The charge on any one of the 2,mu F capacitor | vedclass

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(D) The circuit consists of two parallel branches connected to a $2\,V$ battery.Branch $1$ contains a $1\,\mu F$ capacitor. The potential difference a

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$2, 2$

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(D) The circuit consists of two parallel branches connected to a $2\,V$ battery.Branch $1$ contains a $1\,\mu F$ capacitor. The potential difference across this branch is $2\,V$. Therefore, the charge on the $1\,\mu F$ capacitor is $Q_1 = C_1 V = (1\,\mu F) 	imes (2\,V) = 2\,\mu C$.Branch $2$ contains two $2\,\mu F$ capacitors in series. The equivalent capacitance of this branch is $C_{eq} = \frac{2\,\mu F 	imes 2\,\mu F}{2\,\mu F + 2\,\mu F} = 1\,\mu F$. The potential difference across this branch is also $2\,V$. Therefore, the charge on each of the $2\,\mu F$ capacitors is $Q_2 = C_{eq} V = (1\,\mu F) 	imes (2\,V) = 2\,\mu C$.Thus, the charge on any one of the $2\,\mu F$ capacitors is $2\,\mu C$ and the charge on the $1\,\mu F$ capacitor is $2\,\mu C$. The correct option is $D$.

The charge on any one of the $2\,\mu F$ capacitors and the $1\,\mu F$ capacitor will be given respectively (in $\mu C$) as

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