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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 .

asked Feb 9, 2017 in Physics Problems by koolman (4,196 points)
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

1 Answer

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Best answer

you are correct so far;
$-lnI/I_0=(R/L)T$
now take minus sign to the power
this would just cause reciprocal of $I / I_0$
$lnI_0/I=(R/L)T$ (after you consider minus on power)
so $ln2=(R/L)T$
so T =0.693L/R
now find charge with the time

answered Feb 9, 2017 by physicsapproval (2,290 points)
selected Feb 9, 2017 by koolman
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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