Circuits and Ohm’s Law

SPH4UW

Electric TerminologyCurrent: Moving Charges

Symbol: IUnit: Amp Coulomb/secondCount number of charges which pass point/sec

Power: Energy/Time Symbol: PUnit: Watt Joule/second = Volt Coulomb/secP = IV

Physical Resistor

Resistance: Traveling through a resistor, electrons bump into things which slows them down. R = r L /A r: L: A:

Ohms Law I = V/RDouble potential difference double current

A

L

Resistivity (E/J or Ωm)

Length

Area

UnderstandingTwo cylindrical resistors are made from the same material. They are of equal length but one has twice the diameter of the other.

1. R1 > R2

2. R1 = R2

3. R1 < R2

21

LR

A

Simple Circuit

Practice…Calculate I (current) when e = 24 Volts and

R = 8 W

Re

I

I

Ohm’s Law: V =IR24

83

V

VI

R

A

Resistors in Series

One wire: Effectively adding lengths:

Req=r(L1+L2)/A

Since R L, add resistance:R

R

= 2RReq = R1 + R2

Resistors in Series

Resistors connected end-to-end:If current goes through one resistor,

it must go through other.

I1 = I2 = Ieq

Both have voltage drops:

V1 + V2 = VeqR1

R2

Req

Req = Veq = V1 + V2 = R1 + R2 Ieq Ieq

UnderstandingCompare I1 the current through R1, with I10 the current through R10.

1. I1 < I10

2. I1 = I10

3. I1 > I10

R1=1W

e0 R10=10W

Compare V1 the voltage across R1, with V10 the voltage across R10.

1. V1>V10 2. V1=V10 3. V1<V10

ACT: Series Circuit

Note: I is the same everywhere in this circuit!

V1 = I1 R1 = 1 x I

V10 = I10 R10 = 10 x I

Practice:Resistors in Series

Calculate the voltage across each resistor if the battery has potential V0 = 22 volts.

• R12 =

• V12 =

• I12 =

R12e0

Expand:• V1 =

• V2 =

R1=1W

e0 R2=10W

Check: V1 + V2 = V12 ?

Simplify (R1 and R2 in series):

R1=1W

e0 R2=10W

R1 + R2= 11 W

V1 + V2= 22 VI1=I2=V12/R12 = 2 Amps

I1R1= 2 x 1 = 2 VoltsI2R2= 2 x 10 = 20 Volts

Yes

Resistors in ParallelTwo wires:

Effectively adding the AreaSince R a 1/A add 1/R:

R R = R/2

1 2

1 1 1

eqR R R

Resistors in Parallel

Both ends of resistor are connected:Current is split between two wires:

I1 + I2 = IeqVoltage is same across each:

V1 = V2 = Veq

ReqR2R11

1 2

211 1 1eq

eq eqeqR R R

I I

V V

UnderstandingWhat happens to the current through R2 when the switch is closed?

1) Increases

2) Remains Same

3) Decreases

What happens to the current through the battery?

(1) Increases

(2) Remains Same

(3) Decreases

ACT: Parallel Circuit

V2 = ε = I2R2

Ibattery = I2 + I3

Practice: Resistors in Parallel

Determine the current through the battery.Let E = 60 Volts, R2 = 20 W and R3=30 W.

R2 R3e

R23e

Simplify: R2 and R3 are in parallel

23 2 3

23 2 3

2323

23

1 1 1

60

R R R

V V V V

VI

R

23

23

1 1 1 1

20 30 12

12

R

R

23

60

125

VI

A

Johnny “Danger” Powers uses one power strip to plug in his microwave, coffee pot, space heater, toaster, and guitar amplifier all into one outlet.

1. The resistance of the kitchen circuit is too high.

2. The voltage across the kitchen circuit is too high.

3. The current in the kitchen circuit is too high.

Toaster Coffee Pot Microwave

10 A 5 A 10 A

25 A

This is dangerous because…(By the way, power strips are wired in parallel.)

Understanding

• Resistors– Physical R = r L/A– Series Req = R1 + R2

– Parallel 1/Req = 1/R1 + 1/R2

– Power P = IV

Summary

Voltage

Current

Resistance

Series Parallel

Summary

Different for each resistor.Vtotal = V1 + V2

IncreasesReq = R1 + R2

Same for each resistorItotal = I1 = I2

Same for each resistor.Vtotal = V1 = V2

Decreases1/Req = 1/R1 + 1/R2

WiringEach resistor on the same wire.

Each resistor on a different wire.

Different for each resistorItotal = I1 + I2

R1 R2

R1

R2

Understanding

Which configuration has the smallest resistance?

1

2

3

1 2 3

Which configuration has the largest resistance?123

R 2R R/2

Parallel + Series Tests

Resistors R1 and R2 are in series if and only if every loop that contains R1 also contains R2

Resistors R1 and R2 are in parallel if and only if you can make a loop that has ONLY R1 and R2

UnderstandingDetermine the voltage and current in each resistor

R3=17W

e0=10V

R4=5W

R2=25W

R1=15WFirst we notice the voltage drop through these two resistor groupings total 10V.

Let’s combine to find REQ

1 2 3

1 1 1 1

1 1 1

15 25 176.04

EQ

EQ

R R R R

R

Let’s now add R4 to this REQ

4

6.04 5

11.04

tot EQR R R

Understanding

R3=17W

e0=10V

R4=5W

R2=25W

R1=15W

e0=10V RT=11.04W 100.906

11.04

V VI A

V IR

R

R4:

4

4 4

0.91

0.906 5

4.5

tot

I A

V I R

A

V

11

1

5.5

150.36

5.5

EQVIR

V

A

V V

22

2

5.5

250.22

5.5

EQVIR

V

A

V V

R1: R2: R3: 33

2

5.5

170.32

5.5

EQVIR

V

A

V V

Since voltage drop has to total

10V, VEQ=10V-4.5V=5.5V

Let’s determine the current, IT

R1, R2, R3, experience the same voltage

This current will flow through each of the resistor groupings

5Ω

7Ω12V 10Ω

The V-I-R Chart

V I RR1R2R3Total

Determine the Voltage, Current, and Resistance

Step 1:

Fill out the table with known resistors and the Total Voltage for circuit

V I RR1 5R2 7R3 10Total 12

5Ω

7Ω12V 10Ω

The V-I-R Chart

V I RR1 5R2 7R3 10Total 12

Determine the Voltage, Current, and Resistance

Step 2:

Using resistor laws, determine total resistance of circuit.

1 1 1

7 10

4.117

eq

eq

R

R

5 4.117

9.117TR

V I RR1 5R2 7R3 10Total 12 9.117

5Ω

7Ω12V 10Ω

The V-I-R Chart

Determine the Voltage, Current, and Resistance

Step 3:

Using Ohm’s law, determine the Current of circuit.

V I RR1 5R2 7R3 10Total 12 9.117

12

9.1171.32

V IR

VI

VR

A

V I RR1 5R2 7R3 10Total 12 1.32 9.117

Step 4:

Since initial current amount will also pass through resistor R1, we can determine its voltage drop.

1.32 5

6.6

A

V IR

V

V I RR1 6.6 1.32 5R2 7R3 10Total 12 1.32 9.117

5Ω

7Ω12V 10Ω

The V-I-R Chart

Determine the Voltage, Current, and Resistance

Step 5:

Since R2 and R3 have the same Voltage drop, we then must have 12V-6.6V=5.4V.

V I RR1 6.6 1.32 5R2 7R3 10Total 12 1.32 9.117

V I RR1 6.6 1.32 5R2 5.4 7R3 5.4 10Total 12 1.32 9.117

5Ω

7Ω12V 10Ω

The V-I-R Chart

Determine the Voltage, Current, and Resistance

Step 6:

Use ohm’s law to find currents

V I RR1 6.6 1.32 5R2 5.4 7R3 5.4 10Total 12 1.32 9.117

V I RR1 6.6 1.32 5R2 5.4 0.77 7R3 5.4 0.54 10Total 12 1.32 9.117

Understanding

R3=17W

e0=10V

R4=5W

R2=25W

R1=15WLet’s follow a conventional current path through R1

0 1 4

10 0.366 15 0.91 5

10 5.5 4.5

0

TV IR IR

V A A

V V V

V

You can pick any path through the circuit and the total voltage increases and decrease will balance

You can reverse the direction of the current and thus the signs, (batteries increase the voltage, resistors drop the voltage) and obtain the same results.

5V

2.5R

5

2.52

VI

RV

A

2 5 )

10

P IV

A V

W

22

5

.5

V

A

I

V

R

2

25

2.510

VP

R

V

W

Let us calculate the Current and the Power (used/generated) by the elements of the following circuit.

What happens to the Power delivery and consumption if another identical bulb is place in parallel or in series with the first?

5V

2.5R

5

1.254

VI

R

A

V

4 5 )

20

P IV

A V

W

5 2.5

2

I

I

I

V R

V

A

2

25

2.510

VP

R

V

W

Let us calculate the Current and the Power (used/generated) by the elements of the following circuit when bulbs are in parallel.

2A

2A

4A

2.5R

Because the bulbs (resistors) are in parallel, we use the parallel law to determine total resistance of the circuit.

5V

2.5R

5

2.52

VI

RV

A

2 5 )

10

P IV

A V

W

22

5

.5

V

A

I

V

R

2

25

2.510

VP

R

V

W

5V

2.5R

51

5

VI

RV

A

5

5

1 )

P IV

A V

W

2.51

2.5

V IR

A

V

2

22.5

2.52.5

VP

R

V

W

Let us calculate the Current and the Power (used/generated) by the elements of the following circuit when bulbs are in series.

1A

2.5R

Because the bulbs (resistors) are in series, we use the series law to determine total resistance of the circuit.

5V

2.5R

5

2.52

VI

RV

A

2 5 )

10

P IV

A V

W

22

5

.5

V

A

I

V

R

2

25

2.510

VP

R

V

W

Understanding

v

A

a

bJ

A voltmeter is a device that’s used to measure the voltage between two points in a circuit. An ammeter is used to measure current. Determine the readings on the voltmeter and the ammeter

2400V

Understanding

v

A

a

bJ

Let’s first combine the 2 parallel resistors

1 1 1

600 300

200

eq

eq

R

R

Now we have 3 resistors in series

50 200 150

400tot

eq

R

R

Thus we can determine the

current supplied by the battery2400

6400R

VAI

2400V

Understanding

v

A

a

bJ

At junction J this 6A current splits. Since the top resistor is twice the bottom resistor, half as much current will flow through it.

Therefore the current through the top resistor is 2A and the bottom resistor is 4A

6A

6A

6A

6A

2A

4A

Thus the reading of the ammeter is 4A

2400V

Understanding

v

A

a

bJ

6A

6A

6A

6A

2A

4A

The voltmeter will read a voltage drop of : 3 3 6 150 900AI R V

Between points a and b

Therefore the potential at point b is 900V lower than at point a

2400V