- #1
- 57
- 0
I think you need to know the frequency of the supply to get started.... do you have it?
Assuming the switch is closed and the capacitor creates an open circuit... I'd have the source with the 10 and 60 ohm resistors... Using voltage divider I find that
Vc(0-) = Vc(0+) = 25.71 V, right?
Alright.. sounds good.. I'll write what I have.. can you comment/guide me through it and point out any errors? Please...
Now as t = infinity...
i(inf) = 0... right? since the capacitor will act as an open circuit...
Now as for using KVL... after the initial differentiation of the KVL I get...
L d2i(t)/dt2 + R di(t)/dt + i(t)/C = 0
2.5 d2i(t)/dt2 + 120 di(t)/dt + i(t)/1m = 0
Now.. if I divide the equation by 2.5 I get..
d2i(t)/d2t + 48 di(t)/dt + 400 i(t) = 0
damping ratio = (48/2)/sqrt(400) = 1.2
since damping ratio > 1... circuit is overdamped...
s1 = -10.73
s2 = -37.26
i(t) = K1 e^(-10.73 t) + K2 e^(-37.26 t)
Ok... how about now? Any errors...??
RightOkay.. btw, you meant that K1 = - K2.. right?
so the initial rate of change of the current...
take the derivative of
i(t) = K1 e^(s1 * t) + K2 e^(s2 * t)
at t = 0.. gives
s1K1 + s2K2 = 0... which is what I had before and it was wrong... any other hint..?
The voltage of the inductor at t=0+ should be equal to:
VL(0+) = - Vc(0+) = -25.71, right?
Which implies that
di(0+)/dt = -25.71, Thus we can write that:
s1K1 + s2K2 = -25.71
K1 = -K2
K1 = -0.97 and K2 = 0.97, yes/no/maybe?
Sorry, I'm really tired.. >.<
dI/dt = Vc/L ?
dI/dt = 25.71/2.5 ?
dI/dt = 10.28?
and s1K1 + s2K2 = 10.286....
K1 = 0.388 and K2 = -0.388
Therefore...
i(t) = 0.388e^(-10.73 * t) + -0.388e^(-37.26 * t)
? =0
Looks good! Bravo!
Looks okay. I think if you made the maximum timestep a bit smaller you might resolve the peak a bit better (eliminate the "flat bottom".
Here's the plot of the mathematical version:
![]()
I've negated the current values ( -I(t) ) in order to match your Spice model current direction.
Perfect... thank you so much... I owe you big time... If I can repay you with something (except money.. I'm poor!) let me know!! :)
You're quite welcome. Just go get an "A" in your course!