kirchhoff's Law - The Burns Home Page

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kirchhoff's Law - The Burns Home Page

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

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