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(D) The given circuit consists of two parallel branches connected across a $150 V$ battery. Branch $1$ (top) contains two $20 \mu F$ capacitors in ser

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$3 	imes 10^{-3} C$

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(D) The given circuit consists of two parallel branches connected across a $150 V$ battery. Branch $1$ (top) contains two $20 \mu F$ capacitors in series. The equivalent capacitance $C_1$ is given by: $\frac{1}{C_1} = \frac{1}{20} + \frac{1}{20} = \frac{2}{20} = \frac{1}{10} \implies C_1 = 10 \mu F$. Branch $2$ (bottom) also contains two $20 \mu F$ capacitors in series. The equivalent capacitance $C_2$ is: $\frac{1}{C_2} = \frac{1}{20} + \frac{1}{20} = \frac{1}{10} \implies C_2 = 10 \mu F$. Since these two branches are in parallel, the total equivalent capacitance $C_{eq}$ is: $C_{eq} = C_1 + C_2 = 10 \mu F + 10 \mu F = 20 \mu F$. The total charge $Q$ stored is given by $Q = C_{eq} V$: $Q = 20 	imes 10^{-6} F 	imes 150 V = 3000 	imes 10^{-6} C = 3 	imes 10^{-3} C$.

Five identical capacitors, each of capacitance $20 \mu F$, are connected to a battery of $150 V$ in the combination shown in the diagram. What is the total amount of charge stored?

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