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(B) To obtain an equivalent capacitance of $5 \ \mu F$ using $2 \ \mu F$ capacitors,we can arrange them in a combination of series and parallel circui

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$4$

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(B) To obtain an equivalent capacitance of $5 \ \mu F$ using $2 \ \mu F$ capacitors,we can arrange them in a combination of series and parallel circuits.$1$. Connect two $2 \ \mu F$ capacitors in series. The equivalent capacitance $C_s$ is given by $\frac{1}{C_s} = \frac{1}{2} + \frac{1}{2} = 1 \ \mu F$.$2$. Connect two $2 \ \mu F$ capacitors in parallel. The equivalent capacitance $C_p$ is given by $C_p = 2 + 2 = 4 \ \mu F$.$3$. Now,connect the series combination $(1 \ \mu F)$ and the parallel combination $(4 \ \mu F)$ in parallel. The total equivalent capacitance $C_{eq} = 1 \ \mu F + 4 \ \mu F = 5 \ \mu F$.$4$. The total number of capacitors used is $2$ (in series) $+ 2$ (in parallel) $= 4$ capacitors.Thus,the minimum number of capacitors required is $4$.

The minimum number of capacitors each of $2 \ \mu F$ required to make a circuit with an equivalent capacitance of $5 \ \mu F$ is

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