# LC Circuit

1. Feb 25, 2008

### AiRAVATA

Hello guys. I have a simple question regarding an LC circuit.

Imagine a voltage source $V_0$, a capacitor $C$ and an inductor $L$, all hooked up in series. I know that the equation governing the behvior of the system is

$$V_0=\frac{1}{C}q(t)+L\ddot{q}(t),$$

and hence

$$q(t)=A\cos \omega t + B\sin \omega t + CV_0.$$

What I'm having trouble with is the initial conditions. Is it fair to assume that in $t=0$ there is no charge nor current in the system?

If I put a switch in the system, how would the initial conditions change (assuming is open in $t=0$ and closed in $t>0$)?

2. Feb 25, 2008

### John Creighto

Yes because the voltage across a capacitor cannot change instantaneously and then current though an inductor cannot change instantaneously.

3. Feb 25, 2008

### AiRAVATA

$$q(t)=CV_0 (1-\cos \omega t)$$

no matter if I have a switch or not?

4. Feb 26, 2008

### AiRAVATA

Well, in case you have been wondering, It's all wrong!

What I have to do is imagine a RLC ciruit, solve it with conditions $i(0)=V_0/R, \, i'(0)=0$, integrate in $t$, divide by $C$ and then take the limit as $R \rightarrow 0$. Then I'll know what's the voltage passing trough the capacitor on my original LC circuit!

Yeah!

5. Feb 26, 2008

### John Creighto

Well, you didn't give us that initial set of conditions.

6. Feb 27, 2008

### AiRAVATA

I know, I know. It was exactly that what made me realize my minstake. Thanks for the input tough, you really got me thinking.