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Question

How do you verify superposition theorem?


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Solution

Superposition theorem

  1. According to the superposition theorem, the response across any element in a linear bilateral network with multiple sources equals the sum of the responses obtained from each source when it is analyzed independently, while the other sources are substituted by their internal resistances.
  2. Networks with two or more sources are solved using the superposition theorem. Only circuits that follow Ohm's Law are capable of applying the superposition theorem.

Step 1: Draw the circuit diagram

The resultant current in any branch of a linear network with multiple voltages or current sources acting simultaneously is equal to the algebraic sum of the currents that would be generated in that branch if each source acted independently, replacing all other independent sources with their internal resistances.

Here, R1,R2,&R3 are the resistance, V1&V2 are given two voltage source, I1' is current across the resistance R1, I2' is current across the resistance R2, and I3' is current across the resistance R3.

Step 2: Apply the Superposition theorem in figure 1

Knows, resistors are connected in series when they are connected one after the other, so Rtotal=R1+R2+R3+..., and the combined overall resistance of two parallel-connected resistors: Rtotal=R1×R2R1+R2.

So, in figure 1: Rtotal=R2×R3R2+R3+R1

Applying the superposition theorem to the given figure 1, let's first take the sources V1 alone and then short out V2 as shown in figure 2.

I1'=V1R2×R3R2+R3+R1I2'=I1'×R3R2+R3I3'=I1'-I2'

Step 3: Draw the circuit with only V1 short-circuited

The circuit with only V1 short-circuited is shown as figure 3:

Now here, I1'' is current across the resistance R1, I2'' is current across the resistance R2, and I3'' is current across the resistance R3.

Step 4: When V1 is removed via a short circuit

Note that, in figure 2: Rtotal=R1×R3R1+R3+R2

When V1 is removed via a short circuit, the circuit will only be powered by V2, as seen in figure 3. Then,

I2''=V2R1×R3R1+R3+R2I1''=I2''×R3R1+R3I3''=I2''-I1''

Now, according to the superposition theorem:

I3=I3'+I3''I2=I2'-I2''I1=I1'-I1''

Hence, verified the superposition theorem.


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