Exam Style Question - 3

QuestionA car battery has an e.m.f. of 12 V and an internal resistance of 0.020 Ω. When the starter motor is operated, the battery supplies a current of 150 A.(a) Calculate the terminal potential difference while the starter motor runs.
(b) Calculate the power wasted in the battery and the power delivered to the motor.
(c) Explain why the headlights dim while the starter motor is running.(a)ε = V + Ir, so V = ε − IrV = 12 − (150 × 0.020) = 12 − 3.0 = 9.0 VThe 3.0 V is the lost volts.(b)P_wasted = I²r = 150² × 0.020 = 450 WP_motor = VI = 9.0 × 150 = 1350 WCheck: the total power from the e.m.f. is εI = 12 × 150 = 1800 W = 450 W + 1350 W.(c) The headlights are connected in parallel across the battery terminals, so they receive the terminal p.d. The very large starter current makes the lost volts large, so the terminal p.d. falls from nearly 12 V to 9.0 V. With a smaller p.d. across them, the lamps draw less current and their power falls, so they dim.Exam tip: internal resistance matters most when the current is large. For a headlight current of 0.50 A on its own, the lost volts would be only 0.010 V.