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

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Potential drop across $3 μ F$ capacitor is 10V

Charge on $3 μ F$ capacitor is $21 μ F$

Emf of the cell is 30V

P.D across $12 μ F$ capacitor is 5V

answer is C.

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Detailed Solution

$q_{7} = CV = 42 μ C$

Thus, $q_{3} = 42 μ C$

$\begin{array}{l} V_{3} = \frac{42}{3} = 14 V \\ V_{3.9} = 20 V \\ q_{3.9} = CV = 78 μ C \\ q = 42 + 78 = 120 μ C \\ V_{12} = \frac{q}{C_{12}} = \frac{120}{12} = 10 V \\ E = V_{3.9} + V_{12} = 30 V \end{array}$

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