Question

For the circuit shown in the figure, find the current i(t) and sketch it. You can assume that the solution to the KVL equation will have the form
i(t) = K1 + K2(e^st).

Answer 1

Assume that the only resistance is the 50 Ohm resistor and the inductor is an ideal inductor and refer to this link: http://www.electronics-tutorials.ws/inductor/lr-circuits.html

Answer 2

Hi, thanks for your help! One question though, the time constant is L/R right? So that would mean .1/50 = .002, but you had .02 seconds. Could you explain how you knew to change it from .002 to .02?

Answer 3

Yes, you're correct I made a mistake, thinking the inductor was 1 Henry lol.

Answer 4

Did you follow it, even with that error?

Answer 5

Here is the post revised to correct the time constant value.
Suggest you let the X axis represent t (time) and let each increment represent a value of one time constant R/L (.1/50) or 0.002 seconds. Go from zero to at least 6 increments to .012 seconds. Let the y axis represent current starting from 0 and go to a max of 2 amps or the steady state value. For all practical purposes you can show the current reaching this at 5 time constants. Look at the graph in the link and calculate the y values using the percentages shown for the time constants. At the first time constant .02 the current will be 67% of maximum or 1.34 amps (.67 X 2 Amps) Good luck with this.

Answer 6

Yeah I did, thank you!