Question

State whether the following statement is True or False.
The current in the circuit with inductance $L$ and resistance $R$ and voltage $Esin\omega t$ is given by $L\frac{di}{dt}+Ri=Esin\omega t$.if $i=0$ at $t=0$ then,current is given by $i=\frac{E}{\sqrt{E^2+L^2{\omega }^2}}\left[sin\right(\omega t-\varphi )+e^{-Rt/L}sin\varphi ]$ where $\varphi =ta{n}^{-1}\left(\frac{L\omega }{R}\right)$

• A. True
• B. False

Answer 1

The correct option is A True

The differential equation is $\frac{di}{dt}+\frac{iR}{L}=\frac{E\sin \omega t}{L}$,

This is linear differential equation,

The integrating factor is $e^{\frac{Rt}{L}}$,

The solution is $i{e}^{\frac{Rt}{L}}=\int e^{\frac{Rt}{L}}E\sin \omega tdt$,

The integral can be solved by using integration by parts,

where $u=E\sin \omega tandv=e^{\frac{Rt}{L}}$,

The solution is $i=\frac{E}{\sqrt{E^2+L^2{\omega }^2}}\left(\sin (\omega t-\varphi )+e^{-\frac{Rt}{L}}\right)where\varphi ={\tan }^{-1}\frac{L\omega }{R}$