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Superposition theorem, 5 Ω current

Short answer

The resistor in question is actually 0.5 Ω. Using superposition: with 15 V alone, current through 0.5 Ω is 4.5 A P→Q; with 20 V alone, 8 A Q→P. Net current is 3.5 A from right to left through the 0.5 Ω resistor.

Find the current in 5-ohm resistor in the network shown below by using superposition theorem. [Figure: 15 V source with 1 Ω on the left; 0.5 Ω on top; resistors 1 Ω, 2 Ω, 5 Ω and 1 Ω in a bridge arrangement; 20 V source on the right.]

Reading the figure

The figure contains no 5 Ω resistor. The branch between the two top corners P and Q is 0.5 Ω, and the question is the same circuit as the 2069 Asar question on the 0.5 Ω resistor. So the required current is the current in the 0.5 Ω resistor.

Circuit: node P (top left) has 1 Ω in series with the 15 V battery (+ on top) on the left, and a 1 Ω going to the centre point O. Node Q (top right) has a 2 Ω going to O and a 1 Ω in series with the 20 V battery (+ on the 1 Ω side). The 0.5 Ω joins P and Q. 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 P to ground through the 0.5 Ω: 0.5 + 0.667 = 1.167 Ω.

V_P (1/1 + 1/1 + 1/1.167) = 15/1 → V_P = 15/2.857 = 5.25 V V_Q = 5.25 × 0.667/1.167 = 3.0 V

I′ = (5.25 − 3.0)/0.5 = 4.5 A from 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/(0.5 + 0.5)) = 20/1 → V_Q = 20/2.5 = 8 V V_P = 8 × 0.5/1.0 = 4 V

I″ = (8 − 4)/0.5 = 8 A from Q to P.

Total

The two currents oppose, so

I = 8 − 4.5 = 3.5 A from Q to P (right to left)

Check by solving the full circuit: V_P = 9.25 V and V_Q = 11 V, so I = 1.75/0.5 = 3.5 A.

Answer: I(0.5 Ω) = 3.5 A, from the right node to the left node.

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