which inductor current waveform is correct?

Hi

Please tell me which current waveform is correct. Thank you for the help.

Regards
PG

Hi PG,

The waveform you have drawn as "waveform 2" does not look correct because it is drawn to look like a triangle.

Since during the first interval the voltage is a ramp, that means the current for that interval will be a parabola involving t^2, so it wont be a ramp with a straight line but the line will be curved gently upwards until the end of the interval.

For the second interval the voltage is a negative pulse so for that section the current will be a negative slope ramp.

Is that enough help or do you want to see the exact solution?

Thanks a lot, MrAl.

Seriously, I don't get it. Did I do the integration the correct way? Was there something wrong with the formula used? I'm not well versed in calculus but I'm learning it.

The guy whose work I have with me has told me that it was this way the instructor solved it. My answer, 2 mA, for 0< t <0.1 is the same as of that someone else. The guy has the expression i(t)=4mA - 20mA(t) for the interval 0.1< t <0.2, which to me seems wrong.

I don't know how else I can integrate the v(t)=-2/1000 over the interval 0.1< t <0.2. Could you please help me with it?

Thanks a lot.

Hi again,

I'll start and see if you can get the rest.

For the first interval, we have a voltage ramp that goes from 0 to 4v in time 0 to 0.1 seconds.
The slope therefore is 4/0.1=40 as you noted.
This makes the equation for voltage:
V(t)=40*t

Integrating that with respect to t we get;
40*t^2/2+K=20*t^2+K
with K=0 we have:
20*t^2.

This is a parabola with center at the origin.

v=L*di/dt
v dt=L di
1/L*v dt=di
1/L*40*t dt=di
1/L*20*t^2=i
with L=100H and t=0.1 sec, we get:
i=2ma

So you got the right answer i think, but your writing is hard to read, sorry. If you type out your results i can see it better.

You want to do the second interval or you want to see that too?
Intuitively when you see a ramp voltage you expect a parabola current, when you see a pulse voltage you expect a ramp current (inductor).