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Byju's Answer

Standard XII

Physics

Kirchhoff's Voltage Law

In the circui...

Question

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 V1 when only S1 is closed, V2 when only S2 is closed and V3 when both S1 and S2 are closed. Choose the correct relation between V1, V2 and V3.

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