The effective capacitance between points X an | vedclass

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(C) Given, $C_1 = C_2 = C_3 = C_4 = C_5 = 4 \mu F$. Looking at the circuit, we can identify that the capacitors $C_1, C_2, C_3, C_4, C_5$ form a bridg

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$4$

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(C) Given, $C_1 = C_2 = C_3 = C_4 = C_5 = 4 \mu F$. Looking at the circuit, we can identify that the capacitors $C_1, C_2, C_3, C_4, C_5$ form a bridge network. Let the nodes be $X$ (input), $Y$ (output), and intermediate nodes $A, B, C$. By analyzing the potential distribution, we can see that the circuit is equivalent to a parallel combination of two branches. Specifically, the branch containing $C_1$ and $C_2$ in series is in parallel with the branch containing $C_4$ and $C_5$ in series, and so on. However, a simpler way to view this is to recognize that the circuit simplifies to two parallel branches, each consisting of two capacitors in series. $C_{eq} = (C_1 	ext{ in series with } C_2) + (C_4 	ext{ in series with } C_5)$. Since all $C = 4 \mu F$, the series combination of two $4 \mu F$ capacitors is $\frac{4 	imes 4}{4 + 4} = 2 \mu F$. Thus, $C_{eq} = 2 \mu F + 2 \mu F = 4 \mu F$.

The effective capacitance between points $X$ and $Y$ in the figure shown below is. (Assume all the capacitors are $4 \mu F$ each) (in $\mu F$)

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