Question

An electromotive force 120, 0<t<20, E(t) = (o, t >20, is applied to an LR-series circuit in which the inductance is 20 henries and the resistance is 2 ohms. Find the current it) if i(0)-0.

Answer 1

$\text{For a series circuit containing only a resistor and an inductor} \\[1 em] \text{Kirchhoff's second law states that the sum of the voltage } \\[1 em] \text{ drop across the inductor } L \frac{d i}{d t} \\[1 em] \text{ and the voltage drop across the resistor}~(iR) \\[1 em] \text{ is the same as the impressed voltage on the circuit } E(t) \\[1 em] L \frac{d i}{d t}+R i=E(t) \\[1 em] \because \quad L=0.1 \mathrm{H} \quad \text { and } \quad R=50 \Omega \quad \text { and } \quad E=20 \mathrm{V} \\[1 em] \therefore \quad 0.1 \frac{d i}{d t}+50 i=20\quad \\[1 em] \therefore \frac{d i}{d t}+500 i=200\\[1 em] \text{ linear differential Equation} \\[1 em] \text{ The integrating factor is}~ e^{\int 500~ d t}=e^{500 t}\\[1 em] \therefore i ~e^{500 t} =\int 200 e^{500 t} d t = \frac{2}{5} e^{500 t}+c \\[1 em] \therefore i(t) = \frac{2}{5} +c~ e^{-500 t} \\[1 em] \text{ Applying the initial condition i(0)=0 } \\[1 em] 0=\frac{2}{5}+c~e^0 \quad \rightarrow \quad c=-\frac{2}{5} \\[1 em] \text{ Substitute in i(t)} \\[1 em] \therefore \text{ The current i(t) if i(0) = 0 ~is }i(t)=\frac{2}{5}-\frac{2}{5} e^{-500 t} ~~~(1) \\[1 em] \therefore \lim _{t \rightarrow \infty} i(t)=\lim _{t \rightarrow \infty}\left(\frac{2}{5}-\frac{2}{5} e^{-500 t}\right) =\frac{2}{5} -\frac{2}{5} (0)=\frac{2}{5}~~~~~~~(2)$