percibaa8

2021-12-31

A 8 volt battery is connected to a series circuit in which the inductance is 4 henry and the resistance is 5 ohms. Find the current if the initial current is zero. What will the current be after a very long time?

levurdondishav4

Let the current flows in the circuit is (t)
According to KVL,
$-8+{v}_{L}+{v}_{R}=0$
${V}_{L}+{V}_{R}=8$
$L\frac{di}{dt}+i×R=8$
$4\frac{di}{dt}+5i=8$
$4\frac{di}{dt}=8-5i$
$\frac{di}{8-5i}=\frac{dt}{4}$
Integrate both side
$\int \frac{di}{8-5i}=\int \frac{dt}{4}$
Let $8-5i=x$
$di=-\frac{dx}{5}$
$-\int \frac{dx}{5x}=\int \frac{dt}{4}$
$-\frac{1}{5}\mathrm{ln}x=\frac{t}{4}+c$
$-\frac{1}{5}\mathrm{ln}\left(8-5i\right)=\frac{t}{4}+c$
At $t=0;i=0$
$-\frac{1}{5}\mathrm{ln}8=c$
$-\frac{1}{5}\mathrm{ln}\left(8-5i\right)=\frac{t}{4}-\frac{1}{5}\mathrm{ln}8$
$\frac{1}{5}\mathrm{ln}8-\frac{1}{5}\mathrm{ln}\left(8-5i\right)=\frac{t}{4}$
$\frac{1}{5}\mathrm{ln}\left(\frac{8}{8-5i}\right)=\frac{t}{4}\left(\therefore \mathrm{ln}\frac{a}{b}=\mathrm{ln}a-\mathrm{ln}b\right)$
$\mathrm{ln}\left(\frac{8}{8-5i}\right)=\frac{5t}{4}$
$\frac{8}{8-5i}={e}^{5\frac{t}{4}}$
$8-5i=8{e}^{-5\frac{t}{4}}$
$5i=8-8{e}^{-5\frac{t}{4}}$

usaho4w

Let I be current at ant time t
EMF $E=5$ volts, resistance $R=50$ ohms and inductance $L=1$ henry
The basic equation of current for RL circuit is $\frac{dI}{dt}+\frac{R}{L}I=\frac{E}{L}$
$⇒\frac{dI}{dt}+50I=5$
This is a linear equation and the solution of it is $I=c{e}^{-50t}+\frac{1}{10}$
At $t=0,I=0$. So we get $c=-\frac{1}{10}$.
Therefore the equation for current will become $I=-\frac{1}{10}{e}^{-50t}+\frac{1}{10}$
As t tends to infinite , the value of I tends to $\frac{1}{10}$

Vasquez

$\begin{array}{}L=20H\\ R=10\mathrm{\Omega }\\ i\left(t\right)=\frac{V}{R}\left[1-{e}^{-\frac{RT}{L}}\right]\\ \frac{di\left(t\right)}{dt}|t=0=\frac{R}{L}\frac{V}{R}\left[1-{e}^{-\frac{RT}{L}}\right]\\ \frac{1}{2}×\frac{5}{10}{e}^{0}\\ =\frac{1}{4}⇒0.25Amp/s\end{array}$