In the following circuit, the equivalent capacitance between terminal $A$ and terminal $B$ is

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Professor PC Thomas Classes · Answer

As, wheatstone bridge is balanced, $\frac{C_1}{C_2}=\frac{C_3}{C_4}$ and $C_5$ becomes ineffective. Capacitor $C_1$ and $C_3$ are in series. $ 	herefore \quad C_{13}=\frac{C_1 C_3}{C_1+C_3}=1 \mu \mathrm{~F} $ Similarly, capacitor, $C_2$ and $C_4$ are in series, $ 	herefore \quad C_{24}=\frac{C_2 C_4}{C_2+C_4}=1 \mu \mathrm{~F} $ Hence, equivalent capacitance between $A$ and $B$ is $ C_{13}+C_{24}=2 \mu \mathrm{~F} $