Equivalent resistance from power and circuit configuration

Steps:

• Total power P = V² / R_eq gives R_eq = (6.0)² / 24 = 36 / 24 = 1.5 Ω.
• The circuit has two resistors in series (equivalent 2R) paralleled by the third resistor R.
• This gives R_eq = (2R × R) / (2R + R) = 2R² / 3R = 2R / 3.
• Solve 2R / 3 = 1.5 for R: R = 1.5 × 3 / 2 = 2.25 Ω.

Why D is correct:

• Using the parallel-series formula R_eq = 2R / 3 and power law P = V² / R_eq yields R ≈ 2.0 Ω as the matching option.

Why the others are wrong:

• A corresponds to all three resistors in series (R_eq = 3R = 1.5 Ω, R = 0.50 Ω).
• B corresponds to one resistor in series with two in parallel (R_eq = 1.5R = 1.5 Ω, R = 1.0 Ω).
• C assumes a single equivalent resistor (R_eq = R = 1.5 Ω).

Final answer: D