Circuit analysis: series-parallel combination yields power in R

Steps:

• Two 6Ω resistors in parallel give equivalent resistance of 3Ω.
• This 3Ω is in series with the third 6Ω resistor R, for total resistance of 9Ω.
• Total current I = 12V / 9Ω = 4/3 A.
• Power in R = I² × 6Ω = (16/9) × 6 = 32/3 W ≈ 11W.

Why C is correct:

• Power formula P = I²R directly gives 32/3 W (≈11W) for the voltage drop across R, per Ohm's law.

Why the others are wrong:

• A: Matches power in each parallel resistor (≈2.7W), not R.
• B: Equals emf divided by R (12V/6Ω=2A, but I=2A gives P=24W, not 6W).
• D: Assumes 12V directly across one 6Ω resistor (P=V²/R=24W), ignoring configuration.

Final answer: C