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  4. In the circuit shown, the electromotive force of the battery is 9 V and its internal resistance is 15 Ω. The two identical voltmeters can be considered ideal. Let V1 and V1' be the reading of 1 st voltmeter when switch is open and closed, respectively. Similarly, let V2 and V2' be the reading of 2 nd voltmeter when switch is open and closed, respectively. Then (V2'-V2/V1-V1')=

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

Current Electricity

Solution:

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