In the given capacitive network,the resultant | vedclass

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(B) $1$. Capacitors $C_3$ and $C_4$ are in series. Their equivalent capacitance $C_S$ is given by:$\frac{1}{C_S} = \frac{1}{C_3} + \frac{1}{C_4} = \fr

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(B) $1$. Capacitors $C_3$ and $C_4$ are in series. Their equivalent capacitance $C_S$ is given by:$\frac{1}{C_S} = \frac{1}{C_3} + \frac{1}{C_4} = \frac{1}{8} + \frac{1}{8} = \frac{2}{8} = \frac{1}{4} \implies C_S = 4 \mu F$.$2$. Capacitors $C_5$ and $C_6$ are in parallel. Their equivalent capacitance $C_P$ is:$C_P = C_5 + C_6 = 4 \mu F + 4 \mu F = 8 \mu F$.$3$. Now,$C_2$ and $C_P$ are in series. Their equivalent capacitance $C_{2P}$ is:$C_{2P} = \frac{C_2 \cdot C_P}{C_2 + C_P} = \frac{8 \cdot 8}{8 + 8} = \frac{64}{16} = 4 \mu F$.$4$. This $C_{2P}$ is in parallel with $C_S$. Their equivalent capacitance $C_{total}'$ is:$C_{total}' = C_{2P} + C_S = 4 \mu F + 4 \mu F = 8 \mu F$.$5$. Finally,$C_1$ and $C_{total}'$ are in series. The resultant capacitance $C_{AB}$ is:$C_{AB} = \frac{C_1 \cdot C_{total}'}{C_1 + C_{total}'} = \frac{8 \cdot 8}{8 + 8} = \frac{64}{16} = 4 \mu F$.

In the given capacitive network,the resultant capacitance between points $A$ and $B$ is: (in $\mu F$)

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