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The potential drop across $7 μ F$ capacitor is $6 V$ . Then

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Charge on $3 μ F$ capacitor is $42 μ C$

P.D across $12 μ F$ capacitor is $10 V$

emf of the cell is $30 V$

Potential drop across $3 μ F$ capacitor is $14 V$

answer is A, B, C, D.

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Detailed Solution

$C_{e q} = 4 μ F$ ; Charge on $7 μ F = 6 \times 7 = 42 μ C$
Charge on $3 μ F$ is same.
Hence pd across it $= \frac{42}{3} V_{2} = 14 V$
Hence pd across $3.9 μ F$ is $= 6 + 14 = 20 V$ .
$20 = \frac{12}{12 + 6} \times V \Rightarrow V = 30 V$

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