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Maximum power in bridge load RL

Short answer

Thevenin voltage across the bridge is 8.33 V and Thevenin resistance is 20.83 Ω. Maximum power transfer requires RL = 20.83 Ω, and the maximum power is 0.833 W.

Determine the value of load resistance RL to receive maximum power from the source. Also find the maximum power delivered to the load in the circuit shown in figure below: [Figure: 100 V source feeding a bridge; upper arms 10 Ω and 20 Ω, lower arms 30 Ω and 40 Ω; variable load RL connected across the bridge between the middle nodes.]

Reading the figure

The 100 V source (+ on top) feeds a bridge. Top vertex T and bottom vertex B are joined to the source. Upper arms: 10 Ω (T to left node X) and 20 Ω (T to right node Y). Lower arms: 30 Ω (X to B) and 40 Ω (Y to B). R_L is connected between X and Y.

          T (+100 V)
         /  \
      10 Ω   20 Ω
       /      \
     X --[R_L]-- Y
       \      /
      30 Ω   40 Ω
         \  /
          B (0 V)

Step 1: Thevenin voltage across X-Y (R_L removed)

V_X = 100 × 30/(10 + 30) = 75 V V_Y = 100 × 40/(20 + 40) = 66.67 V

V_th = V_X − V_Y = 8.33 V

Step 2: Thevenin resistance (source shorted, T and B joined)

From X: 10 Ω ∥ 30 Ω = 7.5 Ω From Y: 20 Ω ∥ 40 Ω = 13.33 Ω

R_th = 7.5 + 13.33 = 20.83 Ω

Step 3: Maximum power

R_L = R_th = 20.83 Ω

P_max = V_th²/(4 R_th) = 8.333²/(4 × 20.833) = 69.44/83.33 = 0.833 W

Answer: R_L = 20.83 Ω; P_max = 0.833 W

✓ Answer checked by Claude Opus 5.5 (high effort)

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