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Parallel and Series Combination of Capacitors

Question

The potential difference across $\frac{C (V_{1}^{2} - V_{2}^{2})}{L}$ condenser is :

$40 V o l t$
$80 V o l t$
$60 V o l t$
$120 V o l t$

A

60 V o l t

B

40 V o l t

C

80 V o l t

D

120 V o l t

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Solution

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As both the capacitor are in series the charge flowing through it is same,
$Q_{1} = Q_{2}$
Now if we consider potential across $3 μ F$ capacitor as V then potential across $6 μ F$ capacitor is given as $120 - V$ ,
$(120 - V) 6 μ F = V (3 μ F)$
$720 μ F = V (9 μ C)$
$V = 80 V$

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