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Question: Answered & Verified by Expert
A 2 μF capacitor is charged as shown in the figure. The percentage of its stored energy dissipated as heat after the switch S is turned to position 2 is

D:\Session 2018-2019\Online Test Paper\Online CBT Series-2018\Phase-II\Test-11_(07-03-2019)\Images\q18.jpg
PhysicsElectrostaticsNEET
Options:
  • A 0 %
  • B 20 %
  • C 75 %
  • D 80 %
Solution:
1444 Upvotes Verified Answer
The correct answer is: 80 %
Einitial=Q022C=12CV02=V02
Final energy E(f)=Q022Ceff
E(f)=12×2v010×2v0=v025
ΔE(loss)=4v025,
% loss = 45v02v02×100
ΔE(loss) %=80%

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