A 5μF capacitor is charged fully by a 220V supply. It is then disconnected from the supply and is connected in series to another uncharged 2.5μF capacitor. If the energy change during the charge redistribution is X100J then value of X to the nearest integer is_______.

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answer is 0004.00.

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Detailed Solution

Energy lost during charge redistribution =12C1C2C1+C2V1−V22=125×2.55+2.5×220−02×10−6∵For uncharged capacitance V=0=56×2.22×10−2=X100⇒x=5×2.226=4.03

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