A 5µF capacitor is charged fully by a 220 V supply. It is then disconnected from the supply and is connected in series to another uncharged 2.5µF capacitor. If the energy change during the charge redistribution is X/100 J then value of X to the nearest integer is:

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5 months ago

Heat loss; ΔH = U? - U? = 1/2 (C? / (C? +C? ) (V? -V? )²
= 1/2 * (5*2.5)/ (5+2.5) (220-0)²µJ
= 5/ (2*3) * 22*22*100*10? J
= 5*11*22/3 * 10? J = 1210/3 * 10? J = 1210/3 * 10? ³ J * 4 * 10? ²
According to questions
x/100 = 4*10? ²
So, x=4
Note: But given answer by JEE Main x=36

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Physics Electrostatic Potential and Capacitance 2025

Physics Electrostatic Potential and Capacitance 2025

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