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Byju's Answer

Standard XII

Physics

Wheatstone Bridge

In the circui...

Question

In the circuit shown, the electromotive force of the battery is 9V and its internal resistance is $15 Ω$ . The two identical voltmeters can be considered ideal. Let $V_{1}$ and $V_{1}^{'}$ be the reading of $1^{s t}$ voltmeter when switch is open and closed, respectively. Similarly, let $V_{2}$ and $V_{2}^{'}$ be the reading of $2^{n d}$ voltmeter when switch is open and closed, respectively. Then $\frac{V_{2}^{^{'}} - V_{2}}{V_{1} - V_{1}^{^{'}}} =$

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Solution

$R_{e q} = 45 Ω$

Current in the circuit,
$I = \frac{9}{45} = \frac{1}{5} A = 0.2 A$

When switch is open :
$V_{1} + V_{2} = I R = 0.2 (30) = 6 V$

Both voltmeter are similar
Hence $V_{1} = V_{2} = 3 V$

When switch is closed :
$V_{1}^{^{'}} = I R = 0.2 (10) = 2 V;$
$V_{2}^{^{'}} = I R = 0.2 (20) = 4 V$

Therefore
$\frac{V_{2}^{^{'}} - V_{2}}{V_{1} - V_{1}^{^{'}}} = \frac{4 - 3}{3 - 2} = 1$

Hence required value = 1

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