Question

The current in a discharging LR circuit is given by $i=i_{0}e-t\tau$ where \tau is the time constant of the circuit. Calculate the rms current for the period t = 0 to $t=\tau$.

Answer 1

Since i = $i_0{e}^{\frac{-t}{\tau }}$

$i_2=\frac{1}{\tau }{\int }_0^{\tau }{i}_0^2{e}^{\frac{-2t}{\tau }}$ dt

=$\frac{i_0^2}{\tau }{\int }_0^{\tau }{e}^{\frac{-2t}{\tau }}$ dt

= $\frac{i_0^2}{\tau }\times {\left[\frac{\tau }{2}{e}^{\frac{-2t}{\tau }}\right]}_0^e$

= - $\frac{i_0^2}{\tau }\times \frac{\tau }{2}\times [e^{-2}-1]$

$\sqrt{i^2}2e^2=\sqrt{\frac{-i_2^0}{2}(\frac{1}{e^2}-1)}$

= $\sqrt{\frac{i_2^0}{2}(e^2-1)}$

= $\frac{i_0}{e}\sqrt{\frac{e^2-1}{2}}$