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Find R for maximum power, 10 A and 48 V

Short answer

Maximum power transfer occurs when R equals the Thevenin resistance, 10.67 Ω. The maximum power delivered is 13.5 W.

What is the value of R such that maximum power is transferred to it? Find the value of this maximum power for the circuit shown below. [Figure: 10 A current source on the left; 4 Ω and 10 Ω shunt resistors; R in the top branch; 20 Ω resistor and 48 V source on the right.]

Reading the figure

Left node A: 10 A source (up) into A and a 4 Ω shunt to ground. R joins A to node B. Node B: 10 Ω shunt to ground, and 20 Ω in series with the 48 V battery (+ on top) to ground.

R is the load; find its Thevenin equivalent.

Step 1: Open-circuit voltage (R removed)

  • V_A = 10 × 4 = 40 V (all the 10 A flows through the 4 Ω).
  • V_B = 48 × 10/(10 + 20) = 16 V (voltage divider).

V_th = V_A − V_B = 40 − 16 = 24 V

Step 2: Thevenin resistance (sources off)

10 A opened, 48 V shorted: R_th = 4 + (10 ∥ 20) = 4 + 6.667 = 10.67 Ω

Step 3: Maximum power

R = R_th = 10.67 Ω (32/3 Ω)

P_max = V_th²/(4 R_th) = 24²/(4 × 10.667) = 576/42.67 = 13.5 W

Check: current in R = 24/(2 × 10.667) = 1.125 A; P = 1.125² × 10.667 = 13.5 W.

Answer: R = 10.67 Ω; P_max = 13.5 W

✓ Answer checked by Claude Opus 5.5 (high effort)

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