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Maximum power transfer, find R

Short answer

Thevenin voltage V_th = 11 V, Thevenin resistance R_th = 2.625 Ω. For maximum power transfer, R = R_th = 2.625 Ω. Maximum power P_max = V_th²/(4R_th) = 11.52 W.

Find the value of Resistance R to have maximum power transfer in the circuit as shown in below figure. Also obtain the amount of maximum power. [Figure: 2 A current source on the left; 6 Ω in series with a 6 V battery on the top; 15 Ω, 3 Ω and R in vertical branches; 8 V battery in the bottom branch between the 3 Ω and R branches.]

Reading the figure

Bottom rail = ground. Left node a: 2 A source (up) into a, 15 Ω from a to ground, and 6 Ω in series with the 6 V battery going to node b (the battery's + terminal is on the 6 Ω side). Node b: 3 Ω to ground. The load R is connected between b and the bottom of the 8 V battery (+ terminal on the ground side), so the 8 V battery is in series with R.

The load R is taken between terminals b and c, where c is the lower end of the R branch (−8 V relative to ground).

Step 1: Open-circuit voltage (R removed)

Node a: 2 = V_a/15 + (V_a − 6)/(6 + 3) Multiply by 45: 90 = 3 V_a + 5 V_a − 30, so V_a = 15 V Current in the 6 Ω and 3 Ω = (15 − 6)/9 = 1 A V_b = 1 × 3 = 3 V

V_c = −8 V (battery + is on the ground side), so

V_th = V_b − V_c = 3 − (−8) = 11 V

Step 2: Thevenin resistance (sources off)

2 A opened, batteries shorted: from b, the 3 Ω is in parallel with (6 Ω + 15 Ω).

R_th = 3 × 21/(3 + 21) = 2.625 Ω

Step 3: Maximum power

R = R_th = 2.625 Ω

P_max = V_th²/(4 R_th) = 11²/(4 × 2.625) = 121/10.5 = 11.52 W

Answer: R = 2.625 Ω; P_max ≈ 11.52 W

Assumption: plate polarities are read as 6 V (+ towards the 6 Ω resistor) and 8 V (+ on the ground side).

✓ Answer checked by Claude Opus 5.5 (high effort)

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