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.
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\u00d72}{4}}"
"I=9.837A"
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