Question

The current in a series $RL$ circuit decays as $I=I_0{e}^{\frac{-t}{\tau }}$. Obtain the rms current in the interval $0\le t\le \tau$.

• A. $\frac{I_0}{2}$
• B. $\frac{I_0}{e}$
• C. $\frac{I_0}{\sqrt{2}e}$
• D. $\frac{I_0}{e}\sqrt{\frac{e^2-1}{2}}$

Answer 1

The correct option is D $\frac{I_0}{e}\sqrt{\frac{e^2-1}{2}}$

$I_{rms}^2=\frac{1}{\tau }\underset{0}{\overset{t}{\int }}{I}^{2}dt=\frac{1}{\tau }\underset{0}{\overset{\tau }{\int }}{I}_0^2{e}^{\frac{-2t}{\tau }}dt={\begin{array}{c}\frac{-I_0^2{e}^{\frac{-2t}{\tau }}}{\tau \frac{2}{\tau }}\end{array}|}_0^{\tau }\Rightarrow I_{rms}=\frac{I_0}{e}\sqrt{\frac{e^2-1}{2}}$
$\therefore \left(d\right)$ is the correct answer.