A2 Daily A Level Physics question
A 6.0 V cell with internal resistance 1.0 Ω powers a single 5.0 Ω lamp. A second identical lamp is then connected in parallel across the first, using the same cell. Assuming the lamps behave ohmically over this range, what is the power in each lamp compared to the single-lamp case? Choose the nearest percentage.
Answer
The correct answer is D.
Correct: D — Decreases to about 74% of the original power in each lamp. With one lamp: current = 6.0 V / (5.0 Ω + 1.0 Ω) = 1.0 A, so P1 = I^2R = 5.0 W. With two lamps in parallel: load is 2.5 Ω, total is 2.5 Ω + 1.0 Ω = 3.5 Ω, so terminal p.d. ≈ 6.0 × (2.5/3.5) ≈ 4.29 V; power per lamp ≈ (4.29^2)/5.0 ≈ 3.7 W ≈ 74% of 5.0 W. A assumes lower external resistance necessarily boosts each lamp’s power; it ignores the increased internal voltage drop—calculation gives about 74%, not an increase to 150%. B treats power as being simply “shared” between lamps; in parallel, each lamp sees the same terminal p.d., which falls but not to the extent of halving power. C would be true only for an ideal source with zero internal resistance; here increased current raises the internal drop, reducing terminal p.d. and hence power. D matches the quantitative result when accounting for the cell’s internal resistance.