The potential difference across the 4 mu F ca | vedclass

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(C) $1$. First,calculate the equivalent capacitance of the parallel combination of the $2 \mu F$ and $4 \mu F$ capacitors.$C_p = 2 \mu F + 4 \mu F = 6

Vedclass · Accepted Answer

$5.4$

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(C) $1$. First,calculate the equivalent capacitance of the parallel combination of the $2 \mu F$ and $4 \mu F$ capacitors.$C_p = 2 \mu F + 4 \mu F = 6 \mu F$.$2$. Now,the circuit consists of a $4 \mu F$ capacitor in series with the equivalent $6 \mu F$ capacitor,connected to a $9 V$ battery.$3$. The equivalent capacitance of the entire circuit $(C_{eq})$ is given by:$\frac{1}{C_{eq}} = \frac{1}{4 \mu F} + \frac{1}{6 \mu F} = \frac{3 + 2}{12 \mu F} = \frac{5}{12 \mu F}$.$C_{eq} = \frac{12}{5} \mu F = 2.4 \mu F$.$4$. The total charge $(Q)$ supplied by the battery is:$Q = C_{eq} 	imes V = 2.4 \mu F 	imes 9 V = 21.6 \mu C$.$5$. Since the $4 \mu F$ capacitor is in series with the parallel combination,the same charge $Q = 21.6 \mu C$ flows through it.$6$. The potential difference $(V_1)$ across the $4 \mu F$ capacitor is:$V_1 = \frac{Q}{C_1} = \frac{21.6 \mu C}{4 \mu F} = 5.4 V$.

The potential difference across the $4 \mu F$ capacitor in series with the parallel combination in the following circuit is: (in $V$)

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