When the lights of a car are switched on, an ammeter in series with them reads10.0 Aand a voltmeter connected across them reads12.0 V(Fig. 27-60). When the electric starting motor is turned on, the ammeter reading drops to8.00 Aand the lights dim somewhat. If the internal resistance of the battery is0.0500 ohmand that of the ammeter is negligible, what are (a) the emf of the battery and (b) the current through the starting motor when the lights are on?

Ammeter reading of the car lights in series, i=10 A

Voltmeter reading, V=12 V

After the starting motor, the ammeter reading becomes i_A^'=8 A

The internal resistance of the battery, ${\mathrm{R}}_{\mathrm{int}}=0.05\mathrm{\Omega }$

Kirchhoff's voltage law gives the voltage equation that is used to calculate the emf before and after the start of the motor. Again, for the given condition, the internal resistance plays the role after the start of the motor.

Formulae:

The voltage equation using Ohm’s law, $\mathrm{V}=\mathrm{IR}$ (1)

Kirchhoff’s voltage law, $\sum _{\mathrm{closed}\mathrm{loop}}\mathrm{V}=0$ (2)

The emf of the battery can be given using equations (1) and (2) as follows:

$\begin{array}{rcl}\mathrm{\epsilon }& =& \mathrm{V}+\mathrm{ir}\\ & =& 12\mathrm{V}+\left(10\mathrm{A}\right)\left(0.05\mathrm{\Omega }\right)\\ & =& 12.5\mathrm{V}\end{array}$

Hence, the emf of the battery is 12.5 V.

Now, after the motor is started, the new voltmeter reading can be given using equation (1) as follows:

$\begin{array}{rcl}{\mathrm{V}}^{\text{'}}& =& {\mathrm{i}}_{\mathrm{A}}^{\text{'}}{\mathrm{R}}_{\mathrm{light}}\\ & =& \left(8.00\mathrm{A}\right)\left(12.0\mathrm{V}/10\mathrm{A}\right)\\ & =& 9.60\mathrm{V}\end{array}$

Now, the emf equation of the battery can be given using equation (1) in equation (2) and thus the current of the motor can be calculated as follows:

$\begin{array}{rcl}\mathrm{\epsilon }& =& \mathrm{V}+\left({\mathrm{i}}_{\mathrm{motor}}+8\mathrm{A}\right)\mathrm{r}\\ {\mathrm{i}}_{\mathrm{motor}}& =& \frac{\mathrm{\epsilon }-{\mathrm{V}}^{\text{'}}}{\mathrm{r}}-8\mathrm{A}\\ {\mathrm{i}}_{\mathrm{motor}}& =& \frac{12.5\mathrm{V}-9.60\mathrm{V}}{0.05\mathrm{\Omega }}-8\mathrm{A}\\ {\mathrm{i}}_{\mathrm{motor}}& =& 50.0\mathrm{A}\end{array}$

Hence, the current value is 50.0 A.