In the circuit shown in the figure,the ratio | vedclass

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(C) The $4 \mu F$ capacitor is connected directly across the $6 	ext{ V}$ battery,so the potential difference across it is $V_{4} = 6 	ext{ V}$.Thus

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$3/8$

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(C) The $4 \mu F$ capacitor is connected directly across the $6 	ext{ V}$ battery,so the potential difference across it is $V_{4} = 6 	ext{ V}$.Thus,the charge on the $4 \mu F$ capacitor is $Q_{4} = C_{4} 	imes V_{4} = 4 \mu F 	imes 6 	ext{ V} = 24 \mu C$.The $3 \mu F$ capacitor is in series with the parallel combination of $2 \mu F$ and $5 \mu F$ capacitors. The equivalent capacitance of the parallel part is $C_{p} = 2 \mu F + 5 \mu F = 7 \mu F$.The total capacitance of this branch is $\frac{1}{C_{branch}} = \frac{1}{3 \mu F} + \frac{1}{7 \mu F} = \frac{10}{21 \mu F}$,so $C_{branch} = 2.1 \mu F$.The total charge flowing through this branch is $Q_{branch} = C_{branch} 	imes 6 	ext{ V} = 2.1 \mu F 	imes 6 	ext{ V} = 12.6 \mu C$.The potential difference across the parallel combination ($2 \mu F$ and $5 \mu F$) is $V_{p} = \frac{Q_{branch}}{C_{p}} = \frac{12.6 \mu C}{7 \mu F} = 1.8 	ext{ V}$.Since the $5 \mu F$ capacitor is in parallel,the potential difference across it is $1.8 	ext{ V}$.The charge on the $5 \mu F$ capacitor is $Q_{5} = 5 \mu F 	imes 1.8 	ext{ V} = 9 \mu C$.The ratio of charges is $\frac{Q_{5}}{Q_{4}} = \frac{9 \mu C}{24 \mu C} = \frac{3}{8}$.

In the circuit shown in the figure,the ratio of charges on $5 \mu F$ and $4 \mu F$ capacitors is:

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