
Heat Loss is given by,
H=2(C1+C2)C1C2(V1−V2)2
V1 is initial voltage of capacitor C=V0
V2 is the initial voltage of capacitor 2C=0
=2(c+2c)c×2c(V0−0)2=6CV02
H=61C0V02
JEE Main 2020 — Physics Electromagnetism
A capacitor C is fully charged with voltage V0. After disconnecting the voltage source, it is connected in parallel with another uncharged capacitor of capacitance 2C. The energy loss in the process after the charge is distributed between the two capacitors is :
Held on 4 Sept 2020 · Verified 6 Jul 2026.
21CV02
31CV02
41CV02
61CV02
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