Physics, asked by bidiyasarmanish5540, 1 year ago

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

Answers

Answered by shaina55
8
By using
v=voe = t/cr=>40=50e-2/cr=>e-1/cr=4/5
potential difference after

2sec
v=ve-2/cr=50(e-1/cr)2=50(4/5)2=32v

fraction of energy after 1

sec

1/2c(vf)2
_______ =(40/50)2=16/25
1/2c(vi)2

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