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Maximum power transfer, find R

Short answer

Thevenin voltage V_th = 13.33 V, Thevenin resistance R_th = 5 Ω∥10 Ω = 3.33 Ω. R should equal R_th = 3.33 Ω for maximum power; maximum power P_max = V_th²/(4R_th) = 13.33 W.

Calculate the value of 'R' such that maximum power will be absorbed by it in the given circuit. [Figure: 5 Ω and 10 Ω resistors in the top branch; 10 V source on the left; R in the middle vertical branch; 20 V source on the right.]

Reading the figure

Left branch: 10 V battery (+ on top). Right branch: 20 V battery (+ on top). Top: 5 Ω from the left top node to the middle node M, and 10 Ω from M to the right top node. R is the vertical branch at M. Bottom rail common.

So R is connected at M to two source branches: (10 V, 5 Ω) and (20 V, 10 Ω).

Step 1: Thevenin voltage (R removed)

V_th = V_M = (10/5 + 20/10)/(1/5 + 1/10) = 4/0.3 = 13.33 V

Step 2: Thevenin resistance

V sources shorted: R_th = 5 ∥ 10 = 50/15 = 3.33 Ω

Step 3: Maximum power

R = R_th = 3.33 Ω

P_max = V_th²/(4 R_th) = 13.33²/(4 × 3.333) = 177.8/13.33 = 13.33 W

Check: current in R = 13.33/(2 × 3.333) = 2 A, so P = 2² × 3.333 = 13.33 W.

Answer: R = 3.33 Ω (10/3 Ω); P_max = 13.33 W

✓ Answer checked by Claude Opus 5.5 (high effort)

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