Three capacitors each of $4\,\,\mu F$  are to be connected in such a way that the effective capacitance is $6\,\,\mu F.$  This can be done by connecting thern
JEE MAIN 2016, Diffcult
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To get effective capacitance of $6\,\mu F$ two capacitors of $4\, \mu \mathrm{F}$ each connected in series and one of $4\, \mu \mathrm{F}$ capacitor in parallel with them.

Two capacitances in series

$\therefore \frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$

$1$ capacitor in parallel

$\therefore $ $C_{\mathrm{eq}}=C_{3}+C=4+2=6\, \mu \mathrm{F}$

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