bulb is added to each circuit (3 bulbs total) how does the brightness of the bulbs change Or not

In the series circuit the bulbs DIM WHY

( )22 2singlebulb 3 1

9 9PVV VP

R R R= = = =

1 2 3V V V V= + +In series each of the three equal bulbs gets one third of the Voltage

(V3) that a single bulb would get

Note P=VI but I is due to the equivalent Resistance I = VRs =V3RSo the Current through each is 13 the current through a single bulb andP=VI=V3 x I3 = VI9 = P9 The bulbs burn 19 as bright

Circuits Problem3 Bulbs in ParallelIf one more bulb is added to each circuit (3 bulbs total) how does the brightness of the bulbs change Or not

In the parallel circuit the bulbs DO NOT DIM WHY

2

singlebulbVP PR

= =

In parallel each of the three equal bulbs gets the full voltage of the battery source

Is this getting something for nothing

NO Parallel circuits drain the battery faster

Parallel CircuitsbullAs the number of branches is increased the overall resistance of the circuit is DECREASED

bullOverall resistance is lowered with each added path between any two points of the circuit

bullThis means the overall resistance of the circuit is less than the resistance of any one of the branches (Weird)

bullAs overall resistance is lowered more current is drawn This is how you blow fuses

FusesIf the current drawn exceeds safe levels the fuse melts and the

circuit lsquobreaksrdquo ndash most house have switches not fuses

Limitations of DC Transmissionbull Large currents in wires produce heat and energy losses by

P = I2Rbull Large expensive conductors would be needed or else very

high voltage drops (and efficiency losses) would resultbull High loads of direct current could rarely be transmitted for

distances greater than one mile without introducing excessive voltage drops

bull Direct current can not easily be changed to higher or lower voltages Separate electrical lines are needed to distribute power to appliances that used different voltages for example lighting and electric motors

Advantages of AC Transmission

bull Alternating Current can be transformed to lsquosteprsquo the voltage up or down with transformers

bull Power is transmitted at great distances at HIGH voltages and LOW currents and then stepped down to low voltages for use in homes (240V) and industry (440V)

bull Convert AC to DC with a rectifier in appliances

AC is more efficient for Transmission amp Distribution of

electrical power than DC

In The FuturehellipLong Distance AC Power Transmission

may not be needed

War of Currents 1880rsquos

Thomas Edison American inventor and businessman pushed for the development of a DC power network

George WestinghouseAmerican entrepreneur and engineer backed financially the development of a practical AC power network

Nikola Tesla Serbian inventor physicist and electro-mechanical engineer was instrumental in developing AC networks

QUESTION

The power rating for two light bulbs read30W and 60W Which bulb has the greatest

resistance at 120V 2 2 P V R R V P= rarr =

2(120 ) 30 480R V W= = Ω2(120 ) 60 240R V W= = Ω

Which burns brighter and why

Quick Quiz 278

For the two lightbulbs shown in this figure rank the current values at the points from greatest to least

Ia = Ib gt Ic = Id gt Ie = If

The 60 W bulb has the lowest resistance and therefore draws the most current

Which light burns the brightest

bull The voltage of each device is the full voltage of the EMF source (the battery)

bull The total current is divided between each path

Parallel Circuits

1 21 2 1 2

1 1( )P

V V VI I I VR R R R R

= + = + = + =

1 2 3

1 1 1 1 PR R R R

= + + +

Equivalent Resistance

bull The current is the same in each devicebull The equivalent resistance of the circuit is the sum

of the individual resistances R=R1+R2

Series Circuits

1 2 1 2 3+ SV V V R R R R= = + + +

1 2 1 2 1 2( + ) + + I R R IR IR V V= = =

total totalV IR=

bull The current is the same in each devicebull The equivalent resistance of the circuit is the sum

of the individual resistances R=R1+R2

Series Circuits

1 2 1 2 3+ SV V V R R R R= = + + +

To find the current use the total voltage and equivalent resistance

S

VIR

=

Circuits ProblemBulbs in Series vs ParallelA circuit contains a 48-V battery and two 240Ω light bulbs

In which circuit does each bulb burn brighterRULE THE MORE POWER DISSIPATED IN A BULB THE

BRIGHTER IT IS

P IVFind the power in each bulb when in series and in parallel

=

Circuits ProblemBulbs in Series vs ParallelIf a bulb burns out - what happens to the other bulb in

each circuit Does it go out Is it brighter Dimmer Or

In the series circuit the burned out bulb will short the circuit and the other bulb will go out

In the parallel circuit the other bulb will have the same brightness

In SeriesThe Voltage is divided in series each bulb gets half V = 24V

2 2(24 ) 24240

V V VP IV V WR R

= = = = =Ω

Circuits ProblemBulbs in Series vs Parallel

In ParallelVoltage is the same in each bulb 48V

2 2(48 ) 96240

V VP WR

= = =Ω

Parallel Bulbs Burn Brighter

Circuits Problem3 Bulbs in SeriesIf one more bulb is added to each circuit (3 bulbs total) how does the brightness of the bulbs change Or not

In the series circuit the bulbs DIM WHY

( )22 2singlebulb 3 1

9 9PVV VP

R R R= = = =

1 2 3V V V V= + +In series each of the three equal bulbs gets one third of the Voltage

(V3) that a single bulb would get

Note P=VI but I is due to the equivalent Resistance I = VRs =V3RSo the Current through each is 13 the current through a single bulb andP=VI=V3 x I3 = VI9 = P9 The bulbs burn 19 as bright

Circuits Problem3 Bulbs in ParallelIf one more bulb is added to each circuit (3 bulbs total) how does the brightness of the bulbs change Or not

In the parallel circuit the bulbs DO NOT DIM WHY

2

singlebulbVP PR

= =

In parallel each of the three equal bulbs gets the full voltage of the battery source

Is this getting something for nothing

NO Parallel circuits drain the battery faster

Parallel CircuitsbullAs the number of branches is increased the overall resistance of the circuit is DECREASED

bullOverall resistance is lowered with each added path between any two points of the circuit

bullThis means the overall resistance of the circuit is less than the resistance of any one of the branches (Weird)

bullAs overall resistance is lowered more current is drawn This is how you blow fuses

FusesIf the current drawn exceeds safe levels the fuse melts and the

circuit lsquobreaksrdquo ndash most house have switches not fuses

Limitations of DC Transmissionbull Large currents in wires produce heat and energy losses by

P = I2Rbull Large expensive conductors would be needed or else very

high voltage drops (and efficiency losses) would resultbull High loads of direct current could rarely be transmitted for

distances greater than one mile without introducing excessive voltage drops

bull Direct current can not easily be changed to higher or lower voltages Separate electrical lines are needed to distribute power to appliances that used different voltages for example lighting and electric motors

Advantages of AC Transmission

bull Alternating Current can be transformed to lsquosteprsquo the voltage up or down with transformers

bull Power is transmitted at great distances at HIGH voltages and LOW currents and then stepped down to low voltages for use in homes (240V) and industry (440V)

bull Convert AC to DC with a rectifier in appliances

AC is more efficient for Transmission amp Distribution of

electrical power than DC

In The FuturehellipLong Distance AC Power Transmission

may not be needed

War of Currents 1880rsquos

Thomas Edison American inventor and businessman pushed for the development of a DC power network

George WestinghouseAmerican entrepreneur and engineer backed financially the development of a practical AC power network

Nikola Tesla Serbian inventor physicist and electro-mechanical engineer was instrumental in developing AC networks

Quick Quiz 278

For the two lightbulbs shown in this figure rank the current values at the points from greatest to least

Ia = Ib gt Ic = Id gt Ie = If

The 60 W bulb has the lowest resistance and therefore draws the most current

Which light burns the brightest

bull The voltage of each device is the full voltage of the EMF source (the battery)

bull The total current is divided between each path

Parallel Circuits

1 21 2 1 2

1 1( )P

V V VI I I VR R R R R

= + = + = + =

1 2 3

1 1 1 1 PR R R R

= + + +

Equivalent Resistance

bull The current is the same in each devicebull The equivalent resistance of the circuit is the sum

of the individual resistances R=R1+R2

Series Circuits

1 2 1 2 3+ SV V V R R R R= = + + +

1 2 1 2 1 2( + ) + + I R R IR IR V V= = =

total totalV IR=

bull The current is the same in each devicebull The equivalent resistance of the circuit is the sum

of the individual resistances R=R1+R2

Series Circuits

1 2 1 2 3+ SV V V R R R R= = + + +

To find the current use the total voltage and equivalent resistance

S

VIR

=

Circuits ProblemBulbs in Series vs ParallelA circuit contains a 48-V battery and two 240Ω light bulbs

In which circuit does each bulb burn brighterRULE THE MORE POWER DISSIPATED IN A BULB THE

BRIGHTER IT IS

P IVFind the power in each bulb when in series and in parallel

=

Circuits ProblemBulbs in Series vs ParallelIf a bulb burns out - what happens to the other bulb in

each circuit Does it go out Is it brighter Dimmer Or

In the series circuit the burned out bulb will short the circuit and the other bulb will go out

In the parallel circuit the other bulb will have the same brightness

In SeriesThe Voltage is divided in series each bulb gets half V = 24V

2 2(24 ) 24240

V V VP IV V WR R

= = = = =Ω

Circuits ProblemBulbs in Series vs Parallel

In ParallelVoltage is the same in each bulb 48V

2 2(48 ) 96240

V VP WR

= = =Ω

Parallel Bulbs Burn Brighter

Circuits Problem3 Bulbs in SeriesIf one more bulb is added to each circuit (3 bulbs total) how does the brightness of the bulbs change Or not

In the series circuit the bulbs DIM WHY

( )22 2singlebulb 3 1

9 9PVV VP

R R R= = = =

1 2 3V V V V= + +In series each of the three equal bulbs gets one third of the Voltage

(V3) that a single bulb would get

Note P=VI but I is due to the equivalent Resistance I = VRs =V3RSo the Current through each is 13 the current through a single bulb andP=VI=V3 x I3 = VI9 = P9 The bulbs burn 19 as bright

Circuits Problem3 Bulbs in ParallelIf one more bulb is added to each circuit (3 bulbs total) how does the brightness of the bulbs change Or not

In the parallel circuit the bulbs DO NOT DIM WHY

2

singlebulbVP PR

= =

In parallel each of the three equal bulbs gets the full voltage of the battery source

Is this getting something for nothing

NO Parallel circuits drain the battery faster

Parallel CircuitsbullAs the number of branches is increased the overall resistance of the circuit is DECREASED

bullOverall resistance is lowered with each added path between any two points of the circuit

bullThis means the overall resistance of the circuit is less than the resistance of any one of the branches (Weird)

bullAs overall resistance is lowered more current is drawn This is how you blow fuses

FusesIf the current drawn exceeds safe levels the fuse melts and the

circuit lsquobreaksrdquo ndash most house have switches not fuses

Limitations of DC Transmissionbull Large currents in wires produce heat and energy losses by

P = I2Rbull Large expensive conductors would be needed or else very

high voltage drops (and efficiency losses) would resultbull High loads of direct current could rarely be transmitted for

distances greater than one mile without introducing excessive voltage drops

bull Direct current can not easily be changed to higher or lower voltages Separate electrical lines are needed to distribute power to appliances that used different voltages for example lighting and electric motors

Advantages of AC Transmission

bull Alternating Current can be transformed to lsquosteprsquo the voltage up or down with transformers

bull Power is transmitted at great distances at HIGH voltages and LOW currents and then stepped down to low voltages for use in homes (240V) and industry (440V)

bull Convert AC to DC with a rectifier in appliances

AC is more efficient for Transmission amp Distribution of

electrical power than DC

In The FuturehellipLong Distance AC Power Transmission

may not be needed

War of Currents 1880rsquos

Thomas Edison American inventor and businessman pushed for the development of a DC power network

George WestinghouseAmerican entrepreneur and engineer backed financially the development of a practical AC power network

Nikola Tesla Serbian inventor physicist and electro-mechanical engineer was instrumental in developing AC networks

bull The voltage of each device is the full voltage of the EMF source (the battery)

bull The total current is divided between each path

Parallel Circuits

1 21 2 1 2

1 1( )P

V V VI I I VR R R R R

= + = + = + =

1 2 3

1 1 1 1 PR R R R

= + + +

Equivalent Resistance

bull The current is the same in each devicebull The equivalent resistance of the circuit is the sum

of the individual resistances R=R1+R2

Series Circuits

1 2 1 2 3+ SV V V R R R R= = + + +

1 2 1 2 1 2( + ) + + I R R IR IR V V= = =

total totalV IR=

bull The current is the same in each devicebull The equivalent resistance of the circuit is the sum

of the individual resistances R=R1+R2

Series Circuits

1 2 1 2 3+ SV V V R R R R= = + + +

To find the current use the total voltage and equivalent resistance

S

VIR

=

Circuits ProblemBulbs in Series vs ParallelA circuit contains a 48-V battery and two 240Ω light bulbs

In which circuit does each bulb burn brighterRULE THE MORE POWER DISSIPATED IN A BULB THE

BRIGHTER IT IS

P IVFind the power in each bulb when in series and in parallel

=

Circuits ProblemBulbs in Series vs ParallelIf a bulb burns out - what happens to the other bulb in

each circuit Does it go out Is it brighter Dimmer Or

In the series circuit the burned out bulb will short the circuit and the other bulb will go out

In the parallel circuit the other bulb will have the same brightness

In SeriesThe Voltage is divided in series each bulb gets half V = 24V

2 2(24 ) 24240

V V VP IV V WR R

= = = = =Ω

Circuits ProblemBulbs in Series vs Parallel

In ParallelVoltage is the same in each bulb 48V

2 2(48 ) 96240

V VP WR

= = =Ω

Parallel Bulbs Burn Brighter

Circuits Problem3 Bulbs in SeriesIf one more bulb is added to each circuit (3 bulbs total) how does the brightness of the bulbs change Or not

In the series circuit the bulbs DIM WHY

( )22 2singlebulb 3 1

9 9PVV VP

R R R= = = =

1 2 3V V V V= + +In series each of the three equal bulbs gets one third of the Voltage

(V3) that a single bulb would get

Note P=VI but I is due to the equivalent Resistance I = VRs =V3RSo the Current through each is 13 the current through a single bulb andP=VI=V3 x I3 = VI9 = P9 The bulbs burn 19 as bright

Circuits Problem3 Bulbs in ParallelIf one more bulb is added to each circuit (3 bulbs total) how does the brightness of the bulbs change Or not

In the parallel circuit the bulbs DO NOT DIM WHY

2

singlebulbVP PR

= =

In parallel each of the three equal bulbs gets the full voltage of the battery source

Is this getting something for nothing

NO Parallel circuits drain the battery faster

Parallel CircuitsbullAs the number of branches is increased the overall resistance of the circuit is DECREASED

bullOverall resistance is lowered with each added path between any two points of the circuit

bullThis means the overall resistance of the circuit is less than the resistance of any one of the branches (Weird)

bullAs overall resistance is lowered more current is drawn This is how you blow fuses

FusesIf the current drawn exceeds safe levels the fuse melts and the

circuit lsquobreaksrdquo ndash most house have switches not fuses

Limitations of DC Transmissionbull Large currents in wires produce heat and energy losses by

P = I2Rbull Large expensive conductors would be needed or else very

high voltage drops (and efficiency losses) would resultbull High loads of direct current could rarely be transmitted for

distances greater than one mile without introducing excessive voltage drops

bull Direct current can not easily be changed to higher or lower voltages Separate electrical lines are needed to distribute power to appliances that used different voltages for example lighting and electric motors

Advantages of AC Transmission

bull Alternating Current can be transformed to lsquosteprsquo the voltage up or down with transformers

bull Power is transmitted at great distances at HIGH voltages and LOW currents and then stepped down to low voltages for use in homes (240V) and industry (440V)

bull Convert AC to DC with a rectifier in appliances

AC is more efficient for Transmission amp Distribution of

electrical power than DC

In The FuturehellipLong Distance AC Power Transmission

may not be needed

War of Currents 1880rsquos

Thomas Edison American inventor and businessman pushed for the development of a DC power network

George WestinghouseAmerican entrepreneur and engineer backed financially the development of a practical AC power network

Nikola Tesla Serbian inventor physicist and electro-mechanical engineer was instrumental in developing AC networks

bull The current is the same in each devicebull The equivalent resistance of the circuit is the sum

of the individual resistances R=R1+R2

Series Circuits

1 2 1 2 3+ SV V V R R R R= = + + +

1 2 1 2 1 2( + ) + + I R R IR IR V V= = =

total totalV IR=

bull The current is the same in each devicebull The equivalent resistance of the circuit is the sum

of the individual resistances R=R1+R2

Series Circuits

1 2 1 2 3+ SV V V R R R R= = + + +

To find the current use the total voltage and equivalent resistance

S

VIR

=

Circuits ProblemBulbs in Series vs ParallelA circuit contains a 48-V battery and two 240Ω light bulbs

In which circuit does each bulb burn brighterRULE THE MORE POWER DISSIPATED IN A BULB THE

BRIGHTER IT IS

P IVFind the power in each bulb when in series and in parallel

=

Circuits ProblemBulbs in Series vs ParallelIf a bulb burns out - what happens to the other bulb in

each circuit Does it go out Is it brighter Dimmer Or

In the series circuit the burned out bulb will short the circuit and the other bulb will go out

In the parallel circuit the other bulb will have the same brightness

In SeriesThe Voltage is divided in series each bulb gets half V = 24V

2 2(24 ) 24240

V V VP IV V WR R

= = = = =Ω

Circuits ProblemBulbs in Series vs Parallel

In ParallelVoltage is the same in each bulb 48V

2 2(48 ) 96240

V VP WR

= = =Ω

Parallel Bulbs Burn Brighter

Circuits Problem3 Bulbs in SeriesIf one more bulb is added to each circuit (3 bulbs total) how does the brightness of the bulbs change Or not

In the series circuit the bulbs DIM WHY

( )22 2singlebulb 3 1

9 9PVV VP

R R R= = = =

1 2 3V V V V= + +In series each of the three equal bulbs gets one third of the Voltage

(V3) that a single bulb would get

Note P=VI but I is due to the equivalent Resistance I = VRs =V3RSo the Current through each is 13 the current through a single bulb andP=VI=V3 x I3 = VI9 = P9 The bulbs burn 19 as bright

Circuits Problem3 Bulbs in ParallelIf one more bulb is added to each circuit (3 bulbs total) how does the brightness of the bulbs change Or not

In the parallel circuit the bulbs DO NOT DIM WHY

2

singlebulbVP PR

= =

In parallel each of the three equal bulbs gets the full voltage of the battery source

Is this getting something for nothing

NO Parallel circuits drain the battery faster

Parallel CircuitsbullAs the number of branches is increased the overall resistance of the circuit is DECREASED

bullOverall resistance is lowered with each added path between any two points of the circuit

bullThis means the overall resistance of the circuit is less than the resistance of any one of the branches (Weird)

bullAs overall resistance is lowered more current is drawn This is how you blow fuses

FusesIf the current drawn exceeds safe levels the fuse melts and the

circuit lsquobreaksrdquo ndash most house have switches not fuses

Limitations of DC Transmissionbull Large currents in wires produce heat and energy losses by

P = I2Rbull Large expensive conductors would be needed or else very

high voltage drops (and efficiency losses) would resultbull High loads of direct current could rarely be transmitted for

distances greater than one mile without introducing excessive voltage drops

bull Direct current can not easily be changed to higher or lower voltages Separate electrical lines are needed to distribute power to appliances that used different voltages for example lighting and electric motors

Advantages of AC Transmission

bull Alternating Current can be transformed to lsquosteprsquo the voltage up or down with transformers

bull Power is transmitted at great distances at HIGH voltages and LOW currents and then stepped down to low voltages for use in homes (240V) and industry (440V)

bull Convert AC to DC with a rectifier in appliances

AC is more efficient for Transmission amp Distribution of

electrical power than DC

In The FuturehellipLong Distance AC Power Transmission

may not be needed

War of Currents 1880rsquos

Thomas Edison American inventor and businessman pushed for the development of a DC power network

George WestinghouseAmerican entrepreneur and engineer backed financially the development of a practical AC power network

Nikola Tesla Serbian inventor physicist and electro-mechanical engineer was instrumental in developing AC networks

bull The current is the same in each devicebull The equivalent resistance of the circuit is the sum

of the individual resistances R=R1+R2

Series Circuits

1 2 1 2 3+ SV V V R R R R= = + + +

To find the current use the total voltage and equivalent resistance

S

VIR

=

Circuits ProblemBulbs in Series vs ParallelA circuit contains a 48-V battery and two 240Ω light bulbs

In which circuit does each bulb burn brighterRULE THE MORE POWER DISSIPATED IN A BULB THE

BRIGHTER IT IS

P IVFind the power in each bulb when in series and in parallel

=

Circuits ProblemBulbs in Series vs ParallelIf a bulb burns out - what happens to the other bulb in

each circuit Does it go out Is it brighter Dimmer Or

In the series circuit the burned out bulb will short the circuit and the other bulb will go out

In the parallel circuit the other bulb will have the same brightness

In SeriesThe Voltage is divided in series each bulb gets half V = 24V

2 2(24 ) 24240

V V VP IV V WR R

= = = = =Ω

Circuits ProblemBulbs in Series vs Parallel

In ParallelVoltage is the same in each bulb 48V

2 2(48 ) 96240

V VP WR

= = =Ω

Parallel Bulbs Burn Brighter

Circuits Problem3 Bulbs in SeriesIf one more bulb is added to each circuit (3 bulbs total) how does the brightness of the bulbs change Or not

In the series circuit the bulbs DIM WHY

( )22 2singlebulb 3 1

9 9PVV VP

R R R= = = =

1 2 3V V V V= + +In series each of the three equal bulbs gets one third of the Voltage

(V3) that a single bulb would get

Note P=VI but I is due to the equivalent Resistance I = VRs =V3RSo the Current through each is 13 the current through a single bulb andP=VI=V3 x I3 = VI9 = P9 The bulbs burn 19 as bright

Circuits Problem3 Bulbs in ParallelIf one more bulb is added to each circuit (3 bulbs total) how does the brightness of the bulbs change Or not

In the parallel circuit the bulbs DO NOT DIM WHY

2

singlebulbVP PR

= =

In parallel each of the three equal bulbs gets the full voltage of the battery source

Is this getting something for nothing

NO Parallel circuits drain the battery faster

Parallel CircuitsbullAs the number of branches is increased the overall resistance of the circuit is DECREASED

bullOverall resistance is lowered with each added path between any two points of the circuit

bullThis means the overall resistance of the circuit is less than the resistance of any one of the branches (Weird)

bullAs overall resistance is lowered more current is drawn This is how you blow fuses

FusesIf the current drawn exceeds safe levels the fuse melts and the

circuit lsquobreaksrdquo ndash most house have switches not fuses

Limitations of DC Transmissionbull Large currents in wires produce heat and energy losses by

P = I2Rbull Large expensive conductors would be needed or else very

high voltage drops (and efficiency losses) would resultbull High loads of direct current could rarely be transmitted for

distances greater than one mile without introducing excessive voltage drops

bull Direct current can not easily be changed to higher or lower voltages Separate electrical lines are needed to distribute power to appliances that used different voltages for example lighting and electric motors

Advantages of AC Transmission

bull Alternating Current can be transformed to lsquosteprsquo the voltage up or down with transformers

bull Power is transmitted at great distances at HIGH voltages and LOW currents and then stepped down to low voltages for use in homes (240V) and industry (440V)

bull Convert AC to DC with a rectifier in appliances

AC is more efficient for Transmission amp Distribution of

electrical power than DC

In The FuturehellipLong Distance AC Power Transmission

may not be needed

War of Currents 1880rsquos

Thomas Edison American inventor and businessman pushed for the development of a DC power network

George WestinghouseAmerican entrepreneur and engineer backed financially the development of a practical AC power network

Nikola Tesla Serbian inventor physicist and electro-mechanical engineer was instrumental in developing AC networks

Circuits ProblemBulbs in Series vs ParallelA circuit contains a 48-V battery and two 240Ω light bulbs

In which circuit does each bulb burn brighterRULE THE MORE POWER DISSIPATED IN A BULB THE

BRIGHTER IT IS

P IVFind the power in each bulb when in series and in parallel

=

Circuits ProblemBulbs in Series vs ParallelIf a bulb burns out - what happens to the other bulb in

each circuit Does it go out Is it brighter Dimmer Or

In the series circuit the burned out bulb will short the circuit and the other bulb will go out

In the parallel circuit the other bulb will have the same brightness

In SeriesThe Voltage is divided in series each bulb gets half V = 24V

2 2(24 ) 24240

V V VP IV V WR R

= = = = =Ω

Circuits ProblemBulbs in Series vs Parallel

In ParallelVoltage is the same in each bulb 48V

2 2(48 ) 96240

V VP WR

= = =Ω

Parallel Bulbs Burn Brighter

Circuits Problem3 Bulbs in SeriesIf one more bulb is added to each circuit (3 bulbs total) how does the brightness of the bulbs change Or not

In the series circuit the bulbs DIM WHY

( )22 2singlebulb 3 1

9 9PVV VP

R R R= = = =

1 2 3V V V V= + +In series each of the three equal bulbs gets one third of the Voltage

(V3) that a single bulb would get

Note P=VI but I is due to the equivalent Resistance I = VRs =V3RSo the Current through each is 13 the current through a single bulb andP=VI=V3 x I3 = VI9 = P9 The bulbs burn 19 as bright

Circuits Problem3 Bulbs in ParallelIf one more bulb is added to each circuit (3 bulbs total) how does the brightness of the bulbs change Or not

In the parallel circuit the bulbs DO NOT DIM WHY

2

singlebulbVP PR

= =

In parallel each of the three equal bulbs gets the full voltage of the battery source

Is this getting something for nothing

NO Parallel circuits drain the battery faster

Parallel CircuitsbullAs the number of branches is increased the overall resistance of the circuit is DECREASED

bullOverall resistance is lowered with each added path between any two points of the circuit

bullThis means the overall resistance of the circuit is less than the resistance of any one of the branches (Weird)

bullAs overall resistance is lowered more current is drawn This is how you blow fuses

FusesIf the current drawn exceeds safe levels the fuse melts and the

circuit lsquobreaksrdquo ndash most house have switches not fuses

Limitations of DC Transmissionbull Large currents in wires produce heat and energy losses by

P = I2Rbull Large expensive conductors would be needed or else very

high voltage drops (and efficiency losses) would resultbull High loads of direct current could rarely be transmitted for

distances greater than one mile without introducing excessive voltage drops

bull Direct current can not easily be changed to higher or lower voltages Separate electrical lines are needed to distribute power to appliances that used different voltages for example lighting and electric motors

Advantages of AC Transmission

bull Alternating Current can be transformed to lsquosteprsquo the voltage up or down with transformers

bull Power is transmitted at great distances at HIGH voltages and LOW currents and then stepped down to low voltages for use in homes (240V) and industry (440V)

bull Convert AC to DC with a rectifier in appliances

AC is more efficient for Transmission amp Distribution of

electrical power than DC

In The FuturehellipLong Distance AC Power Transmission

may not be needed

War of Currents 1880rsquos

Thomas Edison American inventor and businessman pushed for the development of a DC power network

George WestinghouseAmerican entrepreneur and engineer backed financially the development of a practical AC power network

Nikola Tesla Serbian inventor physicist and electro-mechanical engineer was instrumental in developing AC networks

Circuits ProblemBulbs in Series vs ParallelIf a bulb burns out - what happens to the other bulb in

each circuit Does it go out Is it brighter Dimmer Or

In the series circuit the burned out bulb will short the circuit and the other bulb will go out

In the parallel circuit the other bulb will have the same brightness

In SeriesThe Voltage is divided in series each bulb gets half V = 24V

2 2(24 ) 24240

V V VP IV V WR R

= = = = =Ω

Circuits ProblemBulbs in Series vs Parallel

In ParallelVoltage is the same in each bulb 48V

2 2(48 ) 96240

V VP WR

= = =Ω

Parallel Bulbs Burn Brighter

Circuits Problem3 Bulbs in SeriesIf one more bulb is added to each circuit (3 bulbs total) how does the brightness of the bulbs change Or not

In the series circuit the bulbs DIM WHY

( )22 2singlebulb 3 1

9 9PVV VP

R R R= = = =

1 2 3V V V V= + +In series each of the three equal bulbs gets one third of the Voltage

(V3) that a single bulb would get

Note P=VI but I is due to the equivalent Resistance I = VRs =V3RSo the Current through each is 13 the current through a single bulb andP=VI=V3 x I3 = VI9 = P9 The bulbs burn 19 as bright

Circuits Problem3 Bulbs in ParallelIf one more bulb is added to each circuit (3 bulbs total) how does the brightness of the bulbs change Or not

In the parallel circuit the bulbs DO NOT DIM WHY

2

singlebulbVP PR

= =

In parallel each of the three equal bulbs gets the full voltage of the battery source

Is this getting something for nothing

NO Parallel circuits drain the battery faster

Parallel CircuitsbullAs the number of branches is increased the overall resistance of the circuit is DECREASED

bullOverall resistance is lowered with each added path between any two points of the circuit

bullThis means the overall resistance of the circuit is less than the resistance of any one of the branches (Weird)

bullAs overall resistance is lowered more current is drawn This is how you blow fuses

FusesIf the current drawn exceeds safe levels the fuse melts and the

circuit lsquobreaksrdquo ndash most house have switches not fuses

Limitations of DC Transmissionbull Large currents in wires produce heat and energy losses by

P = I2Rbull Large expensive conductors would be needed or else very

high voltage drops (and efficiency losses) would resultbull High loads of direct current could rarely be transmitted for

distances greater than one mile without introducing excessive voltage drops

bull Direct current can not easily be changed to higher or lower voltages Separate electrical lines are needed to distribute power to appliances that used different voltages for example lighting and electric motors

Advantages of AC Transmission

bull Alternating Current can be transformed to lsquosteprsquo the voltage up or down with transformers

bull Power is transmitted at great distances at HIGH voltages and LOW currents and then stepped down to low voltages for use in homes (240V) and industry (440V)

bull Convert AC to DC with a rectifier in appliances

AC is more efficient for Transmission amp Distribution of

electrical power than DC

In The FuturehellipLong Distance AC Power Transmission

may not be needed

War of Currents 1880rsquos

Thomas Edison American inventor and businessman pushed for the development of a DC power network

George WestinghouseAmerican entrepreneur and engineer backed financially the development of a practical AC power network

Nikola Tesla Serbian inventor physicist and electro-mechanical engineer was instrumental in developing AC networks

In SeriesThe Voltage is divided in series each bulb gets half V = 24V

2 2(24 ) 24240

V V VP IV V WR R

= = = = =Ω

Circuits ProblemBulbs in Series vs Parallel

In ParallelVoltage is the same in each bulb 48V

2 2(48 ) 96240

V VP WR

= = =Ω

Parallel Bulbs Burn Brighter

Circuits Problem3 Bulbs in SeriesIf one more bulb is added to each circuit (3 bulbs total) how does the brightness of the bulbs change Or not

In the series circuit the bulbs DIM WHY

( )22 2singlebulb 3 1

9 9PVV VP

R R R= = = =

1 2 3V V V V= + +In series each of the three equal bulbs gets one third of the Voltage

(V3) that a single bulb would get

Note P=VI but I is due to the equivalent Resistance I = VRs =V3RSo the Current through each is 13 the current through a single bulb andP=VI=V3 x I3 = VI9 = P9 The bulbs burn 19 as bright

Circuits Problem3 Bulbs in ParallelIf one more bulb is added to each circuit (3 bulbs total) how does the brightness of the bulbs change Or not

In the parallel circuit the bulbs DO NOT DIM WHY

2

singlebulbVP PR

= =

In parallel each of the three equal bulbs gets the full voltage of the battery source

Is this getting something for nothing

NO Parallel circuits drain the battery faster

Parallel CircuitsbullAs the number of branches is increased the overall resistance of the circuit is DECREASED

bullOverall resistance is lowered with each added path between any two points of the circuit

bullThis means the overall resistance of the circuit is less than the resistance of any one of the branches (Weird)

bullAs overall resistance is lowered more current is drawn This is how you blow fuses

FusesIf the current drawn exceeds safe levels the fuse melts and the

circuit lsquobreaksrdquo ndash most house have switches not fuses

Limitations of DC Transmissionbull Large currents in wires produce heat and energy losses by

P = I2Rbull Large expensive conductors would be needed or else very

high voltage drops (and efficiency losses) would resultbull High loads of direct current could rarely be transmitted for

distances greater than one mile without introducing excessive voltage drops

bull Direct current can not easily be changed to higher or lower voltages Separate electrical lines are needed to distribute power to appliances that used different voltages for example lighting and electric motors

Advantages of AC Transmission

bull Alternating Current can be transformed to lsquosteprsquo the voltage up or down with transformers

bull Power is transmitted at great distances at HIGH voltages and LOW currents and then stepped down to low voltages for use in homes (240V) and industry (440V)

bull Convert AC to DC with a rectifier in appliances

AC is more efficient for Transmission amp Distribution of

electrical power than DC

In The FuturehellipLong Distance AC Power Transmission

may not be needed

War of Currents 1880rsquos

Thomas Edison American inventor and businessman pushed for the development of a DC power network

George WestinghouseAmerican entrepreneur and engineer backed financially the development of a practical AC power network

Nikola Tesla Serbian inventor physicist and electro-mechanical engineer was instrumental in developing AC networks

Circuits Problem3 Bulbs in SeriesIf one more bulb is added to each circuit (3 bulbs total) how does the brightness of the bulbs change Or not

In the series circuit the bulbs DIM WHY

( )22 2singlebulb 3 1

9 9PVV VP

R R R= = = =

1 2 3V V V V= + +In series each of the three equal bulbs gets one third of the Voltage

(V3) that a single bulb would get

Note P=VI but I is due to the equivalent Resistance I = VRs =V3RSo the Current through each is 13 the current through a single bulb andP=VI=V3 x I3 = VI9 = P9 The bulbs burn 19 as bright

Circuits Problem3 Bulbs in ParallelIf one more bulb is added to each circuit (3 bulbs total) how does the brightness of the bulbs change Or not

In the parallel circuit the bulbs DO NOT DIM WHY

2

singlebulbVP PR

= =

In parallel each of the three equal bulbs gets the full voltage of the battery source

Is this getting something for nothing

NO Parallel circuits drain the battery faster

Parallel CircuitsbullAs the number of branches is increased the overall resistance of the circuit is DECREASED

bullOverall resistance is lowered with each added path between any two points of the circuit

bullThis means the overall resistance of the circuit is less than the resistance of any one of the branches (Weird)

bullAs overall resistance is lowered more current is drawn This is how you blow fuses

FusesIf the current drawn exceeds safe levels the fuse melts and the

circuit lsquobreaksrdquo ndash most house have switches not fuses

Limitations of DC Transmissionbull Large currents in wires produce heat and energy losses by

P = I2Rbull Large expensive conductors would be needed or else very

high voltage drops (and efficiency losses) would resultbull High loads of direct current could rarely be transmitted for

distances greater than one mile without introducing excessive voltage drops

bull Direct current can not easily be changed to higher or lower voltages Separate electrical lines are needed to distribute power to appliances that used different voltages for example lighting and electric motors

Advantages of AC Transmission

bull Alternating Current can be transformed to lsquosteprsquo the voltage up or down with transformers

bull Power is transmitted at great distances at HIGH voltages and LOW currents and then stepped down to low voltages for use in homes (240V) and industry (440V)

bull Convert AC to DC with a rectifier in appliances

AC is more efficient for Transmission amp Distribution of

electrical power than DC

In The FuturehellipLong Distance AC Power Transmission

may not be needed

War of Currents 1880rsquos

Thomas Edison American inventor and businessman pushed for the development of a DC power network

George WestinghouseAmerican entrepreneur and engineer backed financially the development of a practical AC power network

Nikola Tesla Serbian inventor physicist and electro-mechanical engineer was instrumental in developing AC networks

Circuits Problem3 Bulbs in ParallelIf one more bulb is added to each circuit (3 bulbs total) how does the brightness of the bulbs change Or not

In the parallel circuit the bulbs DO NOT DIM WHY

2

singlebulbVP PR

= =

In parallel each of the three equal bulbs gets the full voltage of the battery source

Is this getting something for nothing

NO Parallel circuits drain the battery faster

Parallel CircuitsbullAs the number of branches is increased the overall resistance of the circuit is DECREASED

bullOverall resistance is lowered with each added path between any two points of the circuit

bullThis means the overall resistance of the circuit is less than the resistance of any one of the branches (Weird)

bullAs overall resistance is lowered more current is drawn This is how you blow fuses

FusesIf the current drawn exceeds safe levels the fuse melts and the

circuit lsquobreaksrdquo ndash most house have switches not fuses

Limitations of DC Transmissionbull Large currents in wires produce heat and energy losses by

P = I2Rbull Large expensive conductors would be needed or else very

high voltage drops (and efficiency losses) would resultbull High loads of direct current could rarely be transmitted for

distances greater than one mile without introducing excessive voltage drops

bull Direct current can not easily be changed to higher or lower voltages Separate electrical lines are needed to distribute power to appliances that used different voltages for example lighting and electric motors

Advantages of AC Transmission

bull Alternating Current can be transformed to lsquosteprsquo the voltage up or down with transformers

bull Power is transmitted at great distances at HIGH voltages and LOW currents and then stepped down to low voltages for use in homes (240V) and industry (440V)

bull Convert AC to DC with a rectifier in appliances

AC is more efficient for Transmission amp Distribution of

electrical power than DC

In The FuturehellipLong Distance AC Power Transmission

may not be needed

War of Currents 1880rsquos

Thomas Edison American inventor and businessman pushed for the development of a DC power network

George WestinghouseAmerican entrepreneur and engineer backed financially the development of a practical AC power network

Nikola Tesla Serbian inventor physicist and electro-mechanical engineer was instrumental in developing AC networks

Parallel CircuitsbullAs the number of branches is increased the overall resistance of the circuit is DECREASED

bullOverall resistance is lowered with each added path between any two points of the circuit

bullThis means the overall resistance of the circuit is less than the resistance of any one of the branches (Weird)

bullAs overall resistance is lowered more current is drawn This is how you blow fuses

FusesIf the current drawn exceeds safe levels the fuse melts and the

circuit lsquobreaksrdquo ndash most house have switches not fuses

Limitations of DC Transmissionbull Large currents in wires produce heat and energy losses by

P = I2Rbull Large expensive conductors would be needed or else very

high voltage drops (and efficiency losses) would resultbull High loads of direct current could rarely be transmitted for

distances greater than one mile without introducing excessive voltage drops

bull Direct current can not easily be changed to higher or lower voltages Separate electrical lines are needed to distribute power to appliances that used different voltages for example lighting and electric motors

Advantages of AC Transmission

bull Alternating Current can be transformed to lsquosteprsquo the voltage up or down with transformers

bull Power is transmitted at great distances at HIGH voltages and LOW currents and then stepped down to low voltages for use in homes (240V) and industry (440V)

bull Convert AC to DC with a rectifier in appliances

AC is more efficient for Transmission amp Distribution of

electrical power than DC

In The FuturehellipLong Distance AC Power Transmission

may not be needed

War of Currents 1880rsquos

Thomas Edison American inventor and businessman pushed for the development of a DC power network

George WestinghouseAmerican entrepreneur and engineer backed financially the development of a practical AC power network

Nikola Tesla Serbian inventor physicist and electro-mechanical engineer was instrumental in developing AC networks

FusesIf the current drawn exceeds safe levels the fuse melts and the

circuit lsquobreaksrdquo ndash most house have switches not fuses

Limitations of DC Transmissionbull Large currents in wires produce heat and energy losses by

P = I2Rbull Large expensive conductors would be needed or else very

high voltage drops (and efficiency losses) would resultbull High loads of direct current could rarely be transmitted for

distances greater than one mile without introducing excessive voltage drops

bull Direct current can not easily be changed to higher or lower voltages Separate electrical lines are needed to distribute power to appliances that used different voltages for example lighting and electric motors

Advantages of AC Transmission

bull Alternating Current can be transformed to lsquosteprsquo the voltage up or down with transformers

bull Power is transmitted at great distances at HIGH voltages and LOW currents and then stepped down to low voltages for use in homes (240V) and industry (440V)

bull Convert AC to DC with a rectifier in appliances

AC is more efficient for Transmission amp Distribution of

electrical power than DC

In The FuturehellipLong Distance AC Power Transmission

may not be needed

War of Currents 1880rsquos

Thomas Edison American inventor and businessman pushed for the development of a DC power network

George WestinghouseAmerican entrepreneur and engineer backed financially the development of a practical AC power network

Nikola Tesla Serbian inventor physicist and electro-mechanical engineer was instrumental in developing AC networks

Limitations of DC Transmissionbull Large currents in wires produce heat and energy losses by

P = I2Rbull Large expensive conductors would be needed or else very

high voltage drops (and efficiency losses) would resultbull High loads of direct current could rarely be transmitted for

distances greater than one mile without introducing excessive voltage drops

bull Direct current can not easily be changed to higher or lower voltages Separate electrical lines are needed to distribute power to appliances that used different voltages for example lighting and electric motors

Advantages of AC Transmission

bull Alternating Current can be transformed to lsquosteprsquo the voltage up or down with transformers

bull Power is transmitted at great distances at HIGH voltages and LOW currents and then stepped down to low voltages for use in homes (240V) and industry (440V)

bull Convert AC to DC with a rectifier in appliances

AC is more efficient for Transmission amp Distribution of

electrical power than DC

In The FuturehellipLong Distance AC Power Transmission

may not be needed

War of Currents 1880rsquos

Thomas Edison American inventor and businessman pushed for the development of a DC power network

George WestinghouseAmerican entrepreneur and engineer backed financially the development of a practical AC power network

Nikola Tesla Serbian inventor physicist and electro-mechanical engineer was instrumental in developing AC networks

Advantages of AC Transmission

bull Alternating Current can be transformed to lsquosteprsquo the voltage up or down with transformers

bull Power is transmitted at great distances at HIGH voltages and LOW currents and then stepped down to low voltages for use in homes (240V) and industry (440V)

bull Convert AC to DC with a rectifier in appliances

AC is more efficient for Transmission amp Distribution of

electrical power than DC

In The FuturehellipLong Distance AC Power Transmission

may not be needed

War of Currents 1880rsquos

Thomas Edison American inventor and businessman pushed for the development of a DC power network

George WestinghouseAmerican entrepreneur and engineer backed financially the development of a practical AC power network

Nikola Tesla Serbian inventor physicist and electro-mechanical engineer was instrumental in developing AC networks

In The FuturehellipLong Distance AC Power Transmission

may not be needed

War of Currents 1880rsquos

Thomas Edison American inventor and businessman pushed for the development of a DC power network

George WestinghouseAmerican entrepreneur and engineer backed financially the development of a practical AC power network

Nikola Tesla Serbian inventor physicist and electro-mechanical engineer was instrumental in developing AC networks

War of Currents 1880rsquos

Thomas Edison American inventor and businessman pushed for the development of a DC power network

George WestinghouseAmerican entrepreneur and engineer backed financially the development of a practical AC power network

Nikola Tesla Serbian inventor physicist and electro-mechanical engineer was instrumental in developing AC networks

The first electric chair which was used to execute William Kemmler in 1890

Edison wired NYC with DC He carried out a campaign to discourage the use of AC including spreading information on fatal AC accidents killing animals and lobbying against the use of AC in state legislatures Edison opposed capital punishment but his desire to disparage the system of alternating current led to the invention of the electric chair Harold P Brown who was at this time being secretly paid by Edison constructed the first electric chair for the state of New York in order to promote the idea that alternating current was deadlier than DC

Edisons Publicity Campaign

The first electric chair which was used to execute William Kemmler in 1890

Nebraska Only state that requires it 15-second-long jolt of 2450 volts of electricity (~ 8 Amps)

GE amp Edison We bring good things to lightMore than a 1000 killed since 1890

Any practical distribution system will use voltage levels quite sufficient for a dangerous amount of current to flow whether it uses alternating or direct current Ultimately the advantages of AC power transmission outweighed this theoretical risk and it was eventually adopted as the standard worldwide after Nikola Tesla designed the first AC hydroelectric power plant at Niagara Falls New York which started producing electrical power in 1895

Is AC Deadlier than DCThey are BOTH Deadly

Current (A) Effect

0001 Can be felt0005 Painful0010 Causes involuntary muscle spasms0015 Causes loss of muscle control0070 If through heart serious

If current lasts for 1 s - FATAL

Dry Skin Body Resistance 500000 ΩWet Skin Body Resistance 1000 Ω

Electric Shock

Electric Shock

bullElectric Shock occurs when current is produced in the body which is caused by an impressed voltage

bullVoltage is the CAUSE bullCurrent does the DAMAGE

What causes electric Shock in the human body Voltage or Current

Is AC Deadlier than DCbull Low frequency (50 - 60 Hz) AC

currents can be more dangerous than similar levels of DC current since the alternating fluctuations can cause the heart to lose coordination inducing ventricular fibrillation which then rapidly leads to death

bull High voltage DC power can be more dangerous than AC however since it tends to cause muscles to lock in position stopping the victim from releasing the energised conductor once grasped

Frequency Matters

QuestionWhat current would you draw if you were unfortunate to short-circuit a 120 V line with dry

hands Wet handsUse Ohmrsquos Law V = IR

Dry Skin Body Resistance 500000 ΩWet Skin Body Resistance 1000 Ω

QuestionWhat current would you draw if you were unfortunate to short-circuit a 120 V line with dry

hands Wet handsUse Ohmrsquos Law V = IR

DRY I = VR = 120 V500000 Ω = 00024 (live)

WET I = VR = 120 V1000 Ω = 12 (dead)

Electric Shock TherapyELECTRO CONVULSIVE

THERAPYAn electric shock is applied to produce a convulsive seizure

The shock is typically between 140 - 170 volts and lasts between 05 and 1 seconds No explanation of how it works

Used in the treatment of 1Chronic endogenous depression 2Bipolar disorder 3Acute mania 4Certain types of schizophrenia

In the US 33000 - 50000 people receive ECT each year

• Chapter 27 Current amp Resistance
• Current Dead or Alive
• I = Coulombsecond = Ampere
• Engine Current Problem
• Current is charge in motion
• Atomic Vision of Ohmrsquos Law
• Engine Current Problem Again
• Fuse Problem
• Resistivity
• Resistivity Values
• Engine Current Problem
• You Try Problem
• Resistance Question
• Radial Resistance of a Coaxial Cable Leakage
• Resistors
• When are light bulbs more likely to blow When hot or cold
• Ohmic Material
• Nonohmic Material
• Superconductivity
• Superconductivity
• Resistance QUESTION
• What makes the Glow
• Ohmic Heat Loss
• Electrical Power
• If V = 120V What is I USE P = IV=gt I = PV
• If V = 120V What is I USE P = IV=gt I = PV
• Electric Shock
• Electric Shock
• Electric BillCost to run for 1 hr $05 per 1 kw-hr Cost = Power x Time x Rate
• Electric BillCost to run for 1 hr $05 per 1 kw-hr Cost = Power x Time x Rate
• QUESTION
• QUESTION
• QUESTION
• Circuits ProblemBulbs in Series vs Parallel
• Circuits ProblemBulbs in Series vs Parallel
• Circuits ProblemBulbs in Series vs Parallel
• Circuits Problem3 Bulbs in Series
• Circuits Problem3 Bulbs in Parallel
• Parallel Circuits
• Limitations of DC Transmission
• In The FuturehellipLong Distance AC Power Transmission may not be needed
• War of Currents 1880rsquos
• Edisons Publicity Campaign
• Electric Shock
• Electric Shock
• Is AC Deadlier than DC
• Frequency Matters
• QuestionWhat current would you draw if you were unfortunate to short-circuit a 120 V line with dry hands Wet handsUse Ohmrsquo
• QuestionWhat current would you draw if you were unfortunate to short-circuit a 120 V line with dry hands Wet handsUse Ohmrsquo
• Electric Shock TherapyELECTRO CONVULSIVE THERAPY

The first electric chair which was used to execute William Kemmler in 1890

Nebraska Only state that requires it 15-second-long jolt of 2450 volts of electricity (~ 8 Amps)

GE amp Edison We bring good things to lightMore than a 1000 killed since 1890

Any practical distribution system will use voltage levels quite sufficient for a dangerous amount of current to flow whether it uses alternating or direct current Ultimately the advantages of AC power transmission outweighed this theoretical risk and it was eventually adopted as the standard worldwide after Nikola Tesla designed the first AC hydroelectric power plant at Niagara Falls New York which started producing electrical power in 1895

Is AC Deadlier than DCThey are BOTH Deadly

Current (A) Effect

0001 Can be felt0005 Painful0010 Causes involuntary muscle spasms0015 Causes loss of muscle control0070 If through heart serious

If current lasts for 1 s - FATAL

Dry Skin Body Resistance 500000 ΩWet Skin Body Resistance 1000 Ω

Electric Shock

Electric Shock

bullElectric Shock occurs when current is produced in the body which is caused by an impressed voltage

bullVoltage is the CAUSE bullCurrent does the DAMAGE

What causes electric Shock in the human body Voltage or Current

Is AC Deadlier than DCbull Low frequency (50 - 60 Hz) AC

currents can be more dangerous than similar levels of DC current since the alternating fluctuations can cause the heart to lose coordination inducing ventricular fibrillation which then rapidly leads to death

bull High voltage DC power can be more dangerous than AC however since it tends to cause muscles to lock in position stopping the victim from releasing the energised conductor once grasped

Frequency Matters

QuestionWhat current would you draw if you were unfortunate to short-circuit a 120 V line with dry

hands Wet handsUse Ohmrsquos Law V = IR

Dry Skin Body Resistance 500000 ΩWet Skin Body Resistance 1000 Ω

QuestionWhat current would you draw if you were unfortunate to short-circuit a 120 V line with dry

hands Wet handsUse Ohmrsquos Law V = IR

DRY I = VR = 120 V500000 Ω = 00024 (live)

WET I = VR = 120 V1000 Ω = 12 (dead)

Electric Shock TherapyELECTRO CONVULSIVE

THERAPYAn electric shock is applied to produce a convulsive seizure

The shock is typically between 140 - 170 volts and lasts between 05 and 1 seconds No explanation of how it works

Used in the treatment of 1Chronic endogenous depression 2Bipolar disorder 3Acute mania 4Certain types of schizophrenia

In the US 33000 - 50000 people receive ECT each year

• Chapter 27 Current amp Resistance
• Current Dead or Alive
• I = Coulombsecond = Ampere
• Engine Current Problem
• Current is charge in motion
• Atomic Vision of Ohmrsquos Law
• Engine Current Problem Again
• Fuse Problem
• Resistivity
• Resistivity Values
• Engine Current Problem
• You Try Problem
• Resistance Question
• Radial Resistance of a Coaxial Cable Leakage
• Resistors
• When are light bulbs more likely to blow When hot or cold
• Ohmic Material
• Nonohmic Material
• Superconductivity
• Superconductivity
• Resistance QUESTION
• What makes the Glow
• Ohmic Heat Loss
• Electrical Power
• If V = 120V What is I USE P = IV=gt I = PV
• If V = 120V What is I USE P = IV=gt I = PV
• Electric Shock
• Electric Shock
• Electric BillCost to run for 1 hr $05 per 1 kw-hr Cost = Power x Time x Rate
• Electric BillCost to run for 1 hr $05 per 1 kw-hr Cost = Power x Time x Rate
• QUESTION
• QUESTION
• QUESTION
• Circuits ProblemBulbs in Series vs Parallel
• Circuits ProblemBulbs in Series vs Parallel
• Circuits ProblemBulbs in Series vs Parallel
• Circuits Problem3 Bulbs in Series
• Circuits Problem3 Bulbs in Parallel
• Parallel Circuits
• Limitations of DC Transmission
• In The FuturehellipLong Distance AC Power Transmission may not be needed
• War of Currents 1880rsquos
• Edisons Publicity Campaign
• Electric Shock
• Electric Shock
• Is AC Deadlier than DC
• Frequency Matters
• QuestionWhat current would you draw if you were unfortunate to short-circuit a 120 V line with dry hands Wet handsUse Ohmrsquo
• QuestionWhat current would you draw if you were unfortunate to short-circuit a 120 V line with dry hands Wet handsUse Ohmrsquo
• Electric Shock TherapyELECTRO CONVULSIVE THERAPY

Any practical distribution system will use voltage levels quite sufficient for a dangerous amount of current to flow whether it uses alternating or direct current Ultimately the advantages of AC power transmission outweighed this theoretical risk and it was eventually adopted as the standard worldwide after Nikola Tesla designed the first AC hydroelectric power plant at Niagara Falls New York which started producing electrical power in 1895

Is AC Deadlier than DCThey are BOTH Deadly

Current (A) Effect

0001 Can be felt0005 Painful0010 Causes involuntary muscle spasms0015 Causes loss of muscle control0070 If through heart serious

If current lasts for 1 s - FATAL

Dry Skin Body Resistance 500000 ΩWet Skin Body Resistance 1000 Ω

Electric Shock

Electric Shock

bullElectric Shock occurs when current is produced in the body which is caused by an impressed voltage

bullVoltage is the CAUSE bullCurrent does the DAMAGE

What causes electric Shock in the human body Voltage or Current

Is AC Deadlier than DCbull Low frequency (50 - 60 Hz) AC

currents can be more dangerous than similar levels of DC current since the alternating fluctuations can cause the heart to lose coordination inducing ventricular fibrillation which then rapidly leads to death

bull High voltage DC power can be more dangerous than AC however since it tends to cause muscles to lock in position stopping the victim from releasing the energised conductor once grasped

Frequency Matters

QuestionWhat current would you draw if you were unfortunate to short-circuit a 120 V line with dry

hands Wet handsUse Ohmrsquos Law V = IR

Dry Skin Body Resistance 500000 ΩWet Skin Body Resistance 1000 Ω

QuestionWhat current would you draw if you were unfortunate to short-circuit a 120 V line with dry

hands Wet handsUse Ohmrsquos Law V = IR

DRY I = VR = 120 V500000 Ω = 00024 (live)

WET I = VR = 120 V1000 Ω = 12 (dead)

Electric Shock TherapyELECTRO CONVULSIVE

THERAPYAn electric shock is applied to produce a convulsive seizure

The shock is typically between 140 - 170 volts and lasts between 05 and 1 seconds No explanation of how it works

Used in the treatment of 1Chronic endogenous depression 2Bipolar disorder 3Acute mania 4Certain types of schizophrenia

In the US 33000 - 50000 people receive ECT each year

• Chapter 27 Current amp Resistance
• Current Dead or Alive
• I = Coulombsecond = Ampere
• Engine Current Problem
• Current is charge in motion
• Atomic Vision of Ohmrsquos Law
• Engine Current Problem Again
• Fuse Problem
• Resistivity
• Resistivity Values
• Engine Current Problem
• You Try Problem
• Resistance Question
• Radial Resistance of a Coaxial Cable Leakage
• Resistors
• When are light bulbs more likely to blow When hot or cold
• Ohmic Material
• Nonohmic Material
• Superconductivity
• Superconductivity
• Resistance QUESTION
• What makes the Glow
• Ohmic Heat Loss
• Electrical Power
• If V = 120V What is I USE P = IV=gt I = PV
• If V = 120V What is I USE P = IV=gt I = PV
• Electric Shock
• Electric Shock
• Electric BillCost to run for 1 hr $05 per 1 kw-hr Cost = Power x Time x Rate
• Electric BillCost to run for 1 hr $05 per 1 kw-hr Cost = Power x Time x Rate
• QUESTION
• QUESTION
• QUESTION
• Circuits ProblemBulbs in Series vs Parallel
• Circuits ProblemBulbs in Series vs Parallel
• Circuits ProblemBulbs in Series vs Parallel
• Circuits Problem3 Bulbs in Series
• Circuits Problem3 Bulbs in Parallel
• Parallel Circuits
• Limitations of DC Transmission
• In The FuturehellipLong Distance AC Power Transmission may not be needed
• War of Currents 1880rsquos
• Edisons Publicity Campaign
• Electric Shock
• Electric Shock
• Is AC Deadlier than DC
• Frequency Matters
• QuestionWhat current would you draw if you were unfortunate to short-circuit a 120 V line with dry hands Wet handsUse Ohmrsquo
• QuestionWhat current would you draw if you were unfortunate to short-circuit a 120 V line with dry hands Wet handsUse Ohmrsquo
• Electric Shock TherapyELECTRO CONVULSIVE THERAPY

Current (A) Effect

0001 Can be felt0005 Painful0010 Causes involuntary muscle spasms0015 Causes loss of muscle control0070 If through heart serious

If current lasts for 1 s - FATAL

Dry Skin Body Resistance 500000 ΩWet Skin Body Resistance 1000 Ω

Electric Shock

Electric Shock

bullElectric Shock occurs when current is produced in the body which is caused by an impressed voltage

bullVoltage is the CAUSE bullCurrent does the DAMAGE

What causes electric Shock in the human body Voltage or Current

Is AC Deadlier than DCbull Low frequency (50 - 60 Hz) AC

currents can be more dangerous than similar levels of DC current since the alternating fluctuations can cause the heart to lose coordination inducing ventricular fibrillation which then rapidly leads to death

bull High voltage DC power can be more dangerous than AC however since it tends to cause muscles to lock in position stopping the victim from releasing the energised conductor once grasped

Frequency Matters

QuestionWhat current would you draw if you were unfortunate to short-circuit a 120 V line with dry

hands Wet handsUse Ohmrsquos Law V = IR

Dry Skin Body Resistance 500000 ΩWet Skin Body Resistance 1000 Ω

QuestionWhat current would you draw if you were unfortunate to short-circuit a 120 V line with dry

hands Wet handsUse Ohmrsquos Law V = IR

DRY I = VR = 120 V500000 Ω = 00024 (live)

WET I = VR = 120 V1000 Ω = 12 (dead)

Electric Shock TherapyELECTRO CONVULSIVE

THERAPYAn electric shock is applied to produce a convulsive seizure

The shock is typically between 140 - 170 volts and lasts between 05 and 1 seconds No explanation of how it works

Used in the treatment of 1Chronic endogenous depression 2Bipolar disorder 3Acute mania 4Certain types of schizophrenia

In the US 33000 - 50000 people receive ECT each year

• Chapter 27 Current amp Resistance
• Current Dead or Alive
• I = Coulombsecond = Ampere
• Engine Current Problem
• Current is charge in motion
• Atomic Vision of Ohmrsquos Law
• Engine Current Problem Again
• Fuse Problem
• Resistivity
• Resistivity Values
• Engine Current Problem
• You Try Problem
• Resistance Question
• Radial Resistance of a Coaxial Cable Leakage
• Resistors
• When are light bulbs more likely to blow When hot or cold
• Ohmic Material
• Nonohmic Material
• Superconductivity
• Superconductivity
• Resistance QUESTION
• What makes the Glow
• Ohmic Heat Loss
• Electrical Power
• If V = 120V What is I USE P = IV=gt I = PV
• If V = 120V What is I USE P = IV=gt I = PV
• Electric Shock
• Electric Shock
• Electric BillCost to run for 1 hr $05 per 1 kw-hr Cost = Power x Time x Rate
• Electric BillCost to run for 1 hr $05 per 1 kw-hr Cost = Power x Time x Rate
• QUESTION
• QUESTION
• QUESTION
• Circuits ProblemBulbs in Series vs Parallel
• Circuits ProblemBulbs in Series vs Parallel
• Circuits ProblemBulbs in Series vs Parallel
• Circuits Problem3 Bulbs in Series
• Circuits Problem3 Bulbs in Parallel
• Parallel Circuits
• Limitations of DC Transmission
• In The FuturehellipLong Distance AC Power Transmission may not be needed
• War of Currents 1880rsquos
• Edisons Publicity Campaign
• Electric Shock
• Electric Shock
• Is AC Deadlier than DC
• Frequency Matters
• QuestionWhat current would you draw if you were unfortunate to short-circuit a 120 V line with dry hands Wet handsUse Ohmrsquo
• QuestionWhat current would you draw if you were unfortunate to short-circuit a 120 V line with dry hands Wet handsUse Ohmrsquo
• Electric Shock TherapyELECTRO CONVULSIVE THERAPY

Electric Shock

bullElectric Shock occurs when current is produced in the body which is caused by an impressed voltage

bullVoltage is the CAUSE bullCurrent does the DAMAGE

What causes electric Shock in the human body Voltage or Current

Is AC Deadlier than DCbull Low frequency (50 - 60 Hz) AC

currents can be more dangerous than similar levels of DC current since the alternating fluctuations can cause the heart to lose coordination inducing ventricular fibrillation which then rapidly leads to death

bull High voltage DC power can be more dangerous than AC however since it tends to cause muscles to lock in position stopping the victim from releasing the energised conductor once grasped

Frequency Matters

QuestionWhat current would you draw if you were unfortunate to short-circuit a 120 V line with dry

hands Wet handsUse Ohmrsquos Law V = IR

Dry Skin Body Resistance 500000 ΩWet Skin Body Resistance 1000 Ω

QuestionWhat current would you draw if you were unfortunate to short-circuit a 120 V line with dry

hands Wet handsUse Ohmrsquos Law V = IR

DRY I = VR = 120 V500000 Ω = 00024 (live)

WET I = VR = 120 V1000 Ω = 12 (dead)

Electric Shock TherapyELECTRO CONVULSIVE

THERAPYAn electric shock is applied to produce a convulsive seizure

The shock is typically between 140 - 170 volts and lasts between 05 and 1 seconds No explanation of how it works

Used in the treatment of 1Chronic endogenous depression 2Bipolar disorder 3Acute mania 4Certain types of schizophrenia

In the US 33000 - 50000 people receive ECT each year

• Chapter 27 Current amp Resistance
• Current Dead or Alive
• I = Coulombsecond = Ampere
• Engine Current Problem
• Current is charge in motion
• Atomic Vision of Ohmrsquos Law
• Engine Current Problem Again
• Fuse Problem
• Resistivity
• Resistivity Values
• Engine Current Problem
• You Try Problem
• Resistance Question
• Radial Resistance of a Coaxial Cable Leakage
• Resistors
• When are light bulbs more likely to blow When hot or cold
• Ohmic Material
• Nonohmic Material
• Superconductivity
• Superconductivity
• Resistance QUESTION
• What makes the Glow
• Ohmic Heat Loss
• Electrical Power
• If V = 120V What is I USE P = IV=gt I = PV
• If V = 120V What is I USE P = IV=gt I = PV
• Electric Shock
• Electric Shock
• Electric BillCost to run for 1 hr $05 per 1 kw-hr Cost = Power x Time x Rate
• Electric BillCost to run for 1 hr $05 per 1 kw-hr Cost = Power x Time x Rate
• QUESTION
• QUESTION
• QUESTION
• Circuits ProblemBulbs in Series vs Parallel
• Circuits ProblemBulbs in Series vs Parallel
• Circuits ProblemBulbs in Series vs Parallel
• Circuits Problem3 Bulbs in Series
• Circuits Problem3 Bulbs in Parallel
• Parallel Circuits
• Limitations of DC Transmission
• In The FuturehellipLong Distance AC Power Transmission may not be needed
• War of Currents 1880rsquos
• Edisons Publicity Campaign
• Electric Shock
• Electric Shock
• Is AC Deadlier than DC
• Frequency Matters
• QuestionWhat current would you draw if you were unfortunate to short-circuit a 120 V line with dry hands Wet handsUse Ohmrsquo
• QuestionWhat current would you draw if you were unfortunate to short-circuit a 120 V line with dry hands Wet handsUse Ohmrsquo
• Electric Shock TherapyELECTRO CONVULSIVE THERAPY

Is AC Deadlier than DCbull Low frequency (50 - 60 Hz) AC

currents can be more dangerous than similar levels of DC current since the alternating fluctuations can cause the heart to lose coordination inducing ventricular fibrillation which then rapidly leads to death

bull High voltage DC power can be more dangerous than AC however since it tends to cause muscles to lock in position stopping the victim from releasing the energised conductor once grasped

Frequency Matters

QuestionWhat current would you draw if you were unfortunate to short-circuit a 120 V line with dry

hands Wet handsUse Ohmrsquos Law V = IR

Dry Skin Body Resistance 500000 ΩWet Skin Body Resistance 1000 Ω

QuestionWhat current would you draw if you were unfortunate to short-circuit a 120 V line with dry

hands Wet handsUse Ohmrsquos Law V = IR

DRY I = VR = 120 V500000 Ω = 00024 (live)

WET I = VR = 120 V1000 Ω = 12 (dead)

Electric Shock TherapyELECTRO CONVULSIVE

THERAPYAn electric shock is applied to produce a convulsive seizure

The shock is typically between 140 - 170 volts and lasts between 05 and 1 seconds No explanation of how it works

Used in the treatment of 1Chronic endogenous depression 2Bipolar disorder 3Acute mania 4Certain types of schizophrenia

In the US 33000 - 50000 people receive ECT each year

• Chapter 27 Current amp Resistance
• Current Dead or Alive
• I = Coulombsecond = Ampere
• Engine Current Problem
• Current is charge in motion
• Atomic Vision of Ohmrsquos Law
• Engine Current Problem Again
• Fuse Problem
• Resistivity
• Resistivity Values
• Engine Current Problem
• You Try Problem
• Resistance Question
• Radial Resistance of a Coaxial Cable Leakage
• Resistors
• When are light bulbs more likely to blow When hot or cold
• Ohmic Material
• Nonohmic Material
• Superconductivity
• Superconductivity
• Resistance QUESTION
• What makes the Glow
• Ohmic Heat Loss
• Electrical Power
• If V = 120V What is I USE P = IV=gt I = PV
• If V = 120V What is I USE P = IV=gt I = PV
• Electric Shock
• Electric Shock
• Electric BillCost to run for 1 hr $05 per 1 kw-hr Cost = Power x Time x Rate
• Electric BillCost to run for 1 hr $05 per 1 kw-hr Cost = Power x Time x Rate
• QUESTION
• QUESTION
• QUESTION
• Circuits ProblemBulbs in Series vs Parallel
• Circuits ProblemBulbs in Series vs Parallel
• Circuits ProblemBulbs in Series vs Parallel
• Circuits Problem3 Bulbs in Series
• Circuits Problem3 Bulbs in Parallel
• Parallel Circuits
• Limitations of DC Transmission
• In The FuturehellipLong Distance AC Power Transmission may not be needed
• War of Currents 1880rsquos
• Edisons Publicity Campaign
• Electric Shock
• Electric Shock
• Is AC Deadlier than DC
• Frequency Matters
• QuestionWhat current would you draw if you were unfortunate to short-circuit a 120 V line with dry hands Wet handsUse Ohmrsquo
• QuestionWhat current would you draw if you were unfortunate to short-circuit a 120 V line with dry hands Wet handsUse Ohmrsquo
• Electric Shock TherapyELECTRO CONVULSIVE THERAPY

Frequency Matters

QuestionWhat current would you draw if you were unfortunate to short-circuit a 120 V line with dry

hands Wet handsUse Ohmrsquos Law V = IR

Dry Skin Body Resistance 500000 ΩWet Skin Body Resistance 1000 Ω

QuestionWhat current would you draw if you were unfortunate to short-circuit a 120 V line with dry

hands Wet handsUse Ohmrsquos Law V = IR

DRY I = VR = 120 V500000 Ω = 00024 (live)

WET I = VR = 120 V1000 Ω = 12 (dead)

Electric Shock TherapyELECTRO CONVULSIVE

THERAPYAn electric shock is applied to produce a convulsive seizure

The shock is typically between 140 - 170 volts and lasts between 05 and 1 seconds No explanation of how it works

Used in the treatment of 1Chronic endogenous depression 2Bipolar disorder 3Acute mania 4Certain types of schizophrenia

In the US 33000 - 50000 people receive ECT each year

• Chapter 27 Current amp Resistance
• Current Dead or Alive
• I = Coulombsecond = Ampere
• Engine Current Problem
• Current is charge in motion
• Atomic Vision of Ohmrsquos Law
• Engine Current Problem Again
• Fuse Problem
• Resistivity
• Resistivity Values
• Engine Current Problem
• You Try Problem
• Resistance Question
• Radial Resistance of a Coaxial Cable Leakage
• Resistors
• When are light bulbs more likely to blow When hot or cold
• Ohmic Material
• Nonohmic Material
• Superconductivity
• Superconductivity
• Resistance QUESTION
• What makes the Glow
• Ohmic Heat Loss
• Electrical Power
• If V = 120V What is I USE P = IV=gt I = PV
• If V = 120V What is I USE P = IV=gt I = PV
• Electric Shock
• Electric Shock
• Electric BillCost to run for 1 hr $05 per 1 kw-hr Cost = Power x Time x Rate
• Electric BillCost to run for 1 hr $05 per 1 kw-hr Cost = Power x Time x Rate
• QUESTION
• QUESTION
• QUESTION
• Circuits ProblemBulbs in Series vs Parallel
• Circuits ProblemBulbs in Series vs Parallel
• Circuits ProblemBulbs in Series vs Parallel
• Circuits Problem3 Bulbs in Series
• Circuits Problem3 Bulbs in Parallel
• Parallel Circuits
• Limitations of DC Transmission
• In The FuturehellipLong Distance AC Power Transmission may not be needed
• War of Currents 1880rsquos
• Edisons Publicity Campaign
• Electric Shock
• Electric Shock
• Is AC Deadlier than DC
• Frequency Matters
• QuestionWhat current would you draw if you were unfortunate to short-circuit a 120 V line with dry hands Wet handsUse Ohmrsquo
• QuestionWhat current would you draw if you were unfortunate to short-circuit a 120 V line with dry hands Wet handsUse Ohmrsquo
• Electric Shock TherapyELECTRO CONVULSIVE THERAPY

QuestionWhat current would you draw if you were unfortunate to short-circuit a 120 V line with dry

hands Wet handsUse Ohmrsquos Law V = IR

Dry Skin Body Resistance 500000 ΩWet Skin Body Resistance 1000 Ω

QuestionWhat current would you draw if you were unfortunate to short-circuit a 120 V line with dry

hands Wet handsUse Ohmrsquos Law V = IR

DRY I = VR = 120 V500000 Ω = 00024 (live)

WET I = VR = 120 V1000 Ω = 12 (dead)

Electric Shock TherapyELECTRO CONVULSIVE

THERAPYAn electric shock is applied to produce a convulsive seizure

The shock is typically between 140 - 170 volts and lasts between 05 and 1 seconds No explanation of how it works

Used in the treatment of 1Chronic endogenous depression 2Bipolar disorder 3Acute mania 4Certain types of schizophrenia

In the US 33000 - 50000 people receive ECT each year

• Chapter 27 Current amp Resistance
• Current Dead or Alive
• I = Coulombsecond = Ampere
• Engine Current Problem
• Current is charge in motion
• Atomic Vision of Ohmrsquos Law
• Engine Current Problem Again
• Fuse Problem
• Resistivity
• Resistivity Values
• Engine Current Problem
• You Try Problem
• Resistance Question
• Radial Resistance of a Coaxial Cable Leakage
• Resistors
• When are light bulbs more likely to blow When hot or cold
• Ohmic Material
• Nonohmic Material
• Superconductivity
• Superconductivity
• Resistance QUESTION
• What makes the Glow
• Ohmic Heat Loss
• Electrical Power
• If V = 120V What is I USE P = IV=gt I = PV
• If V = 120V What is I USE P = IV=gt I = PV
• Electric Shock
• Electric Shock
• Electric BillCost to run for 1 hr $05 per 1 kw-hr Cost = Power x Time x Rate
• Electric BillCost to run for 1 hr $05 per 1 kw-hr Cost = Power x Time x Rate
• QUESTION
• QUESTION
• QUESTION
• Circuits ProblemBulbs in Series vs Parallel
• Circuits ProblemBulbs in Series vs Parallel
• Circuits ProblemBulbs in Series vs Parallel
• Circuits Problem3 Bulbs in Series
• Circuits Problem3 Bulbs in Parallel
• Parallel Circuits
• Limitations of DC Transmission
• In The FuturehellipLong Distance AC Power Transmission may not be needed
• War of Currents 1880rsquos
• Edisons Publicity Campaign
• Electric Shock
• Electric Shock
• Is AC Deadlier than DC
• Frequency Matters
• QuestionWhat current would you draw if you were unfortunate to short-circuit a 120 V line with dry hands Wet handsUse Ohmrsquo
• QuestionWhat current would you draw if you were unfortunate to short-circuit a 120 V line with dry hands Wet handsUse Ohmrsquo
• Electric Shock TherapyELECTRO CONVULSIVE THERAPY

QuestionWhat current would you draw if you were unfortunate to short-circuit a 120 V line with dry

hands Wet handsUse Ohmrsquos Law V = IR

DRY I = VR = 120 V500000 Ω = 00024 (live)

WET I = VR = 120 V1000 Ω = 12 (dead)

Electric Shock TherapyELECTRO CONVULSIVE

THERAPYAn electric shock is applied to produce a convulsive seizure

The shock is typically between 140 - 170 volts and lasts between 05 and 1 seconds No explanation of how it works

Used in the treatment of 1Chronic endogenous depression 2Bipolar disorder 3Acute mania 4Certain types of schizophrenia

In the US 33000 - 50000 people receive ECT each year

• Chapter 27 Current amp Resistance
• Current Dead or Alive
• I = Coulombsecond = Ampere
• Engine Current Problem
• Current is charge in motion
• Atomic Vision of Ohmrsquos Law
• Engine Current Problem Again
• Fuse Problem
• Resistivity
• Resistivity Values
• Engine Current Problem
• You Try Problem
• Resistance Question
• Radial Resistance of a Coaxial Cable Leakage
• Resistors
• When are light bulbs more likely to blow When hot or cold
• Ohmic Material
• Nonohmic Material
• Superconductivity
• Superconductivity
• Resistance QUESTION
• What makes the Glow
• Ohmic Heat Loss
• Electrical Power
• If V = 120V What is I USE P = IV=gt I = PV
• If V = 120V What is I USE P = IV=gt I = PV
• Electric Shock
• Electric Shock
• Electric BillCost to run for 1 hr $05 per 1 kw-hr Cost = Power x Time x Rate
• Electric BillCost to run for 1 hr $05 per 1 kw-hr Cost = Power x Time x Rate
• QUESTION
• QUESTION
• QUESTION
• Circuits ProblemBulbs in Series vs Parallel
• Circuits ProblemBulbs in Series vs Parallel
• Circuits ProblemBulbs in Series vs Parallel
• Circuits Problem3 Bulbs in Series
• Circuits Problem3 Bulbs in Parallel
• Parallel Circuits
• Limitations of DC Transmission
• In The FuturehellipLong Distance AC Power Transmission may not be needed
• War of Currents 1880rsquos
• Edisons Publicity Campaign
• Electric Shock
• Electric Shock
• Is AC Deadlier than DC
• Frequency Matters
• QuestionWhat current would you draw if you were unfortunate to short-circuit a 120 V line with dry hands Wet handsUse Ohmrsquo
• QuestionWhat current would you draw if you were unfortunate to short-circuit a 120 V line with dry hands Wet handsUse Ohmrsquo
• Electric Shock TherapyELECTRO CONVULSIVE THERAPY

Electric Shock TherapyELECTRO CONVULSIVE

THERAPYAn electric shock is applied to produce a convulsive seizure

The shock is typically between 140 - 170 volts and lasts between 05 and 1 seconds No explanation of how it works

Used in the treatment of 1Chronic endogenous depression 2Bipolar disorder 3Acute mania 4Certain types of schizophrenia

In the US 33000 - 50000 people receive ECT each year

• Chapter 27 Current amp Resistance
• Current Dead or Alive
• I = Coulombsecond = Ampere
• Engine Current Problem
• Current is charge in motion
• Atomic Vision of Ohmrsquos Law
• Engine Current Problem Again
• Fuse Problem
• Resistivity
• Resistivity Values
• Engine Current Problem
• You Try Problem
• Resistance Question
• Radial Resistance of a Coaxial Cable Leakage
• Resistors
• When are light bulbs more likely to blow When hot or cold
• Ohmic Material
• Nonohmic Material
• Superconductivity
• Superconductivity
• Resistance QUESTION
• What makes the Glow
• Ohmic Heat Loss
• Electrical Power
• If V = 120V What is I USE P = IV=gt I = PV
• If V = 120V What is I USE P = IV=gt I = PV
• Electric Shock
• Electric Shock
• Electric BillCost to run for 1 hr $05 per 1 kw-hr Cost = Power x Time x Rate
• Electric BillCost to run for 1 hr $05 per 1 kw-hr Cost = Power x Time x Rate
• QUESTION
• QUESTION
• QUESTION
• Circuits ProblemBulbs in Series vs Parallel
• Circuits ProblemBulbs in Series vs Parallel
• Circuits ProblemBulbs in Series vs Parallel
• Circuits Problem3 Bulbs in Series
• Circuits Problem3 Bulbs in Parallel
• Parallel Circuits
• Limitations of DC Transmission
• In The FuturehellipLong Distance AC Power Transmission may not be needed
• War of Currents 1880rsquos
• Edisons Publicity Campaign
• Electric Shock
• Electric Shock
• Is AC Deadlier than DC
• Frequency Matters
• QuestionWhat current would you draw if you were unfortunate to short-circuit a 120 V line with dry hands Wet handsUse Ohmrsquo
• QuestionWhat current would you draw if you were unfortunate to short-circuit a 120 V line with dry hands Wet handsUse Ohmrsquo
• Electric Shock TherapyELECTRO CONVULSIVE THERAPY

• Chapter 27 Current amp Resistance
• Current Dead or Alive
• I = Coulombsecond = Ampere
• Engine Current Problem
• Current is charge in motion
• Atomic Vision of Ohmrsquos Law
• Engine Current Problem Again
• Fuse Problem
• Resistivity
• Resistivity Values
• Engine Current Problem
• You Try Problem
• Resistance Question
• Radial Resistance of a Coaxial Cable Leakage
• Resistors
• When are light bulbs more likely to blow When hot or cold
• Ohmic Material
• Nonohmic Material
• Superconductivity
• Superconductivity
• Resistance QUESTION
• What makes the Glow
• Ohmic Heat Loss
• Electrical Power
• If V = 120V What is I USE P = IV=gt I = PV
• If V = 120V What is I USE P = IV=gt I = PV
• Electric Shock
• Electric Shock
• Electric BillCost to run for 1 hr $05 per 1 kw-hr Cost = Power x Time x Rate
• Electric BillCost to run for 1 hr $05 per 1 kw-hr Cost = Power x Time x Rate
• QUESTION
• QUESTION
• QUESTION
• Circuits ProblemBulbs in Series vs Parallel
• Circuits ProblemBulbs in Series vs Parallel
• Circuits ProblemBulbs in Series vs Parallel
• Circuits Problem3 Bulbs in Series
• Circuits Problem3 Bulbs in Parallel
• Parallel Circuits
• Limitations of DC Transmission
• In The FuturehellipLong Distance AC Power Transmission may not be needed
• War of Currents 1880rsquos
• Edisons Publicity Campaign
• Electric Shock
• Electric Shock
• Is AC Deadlier than DC
• Frequency Matters
• QuestionWhat current would you draw if you were unfortunate to short-circuit a 120 V line with dry hands Wet handsUse Ohmrsquo
• QuestionWhat current would you draw if you were unfortunate to short-circuit a 120 V line with dry hands Wet handsUse Ohmrsquo
• Electric Shock TherapyELECTRO CONVULSIVE THERAPY