Question

In a series RL circuit, L = 4H, R = 100 ohms and E = 200V . Find the values of current as a function of time. Assume that the initial current is zero. Find the current when t = 2 secons.

Answer 1

Solution;

Using the equation;

$L\frac{di}{dt}+Ri=E$

Rewrite;

$\frac{Di}{dt}+\frac{R}{L}I=\frac EL$

The integrating factor;

$I.F=e^{\int \frac RLdt}=e^{\frac{Rt}{L}}$

I put into the equation;

$e^{\frac{RT}{L}}[\frac{di}{dt}+\frac RLi]=e^{\frac{Rt}{L}}\frac EL$

Integrating respectively;

$I=\frac ER+ce^{\frac{-Rt}{L}}$

At t=0,i=0;

$c=-\frac ER$

$i=\frac ER-\frac ERe^{\frac{-Rt}{L}}$

Substitute the values;

$i=\frac{200}{100}-\frac{200}{100}e^{\frac{-100×2}{4}}$

$I=9.837A$