Kirchhoff’s Laws

SPH4UW

Last Time1 2 3 ...effectiveR R R R • Resistors in series:

• Resistors in parallel:1 2 3

1 1 1 1...

effectiveR R R R

Current thru is same; Voltage drop across is IRi

Voltage drop across is same; Current thru is V/Ri

Last

Lect

ure

• Solved Circuits

• What about this one?To

day

Kirchhoff’s Rules

Kirchhoff’s Voltage Rule (KVR):Sum of voltage drops around a loop is zero.

Kirchhoff’s Current Rule (KCR):Current going in equals current coming out.

Kirchhoff’s Rules

I

a b =-IR

I

a b =IR

=+Ea b

a b =-E

Between a and b

Kirchhoff’s Laws

(1) Label all currents Choose any direction

(2) Choose loop and directionMust start on wire, not element.

(3) Write down voltage drops -Batteries increase or decrease according to which end you encounter first.-Resistors drop if going with current.-Resistors increase if gong against current.

R4

I1

I3I2 I4

R1

E1

R2

R3E2

E3

R5

A

B

e1- I1R1- I2R2-e2=0

For inner loop

PracticeR1=5 W

I

ε1- IR1 - ε2 - IR2 = 050 - 5 I - 10 -15 I = 0I = +2 Amps

e1= 50V

R2=15 We2= 10V

A

B

What if only went from A to B?, Find VB-VA

VB - VA = +IR2 + e2

= 215 + 10 = +40 Volts

VB - VA = e1 - IR1

= 50 - 25 = 40 Volts

Find I:

e2= 10V

R1=5 W

I

e1= 50V

R2=15 WA

B

Label currentsChoose loopWrite KVR

or

Therefore B is 40V higher than A

UnderstandingR1=10 W

E1 = 10 V

IB

I1

E2 = 5 VR2=10 WI2

Resistors R1 and R2 are:

1) in parallel 2) in series 3) neither

+ -

Definition of parallel:

Two elements are in parallel if (and only if) you can make a loop that contains only those two elements.

Definition of series:Two elements are in series if (and only if) every loop that Contains R1 also contains R2

PracticeR1=10 W

E1 = 10 V

IB

I1

R2=10 WI2

1) I1 = 0.5 A

2) I1 = 1.0 A

3) I1 = 1.5 A

How would I1 change if the switch was closed?

E2 = 5 V

1) Increase 2) No change 3) Decrease

Calculate the current through resistor 1.

Understanding: Voltage Law

1 1 1

11

1

0

10

101

I R

IR

V

A

<--Start

slide 7

Understanding

R1=10 W

E1 = 10 V

IB

I1

E2 = 5 VR2=10 WI2

1) I2 = 0.5 A

2) I2 = 1.0 A

3) I2 = 1.5 A

Calculate the current through resistor 2.

Starting at Star and move clockwise around loop

1 2 2 2 0E E I R V

2

2

10 5 10 0

5

100.5

V V I V

VI

A

slide 7

Kirchhoff’s Junction Rule

Current Entering = Current Leaving

I1 I2

I3

I1 = I2 + I3

1) IB = 0.5 A

2) IB = 1.0 A

3) IB = 1.5 A

R=10 W

E1 = 10 V

IB

I1=1.0A

E = 5 V R=10 W

I2=0.5

+ -

Understanding

IB = I1 + I2

= 1.0A + 0.5 A

= 1.5 A

slide 7

Kirchhoff’s Laws

(1) Label all currents Choose any direction

(2) Choose loop and directionYour choice!

(3) Write down voltage dropsFollow any loops

(4) Write down node equationIin = Iout R4

R1

E1

R2

R3E2

E3

I1

I3I2 I4

R5

A

B

You try it!

R1

R2 R3

I1 I3

I2

Loop 1:

1. Label all currents

2. Choose loop and direction

3. Write down voltage drops

Loop 2:

e1

4. Write down node equation

Node:

e2

In the circuit below you are given ε1, ε2, R1, R2 and R3. Find I1, I2 and I3.

(Choose any direction)(Current goes + - for resistor)

(Your choice! Include all circuit elements!)

Loop 1

Loop 2

+ e1 - I1R1 + I2R2 = 0

- I2R2 - I3R3 - e2 = 0

I1 + I2 = I3

3 Equations, 3 unknowns the rest is math!

Calculations

R1

R2 R3

I1 I3

I2

e1

Loop 1

Loop 2

Loop 1:

Loop 2:

+ e1 - I1R1 + I2R2 = 0

Node:

- I2R2 - I3R3 - e2 = 0

I1 + I2 = I3

1

1

2

3

2

10

50

25

100

5

V

R

R

R

V

1

1 2 2 3

1 2 2 1 2

2 1 2

10 50 25 0 25 100 5 0

50 25 10 25 10

10 5 2 20 25

0 5 0

1

I I I I

I I I I

I I

I

I I

1 2

1 2

2

20 25 1

10 5

1

2

7

I I

I I

I A

1

1

25 110

7

9

5

70

0

I

I

A

3 2

3

1

9 1

70 7

1

70I A

I I I

The negatives only indicate that our current direction

choice was wrong.

Practice Circuits

R1=25

R2=100 R3=50

I1

I3I2

e1=12V

In the circuit below you are given ε1, R1, R2 and R3.

a) Determine the total resistance of the circuit

b) Find I1, I2 and I3.

Since R2 and R3 are in parallel

3 2

1 1 1

1 1

50 1003

100100

3

P

P

R R R

R

Now RP and R1 are in serial

1

10025

358.3

T PR R R

This circuit can be broken down into a simple circuit, no need for Kirchhoff

Now: T

VI

R

120.206

58.3

VA

This is the current of I1

The potential, V across R2 and R3 is

1 1

12 25 0.206

6.85

P TV V R I

V A

V

22

6.85

1000.0685

PVIR

V

A

33

6.85

500.137

PVIR

V

A

Therefore:

Practice

1. Label all currents

In the circuit below, find ε1, I2, I3

(Directions are given)

4Ω 6Ω

I1=0.5 I3

I2

e1

Loop 1

Loop 2

12V 4V

2Ω

2. Choose loop and direction (Your choice!)

PracticeIn the circuit below, find ε1, I2, I3

4Ω 6Ω

I1=0.5 I3

I2

e1

Loop 1

Loop 2

12V 4V

2Ω

3. Write down voltage drops

Loop 1: + (0.5A)(2Ω) + ε1- 12V- I2(4Ω) = 0Loop 2: + I2(4Ω) + 12V-4V + I3(6 Ω )= 0

5. Write down node equation

Node: 0.5A + I2 = I3

PracticeIn the circuit below, find ε1, I2, I3

4Ω 6Ω

I1=0.5 I3

I2

Loop 1Loop 2

12V 4V

2Ω

1) + (0.5A)(2Ω) + ε1- 12V - I2(4Ω) = 02) + I2(4Ω) + 12V - 4V + I3(6 Ω )= 03) 0.5A + I2 = I3

3 Equations, 3 Unknownse1

1 2

3

3 2

28 6 4 0

4 0

0

11

.5

V I

V I

I

I A I

220.8 4 056 A IV I

2 28 3 6 4 0V V I I

211 10 0V I 2 1.1I A

111 4 1.1 0V V

1

1

4.4 11

6.6

V V

V

3 1.10.5

0.6

AI A

A

The “-” on the currents indicate that our original direction guess was wrong

PracticeIn the circuit below, find the current in each resistor and the equivalent resistance of the network of five resistors.

2Ω

13V

1Ω

c

1Ω 1Ω

1Ω

I2

I5I4

I3

I1

a b

d

PracticeThis “bridge” network cannot be represented in terms of series and parallel combinations. There are five different currents to determine, but by applying the junction rule to junctions a and b, we can determine then in terms of three unknown currents.

2ΩI1+ I2Loop 1

Loop 3

13V

1Ω

cLoop 2

1Ω 1Ω

1Ω

I2

I2 + I3I1 – I3

I3

I1

a b

d

I5I4

Using the current directions as guides, we will define 3 loops (3 equations for the 3 unknowns)

Practice

2ΩI1+ I2Loop 1

Loop 3+

13V

1Ω

cLoop 2

1Ω 1Ω

1Ω

I2

I2 + I3I1 – I3

I3

I1

a b

d

I5I4

Loop 1: 1 1 313 1 1 0V I I I

Loop 2:

Loop 3:

2 2 313 1 2 0V I I I

1 3 21 1 1 0I I I

This is a set of 3 equations and three unknowns. So let’s solve

PracticeLoop 1: 1 1 313 1 1 0V I I I

Loop 2:

Loop 3:

2 2 313 1 2 0V I I I

1 3 21 1 1 0I I I

From loop 3: 2 1 3I I I

Substitute this into loop 1 and loop 2 (to eliminate I2) 1 313 2 1V I I Loop 1:

Loop 2: 1 313 3 5V I I

Multiply loop 1 by 5 and adding to loop 2 and solving for I1 1

1

78 13

6

V I

I A

thus 2 5I A and 3 1I A

PracticeThe total current is: 1 2 6 5 11I I A A A

The potential drop across this is equal to the battery emf, namely 13V. Therefore the equivalent resistance of the network is:13

1.211eq

VR

A

2ΩI1+ I2Loop 1

Loop 3+

13V

1Ω

cLoop 2

1Ω 1Ω

1Ω

I2

I2 + I3I1 – I3

I3

I1

a b

d

I5I4