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.
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Expert's answer
2021-12-14T07:49:49-0500
"\\text{For a series circuit containing only a resistor and an inductor} \\\\[1 em] \n\\text{Kirchhoff's second law states that the sum of the voltage } \\\\[1 em]\n\\text{ drop across the inductor } L \\frac{d i}{d t} \\\\[1 em]\n\\text{ and the voltage drop across the resistor}~(iR) \\\\[1 em] \n\\text{ is the same as the impressed voltage on the circuit } E(t) \\\\[1 em] \n\nL \\frac{d i}{d t}+R i=E(t) \\\\[1 em]\n\\because \\quad L=0.1 \\mathrm{H} \\quad \\text { and } \\quad R=50 \\Omega \\quad \\text { and } \\quad E=20 \\mathrm{V} \\\\[1 em]\n\\therefore \\quad 0.1 \\frac{d i}{d t}+50 i=20\\quad \\\\[1 em] \n\\therefore \\frac{d i}{d t}+500 i=200\\\\[1 em] \n\n\n\n\n\\text{ linear differential Equation} \\\\[1 em] \n\n \\text{ The integrating factor is}~ e^{\\int 500~ d t}=e^{500 t}\\\\[1 em] \n \n \\therefore i ~e^{500 t} =\\int 200 e^{500 t} d t = \\frac{2}{5} e^{500 t}+c \\\\[1 em] \n\n\\therefore i(t) = \\frac{2}{5} +c~ e^{-500 t} \\\\[1 em] \n\n\\text{ Applying the initial condition i(0)=0 } \\\\[1 em]\n 0=\\frac{2}{5}+c~e^0 \\quad \\rightarrow \\quad c=-\\frac{2}{5} \\\\[1 em]\n\\text{ Substitute in i(t)} \\\\[1 em]\n\\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]\n\\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)"
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