A2 Daily A Level Physics question
A 12 V car battery has an internal resistance of 0.20 Ω. The starter motor can be modelled as a resistor: 0.80 Ω on a warm day, but 1.6 Ω on a cold day. Assuming the battery’s emf and internal resistance do not change, what happens to the potential difference across the motor when its resistance doubles?
Answer
The correct answer is B.
Correct: B — It increases by about 11%. With r = 0.20 Ω, warm: V_motor = 12 × 0.80/(0.80+0.20) = 9.6 V; cold: V_motor = 12 × 1.6/(1.6+0.20) ≈ 10.7 V, so the increase is ≈ (10.7−9.6)/9.6 ≈ 11%. A says it decreases by about 11%, but when the motor resistance increases the current falls, so less voltage is lost inside the battery and the motor’s pd actually rises. B is correct because the share of the emf across the motor increases from 0.80/1.00 to 1.6/1.8, giving about an 11% rise in its pd. C claims it is approximately unchanged, but the internal resistance is not negligible compared to the motor, so the division of the emf changes appreciably (~11%). D suggests it roughly doubles, which is impossible here since the terminal pd cannot exceed the fixed 12 V emf and the change in current is modest.