kirchhoff's Law - The Burns Home Page

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

Your choice!

(3) Write down voltage drops

Follow any loops

Kirchhoff’s **Law**s

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