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(C) $1$. The circuit consists of a $4 \mu	ext{F}$ capacitor in series with a parallel combination of $9 \mu	ext{F}$ and $3 \mu	ext{F}$ capacitors.$

Vedclass · Accepted Answer

$9$

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(C) $1$. The circuit consists of a $4 \mu	ext{F}$ capacitor in series with a parallel combination of $9 \mu	ext{F}$ and $3 \mu	ext{F}$ capacitors.$2$. The equivalent capacitance of the parallel part is $C_p = 9 \mu	ext{F} + 3 \mu	ext{F} = 12 \mu	ext{F}$.$3$. The total equivalent capacitance $C_{eq}$ of the circuit is given by $\frac{1}{C_{eq}} = \frac{1}{4 \mu	ext{F}} + \frac{1}{12 \mu	ext{F}} = \frac{3+1}{12 \mu	ext{F}} = \frac{4}{12 \mu	ext{F}} = \frac{1}{3 \mu	ext{F}}$. Thus, $C_{eq} = 3 \mu	ext{F}$.$4$. The total charge $Q$ supplied by the $12 	ext{ V}$ battery is $Q = C_{eq} 	imes V = 3 \mu	ext{F} 	imes 12 	ext{ V} = 36 \mu	ext{C}$.$5$. This charge $Q$ flows through the $4 \mu	ext{F}$ capacitor. The potential difference $V_1$ across the $4 \mu	ext{F}$ capacitor is $V_1 = \frac{Q}{C_1} = \frac{36 \mu	ext{C}}{4 \mu	ext{F}} = 9 	ext{ V}$.

In the given circuit, the potential difference across the $4 \mu\text{F}$ capacitor is (in $text{ V}$)

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