In the given circuit,C_1=2,mu F, C_2=0.2,mu F | vedclass

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(A) First,simplify the circuit. Capacitors $C_3, C_4,$ and $C_5$ are in series. Their equivalent capacitance $C_{345}$ is given by $\frac{1}{C_{345}}

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(A) First,simplify the circuit. Capacitors $C_3, C_4,$ and $C_5$ are in series. Their equivalent capacitance $C_{345}$ is given by $\frac{1}{C_{345}} = \frac{1}{C_3} + \frac{1}{C_4} + \frac{1}{C_5} = \frac{1}{2} + \frac{1}{4} + \frac{1}{2} = \frac{2+1+2}{4} = \frac{5}{4}$. Thus,$C_{345} = 0.8\,\mu F$.Now,$C_{345}$ is in parallel with $C_2$. Their equivalent capacitance $C_{2345} = C_2 + C_{345} = 0.2 + 0.8 = 1.0\,\mu F$.This combination is in series with $C_1$ and $C_6$. The total equivalent capacitance $C_{eq}$ is $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_{2345}} + \frac{1}{C_6} = \frac{1}{2} + \frac{1}{1} + \frac{1}{2} = 2\,\mu F^{-1}$. So,$C_{eq} = 0.5\,\mu F$.The total charge drawn from the $10\,V$ source is $Q = C_{eq} 	imes V = 0.5\,\mu F 	imes 10\,V = 5\,\mu C$.This charge $Q$ flows through the series combination of $C_1, C_{2345},$ and $C_6$. The potential difference across the parallel combination $(C_{2345})$ is $V_{2345} = \frac{Q}{C_{2345}} = \frac{5\,\mu C}{1\,\mu F} = 5\,V$.The charge on the series branch containing $C_4$ is $Q' = C_{345} 	imes V_{2345} = 0.8\,\mu F 	imes 5\,V = 4\,\mu C$.

In the given circuit,$C_1=2\,\mu F, C_2=0.2\,\mu F, C_3=2\,\mu F, C_4=4\,\mu F, C_5=2\,\mu F, C_6=2\,\mu F$. The charge stored on capacitor $C_4$ is $.....\mu C$.

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