Question #122488
A 20-volt electromotive force is applied to an LR-series circuit in which the inductance is 0.1 henry and the resistance is 50 ohms. Find the current i(t) if i(0) = 0.

i(t = _____

Determine the current as t → ∞.
lim i(t) =
t→∞
1
Expert's answer
2020-06-24T17:58:30-0400

For a series circuit containing only a resistor and an inductorKirchhoff’s second law states that the sum of the voltage  drop across the inductor Ldidt and the voltage drop across the resistor (iR) is the same as the impressed voltage on the circuit E(t)Ldidt+Ri=E(t)L=0.1H and R=50Ω and E=20V0.1didt+50i=20didt+500i=200 linear differential Equation The integrating factor is e500 dt=e500ti e500t=200e500tdt=25e500t+ci(t)=25+c e500t Applying the initial condition i(0)=0 0=25+c e0c=25 Substitute in i(t) The current i(t) if i(0) = 0  is i(t)=2525e500t   (1)limti(t)=limt(2525e500t)=2525(0)=25       (2)\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)

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