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JEE Main 2020 Physics question

PhysicsElectrostatic Potential and CapacitanceJEE Main 2020

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

Answer: -4

Solution

Initial energy = ½·5 μF·(220)² = 0.121 J; the common potential is 220 × 2/3 (as in the paper's working) so

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Question ID VK-012168