Equivalent resistance and power in parallel circuit

Steps:

• Calculate equivalent resistance: $\frac{1}{R_{eq}} = \frac{1}{R} + \frac{1}{2R} = \frac{3}{2R}$, so $R_{eq} = \frac{2R}{3}$.
• Total power: $P = \frac{V^2}{R_{eq}} = \frac{12^2}{(2R/3)} = \frac{144 \times 3}{2R} = \frac{216}{R}$.
• Set $P = 36$ W: $\frac{216}{R} = 36$, so $R = \frac{216}{36} = 6$ Ω.
• Verify: For R=6Ω, $R_{eq} = 4$ Ω, $P = \frac{144}{4} = 36$ W.

Why D is correct:

• R=6Ω yields $R_{eq}=4$ Ω, and $P = V^2 / R_{eq} = 144/4 = 36$ W exactly.

Why the others are wrong:

• A (0.5Ω): $R_{eq} \approx 0.33$ Ω, P≈432 W (too high).
• B (2Ω): $R_{eq} \approx 1.33$ Ω, P≈108 W (too high).
• C (4Ω): $R_{eq} \approx 2.67$ Ω, P≈54 W (too high).

Final answer: D