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Series-parallel resistors, find fourth resistance

A circuit containing three resistors with resistances 12Ω, 18Ω and 36Ω respectively joined in parallel is connected in series with a fourth resistance. The whole circuit is supplied at 60V and it is found that power dissipated in 12Ω resistance is 36watt. Determine the value of fourth resistance and the total power dissipated in the group.

The 12 Ω, 18 Ω and 36 Ω resistors are in parallel, and this group is in series with the unknown resistor R₄ across 60 V.

Step 1: Voltage across the parallel group

P₁₂ = V²/12 = 36, so V² = 432 and V = 20.78 V

Step 2: Branch currents

  • I₁₂ = 20.78/12 = 1.732 A
  • I₁₈ = 20.78/18 = 1.155 A
  • I₃₆ = 20.78/36 = 0.577 A

Total current I = 1.732 + 1.155 + 0.577 = 3.464 A

Step 3: Fourth resistance

Voltage across R₄ = 60 − 20.78 = 39.22 V

R₄ = 39.22/3.464 = 11.32 Ω

Step 4: Power in the group

  • P₁₂ = 36 W
  • P₁₈ = 20.78²/18 = 24 W
  • P₃₆ = 20.78²/36 = 12 W

Power dissipated in the parallel group = 36 + 24 + 12 = 72 W

(Check: the whole circuit takes 60 × 3.464 = 207.8 W, of which R₄ takes 39.22 × 3.464 = 135.8 W.)

Answer: R₄ ≈ 11.32 Ω; power in the group = 72 W

✓ Answer checked by Claude Opus 5.5 (high effort)

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