That looks fine. So when the switch opens the capacitor will have that voltage on it as an initial condition. And the current will still be zero because of the inductor.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?
It's looking good!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?
s1K1 + s2K2 won't be zero. It'll equal the initial rate of change of the current. Go back to the circuit diagram and see if you can't determine what that might be. You've an inductor that has suddenly had a voltage Vc impressed upon it... how will it react?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 equation of interest for the inductor is its defining equation: V = L dI/dt. In this case there's an impressed voltage Vc across inductor L=2.5H. What's dI/dt?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.. >.<
Woohoo... thank you so much!!Looks good! Bravo!
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!! :)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.