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Power in 75Ω across XY

Short answer

The load sees 115 Ω, paralleled with 10 Ω gives 9.2 Ω; voltage across 10 Ω is 37.4 V. Current in 75 Ω is 0.325 A, dissipated power = 0.325² × 75 ≈ 7.93 W.

Calculate the power which would be dissipated in a 75 Ω resistor connected across XY in the network shown below. [Figure: 200 V source; 80 Ω shunt; 20 Ω and 20 Ω series resistors along the top to terminal X; 10 Ω shunt in the middle; 20 Ω and 20 Ω along the bottom to terminal Y; 75 Ω connected across X and Y.]

Reading the figure

The 200 V source feeds a ladder: 80 Ω across the source, then 20 Ω (top) and 20 Ω (bottom) in series, a 10 Ω shunt, another 20 Ω (top) and 20 Ω (bottom), and finally the 75 Ω load across X and Y.

Step 1: Resistance looking from the 10 Ω towards the load

R = 20 + 20 + 75 = 115 Ω

Step 2: Combine with the 10 Ω

10 ∥ 115 = 10 × 115/125 = 9.2 Ω

Step 3: Voltage across the 10 Ω

The 80 Ω shunt is directly across the source, so the rest of the ladder receives 200 V through the first 20 + 20 Ω:

V₁₀ = 200 × 9.2/(9.2 + 40) = 37.40 V

Step 4: Load current and power

I₇₅ = 37.40/115 = 0.3252 A

P = I² R = 0.3252² × 75 = 7.93 W

Answer: Power in the 75 Ω ≈ 7.93 W

✓ Answer checked by Claude Opus 5.5 (high effort)

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