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A parallel plate capacitor  is charge  to a potential  difference  of 50v. It is then discharged through  a resistance  for 2 sec and its potential  drops by 10v.   Calculate  the fraction of energy  stored in the capacitance

A) 0.14

B) 0.50

C) 0.25

D) 0.64

1 Answer

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answered by (7k points)

The correct answer is option (d)

Explanation:

Vi = 50 v
Vf = 40 v 
we know that fraction of stored energy is given by 
[1/2 C(Vi)2] / [1/2 C(Vf)2]

= (40/50)2 = 16/25

= 0.64

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