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(D) The $7 \mu F$ and $3 \mu F$ capacitors are in series. Their equivalent capacitance $C_s$ is given by $\frac{1}{C_s} = \frac{1}{7} + \frac{1}{3} =

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all of the above.

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(D) The $7 \mu F$ and $3 \mu F$ capacitors are in series. Their equivalent capacitance $C_s$ is given by $\frac{1}{C_s} = \frac{1}{7} + \frac{1}{3} = \frac{10}{21} \mu F^{-1}$,so $C_s = 2.1 \mu F$.The charge on the series combination is $Q_2 = C_s 	imes V_s = (2.1 \mu F) 	imes (6 \ V + V_{3 \mu F})$. Since the potential drop across the $7 \mu F$ capacitor is $6 \ V$,the charge $Q_2$ on the $7 \mu F$ capacitor is $Q_2 = (7 \mu F) 	imes (6 \ V) = 42 \mu C$. This charge is the same for the $3 \mu F$ capacitor in series.The potential drop across the $3 \mu F$ capacitor is $V_3 = \frac{Q_2}{3 \mu F} = \frac{42 \mu C}{3 \mu F} = 14 \ V$. Thus,the total potential across the series branch is $V_{branch} = 6 \ V + 14 \ V = 20 \ V$.The charge on the $3.9 \mu F$ capacitor is $Q_1 = (3.9 \mu F) 	imes (20 \ V) = 78 \mu C$.The total charge from the battery is $Q = Q_1 + Q_2 = 78 \mu C + 42 \mu C = 120 \mu C$.The potential drop across the $12 \mu F$ capacitor is $V_{12} = \frac{Q}{12 \mu F} = \frac{120 \mu C}{12 \mu F} = 10 \ V$.The total e.m.f. of the battery is $E = V_{12} + V_{branch} = 10 \ V + 20 \ V = 30 \ V$.Since all statements are correct,the answer is $(D)$.

Four capacitors and a battery are connected as shown. The potential drop across the $7 \mu F$ capacitor is $6 \ V$. Then the:

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