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Find RL for maximum power, 10 V network

Short answer

Find Thevenin equivalent: Vth = 2 V, Rth = 1.2 Ω. For maximum power, RL = Rth = 1.2 Ω. Maximum power is Vth²/(4Rth) = 2²/(4×1.2) = 0.833 W.

Find the value of RL in the network shown below to obtain maximum power in it. Also find the maximum power in RL. [Figure: RL in the top branch; 2 Ω, 2 Ω in the top row; 2 Ω shunt in the middle; 10 V source on the left; 2 Ω in series with a 10 V source on the right.]

Reading the figure

Take the bottom rail as reference. R_L is connected between the top-left node L and the top-right node R.

  • Left: 10 V battery (+ on top) from the bottom rail to L.
  • 2 Ω from L to the middle node M; 2 Ω shunt from M to ground; 2 Ω from M to R.
  • Right: 2 Ω in series with a 10 V battery (+ on top, towards the 2 Ω) from R to ground.

Step 1: Thevenin voltage across L-R (R_L removed)

V_L = 10 V. With R_L removed, the only paths at R are the 2 Ω to M and the right branch (2 Ω with the 10 V battery):

Node M: (V_M − 10)/2 + V_M/2 + (V_M − V_R)/2 = 0 → 3 V_M − V_R = 10 Node R: (V_R − V_M)/2 + (V_R − 10)/2 = 0 → −V_M + 2 V_R = 10

Solving: V_M = 6 V, V_R = 8 V

V_th = V_L − V_R = 10 − 8 = 2 V

Step 2: Thevenin resistance (both batteries shorted)

Shorting the left battery joins L to ground, so R_th is the resistance from R to ground:

  • Right branch: 2 Ω.
  • Path through M: 2 Ω (R to M) in series with M to ground, where the 2 Ω shunt is in parallel with the 2 Ω from M to L (2 ∥ 2 = 1 Ω): 2 + 1 = 3 Ω.
  • R_th = 2 ∥ 3 = 1.2 Ω

Step 3: Maximum power

R_L = R_th = 1.2 Ω

P_max = V_th²/(4 R_th) = 2²/(4 × 1.2) = 4/4.8 = 0.833 W

Answer: R_L = 1.2 Ω; P_max = 0.833 W

✓ Answer checked by Claude Opus 5.5 (high effort)

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