C1=5μFV1=220Volt
C2=2.5μFV2=0
Heat loss;
ΔH=U1−Ut=21C1+C2C1C2(v1−v2)2
=21×(5+2.5)5×2.5(220−0)2μJ
=2×35×22×22×100×10−6J
=35×11×22×10−4J=355×22×10−4J
=31210×10−4J=31210×10−3J=4×10−2
According to questions
100x=4×10−2
so, x=4
JEE Main 2020 — Physics Electromagnetism
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 100XJ then value of X to the nearest integer is :
Held on 2 Sept 2020 · Verified 6 Jul 2026.
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