A capacitor of capacitance C1 = 4 μF is charged to V1 = 80V and another capacitor of capacitance C = 6 μF is charged to V2 = 30V. When they are connected together, the energy lost by the 4 μF capacitor is

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7.8 mJ

4.6 mJ

3.2 mJ

2.5 mJ

answer is A.

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Detailed Solution

V=C1V1+C2V2C1+C2=50 V ∴ Energy lost by 4μF capacitor =12C1V12−12C1V2=7.8 mJ