Equivalence and Linearity: Proof of Linearity and Superposition in Circuit Analysis, Slides of Fundamentals of Electronics

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Lecture Fourteen

Equivalence and Linearity

Equivalence

Table: Equivalent Circuits

Linearity

All resistive circuits (circuits containing resistance elements, independent current and/ or

voltage sources and linear dependent sources) are linear in nature i.e. they are both additive

and homogeneous (scaled).

Proof:

Consider the circuit below:

We wish to calculate Vout = V 2. Assume Vout = α V where α is constant. Then

2 2 4 4

V

I

R R

V 1 can then be calculated as

3 4 1 3 2 2 3 4 4

R R

V R I V R

R R

Hence

1 3 4 1 2 4 2

V R R

I

R R R

Now

2 3 4 0 1 2 4 2

R R R

I I I

R R

And

1 2 3 4 2 3 4 0 1 0 1 2 4

S

R R R R R R R

V V R I V

R R

Therefore the factor α appears in the source voltage. Suppose R 1 = R 2 = R 3 = R 4 = 1 Ω then VS =

5 α V. Now for a = 1, 2, 3... then VS = 5, 10, 15 ... Volts respectively. HENCE PROOVED!

Principle of Superposition

In any linear circuit containing multiple independent sources, either current or

voltage at any point in the network is the algebraic sum of all the individual

contribution of each source acting alone

Proof:

Part A: Circuit containing Voltage Sources

Writing loop equations (^1 1 2 1 2 )

2 2 1 3 2 2

loop R I R I I V

loop R I I R I V

Solving these equations we get

2 3 2 1 2 1 2 2 1 2 2 3 2 1 2 2 3 2

2 3 1 2 2

1 2 3 2 3

2 1 2 2 2 1 2 2 1 2 2 3 2 1 2 2 3 2

2 1 1 2 2

1 2 3 2 3

R R R

I V V

R R R R R R R R R R

R R V R V

R R R R R

R R R

I V V

R R R R R R R R R R

R V R R V

R R R R R

Therefore each current I 1 and I 2 has component due to V 1 and V 2.

For V 1 to act alone, V 2 = 0 i.e. it is replaced with a short circuit, then

1 1 1 1 2 3 2 3 1 2 3

2 3 1 1 2 3 2 3

V V

I

R R R R R

R

R R

R R

V

R R R R R

Now let V 1 = 0 i.e. V 1 is replaced with short circuit,

2 2 2 1 2 3 1 2 3 1 2

1 2 2 1 2 3 1 2

V V

I

R R R R R

R

R R

R R

V

R R R R R

Using current division rule,

2 2 2 1 2 2 2 1 2 1 2 3 1 2 1 2 3 2 3

R R R

I I V V

R R R R R R R R R R R R

Thus

2 3 1 2 2 1 1 2 1 2 3 2 3

R R V R V

I I I

R R R R R

= ′^ + ′′=

On similar reasoning, we can proof that

2 1 1 2 2 2 1 2 3 2 3

R V R R V

I

R R R R R

Part B: Circuit containing current sources

Writing node equations

1 1 2 1 2 1 2 1 2 1 2 2

1 2 2 2 3 1 2 2 3 2 2 3

A

B

V V V R R

node I V V R R R R R

V V V R R

node I V V R R R R R

Solving these equations

1 2 3 1 3 1 1 2 3 1 2 3

1 3 3 1 2 2 1 2 3 1 2 3

A B

A B

R R R R R

V I I

R R R R R R

R R R R R

V I I

R R R R R R

Now letting IA = 0 i.e IA is replaced with open circuit, then I 1 = ‐I 2

R 1 R^3 I

B

V 1 R^2 V 2

I 2

R 1 R^3 I

B

V’ 1 R^2 V^2

I 2