With only the 90 V source, the 8 ohm is in parallel with which resistor, and what is V across the 8 ohm?

12 ohm (4.8 ohm combined), V = 10.588 V

Why are the three currents simply added to get 5.03 A?

All three flow top to bottom in the 8 ohm, so they add in the same direction

Because the sources are identical

2068 Shrawan · 8 marks

Superposition, current in 8Ω

Short answer

By superposition: 90 V source gives 1.3235 A, 60 V gives 2.647 A, 2 A gives 1.059 A, all in the same direction. Total current in the 8 Ω resistor is 5.03 A.

Using superposition theorem, find the current in 8Ω resistor of the network shown below. [Figure: 36 Ω in series with a 90 V source; 12 Ω in series with a 60 V source; 2 A current source; 8 Ω resistor, all in parallel.]

Reading the figure

Four branches are in parallel between the top rail and the bottom rail: 36 Ω in series with the 90 V battery (+ up), 12 Ω in series with the 60 V battery (+ up), the 2 A source (arrow up) and the 8 Ω resistor. All three sources push current into the top rail.

Case 1: 90 V alone (60 V shorted, 2 A opened)

Rest of the circuit across the 8 Ω: 36 Ω in series with the source; the 12 Ω is in parallel with the 8 Ω. 12 ∥ 8 = 4.8 Ω V = 90 × 4.8/(36 + 4.8) = 10.588 V I′ = 10.588/8 = 1.3235 A

Case 2: 60 V alone (90 V shorted, 2 A opened)

36 ∥ 8 = 6.545 Ω V = 60 × 6.545/(12 + 6.545) = 21.176 V I″ = 21.176/8 = 2.647 A

Case 3: 2 A alone (both batteries shorted)

36 ∥ 12 ∥ 8: 1/R = 0.02778 + 0.08333 + 0.125 = 0.23611, so R = 4.235 Ω V = 2 × 4.235 = 8.471 V I‴ = 8.471/8 = 1.059 A

Total (all in the same direction, top to bottom)

I₈ = 1.3235 + 2.647 + 1.059 = 5.03 A

Check by nodal analysis: V = (90/36 + 60/12 + 2)/(1/36 + 1/12 + 1/8) = 9.5/0.2361 = 40.24 V, and I = 40.24/8 = 5.03 A.

Answer: Current in the 8 Ω = 5.03 A

✓ Answer checked by Claude Opus 5.5 (high effort)

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