A2 Daily A Level Physics question
A 12 V camping cooler is rated 48 W when connected directly to a car battery. You instead power it through a long thin lead whose total resistance is 0.50 Ω. Assume the cooler’s resistance is constant. Approximately how much electrical power does the cooler now receive?
Answer
The correct answer is A.
Correct: A — About 35 W. The cooler’s resistance is 12^2/48 = 3 Ω; with 0.50 Ω in series the cooler gets 3/(3.5) of 12 V ≈ 10.3 V, so its power is (10.3)^2/3 ≈ 35 W. A — Uses the correct square dependence of power on the reduced load voltage, matching the calculation. B — Assumes the power rating stays fixed at 12 V regardless of cable losses, ignoring the voltage drop across the 0.50 Ω lead. C — Scales power linearly with voltage (≈0.857×48 ≈ 41 W), but for a fixed resistor power depends on voltage squared. D — Implies a near-halving simply because extra resistance is added; here 0.50 Ω is much smaller than 3 Ω, so the drop cannot be that large.