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Two capacitors of capacities 1µF and C µF are connected in series and the combination is charged to a potential difference of 120 V. If the charge on the combination is 80µC, the energy stored in the capacitor C (in micro joules) is :

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1800

1600

14400

7200

answer is B.

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Detailed Solution

$If potential across 1 μF is 80 V potential of other must be 40 V and in series charges are same Q = C_{1} V_{1} = C_{2} V_{2} so C_{2} must be 2 μF energy stored in C_{2} = \frac{1}{2} 2 \times 10^{- 6} {(40)}^{2} = 1600 μJ$

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