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(D) $1$. The two $4\,\mu F$ capacitors are in parallel,so their equivalent capacitance is $C_p = 4\,\mu F + 4\,\mu F = 8\,\mu F$.$2$. Now,the circuit

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$580\,\mu C;\;145\;V$

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(D) $1$. The two $4\,\mu F$ capacitors are in parallel,so their equivalent capacitance is $C_p = 4\,\mu F + 4\,\mu F = 8\,\mu F$.$2$. Now,the circuit consists of three capacitors in series: $20\,\mu F$,$8\,\mu F$ (the parallel combination),and $12\,\mu F$.$3$. The equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_{eq}} = \frac{1}{20} + \frac{1}{8} + \frac{1}{12} = \frac{6 + 15 + 10}{120} = \frac{31}{120}\,\mu F^{-1}$,so $C_{eq} = \frac{120}{31}\,\mu F$.$4$. The total charge $Q$ supplied by the $300\,V$ source is $Q = C_{eq} 	imes V = \frac{120}{31} 	imes 300 = \frac{36000}{31} \approx 1161.29\,\mu C$.$5$. Since the capacitors are in series,the same charge $Q$ flows through each branch. The charge $Q$ splits equally between the two parallel $4\,\mu F$ capacitors because they have equal capacitance.$6$. Charge on each $4\,\mu F$ capacitor is $q = \frac{Q}{2} = \frac{1161.29}{2} \approx 580.6\,\mu C$.$7$. The potential difference across the $4\,\mu F$ capacitor is $V' = \frac{q}{C} = \frac{580.6}{4} \approx 145.15\,V$. The closest option is $D$.

In the adjoining figure,four capacitors are shown with their respective capacities and the $P.D.$ applied. The charge and the $P.D.$ across the $4\,\mu F$ capacitor will be

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