Question

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 of its initial value is

• A. $\frac{LI}{\sqrt{2}R}$
• B. $\frac{LI}{R}$
• C. $\frac{\sqrt{2}LI}{R}$
• D. $\frac{LI}{2R}$

Answer 1

The correct option is D $\frac{LI}{2R}$

Energy stored in the inductor initially is,

$E_i=\frac{1}{2}L{I}^2....\left(1\right)$

Let the current is $I^{\prime }$ when energy reduces to ${\frac{1}{4}}^{th}$ of $E_i$.

$\frac{E_i}{4}=\frac{1}{2}L(I^{\prime }{)}^2$

Using $\left(1\right)$,

$\frac{1}{4}\times \frac{1}{2}L{I}^2=\frac{1}{2}L(I^{\prime }{)}^2$

$\therefore I^{\prime }=\frac{I}{2}$

Now, current in a decay $L-R$ circuit is given by,

$i=I{e}^{\frac{-t}{\tau }}$

Where, $\tau =\frac{L}{R}$

So, when $i=I^{\prime }=\frac{I}{2}$, we have

$\frac{I}{2}=I{e}^{\frac{-t}{\tau }}$

$\Rightarrow e^{\frac{-t}{\tau }}=\frac{1}{2}$

$\therefore t=\tau \ln 2=\frac{L}{R}\ln 2$

Charge flowing through the resistor at any instant is,

$dq=idt=\left(I{e}^{\frac{-t}{\tau }}\right)dt$

Total charge flown through the resistor from $t=0$ to $t=\frac{L}{R}\ln 2$

$q=\int dq={\int }_0^{\frac{L\ln 2}{R}}I{e}^{\frac{-t}{\tau }}dt$

$=\frac{I{\left[e^{\frac{-t}{\tau }}\right]}_0^{\frac{L\ln 2}{R}}}{\left(\frac{-1}{\tau }\right)}$

$=I\times -\frac{L}{R}[e^{-\ln 2}-e^0][\because \tau =\frac{L}{R}]$

$=-\frac{IL}{R}[\frac{1}{2}-1]=\frac{IL}{2R}$

Hence, $\left(\text{B}\right)$ is the correct answer.