In figure, three capacitors $C_{1}, C_{2} a n d C_{3}$ are joined to a battery. with symbols having their usual meaning, the correct conditions will be

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$Q_{1} = Q_{2} = Q_{3} a n d V_{2} = V_{3} + V$

$Q_{1} = Q_{2} + Q_{3} a n d V = V_{1} + V_{2} + V_{3}$

$Q_{2} = Q_{3} a n d V_{2} = V_{3}$

$Q_{1} = Q_{2} + Q_{3} a n d V = V_{1} + V_{2}$

answer is C.

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Detailed Solution

At junction A, $Q_{1}$ will be divided into $Q_{2} a n d Q_{3} .$ Hence, $Q_{1} = Q_{2} + Q_{3} . C_{2} a n d C_{3}$ are in parallel, so potential on them will be the same, i.e., $V_{2} = V_{3}$

V will be divided into $V_{1} a n d V_{2} (o r V_{3})$

Hence, $V = V_{1} + V_{2} o r V = V_{1} + V_{3}$

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