Five capacitors,each of value 1 mu F,are conn | vedclass

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(B) The circuit consists of two branches connected in parallel between the points $A$ and $B$. Each branch contains two capacitors of $1 \mu F$ connec

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(B) The circuit consists of two branches connected in parallel between the points $A$ and $B$. Each branch contains two capacitors of $1 \mu F$ connected in series. The middle capacitor is connected between the top and bottom wires,but it does not affect the potential difference between $A$ and $B$ in a way that changes the series-parallel configuration of the outer branches. For the left branch,the equivalent capacitance $C_1$ is given by $\frac{1}{C_1} = \frac{1}{1 \mu F} + \frac{1}{1 \mu F} = 2 \mu F^{-1}$,so $C_1 = 0.5 \mu F$. Similarly,for the right branch,the equivalent capacitance $C_2$ is $C_2 = 0.5 \mu F$. Since these two branches are in parallel,the total equivalent capacitance $C_{AB} = C_1 + C_2 = 0.5 \mu F + 0.5 \mu F = 1 \mu F$.

Five capacitors,each of value $1 \mu F$,are connected as shown in the figure. The equivalent capacitance between $A$ and $B$ is (in $\mu F$)

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