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Effective capacitance of parallel combination of two capacitors $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ is $10\; \mu \mathrm{F}$. When these capacitors are individually connected to a voltage source of $1\; \mathrm{V},$ the energy stored in the capacitor $\mathrm{C}_{2}$ is $4$ times that of $\mathrm{C}_{1}$. If these capacitors are connected in series, their effective capacitance will be
$3.2\; \mu \mathrm{F}$
$8.4\; \mu \mathrm{F}$
$1.6\; \mu \mathrm{F}$
$4.2\; \mu \mathrm{F}$
Solution
$\mathrm{C}_{1}+\mathrm{C}_{2}=10$
$\frac{1}{2} \mathrm{C}_{2} \mathrm{V}^{2}=4 \times \frac{1}{2} \mathrm{C}_{1} \mathrm{V}^{2}$
$\therefore \quad \mathrm{C}_{2}=4 \mathrm{C}_{1}$
$\therefore \quad \mathrm{C}_{1}=2 \;and\; \mathrm{C}_{2}=8$
For series combination
$C_{e q}=\frac{C_{1} C_{2}}{C_{1}+C_{2}}=1.6$
