Question

An electromotive force$\mathit{E}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\{\begin{array}{l}\begin{array}{cc}120,& 0\le t\le 20\end{array}\\ \begin{array}{cc}0,& t>20\end{array}\end{array}$

is applied to an LR-series circuit in which the inductance is 20 henries and the resistance is 2 ohms. Find the current i(t) if i(0) - 0.

Short answer

$i\left(t\right)=\left\{\begin{array}{l}\begin{array}{cc}60-60e^{-\frac{1}{10}t},& 0\le t\le 20\\ (60e^2-60)e^{\frac{1}{10}t},& t>20\end{array}\\ \end{array}\right.$

Step-by-step solution

Definition of Series circuits

The linear differential equation for the current i(t),

$L\frac{di}{dt}+Ri=E\left(t\right)$

where L and R are known as the inductance and the resistance, respectively. The current i(t) is also called the response of the system.

Differentiate the equation

We have an electromotive force

$E\left(t\right)=\left\{\begin{array}{l}\begin{array}{cc}120,& 0\le t\le 20\\ 0,& t>20\end{array}\\ \end{array}\right.$

in a$LR$ circuit in series with an inductance$L=20$ henries and resistance$R=2\text{ohms}$ and then we have the differential equation for current $\left(I\right)$as

$L\frac{di}{dt}+Ri=E\left(t\right)$ -------- (1)

and we have to obtain the current$I\left(t\right)$of this circuit if we have the initial current as$I\left(0\right)=0$as the following:

Since we have$L=0.1$ henry,$R=50$ohms but$E$ is variable, then we have the differential equation shown in (1) as

$\begin{array}{c}20\frac{\mathrm{di}}{\mathrm{dt}}+2\mathrm{i}=\mathrm{E}\times \frac{1}{20}\\ \frac{\mathrm{di}}{\mathrm{dt}}+\frac{1}{10}\mathrm{i}=\frac{1}{20}\mathrm{E}\end{array}$

$\frac{\mathrm{di}}{\mathrm{dt}}=\frac{1}{20}\mathrm{E}-\frac{1}{10}\mathrm{i}$ -------- (2)

This equation is a first order linear and separable differential equation, then we can separate variables and then do integration as

Separate the variables

$\begin{array}{c}\frac{di}{\frac{1}{20}E-\frac{1}{10}i}=dt\\ \int \frac{1}{\frac{1}{10}\left(\frac{1}{2}E-i\right)}di=\int dt\\ -10\int \frac{-1}{\frac{1}{2}E-i}di=\int dt\\ -10\ln \left(\frac{1}{2}E-i\right)=t+c_1\end{array}$

$\begin{array}{c}\ln \left(\frac{1}{2}E-i\right)=-\frac{1}{10}t+c_2\\ e^{\ln \left(\frac{1}{2}E-i\right)}=e^{\left(-\frac{1}{10}t+c_2\right)}\\ \frac{1}{2}E-i=e^{c_2}{e}^{-\frac{1}{10}t}\end{array}$

Then we have

$i\left(t\right)=\frac{1}{2}E-e^{c_2}{e}^{\frac{1}{10}t}$

$=\frac{1}{2}E-c{e}^{\frac{1}{10}t}$-------- (3)

Now for the interval $0\le t\le 20$: to find the value of constant $C$we have to apply this point of condition$(i,t)=(0\mathrm{amp},0$ seconds ) into equation (3) as

$0\text{amp}=\frac{1}{2}E-c{e}^0$

Then we have

$c=\frac{1}{2}E$

Substitution

After that, substitute with the value of$c$ into equation (3), then we have

$i\left(t\right)=\frac{1}{2}E-\frac{1}{2}E{e}^{\frac{1}{10}t}$-------- (4)

Then substitute with the value$E=120$volt into equation (4), then we have

$i\left(t\right)=60-60e^{-\frac{1}{10}t}$-------- (5)

is the current of$LR$series circuit for the interval$0\le t\le 20$.Now for the interval$t>20$: we have to find the value of the initial condition of currentfor this interval, (i.e$i\left(20\right)$, by substituting with$t=20$seconds into equation (5) as

$\begin{array}{c}i\left(20\right)=60-60e^{\frac{1}{10}\times 20}\\ =60-60e^{-2}\end{array}$

After that, to find the value of constant$c$,we have to apply this point of condition$(i,t)=(60-60e^{-2}$amp , 20 seconds) into equation (3) as

$\begin{array}{c}60-60e^2=\frac{1}{2}E-c{e}^{-2}\\ c{e}^{-2}=\frac{1}{2}E-60+60e^{-2}\end{array}$

Then we have

$c=\frac{1}{2}{e}^{2}E-60e^2+60$

After that, substitute with the value of$c$ into equation (3), then we have

$i\left(t\right)=\frac{1}{2}E-\left(\frac{1}{2}{e}^{2}E-60e^2+60\right)e^{\frac{1}{10}t}$-------- (6)

Then substitute with the value$E=0$ volt into equation (6), then we have

$i\left(t\right)=0-(0-60e^2+60)e^{\frac{1}{10}t}$

$=(60e^2-60)e^{\frac{1}{10}t}$-------- (7)

is the current of $LR$series circuit for the interval $t>20.$

Finally, from equations (5) and (7), we can obtain the current of series circuit as

$i\left(t\right)=\left\{\begin{array}{l}\begin{array}{cc}60-60e^{-\frac{1}{10}t},& 0\le t\le 20\\ (60e^2-60)e^{\frac{1}{10}t},& t>20\end{array}\\ \end{array}\right.$