# Charge flown in L-R decay circuit

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

Energy in inductor =(1/2)LI$^2$
So final current would be I/2

Using kirchoff equation
$$L\frac{di}{dt} + ir =0$$
Interating it , I got
$$ln(\frac{I}{I_0 }=-\frac{R}{L}T$$
Where I$_0$= I/2
ln(2)=(-R/L)T
How this can be possible for this equation to satisfy T must be negative , that is impossible .

Final current is $I=\frac12 I_0$. You substituted $I_0=\frac12 I$ which is incorrect.

So you get $ln(\frac{I}{I_0})=ln(\frac12)=-ln(2)=-\frac{R}{L}T$.
Thanks for help

You are correct so far.
$-\ln I/I_0=(R/L)T$
Now take minus sign into the power. This would just cause reciprocal of $I / I_0$
$\ln I_0/I=(R/L)T$
so $\ln 2=(R/L)T$
so T =0.693L/R
Now find charge with the time.

answered Feb 9, 2017 by (2,320 points)
edited Mar 13
Then the term of log would be negative that it would be ln(1/2)
I have edited the answer now its more clear
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