Superposition theorem

The superposition theorem is important in circuit analysis as the theorem helps to easily obtain the current or voltage in any branch by any source applied in the circuit. Superposition theorem is used when we have to determine the current in any particular branch of a circuit consisting of voltage or current sources. This theorem helps us to determine current across each branch separately in the circuit. The superposition theorem applies only to linear circuits. It is not valid for bilinear circuits.

Superposition Theorem Statement:

Superposition theorem states that the total voltage or current of a linear or bilateral circuit having more than one source can be the algebraic sum of voltage or current by each source.

In other terms, the superposition theorem is applicable to solving the circuits having multiple voltage or current sources by solving each source separately and replacing other sources by their internal resistance ( short-circuiting voltage source and open circuiting current source).

Steps to solve using superposition theorem:

  1. Select one source and replace the other source (voltage source with short-circuit and current source with the open circuit).
  2. Solve for current in the circuit.
  3. Repeat step 1 & 2 with other remaining sources
  4. Add the current produced by sources individually.

Now let’s solve a question.

Q. Find the current across branch a-b.

Question of superposition theorem

First, we solve for the left-hand side voltage source and replace the other source.

Replace 40V source by short-circuiting wire

Solution for superposition theorem

Now

Req1 = 2 + (4||2)

=  10/3

Ia1 = V/Req1

= 10 / (10/3)

= 3 A

Ia2 = Ia1 * Rbd / Rbc + Rbd …………………… (Current divider rule)

= 3 * (4/6)

= 2 A

Ia3  = Ia1 * Rbc / Rbc + Rbd

= 3* (2/6)

= 1A

Now replace 10V source by short-circuit wire and solve for 40V source.

Answer for superposition theorem

Req2 = 2 + (4||2)

=  10/3 ohm

Ib2  = V/Req2

= 40 / (10/3)

= 12 A

Ib3 = Ib2 * Rab / Rbd + Rab

= 12 * (4/6)

= 8 A

Ib1 = Ib2 * Rbd / Rbd + Rab

= 12 * (2/6)

= 4 A

The total current across branch a-b:

I = Ia1 + Ib1

= 4 + 3

= 7A.

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