Physics 2020 Lectures and Clicker Quizzes

Weeks 5-6

M. GoldmanSpring, 2011

1

Sunday, February 20, 2011

Capacitor: Any two pieces of metal (conductor) brought near each other.

2

Parallel plate capacitor:L

W

d

Area A = L x W

We charge the capacitor by (somehow) putting charge onto the plates (i.e. excess electrons or deficit of electrons).

“Charged Capacitor” has no net charge.

3

+++++++++++++++++++++++

-----------------------------------------

+|Q|

-|Q|

E=0

E=0

ECharges are on the inside surfaces of the conductors due to attraction.

Electric Field (treating plates as infinite) and having area A.

d

vE =E =|σ |

ε0= 1ε0

|Q |A

⎛⎝⎜

⎞⎠⎟

E = 0

E = 0

Voltage and charge-carrying capacity

4

V =ΔV =

vE⋅Δvr =|E |d+++++++++++++++++++++++

-----------------------------------------

+|Q|

-|Q|

Ed

| E |=|σ |ε0

= 1ε0

QA> 0

A

QdV

0ε

C ≡QV

For parallel plate capacitor d

AC 0ε

Note this definitionrequires that bothQ and V be positive.C is always positive!

C ≡|Q ||V |

Clicker Question

5

An electric field of 500 V/m is desired between two parallel plates 10.0 mm apart. How large a Voltage should be applied?

A) 500 VoltsB) 0.5 VoltsC) 2.0 VoltsD) 5.0 VoltsE) None of the above answers

xx

VE ˆ

ΔΔ

rEV

ΔΔ

Clicker Question

6

LL

d

A parallel-plate capacitor has square plates of edge length L, separated by a distance d. If we doubled the dimension L and halve the dimension d, by what factor have we changed the capacitance?

A) C is unchangedB) 0.5xC) 2xD) 4xE) 8x

22 2oo o o

2LA L LC 8

d d d / 2 d

εε ε ε

What are the units of capacitance

7

V

QC

Can think of capacitor as a device for storing charge (and also energy).

Units [Capacitance] = [Coulomb/Volt] = [Farad]

Clicker Question

8

How big is a one Farad Capacitor?

Assume you have two parallel plates that are separated by 1 mm. If you estimate ε0 ~ 10-11, what is the area of the plates to have a 1.0 Farad capacitance?

A)100,000,000 square metersB)1,000 square metersC)1 square meterD) 0.1 square metersE) 0.001 square meters

d

AC 0ε

Real capacitors

9

Multi-farad capacitors in small packages are made by making d very small ~ atomic dimensions.

Normal capacitors are typically F

Capacitor stores both charge and energy.

10

Energy STORED IN CAQPACITOR = work done to put the charges on the capacitor.

+++++++++++++++++++++++

-----------------------------------------

+q

-q

Ed +dq

Suppose the capacitor is not fully charged. How much work is required to move a little bit +dq across?

How much potential energy is stored in a capacitor

11

VdqdUdWWork

dqC

qdU

C

Qqd

C

qdUU

Q

2

2

0

QVCVC

QU

2

1

2

1

2

1 22

Each next piece +dq gets harder and harder.Yes, this is an integral.

NOTE DISTINCTION BETWEEN THIS P.E. AND P.E. OFTEST CHARGE PUT IN BETWEEN CAPACITOR PLATES

Clicker Question

12

A parallel plate capacitor is charged up (+|Q| on one plate and –|Q| on the other). The plates are isolated so the charge Qcannot change. The plates are then pulled apart so that the plate separation d increases. The total electrostatic energy stored in the capacitor?

A) IncreasesB) DecreasesC) Remains constant

+++++++++++++++++++++++

-----------------------------------------

+|Q|

-|Q|

Ed

QVCVC

QU

2

1

2

1

2

1 22

Where is the energy, U?

13

Energy density of E-field =

.Vol

U

volume

energy

=

12

CV 2

Vol.=

12

ε 0Ad

⎛⎝⎜

⎞⎠⎟ |

vE | d( )

2

A ⋅d20 ||

2Eε

Work to charge capacitor is equal to energy stored.

14

QVCVC

QU

2

1

2

1

2

1 22

In general:V

QC

For parallel plate capacitor: d

AC 0ε

Stored energy in a capacitor:

How can you control the voltage or charge on a capacitor?

15

Connect battery with voltage Vto capacitor to put charge Q = CV on each plate Wire makes a

circuit.Wire is equi-potential aftercharges come to rest (neglectresistance).

Battery with voltage Vbetween its ± terminals(standard symbol)

Capacitor, C (standard symbol)

+ -

+|Q| -|Q|

Disconnect battery and charges remain on capacitor plates. Capacitor is now "storing" energy!

+ -

+|Q| -|Q|

What are capacitors good for?

16

Capacitors are good for storing large amounts of energy, which can then be accessed quickly (e.g. camera flash).

Capacitors are also ideal transducers.

Devices that convert physical quantities into electrical signals (e.g. computer keyboard, elevator buttons).

A

d d

AC 0ε

What if the gap between the capacitor plates is not empty?

17

+++++++++++++++++++++++

- - - - - - - - - - - - - - - - - - - - - - -

d

Weakens the field (screening)!

+

-

+

-

+

-

+

-

+

-

+

-

+

-

+

-

+

-

+

-

+

-

+

-

+

-

+

-

+

-

+

-

+

-

+

-

+

-

+

-

+

-

+

-

(Consider top edge of material: row of -q’s next to + plate)

Easy rule to remember… Everywhere there is an ε0, change it to ε0. C =ε0A

d⇒ ε0A

dκ usually > 1, so C increases.

Values and terminology

18

Air = 1.00054Paper = 3.5Water = 80Titanium Ceramic = 130

Kappa = κ = dielectric constantε = κε0 = electrical permittivityε0 = electrical permittivity of free spacek = 1/4πε0 = Coulomb's constant

Clicker Question

19

If we apply a fixed Voltage across the plates of a capacitor, and then we add a dielectric material in the gap, what happens?

A) The charge on the plates remains constant.B) The charge on the plates increases.C) The charge on the plates decreases.

Q = CV SO C AND Q INCREASE

By connecting capacitors C1 and C2 you effectively make a new capacitor with a new capacitance, Ctot

20

Two capacitors in parallel

21 CCCtot

C1

C2

Different rules for connecting them in different ways

21

Two capacitors in series

Ctot =1

1C1

+ 1C2

, or,

1Ctot

= 1C1

+ 1C2

C1 C2

Ctot

Equivalent to

Extend the rules to more than two capacitors

22

4 Capacitors, each with capacitance, C, in series

411111

4321

C

CCCC

Ctot

244

CCCCtot

And in parallel with 4 morecapacitors, C, in series

Clicker Question

23

A bank of three capacitors (C1, C2, C3) is connected as follows.

C1=2 F

C3=1 F

C2=1 F ° What is the effective total capacitance of the bank?

A) C = 0 FaradB) C = 1 FaradC) C = 1.5 FaradD) C = 2.0 FaradE) C = 0.5 Farad

C4 = C2+C3 = 2F, Ctot = 1/(1/2+1/2) = 1F

What happens to charge with parallel capacitors connected to a battery with voltage V?

24

+|Q1|-|Q1|

+|Q2|-|Q2|

Each capacitor has no net charge…

|Qtotal| = |Q1| + |Q2|ΔV

ΔV= V1 = V2

Ctotal = C1 + C2

|Qtotal|/ΔV =|Q1|/ΔV + |Q2|/ΔV

Voltage is same across each capacitorbut charges are not same if C1 ≠ C2

What happens to charge in capacitors in series?

25

+|Q|-|Q|+|Q|-|Q|

Center section has no net charge!(no potential difference along wire between them)

|Q|= (same) magnitude of charge on each capacitorAND on equivalent capacitor!

ΔVtot

| Q |=

ΔV1

| Q |+

ΔV2

| Q |,

1

Ctot

=1

C1

+1

C2

, Ctot =1

1C1

+1

C2

,

C1 C2

Ctot

What kind of device can we use to charge a capacitor?

26

----

++++

How do we get charges to move opposite to the direction that the Coulomb force wants them to move in?

Volta invented the battery in 1800’s.

27

Early on, people knew how to make a brief spark, but not how to keep charges moving until Volta!

----

++++ Liquid or paste inside

battery called electrolyte (often a strong acid)

Two metal plates are called electrodes.

How does a battery work?

28

By using two different metals for the electrodes, we can have two different chemical reactions.

The reactions build up charges creating the electrical potential difference between the metal electrodes.

One unit is called a cell. Many cells put together is called a battery.

Example: Car battery is six 2 V cells = 12 V battery

Car battery

29

Each cell consists of a lead (Pb) electrode and a lead oxide (PbO2) electrode immersed in a solution of water and sulfuric acid (H2SO4).

The lemon battery

30

0.9 V battery

Clicker Question

31

A standard light bulb is designed to light up with a 120 V voltage supply. Roughly how many lemon batteries does it take to light up a standard light bulb?

A) 1B) 15C)150D)1500E)You can’t light a lightbulb with lemon batteries.

Analogy between a battery and an escalator.

32

---------------------

+++++++++++++

++

+ΔVbat

Charging up a capacitor…

33

POTENTIAL DIFFERENCEBETWEEN BATTERY ANDCAPACITOR PLATESDRIVES CURRENT OFELECTRONS

NO MORE POTENTIAL DIFFERENCEBETWEEN BATTERY ANDCAPACITOR PLATES SONO MORE CURRENT

34

ΔΔ

dt

dQ

t

QiI Rate of flow of charge

Units [i]= Coulombs/second= Ampere (A) or “Amp”

Section ofwirecarrying current

ΔQ chargepasses byin time Δt

Note: if ΔQ is negative (electrons moving), current is in negative direction (to left)

Electrons can flow along inside a metal wire if there is an potential drop along the wire to push them.

35

We have alluded to charges moving already.

Push comes from

Didn’t we say |E| = 0 in a metal?Metals in electrostatic equilibrium |E| = 0

In metal wires carrying a current, there is a potential drop along the wire and an associated electric field, E in the wire. E accelerates electrons. However collisions of the electrons with the atoms of the metal keep them from accelerating and enable them to move with an average constant velocity. Collisions = resistance

vF =q

vE =−e

vE

Which way does current go?

36

Electrons flow in metals, not the protons, so the negative electric charges are moving.

-eE EeEqF

Electrons go “upstream” against the electric field vector.

Flow of negative charge in one direction is equivalent toflow of positive charge in the opposite direction

Direction of current, I = ΔQ/Δt produced by electrons moving to right

Direction of motion of electrons

Clicker Question

37

An ideal battery produces a fixed?

A)Current OutputB)Electric Potential Energy OutputC)Power Output D)Electric Potential DifferenceE)None of the Above

How many electrons are flowing by each second when when i = 1.0 A?

38

i

-e

-e

Ce 19106.1

+ direction

ΔNe electronspass by in time Δt

Why is negative flow in one direction equiva-lent to positive flow in opposite direction

• Volta's battery came before discovery of electron. Thought positive and negative fluids could flow as currents

• Didn't know yet that only negative fluid (electrons) could flow as current in metals

• Consider a charged capacitor:o Two equivalent ways that left plate could get

+|Q| when battery is first connected to capacitoro Negative fluid flowing from plate towards +

terminal of battery, ORo Positive fluid flowing from

+ terminal of battery towards plateo Increase in +charge towards plate either

way! Same current!39

+

+|Q| -|Q|

+ -

Earlier

+ -

Or

i

i

i = 10 mA = 0.01 Amps is lethal if it goes through you!But birds can sit on wires without getting electrocuted!

Under certain circumstances I could grab a wire carrying 1000 Amps and live if my body has a much higher electrical resistance than the metal wire. Electrons prefer to mostly flow through the wire (path of least resistance) Don't try it because under other conditions yourbody could provide the pathof least resistance!

How do we attach numbersto the idea of electrical resistance?

Electrocution?

40

Ohmic Materials

41

Most materials are Ohmic.

In Ohmic materials, the current (i) is proportional to the electrical potential difference (V).

High V Low V

E

V =ΔV

| i | ∝ |V | (Ohmic mat.)

constanti

V

42

Resistance, R, of a length of wire (or other material)

constanti

VR

Definition of R “Ohm’s Law” iRV This is not really a physical law.It is a good approximation for "Ohmic" materials.

Resistance has units of Ohms = = Volts/Amp

Clicker Question

43

A light bulb is attached to a battery with constant voltage V. The light filament has resistance R. When the light bulb is first turned on by attaching to the battery, the filament heats up rapidly, and as it heats, its resistance R increases (due to increased scattering of electrons by thermal vibrations). As the light bulb filament heats up, the current i in the filament

A) IncreasesB)DecreasesC)Remains the same.

V stays constant, R increases, so I = V/R decreases

Drift velocity, vd, of electrons in a wire.

44

vd = drift velocity

In metal wires carrying a current, there is a potential drop along the wire and an associated electric field, E, in the wire which accelerates electrons. However collisions of the electrons with the atoms of the metal keep them from accelerating due to E and enable them to move with an average constant velocity, vd.

These same collisions cause the wire to have resistance

Current density, j = current/(unit-area) in wire

45

ΔNe electronspass by in time Δt

A

Δx

ΔNe = [density of electrons] x [volume containing them]

= [ne] x [Δx·A]

Drift velocity of electrons in current in wire

46

Estimate drift velocity, vd, of electrons in a 1 mm diameter copper (Cu) wire carrying current, i =1 Ampere.

1. First need density of copper electrons, nCu 2. Assume one conduction electron per Cu atom so ne = nCu

3. Calculate atom density, nCu, from mass density, µCu

Cu = 9gm / cm3

nCu =Cu6×1023atoms /mol

63g/mol

⎛

⎝⎜

⎞

⎠⎟=8.6×1022Cu atoms / cm3

i = j·A= enCuvd( )· πd4

2⎛

⎝⎜

⎞

⎠⎟

vd = 9 x 10-5 m/s Compare this to thermal velocityof electrons ~ 1x106 m/s

A simple circuit

47

R

V=5V I Ideal wire Rwire = 0.

0Δ wirewire iRV

Ideal battery Rbattery = 0.

Non-ideal Battery has some internal resistance.

Difference between battery connected to capacitor and battery connected to resistor

48

R

V=5V I

Current keeps flowinguntil battery "dies" orwire is disconnected

Initial currentwhile capacitorcharges

No more currentwhen capacitoris fully charged

-+

Quick Overview of Electric Circuit Components:

49

Capacitor – stores charge and potential energy, measured in Farads (F)

Battery – generates a constant electrical potential difference (ΔV) across it. Measured in Volts (V).

Resistor – resists flow of charge due to scattering; dissipates energy. Measured in Ohms ()

First, we will consider circuits with batteries and resistors

50

R

V=5V I

Circuit calculation

51

R

V=5V I Ideal wire Rwire = 0.

0Δ wirewire iRV

Ohm’s Law iRV

= 2

AV

R

Vi 5.2

2

5

Δ0 V

5 V

Voltage drop across resistor5 V5 V5 V

0 V 0 V 0 V

i

What happens inside a resistor when current flows through it?

52

What happens to the resistor (light bulb) when we turn it on?

It emits light. It also gets hot ! Heat is a form of energy.

Power P = Energy/Time = [Joules/second] = [Watts]

Inside a current-carrying resistor, electrostatic potential energy is converted into thermal energy (heat).

Power across a resistor

53

i

V or ΔViVP

R

VRiiVP

22

If the resistor is Ohmic, then V=iR and thus

Power is energy/sec,measured in Watts.1W = 1J/s

Where did the energy go?

54

It cannot increase the drift velocity since that is set by the electric field magnitude. Does this mean Kinetic Energy is unchanged?

Energy goes into heat. Heat is in fact random (thermal) motion of particles, which is microscopic kinetic energy. However, these velocities are random and do not change the overall drift velocity.

Clicker Question

55

“100 Watt” light bulbFor all household appliances, they have:

VV 120Δ

In fact, 95% is heat, and only 5% of energy is in the light.

What is the approximate resistance of the filament (atoperating temperature)?

A)R = 100 OhmsB)R = 144 OhmsC)R = 1250 Ohms D)None of the Above

P =IΔV =ΔV2

R, R=ΔV2

P=(120)2

100

• Resistors in parallel have the same voltage drop across each but generally different currents

• Resistors in series have the same current through each but generally different voltage drops

• Fully-charged capacitors in parallel have same voltage drop across each but generally different |Q| on each

• Fully-charged capacitors in series have same |Q| on each but generally different voltage drops.

56

Junction rule for wire "splitting"

57

Junction Rule

In a steady state, must have i(in) = i(out) at any junction, otherwise charge is building up somewhere, which cannot happen in steady state.Think about water flowing in pipes analogy.

i1

i2

i3

321 iii

Calculations for a circuit

58

RA = 1.0

RB = 2.0

iRV 0Δ wireV iVP Calculate i and V across each resistor and then P.

V = 6 Volts RA RB

iA iB

itotal

iARA = iBRB = V = 6V

Clicker Question

59

Two pieces of wire, one iron and one copper, are connected in series to a voltage source. The copper wire has a much lower resistance than the iron wire. Which section of wire is going to generate the most heat?

(a)Copper(b)Iron(c)Both the same

RA

RB

Ibat IA

IB

Resistance, R, is proportional to resistivity,

60

Resistance R of a piece of conductor depends on1) Material Composition2) Shape and Dimensions

A = Area

L = LengthA

LR

Where is resistivity – measure of internal friction; dependent on material composition.

Clicker Question

61

1 2Two cylindrical resistors are made of the same material (same resistivity ). Resistor 2 is twice as long and has

twice the diameter of resistor 1. What is the ratio R2/R1?

A)2 B) 4 C) 1/2 D) 1/4 E) 1

More about current density, j = current per unit area of wire

62

iRV ΔE = jEquivalent:

Two versions of Ohm’s Law. Resistivity is independent ofarea and length of wire (intensive rather than extensive)

Resistance

Resistivityj=1

E =σE

Conductivity, σ =1/(resistivity)

Clicker Question

63

A copper cylinder is machined to have the following shape. The ends are connected to a battery so that a current flows through the copper.

A C B

Which region A, B, or C has the highest current i?A, B, C, or D) all three have the same

Current in wire is like flow of water through pipes

64

Flow of electrons in wire or resistor is like flow of water in a full pipe (no bubbles or leaks). Voltage is like water pressure!

A C B

2 gallons/min2 gal/min

2 gal/minHigh

PressureLow

Pressure

High Voltage

Low Voltage

ii

i

Clicker Question

65

A copper cylinder is machined to have the following shape. The ends are connected to a battery so that a current flows through the copper.

A C B

Which region A, B, or C has the highest current density j?A, B, C, or D) all three have the same.

Region B has the largest current density j = i/A. All 3 regions have the same current I, so the region with the smallest A has the largest J.

Clicker Question

66

A copper cylinder is machined to have the following shape. The ends are connected to a battery so that a current flows through the copper.

A C B

Which region A, B, or C has thehighest conductivity, σA, B, C, or D) all three have the same.

All three regions have same conductivity σ. Conductivity σ (and resistivity, ) are properties of material, not shape of wire.

Resistors in series: equivalent resistor.

67

itotal

ΔVbat Requiv

ΔVbat = itotalRequiv

ΔVbat

ΔVA

ΔVB

itotal

• Resistors in series have same current, i, through each but generally different voltage differences, ΔV across each resistoro iA = iB = itotal ΔVA ≠ ΔV

• Sum of voltage drops across both resistors = voltage drop across battery: ΔVA + ΔVB = ΔVbat

• Use Ohm's law, iR = ΔV for each resistor and for equiv. resistor

What is the current through the circuit?

68

V

RA

RB

Ibat IA

IB

RA = 1.0

RB = 3.0

V = 6 Volts

ibat =ΔVbat

Requiv

= ΔVbat

RA +RB

=6V4

=1.5A

itotal

ΔVbat Requiv

ΔVbat = itotalRequiv

What is the Voltage change across each resistor?

69

V

RA

RB

Ibat IA

IB

RA = 1.0 iAA=1.5 Amps=1.5 Amps

RB = 3.0 iB = 1.5 Amps

V = 6 Volts

VoltsARiVV AAAA 5.1)0.1)(5.1( ΔVoltsARiVV BBBB 5.4)0.3)(5.1( Δ

ΔVbat

ΔVA

ΔVB

itotal

Another way to think about voltage changes and current changes: go around a loop!

• Total voltage change going around any loop is zero

70

ΔVbat

ΔVA

ΔVB

itotal

PREVIOUSΔVA + ΔVB = ΔVbat

LOOP EQUIVALENTΔVA + ΔVB + ΔVbat = 0

= - ΔVbat

Clicker Question

71

V

RA

RB

Ibat IA

IB

If we consider the total voltage change, starting at Point P and going all the way around the circuit loop back to Point P, what is the total ΔV?A) ΔV = +6.0 VoltsB) ΔV = +1.5 VoltsC) ΔV = -4.5 VoltsD) ΔV = 0.0 VoltsE) ΔV = -6.0 Volts

Point P

How to visualize changes in potential as we go around loop from P1 to P4

72

Clicker Question

73

We start with the left circuit with one lightbulb (A). If we add a second bulb (B) as shown on the right, what happens to bulb A?

V A B

A)Bulb A is equally bright.

B)Bulb A is dimmer than before

C)Bulb A is brighter than before

RA RB

iA iB

itotal

ΔVbat

Resistors in parallel: equivalent resistor• Resistors in parallel have the same voltage difference, ΔV,

across each but generally different currents, i, through eacho ΔVA = ΔVB = ΔVbat iA ≠ iB

• Use junction rule, iA + iB = itotal

• Use Ohm's law, i = ΔV/R for each resistor A and B and for Requiv

74

ΔVA ΔV

B

itotal

ΔVbat Requiv

ΔVbat = itotalRequiv Requiv =

11RA

+ 1RB

1

Requiv

= 1RA

+ 1RB

RA RB

iA iB

itotal

ΔVbat ΔVA ΔV

B

ΔVA+ ΔVB = 0

RA RB

iA iB

itotal

ΔVbat ΔVA ΔV

B

ΔVB+ ΔVbat = 0

Can also go around any of three loops for two resistors in parallel

75

RA RB

iA iB

itotal

ΔVbat ΔVA ΔV

B

Previous treatment

RA RB

iA iB

itotal

ΔVbat ΔVA ΔV

B

ΔVA = ΔVB = ΔVbat

ΔVA+ ΔVbat = 0

ΔVbat = - ΔVbat

ΔVA = - ΔVA

ΔVbat = - ΔVbat

Voltages, resistances and currents in more complicated circuits

• Could keep using equivalent resistor concept over and over again, OR

• Use Kirchoff's laws after selecting junctions and loops

76

Kirchhoff’s Voltage Loop Law:

The change in Voltage around any closed loop must be zero.

Kirchhoff’s Current Junction Law:

In steady state, the current going into a junction (or point) must equal the current going out of that junction (or point).

Resistors in Parallel

77

R1

R2

R3

Find Requiv for three resistors in parallel

78

R1

R2

R3

Equivalent resistor for three resistors in parallel

1

R5

= 1R4

⎡

⎣⎢

⎤

⎦⎥+

1R3

= 1R1

+ 1R2

⎡

⎣⎢

⎤

⎦⎥+

1R3

321

1111

RRR

Req

or,

R1

R2

R3

R1

R2

R3

R5R4

1

R4

= 1R1

+ 1R2

Practice the current junction rule

79

R1

R2

R3

V

What if we apply the current junction rule at this point?

i

i1i2

i3

321 iiii If all three resistors were the same then:

iiii3

1321

Compare resistance of one resistor circuit with equivalent resistance for three in parallel.

80

RRR

VAcross each resistor, voltage = V

R

V

Through each resistor, current = iR = V/R (Ohm’s Law)

Total current = iR & resistance R = V/iR

Total current = 3iR & equivalent resistance = Requiv = V/3iR = R/3

One resistor circuitThree identical resistors in parallel circuit

Clicker Question

81

R1

R2

R3

V

If we connect a battery with V = 6 Volts with three resistors R1 = 10 , R2 = 20 , R3 = 30 ,which of the following is true?A)The current through each resistor

is the same.B)The Voltage change across

each resistor is the same.C)The resistance of each resistor is

the same.D)None of the above are true.

Clicker Question

82

A A

B

We start with the left circuit with one lightbulb (A). The brightness of the bulb directly reflects the power. If we add a second bulb (B) as shown on the right, what happens to the bulb A?

A)Bulb A is equally bright.

B)Bulb A is dimmer than before

C)Bulb A is brighter than before

Which is the best way to wire a house?

83

Clicker Question

84

What is the electric potential difference across the upper bulb (resistor)?

A) |V| = 0 VoltsB) |V| = 6 VoltsC)|V| = 12 VoltsD)|V| = 24 VoltsE)None of the above answers.

Clicker Question

85

If each of these six light bulbs is identical, which bulb is going to be the brightest?

A)Bulb AB)Bulb BC)Bulb CD)Bulb DE)Bulb E

Brightness is proportional to power, P = I(ΔV) = (ΔV)2/R. It is R which is the same for all bulbs, not the power. Only bulb A has the full voltage, ΔVbatt of the battery across it. The other bulbs have smaller potential drops across them and so generate less power

Clicker Question

86

The three light bulbs A, B, and C are identical. How does the brightness of bulbs B and C together compare with the brightness of bulb A?

V=12V

A

B C

A) Total power in B+C = power in A.B) Total power in B+C > power in A.C) Total power in B+C < power in A.

Answer: The voltage drop, ΔV across A is the same as ΔV across the equivalent resistor Requiv = RB + RC = 2RA Use P = (ΔV)2/R separately for A and for the equivalent resistor Requiv. The higher resistance has the lower power so the total power in B+C < power in A.

Logic for previous Clicker Question

87

The three light bulbs A, B, and C are identical. How does the brightness of bulbs B and C together compare with the brightness of bulb A?

V=12V

A

B C

A) Total power in B+C = power in A.B) Total power in B+C > power in A.C) Total power in B+C < power in A.

Logic: Simplify by using equivalent resistor, Requiv, for resistors B and C. Requiv = 2RA Total power dissipated by B and C = power dissipated by Requiv

Which is best formula to use for power? P = iΔV,P = i2R or P = ΔV2/R?.

We know voltages across RA and across Requiv are 12V becausethey are in parallel. Hence use P = ΔV2/R. Compare PA = (12)2/RA with PBC = (12)2/Requiv= (12)2/(2RA)

Requiv

Clicker Question

88

V=12V

1

2

3

In the circuit below, what happens to the brightness of bulb 1, when bulb 2 burns out? (When a bulb burns out there is no current through it acquires an effectively infinite resistance)

A) Bulb 1 gets brighterB) Bulb 1 gets dimmer.C) It's brightness remains the same.(Hint: What happens to the current through bulb 1 if bulb 2 burns out.)

As R2 gets larger,1/Requiv gets smaller so Requiv gets larger and i gets smallerso P1= i2R1 gets smaller

i = 12VR1 +Requiv

, 1Requiv

= 1R3

+ 1R2

Logic for previous clicker Question

89

V=12V

1

2

3

In the circuit below, what happens to the brightness of bulb 1, when bulb 2 burns out? (When a bulb burns out there is no current through it acquires an effectively infinite resistance)

A) Bulb 1 gets brighterB) Bulb 1 gets dimmer.C) It's brightness remains the same.(Hint: What happens to the current through bulb 1 if bulb 2 burns out.)

iwith2 =12V

R1 +Requiv

Logic: Hint asks about current through 1

Compare current without bulb 2 to current with bulb 2

iwithout 2 =12V

R1 +R3 Requiv =1

1R3

+ 1R2

= R3

1+ R3

R2

< R3

How do we use Kirchoff's law to solve for currents, resistances or voltages in complicated circuits

90

Kirchhoff’s Voltage Loop Law:

The change in Voltage around any closed loop must be zero.

Kirchhoff’s Current Junction Law:

In steady state, the current going into a junction (or point) must equal the current going out of that junction (or point).

Example 1: Choose R1 = R2 = R3 = 10 Find the currents, i1, i2, i3

91

V=12V

1

2

3

1. Longer line (+) on battery is higher voltage and shorter line (-) is lower voltage, by convention.

+-

2. We label the current in each section as shown. Note that we can guess the directions. The answer will be independent of our guess but pay attention to the sign.

i1i1

i1

i1

i2

i3i1

Apply Voltage Loop Rule around each possible loop in the circuit.

92

V=12V

1

2

3

+-

i1i1

i1

i1

i2

i3i1

Make sure to label the direction of your loop!

Make sure to pick a starting point to go around the loop.

Then add up the Voltage changes around the loop.

Critical Sign Convention for battery voltages!Critical Sign Convention for battery voltages!

93

- + If your loop goes through a battery from – to + the Voltage increases (e.g. ΔV = +12 V)

- + If your loop goes through a battery from + to – the Voltage decreases (e.g. ΔV = -12 V)

Critical Sign Convention for resistancesCritical Sign Convention for resistances

94

iIf you go across a resistor and the loop direction and guessed current direction are the same, the voltage decreases (e.g. ΔV = -iR)

iIf you go across a resistor and the loop direction and guessed current direction are opposite, the voltage increases (e.g. ΔV = +iR)

Loop rule: sum of voltages differences around any loop equals zero.

95

V=12V

1

2

3

+-

i1i1

i1

i1

i2

i3i1

Starting at Point P go around the loop indicated.

P

VVbat 12Δ333 RiVR Δ

012 1133 Δ RiRiVsum

111 RiVR Δ

Now go around a different closed loop starting at Point P around the new loop

96

V=12V

1

2

3

+-

i1i1

i1

i2

i3i1

P

.

VVbat 12Δ222 RiVR Δ

012 1122 Δ RiRiVsum

111 RiVR Δ

Not enough equations yet

97

012 1122 Δ RiRiVsum

ΔVsum = +12 − i3R3 − i1R1 = 0

For our specific circuit, we used R1 = R2 = R3 = 10

0101012 12 ii

0101012 13 ii

V=12V

1

2

3

i2

i3i3R3 −i2R2 =0

Now use current junction rule to supply third independent equation

98

V=12V

1

2

3

+-

i1i1

i1

i3i1

i2

Q

Apply the current junction rule at point Q below.

132 iii

ii outin

i1

With this added information can now find the currents

99

0101012 12 ii

0101012 13 ii

132 iii

Solve 3 equations and 3 unknowns.

Ampsi

Ampsi

Ampsi

4.0

4.0

8.0

3

2

1

Reality check

100

V=12V

1

2

3

+-

i1i1

i1

i1

i2

i3i1

Ampsi

Ampsi

Ampsi

4.0

4.0

8.0

3

2

1

Does this make sense?i2 = i3 since two parallel, equal resistance paths.Current from R2 and R3 must go into R1.

Clicker Question

101

V=12V

1

2

3

+-

i1i1

i1

i1

i2

i3i1

i1 0.8Ampsi2 0.4Ampsi3 0.4Amps

10

10

10

3

2

1

R

R

R

What is the value of theequivalent resistor that would replace the three resistors below?A)Req = 5 B)Req = 10 C)Req = 15 D)Req = 30 E)None of the Above