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Find RL for maximum power (delta network)

Short answer

Find Thevenin equivalent: Rth = 20 ∥ 90 ∥ 60 = 12.86 Ω, Vth = 10 × Rth/20 = 6.43 V. For maximum power, RL = Rth = 12.86 Ω, and Pmax = Vth²/(4Rth) = 0.804 W.

For the circuit shown in figure below, what will be the value of RL to get the maximum power? What is the maximum power delivered to the load? [Figure: 10 V source; 20 Ω, 60 Ω and 90 Ω resistors arranged in a delta (triangle) network; load RL on the right.]

Reading the figure

All four elements 20 Ω (in series with the 10 V battery), 90 Ω, 60 Ω and R_L connect between the apex node T and the bottom rail. So, as seen by R_L, the rest of the circuit is a 10 V source with 20 Ω in series, in parallel with 90 Ω and 60 Ω (the drawing as a triangle is only the layout).

 T o------+--------+--------+-------+
          |        |        |       |
        [20 Ω]    [90 Ω]  [60 Ω]  [R_L]
         (10 V)    |        |       |
 o-------+---------+--------+-------+

Step 1: Thevenin resistance (source shorted)

R_th = 20 ∥ 90 ∥ 60

1/R_th = 1/20 + 1/90 + 1/60 = 0.05 + 0.01111 + 0.01667 = 0.07778

R_th = 12.86 Ω (90/7 Ω)

Step 2: Thevenin voltage (R_L removed)

V_th = 10 × R_th/20 = 10 × 12.857/20 = 6.43 V

(The 10 V source feeds R_th through 20 Ω as a divider; the 90 Ω and 60 Ω have no source.)

Step 3: Maximum power

R_L = R_th = 12.86 Ω

P_max = V_th²/(4 R_th) = 6.4286²/(4 × 12.857) = 41.33/51.43 = 0.804 W

Answer: R_L = 12.86 Ω; P_max = 0.804 W

✓ Answer checked by Claude Opus 5.5 (high effort)

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