A circuit has three resistors: 4 Ω and 8 Ω in parallel, connected in series with a 6 Ω resistor and a 12 V battery. The power dissipated in the 6 Ω resistor is:
Correct answer: C. 12W
- A. 6W
- B. 8W
- C. 12W
- D. 18W
Explanation
Parallel resistance: 1/Rₚ = 1/4 + 1/8 = 3/8, so Rₚ = 8/3 Ω. Total resistance = 8/3 + 6 = 26/3 Ω. Current I = 12/(26/3) = 18/13 A. Power in 6 Ω = I²R = (18/13)² × 6 ≈ 12 W.
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