- 2069 Asar · 6 marks
Superposition, current in 0.5Ω
Short answer
With 15 V source alone, current is 4.5 A from P to Q; with 20 V source alone, it is 8 A from Q to P. Superposition gives net current of 3.5 A from Q to P in the 0.5 Ω resistor.
State superposition theorem. Find the current in 0.5Ω resistor in the following network shown, by using superposition theorem. [Figure: RL = 0.5 Ω in the top branch; 1 Ω with a 15 V source on the left; 1 Ω and 2 Ω in the middle; 1 Ω with a 20 V source on the right.]
Statement
Superposition theorem: in a linear bilateral network with more than one independent source, the response (current or voltage) in any element is the algebraic sum of the responses due to each source acting alone, with the other voltage sources replaced by short circuits and the other current sources by open circuits.
Reading the figure
Terminals P (top left) and Q (top right) are joined by R_L = 0.5 Ω. At P: 1 Ω in series with the 15 V battery (+ on top) and a 1 Ω to the centre point O. At Q: 2 Ω to O and 1 Ω in series with the 20 V battery (+ on the 1 Ω side). O and the bottom rail are the reference.
Case 1: 15 V alone (20 V shorted)
Right side from Q to ground: 2 Ω ∥ 1 Ω = 0.667 Ω. Path from P: 0.5 + 0.667 = 1.167 Ω.
V_P (1/1 + 1/1 + 1/1.167) = 15 → V_P = 5.25 V V_Q = 5.25 × 0.667/1.167 = 3.0 V I′ = (5.25 − 3)/0.5 = 4.5 A (P to Q)
Case 2: 20 V alone (15 V shorted)
Left side from P to ground: 1 Ω ∥ 1 Ω = 0.5 Ω.
V_Q (1/1 + 1/2 + 1/1.0) = 20 → V_Q = 8 V V_P = 8 × 0.5/1.0 = 4 V I″ = (8 − 4)/0.5 = 8 A (Q to P)
Total
The two currents are opposite:
I = 8 − 4.5 = 3.5 A from Q to P
Answer: Current in the 0.5 Ω = 3.5 A (from the right node to the left node)
✓ Answer checked by Claude Opus 5.5 (high effort)
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