Question 50 of 180

Effective capacitance of parallel combination of two capacitors ${C}_{1}$ and ${C}_{2}$ is $10\mu F.$ When these capacitors are individually connected to a voltage source of $1V,$ the energy stored in the capacitor ${C}_{2}$ is $4$ times that of ${C}_{1}.$ If these capacitors are connected in series, their effective capacitance will be:

A$4.2\mu F$B$3.2\mu F$C$1.6\mu F$D$8.4\mu F$

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Answer: (C) $1.6\mu F$

Given ${C}_{1}+{C}_{2}=10\mu F$ ...(i)
$4\left(\dfrac{1}{2}{C}_{1}{V}^{2}\right)=\dfrac{1}{2}{C}_{2}{V}^{2}$
$\Rightarrow 4{C}_{1}={C}_{2}$...(ii)
From equation (i) and (ii)
${C}_{1}=2\mu F$
${C}_{2}=8\mu F$
If they are in series
${C}_{eq}=\dfrac{{C}_{1}{C}_{2}}{{C}_{1}{+}_{2}}=1.6\mu F$