Superposition Theorem

The superposition theorem is a fundamental principle in linear circuit analysis that allows the response due to multiple independent sources to be calculated one source at a time and then added together. It directly follows from the linearity property of circuits and is widely used in GATE problems involving multiple voltage and current sources. Understanding when superposition applies and when it does not is equally important for both analysis and examination.

Core Concept: Linearity and Superposition

The superposition theorem applies exclusively to linear circuits. A circuit is linear if it satisfies two properties: homogeneity (scaling an input by a constant scales the output by the same constant) and additivity (the response to a sum of inputs equals the sum of individual responses). Resistors, capacitors, and inductors are linear elements, but diodes and transistors operating in nonlinear regions are not.

The theorem states: in a linear circuit with multiple independent sources, the response (voltage or current) at any branch equals the algebraic sum of the responses produced by each source acting alone, while all other independent sources are deactivated. Deactivating a voltage source means replacing it with a short circuit. Deactivating a current source means replacing it with an open circuit.

Dependent sources are never deactivated during superposition. They remain active in every sub-circuit because they are controlled by circuit variables, not by independent inputs. This is a very common source of errors in GATE solutions.

Mathematical Expression

If a linear circuit has n independent sources S1, S2, ..., Sn, and the response at a particular branch is R, then the superposition theorem expresses the total response as:

R_total equals R1 plus R2 plus ... plus Rn, where Ri is the response due to source Si alone with all other independent sources deactivated. Mathematically, this is a direct consequence of writing the circuit equations as a linear system [A][x] = [b], where [b] is the source vector. By linearity, the solution vector [x] is the sum of solutions to each component of [b] acting individually.

Note that superposition does NOT apply to power. Power is proportional to the square of voltage or current, and squaring destroys additivity. The total power dissipated in a resistor is not the sum of powers due to individual sources. This is one of the most tested GATE traps.

Practical Understanding

Superposition is most useful when a circuit has sources of different types (AC and DC, or multiple AC signals at different frequencies). For example, an amplifier circuit with a DC bias source and an AC signal source can be analysed separately: the DC analysis sets the operating point, and the AC small-signal analysis determines the gain. Combining both gives the complete response.

In power systems, superposition helps analyse the contribution of each generator to the total current at a bus. In signal processing, it justifies the use of Fourier analysis: any periodic signal can be decomposed into sinusoids, and the system response to each sinusoid is found independently and then superposed.

Solved Numerical Example

Consider a series circuit with a 10 V voltage source V1, a 6 V voltage source V2, and a 2 ohm resistor R all in a single loop. V1 and V2 are series opposing (pointing in opposite directions). Using superposition, the current due to each source alone is found and then algebraically added.

Given:
V1 = 10 V, V2 = 6 V (series opposing in a single loop), R = 2 Ω

Why this formula applies:
Superposition: find current due to each source independently, then add algebraically.

Formula:
I_total = I_due_to_V1 + I_due_to_V2

Step 1 - V1 active, V2 shorted (replaced by wire):
I1 = V1 / R = 10 / 2 = 5 A  (clockwise direction)

Step 2 - V2 active, V1 shorted:
I2 = V2 / R = 6 / 2 = 3 A  (counterclockwise, opposing I1)

Substitution:
I_total = I1 - I2 = 5 - 3 = 2 A

Calculation:
I_total = 2 A (in the direction of V1's current)

Final Answer:
Net current through resistor = 2 A
Verification using KVL: V1 - V2 = I*R → 10 - 6 = 2*2 = 4 V ✓

Exam Tip: Superposition cannot be applied to find total power dissipation. Power is nonlinear (P = I²R), so P_total is NOT equal to P1 + P2 from individual sources. Always compute total current or voltage first, then find power from the combined response. Also, dependent sources are NEVER turned off during superposition.

Quick Revision

• Superposition applies only to linear circuits. Nonlinear elements (diodes, transistors in nonlinear region) invalidate it.
• Each independent source is activated one at a time. All other voltage sources are short-circuited; all other current sources are open-circuited.
• Dependent sources are NEVER deactivated. They remain in every sub-circuit.
• Total voltage or current = algebraic sum of individual contributions. Signs must be tracked carefully.
• Superposition does NOT apply to power. Compute net current/voltage first, then calculate power.
• Useful for AC+DC circuits: solve DC and AC separately using superposition, then combine for complete response.
• GATE trap: applying superposition for power or forgetting to keep dependent sources active in sub-circuits.