For the circuit shown in figure, which of the following statement is true?

with

S_{1}

closed

V_{1} = 15 V

V_{2} = 20 V

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with

S_{3}

closed

V_{1} = V_{2} = 25 V

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with

S_{1}

and

S_{2}

closed

V_{1} = V_{2} = 0

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with

S_{1}

and

S_{3}

closed

V_{1} = 30 V, V_{2} = 20 V

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Solution

The correct option is D with $S_{1}$ and $S_{3}$ closed $V_{1} = 30 V, V_{2} = 20 V$
charges on the capacitor are
$q_{1} = (30) (2) = 60 μ F$
$q_{2} = (20) (3) = 60 μ F$
$\Rightarrow q_{1} = q_{2} = q (l e t)$
When $s_{3}$ is closed, due to attraction with opposite charge, no flow of charge take place through $s_{3}$ / Therefore, potential difference across capital plates remain unchanged i.e. $v_{1} = 30 v$ and
$v_{2} = 20 v$

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Similar questions

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 $V_{3}$ .