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In a L-R decay circuit, the initial curr...

In a L-R decay circuit, the initial current at t=0 is I. The total charge that has flown through the resistor till the energy in the inductor has reduced to one-fourth its initial value, is

Text Solution

Verified by Experts

The correct Answer is:
`LI//2R`
Initial energy `E_(1)=1/2LI^(2) …(1)`
Final energy `E_(2)=1/2LI_(2)^(2) …(2)`
given that `E_(2)=1/4 E_(1) …(3)`
From `(1), (2)` and `(3)`
`1/2LI_(2)^(2)=1/4xx1/2LI^(2)implies I_(2)=I/2 …(4)`
For decaying `I=I_(0)e^(-t//tau) …(5)`
From `(4)` and `(5)`
`t=taulm2 ...(6)`
`Deltaq=int_(0)^(t)-I e^(-t//tau)dt=Itau[e^(-t//tau)]_(0)^(tau)=IL/R[e^(-ln2)-1]=(LI)/(2R)`
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