Question

In the circuit shown, the electromotive force of the battery is $9V$ and its internal resistance is $15\Omega$. The two identical voltmeters can be considered ideal. Let $V_1$ and $V_1^{\prime }$ be the reading of $1$ st voltmeter when switch is open and closed, respectively. Similarly, let $V_2$ and $V_2^{\prime }$ be the reading of $2$ nd voltmeter when switch is open and closed, respectively. Then $\frac{V_2^{\prime }-V_2}{V_1-V_1^{\prime }}=$

Answer 1

Answer: 1

Current in the circuit,
$I=\frac{9}{45}=\frac{1}{5}A=0.2A$
When switch is open:
$V_1+V_2=6V$
$V_1=V_2=3V$
When switch is closed:
$V_1^{\prime }=2V;V_2^{\prime }=4V$
$\frac{V_2^{\prime }-V_2}{V_1-V_1^{\prime }}=\frac{4-3}{3-2}=1$