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Superposition theorem and maximum power to RL

Short answer

Superposition theorem states that in a linear bilateral network, the current or voltage in any element equals the algebraic sum of contributions from each independent source acting alone with all other independent sources replaced by their internal resistances (voltage sources shorted, current sources opened). For the given circuit, maximum power transfer occurs when RL = Rth = 13.33 Ω, delivering Pmax = 83.33 W.

Explain superposition theorem with suitable example. Find the maximum power that can be delivered to the load resistor RL of the circuit shown in the figure below. [Figure: a 20 V source (+ at top) on the left; from its top terminal a 5 Ω series resistor leads to node A; a 10 Ω resistor connects node A to the bottom (reference) line; a 10 Ω series resistor connects node A to node B; a 4 A current source between the bottom line and node B, arrow pointing up (current into node B); the variable load resistor RL connects node B to the bottom line.]

Superposition theorem

Statement: In a linear, bilateral network with more than one independent source, the current (or voltage) in any element equals the algebraic sum of the currents (or voltages) produced by each source acting alone, with all other independent sources replaced by their internal resistances.

  • An ideal voltage source is replaced by a short circuit.
  • An ideal current source is replaced by an open circuit.
  • Dependent sources are left in the circuit.
  • It applies to currents and voltages, not to power (power is not linear).

Example: A 10 V source in series with 2 Ω feeds node X; a 2 A current source also feeds node X; a 3 Ω resistor connects X to ground. Find the current in 3 Ω.

10 V alone (2 A open): I1 = 10/(2+3)      = 2 A
2 A alone (10 V short): I2 = 2 × 2/(2+3)  = 0.8 A
Total: I = I1 + I2 = 2 + 0.8              = 2.8 A

Maximum power to RL

Remove RL and find the Thevenin equivalent across node B and ground.

Thevenin resistance (20 V source shorted, 4 A source opened):

5 Ω || 10 Ω = (5×10)/(5+10) = 3.333 Ω
Rth = 10 + 3.333            = 13.333 Ω

Thevenin voltage (RL open): the 4 A source current has no path except through the 10 Ω (A–B) into node A. KCL at node A (currents leaving A):

(VA − 20)/5 + VA/10 − 4 = 0
0.2VA − 4 + 0.1VA − 4   = 0
0.3VA                   = 8
VA                      = 26.667 V
Vth = VB = VA + 4 × 10  = 26.667 + 40 = 66.667 V

(The 4 A flows from B to A through the 10 Ω, so B is 40 V above A.)

Maximum power transfer: power is maximum when RL = Rth = 13.333 Ω.

Pmax = Vth² / (4 Rth)
     = (66.667)² / (4 × 13.333)
     = 4444.4 / 53.333
     = 83.33 W

Answer: RL = 13.33 Ω, Pmax = 83.33 W (load current = 66.667/26.667 = 2.5 A).

✓ Answer checked by Claude Opus 5.5 (high effort)

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