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RLC circuit with Battery

B

BehindEnemyLines

Senior member
For RLC circuits, the Kirchoff's theorem tells (for my particular circuit) as q/C + IR + L(dI/dt) = 0 so L(d^2q/dt^2) + R(dq/dt) + q/c = 0. So the charge as a function of time is defined as: q(t) = Qe^(-t/2T)*cos(?t+F), where Q = q maximum, T = inductance time constant, ? = angular frequency, and t = time (seconds).

EDITED: Note above that the capacitor is first charged to Q and then place connected with a resistor and inductor together in series.

However, what would happen to the q(t) if there's a fixed applied voltage (DC voltage) in the circuit (RLC + battery)? Again, using this loop theorem: L(dI/dt) + IR + q/C - e = 0 ==> L(dI/dt) + IR + (1/C)q = e. If I then differentiate: L(d^2q/dt^2) + R(dq/dt) + (1/C)q = 0 since e is a constant. However, this gives me the q(t) like a LC circuit. But I'm stuck in solving for q(t) with the additional constant emf battery remains connected in the circuit. I'm not sure how this would change the overall equation of q(t), if any???
 
So you're assuming an underdamped LC circuit? When solving the second order differential equation, you should have two constants that require boundary conditions to solve. One of the boundary conditions (in this case) is q(infinity) = epsilon*C instead of 0 which your q(t) solution assumes... I believe.
 
EDITED: I incorrectly gave the wrong information.

The RLC circuit has a damped oscillations. The capacitor is first charged with a battery and then removed from the capacitort. The ideal capacitor is then connected to a resistor and an ideal inductor (solenoid) so you've an oscillations of charge on the capacitor depending on time.

But the latter part has the battery remain connected so that @ t = 0s (when switch is closed), Io = 0.0A (initial current), e = a constant # volts, qo (initial charge) = say -900microCoulomb, L = #mH, C = #microF, and R = #Ohms, where # indicates a postive number for the inital conditions? Is this what you mean by initial conditions? Yes, q(infinity) = e*C or q maximum is = e*C. But that is as t approaches infinity. However, what is q(t) for a given time in the circuit with the e remains connected? The second RLC circuit is damped. I think the problem is I don't know what to do with this loop theorem: L(dI/dt) + IR + q/C - e = 0 in solving for q(t) with e in it.
 
Originally posted by: BehindEnemyLines
I forgot to mention that the LC circuit has a harmonic oscillations with no change in amplitude (like a sine wave). This is assuming zero resistance and no radiation so that the oscillation is indefinite (an ideal LC circuit). The capacitor is first charged with a battery and then removed from the capacitort. The ideal capacitor is then connected to an ideal inductor (solenoid) so you've an oscillations of charge on the capacitor depending on time.

But the latter part has the battery remain connected so that @ t = 0s (when switch is closed), Io = 0.0A (initial current), e = a constant # volts, qo (initial charge) = say -900microCoulomb, L = #mH, C = #microF, and R = #Ohms, where # indicates a postive number for the inital conditions? Is this what you mean by initial conditions? Yes, q(infinity) = e*C or q maximum is = e*C. But that is as t approaches infinity. However, what is q(t) for a given time in the circuit with the e remains connected? The second RLC circuit is damped. I think the problem is I don't know what to do with this loop theorem: L(dI/dt) + IR + q/C - e = 0 in solving for q(t) with e in it.
Click to expand...

But

q(t)=Qe^(-t/2T)*cos(?t+F)

Is a damped oscillating equation? I guess you have to make T = infinity?

As for solving the 2nd order differential equation with the epsilon in it, I don't remember the name but I believe it's called a non-homogeneous differential equation with a solution with two parts. The first part is ignoring the constant for now and the other uses a 'guess' and is put back into the original equation and solved with the initial conditions.
 
EDITED: I made a grave error in previous posts. Sorry TuxDave, maybe I got you confused of the LC circuit. Both circuits are RLC except one has the battery remain connected. Basically, the circuit was NOT LC but RLC without a battery connected. The second HAS a battery remain connected.

The equation q(t)=Qe^(-t/2T)*cos(?t+F) IS damped for the battery disconnected. Hopefully, it makes more sense now.
 
Crap, I should be able to do this. I had Diff Eq and physics 2 last semester and this stuff looks very familar to a lot of the problems I had to do. But I can't remember much, maybe I'll give it a shot later.
 
Do you still need an answer?

After reading through it a couple of times, yeah just do like TuxDave said to find your particular solution. Doing a homogenous is easy (I assume that's what's your first problem w/o battery is), but for a non-homogenous eqn (when the d.e. doesn't equal 0) you gotta do more work. I was never that good at them so I'm not sure if I can really help you but try googling 'variation of parameters' or 'method of undetermined coeffecients' which I think are what TuxDave mentioned.
 
duragezic, not really need it but I'm curious to a solution. If anyone has the time, feel free to work with it. Otherwise, it's not necessary really. Maybe with a bit googling, I can solve it myself.
 
Originally posted by: BehindEnemyLines
duragezic, not really need it but I'm curious to a solution. If anyone has the time, feel free to work with it. Otherwise, it's not necessary really. Maybe with a bit googling, I can solve it myself.
Click to expand...

Sorry, I almost forgot about this thread, if you want a solution can you describe the entire circuit and the switches? I'm currently seeing this. You have a battery (with voltage V0) to switch (which is open), to capacitor (charged to V1) to inductor to resistor to ground. then you close the switch and you're trying to derive the equation for charge across the capacitor? Is that right?
 
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