Homework Help: LC-circuit problem

1. Nov 7, 2011

atlantic

An LC-circuit has L = 64mH, C = 121nF.
At the time t=0, the charge on the condensator is 10μC and the current in the inductor is 0.3A. What is the current in the inductor as a function of time?

With:$q = Q_mcos(\omega_0t+\phi)$ we get that: $I = Q_m\omega_0cos(\omega_0t+\phi + \pi/2)$ (because I = dq/dt), where $\omega_0 = 1/(√LC)$

I thought that the initial conditions would mean that I have to solve:
$q(0) = 10*10^{-6} =Q_mcos(\phi)$ and $I(0) = 0.3 = Q_m\omega_0cos(\phi + \pi/2)$. But these equations have no solution(!)

2. Nov 7, 2011

rude man

Here's what I do with a problem like this:

1. write the differential equation without regard to initial conditions.
2. s-transform the equation term-by-term, paying attention to
f'(t) <--> sF(s) - f(0+) and f''(t) <--> s2F(s) - sf(0+) - f'(0+).

The you just invert the ensuing transfer function back to the time domain, and you get all the i.c.'s included.

(If you haven't had the Laplace transform yet I don't know what to tell you.)

3. Nov 8, 2011

atlantic

No, I don't know Laplace:uhh:

4. Nov 8, 2011

rude man

OK, then you have to use standard differential equations for the network, and solve in the traditional diff e manner including initial conditions.

I didn't follow you argument but I would write

i = -CdV/dt (i is + if flowing out of C)
V = Ldi/dt (i is + if flowing into L)

So i = -C(d/dt)Ldi/dt = -LCd2i/dt2 or

LCd2i/dt2 + i = 0

I.C. 1: i(0) = 0.3A
I.C. 2: di/dt(0+) = V0/L but V0 = Q0/C = 1e-5/C so di/dt(0+) = Q0/LC

So now just solve the 2nd order diff eq. with those two initial conditions.