Ohm’s Law

V = I R

+ _

V

IR

Ohm’s Law: V = IR

• A fundamental relationship in electric circuits. • Describes how much potential difference is

required to move charges through a resistance at a given current.

• Materials that have constant resistances are said to obey Ohm’s Law

Power and Energy in Electric Circuits

• Rate at which energy is supplied to a circuit is Power

P = VI measured in WattsCan also be stated as:

P = VI = (IR) I = I2Ror

P = VI = V (V/R) = V2/R

EXAMPLES1) Calculate the rate at which energy is supplied

by a 120 V source to a circuit if the current in the circuit is 5.5 A

Solution: P=VI = (120V)(5.5 A) = 660 W

2) A 150 Ω resistor carries a current of 2.0 A. Calculate the rate at which heat energy is produced in resistor

Solution: Pheat = I2R = (2.0 A)2(150 Ω) = 600 W

Energy (W)

• Recall: Power = Work / timeTherefore

Work (energy) = Power x time

W = Pt = VIt = I2RT = V2t / R

Unit is joules

Example

• How much energy is produced by a 50 V source that generates a current of 5.0 A for 2 minutes?

Solution: Don’t forget time must be in seconds!

W = VIt = (50 V)(5.0 A)(120 s) = 30,000 J = 30 kJ

Series Circuits

Or why old Christmas lights used to all go out when only one bulb was

broken

Series Circuit

• Has only one current path and if that path is interrupted, the entire circuit ceases to operate.

• The diagram represents a circuit containing three resistors in series with meters placed to measure various characteristics of the circuit

• - A - represents an ammeter, a very low resistance device that measures current in a circuit.

• - V - represents a voltmeter, a very high resistance device that measures potential difference across a circuit

In Series circuits…

• Current throughout the circuit is constant; therefore ammeter can be placed at any position.

• Potential Difference is equal to the SUM of the potential differences across all resistances– Known as Kirchhoff’s first rule (or simply the loop

rule)

• Ohm’s law holds for each resistance

For series circuits

• I = I1 = I2 = I3 = …

• V = V1 + V2 + V3 + …

• Req = R1 + R2 + R3 + … - this is the equivalent resistance of the circuit

Example – calculate the meter readings

24 V

3Ω 6Ω 9Ω

Vt

Solution

• First find equivalent resistanceReq = R1 + R2 + R3 + … = 3Ω + 6Ω + 9Ω = 18Ω

• The total potential difference, VT = 24 V since the source supplies the entire circuit

• The current through circuit (I) isV=I Req 24 V = I (18Ω) I = 1.33 A

Solution cont’d

• Potential difference across each resistance can be found using Ohm’s Law V=IR

V1 = (1.33 A) (3Ω) = 4 V

V2 = (1.33 A) (6Ω) = 8 V

V3 = (1.33 A) (9Ω) = 12 V

Important Fact!

• As the number of resistances in a series circuit increases, the equivalent resistance, of the circuit increases and the current through the circuit decreases.

Example

• Suppose a fourth resistance of 18Ω is added to the series circuit.

Calculate (a) equivalent resistance of circuitReq = R1 + R2 + R3 + …

= 3Ω + 6Ω + 9Ω + 18Ω = 36Ω(b) the current through the circuit

V=I Req 24 V = I (36Ω) I = 0.67 A

Parallel Circuits

Parallel Circuits

Parallel Circuits

• Have more than one current path.• If a segment of a // circuit is interrupted, the

result will not necessarily be that the entire circuit ceases to operate.

• House wiring is in //.

Parallel circuits

• Current separates into more than one path.• The point where separation occurs is known

as a junction • The sum of the currents entering a junction

must equal the sum of the currents leaving the junction

• This is Kirchoff’s second rule (or simply the junction rule)

example

• In the diagram below, what are the magnitude and the direction of the current in wire X?

1A

2A

4A

X

For any // circuit

• V = V1 = V2 = V3 = … = Vn voltage is constant

• I = I1 + I2 + I3 + … + In current through entire circuit is equal to the sum of the currents through all resistances (Kirchoff’s 2nd rule)

• Vn = InRn

• neq RRRRR

1...

1111

321

example

• Calculate (a) the equivalent resistance (b) Currents I1 and I2

(c) Total currentof the following circuit.

3 Ω

6 Ω

24 V

I1

IT

I2

Solution

a) To find equivalent resistance:

2

2

1

6

1

3

11

1...

1111

321

eq

eq

neq

R

R

RRRRR

Solution (cont’d)

b) To find I1 and I2 and c) IT

3 Ω

6 Ω

24 V

I1

IT

I2AAAIII

AV

R

VI

AV

R

VI

T 1248

46

24

83

24

21

22

11

Important Note• The equivalent resistance is less than any single

resistance in the circuit.• If more resistance is added in //, the equivalent

resistance decreases and the total current increases.– The result is roughly equivalent to increasing the cross-

sectional area of a conductor.• This is why overloading a household circuit by

connecting too many electrical appliances is dangerous. – As current increases, the amount of heat energy also

increases. This can lead to fires– Fuses and circuit breakers are designed to prevent such

fires from occurring.

example• A 2 – ohm resistor is added in parallel to the

previous circuit. Calculate (a) the equivalent resistance and (b) the total current of the altered circuit.

3 Ω

6 Ω

24 V

I1

IT

I2

2 ΩI(2Ω)

AV

R

VIb

R

R

RRRRRa

eq

eq

eq

neq

241

24)(

1

1

1

2

1

6

1

3

11

1...

1111)(

321