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Superposition theorem, current in RL

Short answer

With the 10 V source alone, current through the 2 Ω load is 2.5 A; with the 5 A source alone, it is 0.625 A. The total current using superposition is 3.125 A.

Find current on load resistor RL, if its resistance is 2 Ω, using superposition theorem. [Figure: 2 Ω resistor along the top; 10 V source on the left with 2 Ω and 2 Ω resistors; load RL in the middle vertical branch; 5 A current source; 2 Ω resistor on the right.]

Reading the figure

Top: 2 Ω from the left node L to the right node R. Middle row: 2 Ω from L to the middle node M, and 2 Ω from M to R. R_L = 2 Ω from M to the bottom rail. Left: 10 V battery (+ on top) between L and ground. Right: 5 A source (arrow up) into R, and a 2 Ω shunt from R to ground.

Case 1: 10 V alone (5 A source opened)

Node M: (V_M − 10)/2 + V_M/2 + (V_M − V_R)/2 = 0 → 3 V_M − V_R = 10 Node R: (V_R − 10)/2 + (V_R − V_M)/2 + V_R/2 = 0 → 3 V_R − V_M = 10

So V_M = V_R = 5 V and I′ = 5/2 = 2.5 A (downwards through R_L).

Case 2: 5 A alone (10 V source shorted)

L is at 0 V. Node M: V_M/2 + V_M/2 + (V_M − V_R)/2 = 0 → 3 V_M = V_R Node R: V_R/2 + (V_R − V_M)/2 + V_R/2 = 5 → 3 V_R − V_M = 10

3 (3 V_M) − V_M = 10, so V_M = 1.25 V and I″ = 1.25/2 = 0.625 A (downwards through R_L).

Total

Both currents flow in the same direction, so

I_L = I′ + I″ = 2.5 + 0.625 = 3.125 A

Answer: Current in R_L = 3.125 A

✓ Answer checked by Claude Opus 5.5 (high effort)

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