When the lights of an automobile are switched on, an ammeter in series with them reads 10.0 A and a voltmeter connected across them reads 12 V (see figure). When the electric starting motor is switched on, the ammeter reading drops to 8.00 A and the light dims somewhat. If the internal resistance of the battery is 50.0 m and that of the ammeter is negligible, we have to find (a) the emf of the battery and (b) the current through the starting motor when the lights are on.

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(a)

Let us consider the circuit shown above in the situation when the lights have been switched on. It is given that the ammeter reading is 10.0 A and the voltmeter reading is 12.0 V. Applying Ohm’s law, we note that the resistance of the light bulbs is

We next calculate the emf of the battery. As 10.0 A of current flows through the internal resistance of the battery,

,

we have

(b)

When the motor is switched on and the lights are also on, the ammeter reading is 8.0 A. Let the current drawn by the motor be A. In this situation the total current flowing through the battery will be

.

Applying Kirchoff’s law to the loop that includes the battery and the lights but excludes the motor, we have

Solving this equation, we find that the current drawn by the motor when the lights are also switched on will be