# kirchhoff's Law - The Burns Home Page kirchhoff's Law - The Burns Home Page

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SPH4UW

Kirchhoff’s Laws

Last Time

• Resistors in series: Reffective

R R R

• Resistors in parallel:

1 2 3

...

Current thru is same; Voltage drop across is IR i

1 1 1 1

...

R R R R

effective

1 2 3

Voltage drop across is same; Current thru is V/R i

• Solved Circuits

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

a b =-IR

I

a b =IR

I

a

b =+E

a b =-E

1

(1) Label all currents

Choose any direction

(2) Choose loop and direction

Must start on wire, not element.

Kirchhoff’s Laws

(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.

e 1 - I 1 R 1 - I 2 R 2 -e 2 =0

For inner loop

A

R 1 I 1

R 2

E 3

B

E 1 I 2 I 3

E2 R 3

R 5

I4

R 4

Label Choose Write KVR currents loop

or

Find I:

ε 1 - IR 1 - ε 2 - IR 2 = 0

50 - 5 I - 10 -15 I = 0

I = +2 Amps

Practice

e 1 = 50V

A

R 1 =5 W

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

V B - V A = e 1 - IR 1

= 50 - 25 = 40 Volts

V B - V A = +IR 2 + e 2

= 215 + 10 = +40 Volts

Therefore B is 40V higher than A

e 1 = 50V

A

R 2 =15 W

R 1 =5 W

R 2 =15 W

I

I

B

e 2 = 10V

B

e 2 = 10V

Understanding

Practice

slide 7

Resistors R 1 and R 2 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.

I 1

E 2 = 5 V

I 2

I B

R 1 =10 W

R 2 =10 W

+ -

E 1 = 10 V

Definition of series:

Two elements are in series if (and only if) every loop that

Contains R 1 also contains R 2

Calculate the current through resistor 1.

1) I 1 = 0.5 A

2) I 1 = 1.0 A

3) I 1 = 1.5 A

e IR

0

1 1 1

e1

I1

R

I 1 R 1 =10 W

E 2 = 5 V

I 2 R 2 =10 W

I B

Understanding

slide 7

Kirchhoff’s Junction Rule

slide 7

Calculate the current through resistor 2.

Current Entering = Current Leaving

I 1

R 1 =10 W

I 1 = I 2 + I 3

I 1 I 2

1) I 2 = 0.5 A

2) I 2 = 1.0 A

3) I 2 = 1.5 A

E 2 = 5 V

I 2

R 2 =10 W

Understanding

I 3

I 1 =1.0A R=10 W

I B

E1 E2 I2R2 0V

10V 5V 10W I 2

0V

5V

I2

10W

0.5A

E 1 = 10 V

Starting at Star and move

clockwise around loop

1) I B = 0.5 A

2) I B = 1.0 A

3) I B = 1.5 A

I B = I 1 + I 2

= 1.0A + 0.5 A

= 1.5 A

E = 5 V

R=10 W

I 2 =0.5

I B

+ -

E 1 = 10 V

(1) Label all currents

Choose any direction

(2) Choose loop and direction

(3) Write down voltage drops

Kirchhoff’s Laws

R 2

(4) Write down node equation

E 1 I 2

I4

I in = I out

R 3 R 4

A

R 1

I 1

I 3

E 3

B

E2

R 5

You try it!

In the circuit below you are given ε 1 , ε 2 , R 1 , R 2 and R 3 . Find I 1 , I 2 and I 3 .

1. Label all currents

2. Choose loop and direction

3. Write down voltage drops

Loop 1:

Loop 2:

4. Write down node equation

Node:

+ e 1 - I 1 R 1 + I 2 R 2 = 0

- I 2 R 2 - I 3 R 3 - e 2 = 0

I 1 + I 2 = I 3

(Choose any direction)

(Current goes + - for resistor)

(Your choice! Include all circuit elements!)

e 1

R 1 I 1 I 3

I 2

R 2 R3

Loop 1

Loop 2

3 Equations, 3 unknowns the rest is math!

e 2

3

Loop 1:

Loop 2:

Node:

Calculations

e1

10V

R1

50W

R2

25W

+ e 1 - I 1 R 1 + I 2 R 2 = 0

R3

100W

e

2

5V

- I 2 R 2 - I 3 R 3 - e 2 = 0

I 1 + I 2 = I 3

10 50I1 25I2 0 25I2 100I3

5 0

50I1 25I2 10 25I2 100I1 I2

5 0

10I

5I

2 20I

25I

1

1

20I

25I

1

1 2

10I

5I

2

1 2

1

I2

A

7

2 1 2

I3 I1

I2

25 1

I1

10

9 1

7 50

70 7

9

1

I1

A I3

A

70

70

e 1

Loop 1

R 1 I 1 I 3

I 2

R 2 R3

Loop 2

The negatives only

indicate that our

current direction

choice was wrong.

Practice Circuits

In the circuit below you are given ε 1 , R 1 , R 2 and R 3 .

a) Determine the total resistance of the circuit

b) Find I 1 , I 2 and I 3 .

This circuit can be broken down into a

simple circuit, no need for Kirchhoff

e 1 =12V

R 1 =25

I 1

R 2 =100

I 2 R 3 =50 I 3

Since R 2 and R 3 are Now R P and R 1 are in serial Now: V

in parallel

I

1 1 1

RT

RP

R

RT

1

12V

RP

R3 R

100

2

0.206A

W 25W

58.3W

1 1

3

This is the current of I

50W

100W

58.3W

1

3

V

P

V

100W

The potential, V across R 2 and R 3 is I2

I3

P

R2

R3

100 VP

VT

R1I

1

RP

W

Therefore: 6.85V

6.85V

3

12V

25W0.206

A

100W

50W

6.85V

0.0685A

0.137A

Practice

Practice

In the circuit below, find ε 1, I 2 , I 3

1. Label all currents

(Directions are given)

2. Choose loop and direction (Your choice!)

3. Write down voltage drops

Loop 1: + (0.5A)(2Ω) + ε 1 - 12V- I 2 (4Ω) = 0

Loop 2: + I 2 (4Ω) + 12V-4V + I 3 (6 Ω )= 0

I 2

e 1

Loop 1

12V

4V

Loop 2

In the circuit below, find ε 1, I 2 , I 3

I 2

12V

4V

e 1

Loop 1

Loop 2

I 1 =0.5 I 3

I 1 =0.5 I 3

5. Write down node equation

Node: 0.5A + I 2 = I 3

4

Practice

In the circuit below, find ε 1, I 2 , I 3 I

e

2

1

3 Equations, 3 Unknowns

12V 4V

Loop 1

1) + (0.5A)(2Ω) + ε 1 - 12V - I 2 (4Ω) = 0

2) + I

2 (4Ω) + 12V - 4V + I 3 (6 Ω )= 0

4Ω Loop 2

3) 0.5A + I 2 = I 3

I 1 =0.5 I 3

11V

e1 4I2

0

11V

e1

41.1V

0

8V

6I3

4I2

0

e1

4.4V

11V

I3 0.5A I I3 0.5A

A 1.1

2

e1

6.6V

8V

6 0. 5A

I 4I

0

0.6A

2 2

8V 3V 6I 4I

0

2 2

11V

10I2

0

I2 1.1A

The “-” on the

currents indicate that

our original direction

guess was wrong

Practice

In the circuit below, find the current in each resistor and the equivalent

resistance of the network of five resistors.

13V

a

I 1

c

I 3

I 4

d

I 2

I 5

b

Practice

This “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.

Loop 2

13V

I 1+ I 2

Loop 1

a

I 1

c

Loop 3 1Ω

I 3

I 1 –

I 4 I 3

d

I 2

I 2 + I 3

I 5

b

Using the

current

directions as

guides, we will

define 3 loops

(3 equations

for the 3

unknowns)

Loop 2

13V

+

I 1+ I 2

Loop 1

Practice

Loop 1: V I1 I1 I3

Loop 2:

Loop 3:

a

I 1

c

Loop 3

1Ω 1Ω

I 3

I 1 –

I 4 I 3

13 1W 1W 0

13V I 1W I I 2W 0

2 2 3

I 1W I 1W I 1W 0

1 3 2

d

I 2

I 2 + I 3

I 5

b

This is a set of 3

equations and

three unknowns.

So let’s solve

5

Practice

Loop 1: V I1 I1 I3

Loop 2:

Loop 3:

13 1W 1W 0

13V I 1W I I 2W 0

2 2 3

I 1W I 1W I 1W 0

1 3 2

From loop 3: I2 I1 I3

Substitute this into loop 1 and loop 2 (to eliminate I 2 )

Loop 1: 13V I12W I31W

Loop 2: 13V I13W I35W

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

78V

I 13

I 6A

1

1 W

thus I 2

5A

and I3 1A

Practice

The total current is: I 1

I 2

6A5A 11A

The potential drop across this is equal to the battery

emf, namely 13V. Therefore the equivalent resistance of

the network is: 13V

Req

1.2W

11A

Loop 2

13V

+

I 1+ I 2

Loop 1

a

I 1

c

Loop 3

1Ω 1Ω

I 3

I 1 –

I 4 I 3

d

I 2

I 2 + I 3

I 5

b

6

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