1

Circuits and Ohm’s Law

SPH4UWElectric Terminology

Current: Moving Charges

Symbol: I

Unit: Amp Coulomb/second

Count number of charges which pass point/sec

Power: Energy/Time

Symbol: P

Unit: Watt Joule/second = Volt Coulomb/sec

P = 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/R

Double 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

r

2

Simple Circuit

Practice…

Calculate I (current) when e = 24 Volts and

R = 8 W

Re

I

I

Ohm’s Law: V =IR24

8

3

V

VI

R

A

W

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

e0R10=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

3

Practice:

Resistors in SeriesCalculate 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 V

I1=I2=V12/R12 = 2 Amps

I1R1= 2 x 1 = 2 Volts

I2R2= 2 x 10 = 20 Volts

Yes

Resistors in Parallel

Two wires:

Effectively adding the Area

Since 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 = Ieq

Voltage 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

4

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

W W W

W

23

60

12

5

VI

A

W

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

SeriesParallel

Summary

Different for each resistor.

Vtotal = V1 + V2

Increases

Req = R1 + R2

Same for each resistor

Itotal = I1 = I2

Same for each resistor.

Vtotal = V1 = V2

Decreases

1/Req = 1/R1 + 1/R2

WiringEach resistor on the

same wire.

Each resistor on a

different wire.

Different for each resistor

Itotal = I1 + I2

R1 R2

R1

R2

5

Understanding

Which configuration has the smallest resistance?

1

2

3

1 2 3

Which configuration has the largest resistance?

1

2

3

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

Understanding

Determine 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 17

6.04

EQ

EQ

R R R R

R

W W W

W

Let’s now add

R4 to this REQ

4

6.04 5

11.04

tot EQR R R

W W

W

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

W

R4:

4

4 4

0.91

0.906 5

4.5

tot

I A

V I R

A

V

W

1

1

1

5.5

15

0.36

5.5

EQVI

R

V

A

V V

W

2

2

2

5.5

25

0.22

5.5

EQVI

R

V

A

V V

W

R1: R2: R3:3

3

2

5.5

17

0.32

5.5

EQVI

R

V

A

V V

W

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

6

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

W W

W5 4.117

9.117

TR W W

W

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

9.117

1.32

V IR

VI

V

R

A

W

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

W

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

7

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

e

W W

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.

5Ve

2.5R W

5

2.5

2

VI

R

V

A

W

2 5 )

10

P IV

A V

W

22

5

.5

V

A

I

V

R

W

2

25

2.5

10

VP

R

V

W

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?

5Ve

2.5R W

5

1.25

4

VI

R

A

V

W

4 5 )

20

P IV

A V

W

22

5

.5

V

A

I

V

R

W

2

25

2.5

10

VP

R

V

W

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 W

Because the bulbs

(resistors) are in parallel,

we use the parallel law to

determine total resistance

of the circuit.

5Ve

2.5R W

5

2.5

2

VI

R

V

A

W

2 5 )

10

P IV

A V

W

22

5

.5

V

A

I

V

R

W

2

25

2.5

10

VP

R

V

W

W

8

5Ve

2.5R W

5

1

5

VI

R

V

A

W

5

5

1 )

P IV

A V

W

2.51

2.5

V IR

A

V

W

2

22.5

2.5

2.5

VP

R

V

W

W

Let us calculate the Current and the Power (used/generated) by

the elements of the following circuit when bulbs are in series.

1

A

2.5R W

Because the bulbs

(resistors) are in series,

we use the series law to

determine total resistance

of the circuit.

5Ve

2.5R W

5

2.5

2

VI

R

V

A

W

2 5 )

10

P IV

A V

W

22

5

.5

V

A

I

V

R

W

2

25

2.5

10

VP

R

V

W

W

Understanding

SPH4UW, Ohm’s Law, Slide 30

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

W W

W

Now we have 3

resistors in series

50 200 150

400

tot

eq

R

R

W W W

WThus we can determine the

current supplied by the battery

24006

400R

VAI

e

W

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

9

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 W

Between points a and b

Therefore the

potential at point b

is 900V lower than

at point a

2400V