Edited for assignment purposes...
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.. >.<
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.