In the circuit shown in the figure, reading of voltmeter is $V_{1}$ when only $S_{1}$ is closed, $V_{2}$ when only $S_{2}$ is closed and $V_{3}$ when both $S_{1}$ and $S_{2}$ are closed. Choose the correct relation between $V_{1}$ , $V_{2}$ and $V_{3}$ .

$V_{2} > V_{1} > V_{3}$

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$V_{3} > V_{2} > V_{1}$

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$V_{3} > V_{1} > V_{2}$

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$V_{1} > V_{2} > V_{3}$

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Solution

The correct option is A
$V_{2} > V_{1} > V_{3}$

When only $S_{1}$ is closed,
$R_{e q} = R + 3 R = 4 R$
$I = \frac{E}{R_{e q}} = \frac{E}{4 R}$
Reading of the voltmeter $V_{1} = \frac{E}{4 R} \times 3 R = \frac{3 E}{4}$

When only $S_{2}$ is closed,
$R_{e q} = R + 6 R = 7 R$
$I = \frac{E}{R_{e q}} = \frac{E}{7 R}$
Reading of the voltmeter $V_{2} = \frac{E}{7 R} \times 6 R = \frac{6 E}{7}$

When both $S_{1}$ and $S_{2}$ are closed,
$R_{e q} = R + \frac{3 R \times 6 R}{3 R + 6 R} = R + 2 R = 3 R$
$I = \frac{E}{R_{e q}} = \frac{E}{3 R}$
Reading of the voltmeter $V_{3} = \frac{E}{3 R} \times 2 R = \frac{2 E}{3}$

Hence, $V_{2} > V_{1} > V_{3}$

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In the circuit shown in the figure, reading of voltmeter is $V_{1}$ when only $S_{1}$ is closed, $V_{2}$ when only $S_{2}$ is closed and $V_{3}$ when both $S_{1}$ and $S_{2}$ are closed. Choose the correct relation between $V_{1}$ , $V_{2}$ and $V_{3}$ .