Three capacitors 3 mu F, 6 mu F, and 6 mu F a | vedclass

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(D) For capacitors connected in series, the equivalent capacitance $C_{eq}$ is given by:$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C

Vedclass · Accepted Answer

$60$

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(D) For capacitors connected in series, the equivalent capacitance $C_{eq}$ is given by:$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$$\frac{1}{C_{eq}} = \frac{1}{3} + \frac{1}{6} + \frac{1}{6} = \frac{2+1+1}{6} = \frac{4}{6} = \frac{2}{3} \mu F^{-1}$So, $C_{eq} = 1.5 \mu F$.The total charge $q$ flowing through the circuit is:$q = C_{eq} 	imes V = 1.5 \mu F 	imes 120 V = 180 \mu C$.In a series circuit, the charge on each capacitor is the same. Therefore, the charge on the $3 \mu F$ capacitor is $q = 180 \mu C$.The potential difference $V_1$ across the $3 \mu F$ capacitor is:$V_1 = \frac{q}{C_1} = \frac{180 \mu C}{3 \mu F} = 60 V$.

Three capacitors $3 \mu F$, $6 \mu F$, and $6 \mu F$ are connected in series to a source of $120 V$. The potential difference, in volts, across the $3 \mu F$ capacitor will be:

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