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Equivalent resistance and power in 5Ω

Short answer

Combine middle branch 3+4=7 Ω, parallel group 2,7,5 Ω gives 1.186 Ω, then add series 1 and 6 Ω: R_eq ≈ 8.19 Ω. Voltage across parallel group = 0.1449 V, power in 5 Ω: P = V²/5 = 0.0042 V² W (42 W for 100 V).

Find the equivalent resistance in the figure shown, and power dissipated in the 5Ω resistor. [Figure: supply voltage V; 1 Ω in series, then a parallel combination of: 2 Ω on top; 3 Ω in series with 4 Ω in the middle; 5 Ω at the bottom; followed by 6 Ω in series.]

The supply voltage V is not given on the scan, so the answer for the power is given in terms of V.

        +--[2 Ω]------+
 o--[1]-+--[3]--[4]---+--[6]--o
        +--[5 Ω]------+

Equivalent resistance

Middle branch: 3 + 4 = 7 Ω

Parallel group: 1/R_p = 1/2 + 1/7 + 1/5 = 0.5 + 0.1429 + 0.2 = 0.8429 R_p = 1.186 Ω (70/59 Ω)

R_eq = 1 + 1.186 + 6 = 8.186 Ω

Power in the 5 Ω resistor

Supply current: I = V/8.186 = 0.1222 V Voltage across the parallel group: V_p = I × 1.186 = 0.1449 V P₅ = V_p²/5 = (0.1449 V)²/5 = 0.0042 V² watts

Example: for V = 100 V, I = 12.2 A, V_p = 14.49 V and P₅ = 42.0 W.

Answer: R_eq ≈ 8.19 Ω; P₅ ≈ 0.0042 V² W (42 W for a 100 V supply)

✓ Answer checked by Claude Opus 5.5 (high effort)

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