LC Circuit: Initial Conditions and Switch

In summary, the conversation discusses an LC circuit with a voltage source, capacitor, and inductor in series. The equation governing the behavior of the system is V_0=\frac{1}{C}q(t)+L\ddot{q}(t), and the initial conditions are discussed. It is fair to assume no charge or current in the system at t=0. The effect of a switch in the system is also considered. The correct approach to solving for the voltage passing through the capacitor is to imagine a RLC circuit and take the limit as R \rightarrow 0. The conversation ends with the realization of a mistake and gratitude for the input.
  • #1
AiRAVATA
173
0
Hello guys. I have a simple question regarding an LC circuit.

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

[tex]V_0=\frac{1}{C}q(t)+L\ddot{q}(t),[/tex]

and hence

[tex]q(t)=A\cos \omega t + B\sin \omega t + CV_0.[/tex]

What I'm having trouble with is the initial conditions. Is it fair to assume that in [itex]t=0[/itex] 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 [itex]t=0[/itex] and closed in [itex]t>0[/itex])?
 
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  • #2
AiRAVATA said:
Hello guys. I have a simple question regarding an LC circuit.

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

[tex]V_0=\frac{1}{C}q(t)+L\ddot{q}(t),[/tex]

and hence

[tex]q(t)=A\cos \omega t + B\sin \omega t + CV_0.[/tex]

What I'm having trouble with is the initial conditions. Is it fair to assume that in [itex]t=0[/itex] 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 [itex]t=0[/itex] and closed in [itex]t>0[/itex])?

Yes because the voltage across a capacitor cannot change instantaneously and then current though an inductor cannot change instantaneously.
 
  • #3
So the answer is

[tex]q(t)=CV_0 (1-\cos \omega t)[/tex]

no matter if I have a switch or not?
 
  • #4
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 [itex]i(0)=V_0/R, \, i'(0)=0[/itex], integrate in [itex]t[/itex], divide by [itex]C[/itex] and then take the limit as [itex]R \rightarrow 0[/itex]. Then I'll know what's the voltage passing trough the capacitor on my original LC circuit!

Yeah!
 
  • #5
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 [itex]i(0)=V_0/R, \, i'(0)=0[/itex], integrate in [itex]t[/itex], divide by [itex]C[/itex] and then take the limit as [itex]R \rightarrow 0[/itex]. 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.
 
  • #6
I know, I know. It was exactly that what made me realize my minstake. Thanks for the input tough, you really got me thinking.
 

Related to LC Circuit: Initial Conditions and Switch

1. What is an LC circuit?

An LC circuit, also known as a tank circuit, is a type of electrical circuit that consists of an inductor (L) and a capacitor (C) connected together. This circuit is used to store and release energy in the form of oscillations or vibrations.

2. What are the initial conditions of an LC circuit?

The initial conditions of an LC circuit refer to the voltage and current values at the start of the circuit's operation. These values are influenced by the properties of the inductor and capacitor, as well as any external influences such as switches or power sources.

3. How do initial conditions affect the behavior of an LC circuit?

The initial conditions play a crucial role in determining the behavior of an LC circuit. They determine the amplitude and frequency of the oscillations, as well as the time it takes for the circuit to reach a steady state.

4. What happens to the initial conditions when a switch is opened or closed in an LC circuit?

When a switch is opened or closed in an LC circuit, the initial conditions are affected. Opening a switch will cause an interruption in the circuit, leading to a change in the initial conditions. Closing a switch will allow the circuit to continue operating, but the initial conditions may still be altered depending on the properties of the switch.

5. How do you calculate the initial conditions of an LC circuit?

The initial conditions of an LC circuit can be calculated using equations that relate the voltage and current values at different points in the circuit. These equations can be derived from the equations that govern the behavior of inductors and capacitors. Alternatively, simulation software can be used to calculate the initial conditions and predict the behavior of an LC circuit.

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