Superposition Theorem With 2 Voltage Sources | Superposition Theorem Example

Superposition Theorem With 2 Voltage Sources | Superposition Theorem Example

Superposition Theorem With 2 Voltage Sources | Superposition Theorem Example


Analysis Of Superposition Theorem Example Which Circuit With 2 Voltage Sources And Superposition Theorem Steps And How We Apply Theorem In The Electrical Circuit






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The superposition theorem or superposition principle is

" The Superposition Theorem states that the response in any element of a linear, bilateral network containing more than one independent sources can be obtained as the algebraic sum of the responses obtained by each independent source acting separately at a time and with all other independent source set equal to zero."

While evaluating the response from one independent source other source can not be removed bodily from the circuit.

Instead , you suppress all other Independent Source by setting their values to zero. 




The suppressed source then behave like the following replacement.


  1. Ideal independent voltage source - short circuit
  2. Ideal independent current source - open circuit

Special care must be taken if the circuit also include the Controlled Sources.


In general a controlled source affect the individual contribution of each independent source,  Consequently Controlled Source are not suppressed during analysis by super position.

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Example of Superposition Theorem With 2 Voltage Sources  calculation:

Find The Current In 20 ohm Resistance In The Circuit Shown In Below Figer (Superposition Theorem Example Circuit - 1 ) By Using Superposition Theorem .

Superposition Theorem Example Circuit - 1

Answer :

Step : 1

Find Current I'¹  Due To 90 V Source Only :

Superposition Theorem Example Circuit - 1(A) 

Superposition Theorem Example Circuit - 1(B) 

From Figer [ Superposition Theorem Example Circuit - 1(B) ] We Get

R(t) = 20 + 4.8  = 24.8 ohm ;

I'¹ = 90/24.8  = 3.63 Amp ;


 Step : 2

Find Current I''¹  Due To 60V Source Only :

Superposition Theorem Example Circuit - 1(C) 

Superposition Theorem Example Circuit - 1(D) 

From Figer [ Superposition Theorem Example Circuit - 1(D) ]  We Get 

R(t) = 12 + 5.71 = 17.71 ohm; 

I(2) = 60 / 17.71 = 3.39 Amp; 

I''¹ = ( 8/28 ) * ( 3.39 )
I''1 = 0.97 Amp ;

Step : 3

Find Current I'''1 Due To 2A Source Only :

Superposition Theorem Example Circuit - 1(E)

Superposition Theorem Example Circuit - 1(F)

From Figer [ Superposition Theorem Example Circuit - 1(F)  ]  We Get 

I'''¹ = ( 4.8 * 2 ) / ( 20 + 4.8 )  = 0.39 Amp;

Step : 4

Calculate I(1) Current Through  20 ohm Resistance  :

I(1) = I'¹ + I''¹ + I'''¹
       = 3.63 - 0.97 - 0.39
       = 2.27 Amp

So In Given Example  The Current Flowing Through 20 ohm Resistance Is 2.27 Ampere. 




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