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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Equivalent capacitance of network is $C_{e q}=4 \mu F$
charge $Q=C_{e q} V=4 V \mu C$
The charge on the $7 \mu F$ or $3 \mu F$ capacitor is
$Q_{2}=(7 \mu F)(6 V)=42 \mu C$
$\frac{Q_{2}}{2.1 \mu F}=\frac{Q_{1}}{12 \mu F}$ or $Q_{1}=(42 \mu C) \frac{(3.9 \mu F)}{(2.1 \mu F)}=78 \mu C$
$Q=Q_{1}+Q_{2}=(78 \mu C+42 \mu C)=120 \mu C=4 V \mu C$
Emf of the battery is $V=30 V$
The potential drop across $12 \mu F$ capacitor is
$\frac{Q}{12 \mu F}=\frac{120 \mu C}{12 \mu F}=10 V$
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