Capacitor, Inductor and a DC power supply in series

May 2009
6
0
A circuit has a battery(V), Capacitor(C) and Inductor(L) connected in series. At t < 0, the switch is open and the capacitor has an initial charge of -800uC and Io = 0.0Amps.. When the switch is closed at t = 0, I want to know what happens here. I only understand an LC circuit where there is only a capacitor and an inductor, but the addition of a battery in this case makes it a little conplicated for me.


V - Ldi/dt - q(t)/C = 0; q(t)= qmax*cos(wt + phi) + VC; I(t)= -w*qmax*sin(wt + phi).



This is what I think: At t = 0 when the switch is closed, the capacitor starts discharging through the inductor and current increases in the circuit. But the battery is also connected in the circuit, so what is the battery doing? Is it continuosly charging the capacitor? I am confused!
 

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Dec 2009
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A circuit has a battery(V), Capacitor(C) and Inductor(L) connected in series. At t < 0, the switch is open and the capacitor has an initial charge of -800uC and Io = 0.0Amps.. When the switch is closed at t = 0, I want to know what happens here. I only understand an LC circuit where there is only a capacitor and an inductor, but the addition of a battery in this case makes it a little conplicated for me.


V - Ldi/dt - q(t)/C = 0; q(t)= qmax*cos(wt + phi) + VC; I(t)= -w*qmax*sin(wt + phi).


This is what I think: At t = 0 when the switch is closed, the capacitor starts discharging through the inductor and current increases in the circuit. But the battery is also connected in the circuit, so what is the battery doing? Is it continuosly charging the capacitor? I am confused!

The circuit you have attached to your question is a simple oscillator circuit. When the switch is open the capacitor has a voltage across it of q/C where q is the charge on the capacitor plate. You may know the amount and sign of the charge on one of the plates, but you do not have the capacitance of the capacitor, the inductance of the inductor, nor the value of the voltage E in front of the switch, so this whole problem has to be looked at in variable form.


With the charge on the capacitor and an unknown capacitance, you do not know which is larger, the voltage across the capacitor or the voltage inputted, E before the switch is even closed. Also the initial current (Io) that flows just after the switch is closed is always going to be 0 for a circuit like this because the inductor cannot change its current instantly, but must initially exponentially increase from 0 and oscillate for your circuit.

If the capacitor is initially charged with a voltage that is greater the voltage E, when the switch is closed, it will, only ideally, discharge enough current instantly into the battery E until they are of the same value, but with the inductor there, it cannot change instantly so it would actually approach an infinitely large sized discharge current into the battery E at which point the inductors magnetic field will collapse over it causing the current to keep flowing up through some negative value to zero, and then race back to a positive value; this is called ringing, and with no real resistance within the circuit it will go on forever. Only because this circuit is an ideal circuit with no real resistances and two reactive elements in series with each other, there will be a constant back and forth of current flow in the circuit such that it will oscillate at a rate equal to 1/sqrt(LC); that's actually its radial frequency, but just divide it by 2pi if you want the oscillating frequency in cycles per second.

The same exact thing will occur if the battery E is initially larger than the voltage across the capacitor before the switch is closed. The only thing that would be different is the initial direction of the oscillation within the circuit.

These are neat little oscillator circuits, but there's always some real resistance in a circuit so the oscillations eventually will die out and the whole thing will come to a rest; with the capacitor voltage being equal and opposed by the battery E. Your circuit has a battery, so any energy that is lost to thermal energy in heating up the real resistances in a circuit, your battery can replenish any lost energy within the circuit; if the battery is inputing energy at a time rate which is also the resonant frequency, in particular, for this circuit; you can make it osciallate at any frequency, but it will really "go" at its resoanat freqeuncy. So this circuit will work as an oscillator in real life also, but only with the battery changing its input voltage with time particualrly at the circuits natural frequency is when its ability to tune certain frequency content from other frequency content, becomes useful; this is how older radios and TV's used to "tune" in a channel. They would use an oscillator circuit like this one and adjust the capacitor so that the resoant frequency for the circuit is that of the radio or TV band they wish to tune in to. The circuit is injected with energy it recieves from its antenna, and at a certain frequency, energy is pushed back and forth between the two elements at very greater amplitudes than energy of other frequencies can push the oscialltions at; because the energy that is causing it to osciallte greatly is at that circuits resoant frequency, so it can be used to pick out energy at certain frequencies from other frequencies. Ideally it would work without the battery forever at some frequency with an initial charge across the capacitor when the switch is closed; again ideally with no real resistances to dampen out any ringing, it will just keep right on ringing around some average value; most likely around the initial voltage that was across the capacitor before the switch was closed; again with no battery. You have an ideal oscillator circuit; ideal though, it will oscillate in real life but not nearly as long as forever as it ideally does. Or stated differently, it will eventually come to a stop around the battery voltage E, for a real circuit that has some resistance. Since nothing was said about accounting for any real resistance then this circuit can be considered an ideal circuit that will ring around the battery voltage E forever, with nothing to dampen the ringing.

Many Smiles,
Craig (Smile)
 
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