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(B) In the given circuit,the capacitors of $1 \mu F$ and $2 \mu F$ are connected in parallel between points $X$ and $Y$.Therefore,their equivalent cap

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$3$

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(B) In the given circuit,the capacitors of $1 \mu F$ and $2 \mu F$ are connected in parallel between points $X$ and $Y$.Therefore,their equivalent capacitance is $C_p = 1 \mu F + 2 \mu F = 3 \mu F$.The capacitors of $4 \mu F$ (connected to $X$) and $4 \mu F$ (connected to $Y$) are connected in series with the parallel combination of $1 \mu F$ and $2 \mu F$ capacitors.However,the question asks for the equivalent capacitance between points $X$ and $Y$. Since the $1 \mu F$ and $2 \mu F$ capacitors are directly connected in parallel between $X$ and $Y$,the equivalent capacitance between $X$ and $Y$ is simply the parallel combination of these two capacitors.Thus,$C_{XY} = 1 \mu F + 2 \mu F = 3 \mu F$.

In the circuit shown in the figure,the equivalent capacitance between the points $X$ and $Y$ is ...... $\mu F$.

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