Skip to main content

Find RL for maximum power and Pmax

Short answer

Thevenin voltage is 0.8 V and Thevenin resistance is 6.4 Ω. For maximum power, set RL = 6.4 Ω; the maximum power is 0.025 W.

Find the value of RL for which the maximum power is transferred in the load resistance RL. Also find the maximum power that can be transferred to the load resistance RL. [Figure: 8 V source with 2 Ω shunt on the left; 4 Ω in the top branch; 6 Ω shunt; 4 Ω in series with a 4 V source; load RL on the right.]

Reading the figure

Left: 8 V battery (+ on top) with a 2 Ω shunt at node P. 4 Ω from P to node M. 6 Ω shunt at M. Then 4 Ω in series with the 4 V battery to the load R_L. The 4 V battery has its + terminal on the 4 Ω (left) side (the longer plate is on the left).

Step 1: Thevenin voltage (R_L removed)

No current flows in the last 4 Ω or the 4 V battery.

V_M = 8 × 6/(4 + 6) = 4.8 V (the 2 Ω is directly across the 8 V source and does not affect V_M)

The 4 V battery opposes V_M (its + is on the M side), so

V_th = 4.8 − 4 = 0.8 V

Step 2: Thevenin resistance (sources off)

8 V shorted (P to ground, the 2 Ω is then in parallel with the short), 4 V shorted:

  • At M: 6 Ω ∥ 4 Ω = 2.4 Ω
  • In series with the 4 Ω: R_th = 2.4 + 4 = 6.4 Ω

Step 3: Maximum power

R_L = R_th = 6.4 Ω

P_max = V_th²/(4 R_th) = 0.8²/(4 × 6.4) = 0.64/25.6 = 0.025 W (25 mW)

Answer: R_L = 6.4 Ω; P_max = 0.025 W

Note: if the 4 V battery polarity were reversed, V_th = 8.8 V and P_max = 3.03 W; the value above uses the polarity read from the figure.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Basic Electrical Engineering chapter 3 questions →