What does the superposition theorem state?

Response in any element is the algebraic sum of the responses due to each source acting alone

Response is the product of the responses of each source

Only the largest source matters

With E1 alone, how is the RL current found?

R2 ∥ RL in series with R1 gives the total current, then a current divider gives the RL share: 0.2203 A

Asked 3/15
2067 Chaitra · 5 marks

Superposition: 90, 50, 130 ohm circuit

Short answer

Superposition theorem states that in a linear network with multiple sources, the response in any element is the algebraic sum of the responses caused by each source acting alone (other voltage sources shorted). For this circuit, analyzing each source separately and summing gives I = 0.458 A through RL.

Describe the principle of superposition theorem by solving the following problem. [Figure: E1 = 100 V with R1 = 90 Ω in series on the left; R2 = 50 Ω with E2 = 60 V in the middle branch; load RL = 130 Ω.] Find the current I through RL.

Superposition theorem: in a linear bilateral network with more than one independent source, the response (current or voltage) in any element equals the algebraic sum of the responses caused by each source acting alone, with the other independent voltage sources replaced by short circuits and current sources by open circuits.

Assumption: both batteries have their positive ends toward the top node, so both drive current through RL. E1 = 100 V, R1 = 90 Ω, E2 = 60 V, R2 = 50 Ω, RL = 130 Ω.

Case 1: E1 acting alone (E2 shorted)

R2 ∥ RL = (50 × 130)/180 = 36.11 Ω R_total = 90 + 36.11 = 126.11 Ω I1 (from E1) = 100/126.11 = 0.793 A Current in RL (current divider): I' = 0.793 × 50/(50 + 130) = 0.2203 A

Case 2: E2 acting alone (E1 shorted)

R1 ∥ RL = (90 × 130)/220 = 53.18 Ω R_total = 50 + 53.18 = 103.18 Ω I2 (from E2) = 60/103.18 = 0.5815 A Current in RL: I'' = 0.5815 × 90/(90 + 130) = 0.2379 A

Both currents flow in the same direction through RL.

Total

I = I' + I'' = 0.2203 + 0.2379

Answer: I = 0.458 A through RL (130 Ω)

✓ Answer checked by Claude Opus 5.5 (high effort)

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Superposition: 90, 50, 130 ohm circuit

Case 1: E1 acting alone (E2 shorted)

Case 2: E2 acting alone (E1 shorted)

Total

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