A2 Daily A Level Physics question
Which statement must be true when a 6.0 V cell with internal resistance 0.50 Ω powers a single 5.0 Ω resistor, and then a second identical 5.0 Ω resistor is connected in parallel across the first? Assume connecting leads have negligible resistance.
Answer
The correct answer is C.
Correct: C — Terminal p.d. falls from about 5.46 V to 5.0 V (≈8% decrease); the power in each 5.0 Ω resistor falls to about 5 W, while the total power in the two resistors increases. With one 5.0 Ω, total R = 5.5 Ω so I ≈ 6/5.5 = 1.09 A and terminal p.d. ≈ 5.46 V; with two in parallel the load is 2.5 Ω, total R = 3.0 Ω, I = 2.0 A, and the larger drop across 0.50 Ω (Ir = 1.0 V) leaves 5.0 V at the terminals; each 5.0 Ω then has P ≈ 5.0 W, while total load power rises to about 10 W. A Terminal p.d. cannot rise: adding a parallel resistor increases current, increasing the internal voltage drop, so the terminal p.d. must fall, not rise to ~5.6 V. B Although the terminal p.d. falls slightly, the power in each identical resistor cannot rise because each sees a lower voltage (P ∝ V^2/R for fixed R). C Matches the correct direction and scale (≈8% drop in terminal p.d.) and correctly distinguishes per‑resistor power (down to ~5 W) from total load power (up). D Assumes an ideal source (r = 0), which is not the case; moreover, even with an ideal source, each resistor’s power would remain the same as the single‑resistor case, not halve.