The minimum number of capacitors each of $2$ $\mu F$ required to make a circuit with an equivalent capacitance $5$ $\mu F$ is
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if two capacitors will be in series and two capacitors will be in parallel as shown in figure then, equivalent capacitance will 5 microfarad.
Let see $AB$ and $BC$ capacitors are in series so, equivalent capacitance $\{ X \}$ of it $=2 microfarad / 2=1 microfarad \{$ in series $1 / Ceq =1 / C 1+1 / C 2\}$
DE and GF capacitors are in parallel, so, equivalent capacitance $\{ Y \}$ of it $=2+2=4 microfarad \{$ in parallel, $Ceq = C 1+ C 2\}$
now, $X$ and $Y$ capacitors are in parallel then, Ceq $=4+1=5$ microfarad.
hence, minimum number of capacitors = 4
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