Circuit Theorems: Thevenin's, Norton's, Superposition, and Maximum Power Transfer, Lecture notes of Circuit Theory

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Outline

• Introduction
• Linearity Property
• Superposition
• Source Transformations
• Thevenin’s Theorem
• Norton’s Theorem
• Derivations of Thevenin’s and Norton’s

Theorems

• Maximum Power Transfer

Linearity Property

Homogeneity property (Scaling)

i  v  iR

ki  kv  kiR

Additivity property 1 1 1 2 2 2

i v i R i v i R

i 1  i 2  ( i 1  i 2 ) R  i 1 R  i 2 R  v 1  v 2

Linear Circuit

• Linear circuit
  • A circuit whose output is linearly related (or directly proportional) to its input.

Example 1

1 2 1 2 1 1 2

12 4 0 4

KVL at loop 1: KVL at lo 16

( ) (

op 3 0;( 2 ) 10 16

2 )

: 0

s x s x s

i i v i i

a b

v v v i i i v

             

(^762 )

( ) (^0 )

) :

6

5 ( 6 i vs^ v^ s

a i

b    

  

Showing that when the source value is doubled, I 0 doubles.

0 2 0 2

When 12V: 12 A 76 When 24V: 24 A 76

s s

v I i v I i

 (^)       ^ ^ 

Q: Find I 0 when v s=12V and v s=24V.

Sol:

Example 2

Q: Assume I 0 = 1 A and use linearity to find the actual value of I 0.

0 1 0 1 1

If 1A, then (3 5 ) 8V

/ 4 2 A,

I V I

I V

(^) 8 V

2 A

 I 2  I 1  I 0  3A

2 1 2 2 8 6 V^32 A

V  V  I    14 , I  V  2

3 A 14 V 2 A

Sol:

Superposition

• Superposition Principle
  • The voltage across (current through) an element in a linear circuit is the algebraic sum of the voltages across (currents through) that element due to each independent source acting alone.
• Turn off, killed, inactive source:
  • Independent voltage source: 0 V ( short circuit )
  • Independent current source: 0 A ( open circuit )
• Dependent sources are left intact.

Superposition Steps

• Steps to apply superposition principle
  1. Turn off all independent sources except one source. Find the output (voltage or current) due to that active source using nodal or mesh analysis.
  2. Repeat step 1 for each of the other independent sources.
  3. Find the total contribution by adding algebraically all the contributions due to the independent sources.

Example 3

Q: Use the superposition theorem to find v in the circuit.

1

Turn off current source 3A, and usevoltage divis (^4 ) 4 8 2

on v  (^)    V^3 2 3

curren

Turn off t divison

voltage source 6V, and use (^8) ( 8 V

4 ) 8

2A v 4

i i

 (^)     

1 2 2 8 1 V 0

v  v  v   

Sol:

Example 4

Q: Find i 0 in the following circuit using superposition.

' '' ' ''

4-A current sour

Let ; is due to is due to

ce 20-V voltage source

o (^) o o o o

i i i i i

   

Sol:

4 5 '' 4 '' 4 5 '' 4 ''

Loop 4: 6 5 0 6 4 0 Loop 5: 10 5 20 0 5 20

o o o o

i i i i i i i i i i

Example 4 ( cont. )

For io ''^ : i 5   io '';Turn off 4-A source

'' 60  i o  1 7 A ' '' 52 60 8 io io io (^) 16 1 7 17 0 .4 06 7 A      ^       



Source Transformation

• A source transformation is the process of

replacing a voltage source vs in series with a

resistor R by a current source is in parallel with

a resistor R , or vice versa

Source Transformation ( cont. )

• The source transformation is also applicable to

dependent sources.

Equivalent Circuit

i (^) _i

• +_

-^ -

v v

v

i

- i^ vs s R

v R

i v

v iR v s

s  

 