Question

The current in a series RL circuit decays as $I=I_0{e}^{-t/\tau }$. Obtain the $\text{rms current}$ 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 $\frac{I_0}{e}\sqrt{\frac{e^2-1}{2}}$

$i_{rms}^2=\frac{1}{\tau }{\int }_0^{\tau }{I}^{2}dt-\frac{1}{\tau }{\int }_0^{\tau }{I}_0^2{e}^{-2i/\tau }dt$
$=\frac{-I_0^2{e}^{-2i/\tau }}{\tau 2/\tau }$
$=i_{rms}=\frac{I_0}{e}\sqrt{\frac{e^2-1}{2}}$