The charge (in mu C) on any one of the 2,mu F | vedclass

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(D) From the circuit diagram,the two $2\,\mu F$ capacitors in the upper arms are in series with each other. Their equivalent capacitance is $C_s = \fr

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$2, 2$

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(D) From the circuit diagram,the two $2\,\mu F$ capacitors in the upper arms are in series with each other. Their equivalent capacitance is $C_s = \frac{2 	imes 2}{2 + 2} = 1\,\mu F$.This $C_s$ is in parallel with the $1\,\mu F$ capacitor. The total equivalent capacitance is $C_{eq} = 1\,\mu F + 1\,\mu F = 2\,\mu F$.The total charge supplied by the $2\,V$ battery is $Q_{total} = C_{eq} 	imes V = 2\,\mu F 	imes 2\,V = 4\,\mu C$.The voltage across the parallel branches is $2\,V$. For the $1\,\mu F$ capacitor,the charge is $Q_1 = C 	imes V = 1\,\mu F 	imes 2\,V = 2\,\mu C$.For the series branch containing two $2\,\mu F$ capacitors,the equivalent capacitance is $1\,\mu F$,so the charge on this branch is $Q_s = 1\,\mu F 	imes 2\,V = 2\,\mu C$. Since they are in series,each $2\,\mu F$ capacitor carries the same charge,$2\,\mu C$.Thus,the charge on any one $2\,\mu F$ capacitor is $2\,\mu C$ and on the $1\,\mu F$ capacitor is $2\,\mu C$.

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

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