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

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Aicharya · Accepted Answer

Given, $C_1=C_2=C_3=C_4=C_5=4 \mu \mathrm{F}$

We can redraw circuit as;

As it forms a balanced wheatstone bridge, $C_2$ does not participate in the circuit. So, $C_{X Y}=\left(C_4\right.$ series $\left.C_3\right) \|\left(C_1\right.$ series $\left.C_5\right)$ $\begin{aligned} & =\left(C_4+C_3\right) \|\left(C_1+C_5\right) \\ & =(4+4)\|(4+4)=8\| 8 \\ & =\frac{8 \times 8}{8+8}=4 \mu \mathrm{F} \\ & =2+2=4 \mu \mathrm{F}\end{aligned}$

The effective capacitance between points $X$ and $Y$ in the figure shown below is. (Assume all the…

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