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Two capacitors $C_{1} = 2 μ F$ and $C_{2} = 8 μ F$ are connected in series across a $300 V$ source. Then

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the potential difference across $C_{2}$ is $240 V$

the charge on each capacitor is $4.8 \times 10^{- 4} C$

The potential difference across $C_{1}$ is $60 V$

the energy stored in the system is $7.2 \times 10^{- 2} J$

answer is A, D.

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Detailed Solution

$Equivalent capacitance \frac{8 \times 2}{10} μF = 1.6 μF charge on each capcitor = 1.6 \times 300 = 4.8 \times 10^{- 4} C Energy stored = \frac{1}{2} \times 1.6 \times 10^{- 6} \times 9 \times 10^{4} = 7.2 \times 10^{- 2} J$

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