LC Circuit: Initial Conditions and Switch

AI Thread Summary
In an LC circuit with a voltage source, capacitor, and inductor, the governing equation is V_0 = (1/C)q(t) + Lq''(t). The initial conditions are crucial, as the voltage across a capacitor and current through an inductor cannot change instantaneously. It is incorrect to assume there is no charge or current at t=0 without considering the circuit's configuration. When a switch is included, the initial conditions must be re-evaluated, particularly if the switch is open at t=0 and closed afterward. Ultimately, analyzing the circuit as an RLC circuit helps clarify the initial conditions and their impact on the voltage across the capacitor.
AiRAVATA
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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)?
 
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AiRAVATA said:
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)?

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

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

no matter if I have a switch or not?
 
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!
 
AiRAVATA said:
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!

Well, you didn't give us that initial set of conditions.
 
I know, I know. It was exactly that what made me realize my minstake. Thanks for the input tough, you really got me thinking.
 
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