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(A) From the given circuit,the capacitors $2.5\,\mu F$ and $1\,\mu F$ are connected in parallel.Their equivalent capacitance $C_p$ is given by $C_p =

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$1.4$

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(A) From the given circuit,the capacitors $2.5\,\mu F$ and $1\,\mu F$ are connected in parallel.Their equivalent capacitance $C_p$ is given by $C_p = 2.5\,\mu F + 1\,\mu F = 3.5\,\mu F$.Now,this parallel combination is in series with the capacitor $C$.The total equivalent capacitance $C_{eq}$ between $A$ and $B$ is given by $\frac{1}{C_{eq}} = \frac{1}{C} + \frac{1}{C_p}$.Given $C_{eq} = 1\,\mu F$,we have $\frac{1}{1} = \frac{1}{C} + \frac{1}{3.5}$.$1 = \frac{1}{C} + \frac{1}{3.5} \implies \frac{1}{C} = 1 - \frac{1}{3.5} = \frac{2.5}{3.5}$.$C = \frac{3.5}{2.5} = 1.4\,\mu F$.

The equivalent capacitance between $A$ and $B$ in the figure is $1\,\mu F$. Then the value of capacitance $C$ is.....$\mu F$

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