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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:
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Correct option is C. $$1.6\mu F$$
Given that $$C_1 + C_2 = 10\mu F$$ (Parallel combination) ...(1)
They are now connected to the same voltage source of $$1V$$.
The energy stored in the two capacitors are given by
$$E_1 = \dfrac{1}{2}C_1V^2$$ $$E_2 = \dfrac{1}{2}C_2V^2$$
Given that $$E_2 = 4E_1$$
$$\therefore \dfrac{1}{2} C_2V^2 = 4 \times C_1V^2$$
$$\therefore C_2 = 4C_1$$ ....(ii)
From (i) and (ii),
$$C_1 + C_2 = C_1 + 4C_1 = 10\mu F$$.
$$5C_1 = 10\mu F$$, $$C_1 = 2\mu F$$
$$C_2 = 10\mu F-C_1 = 10 \mu F - 2\mu F = 8\mu F$$.
When $$C_1$$ and $$C_2$$ are connected in series, the equivalent capacitance of the combination is given by
$$C_{series} = \dfrac{C_1C_2}{C_1+C_2} = \dfrac{2\times 8}{2+8}\mu F = 1.6\mu F$$
They are now connected to the same voltage source of $$1V$$.
The energy stored in the two capacitors are given by
$$E_1 = \dfrac{1}{2}C_1V^2$$ $$E_2 = \dfrac{1}{2}C_2V^2$$
Given that $$E_2 = 4E_1$$
$$\therefore \dfrac{1}{2} C_2V^2 = 4 \times C_1V^2$$
$$\therefore C_2 = 4C_1$$ ....(ii)
From (i) and (ii),
$$C_1 + C_2 = C_1 + 4C_1 = 10\mu F$$.
$$5C_1 = 10\mu F$$, $$C_1 = 2\mu F$$
$$C_2 = 10\mu F-C_1 = 10 \mu F - 2\mu F = 8\mu F$$.
When $$C_1$$ and $$C_2$$ are connected in series, the equivalent capacitance of the combination is given by
$$C_{series} = \dfrac{C_1C_2}{C_1+C_2} = \dfrac{2\times 8}{2+8}\mu F = 1.6\mu F$$
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