Power dissipated over 1 cycle of a AC source-free RLC series circuit?

AI Thread Summary
The discussion focuses on calculating power dissipation in an underdamped RLC series circuit over one cycle. The original poster is trying to derive the current function and faces challenges in determining coefficients A1 and A2 for the equation of current. Participants suggest that while some calculations are necessary, a simpler approach may exist, particularly for parts of the problem related to energy conservation. The conversation highlights the importance of understanding RMS values and the behavior of the circuit at t=0, noting that all charge initially resides in the capacitor. The final answer for energy dissipated is given as 6.9e-4 J, emphasizing the need for a clear understanding of the circuit dynamics.
testobesto
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Homework Statement
382* A 0:01 µF capacitor, a 0.1 H inductor whose resistance is 1000 Ohms, and a switch are connected in a series circuit. The capacitor is initially charged to a potential difference of 400 V.
The switch is then closed.
(b)/ (c) How much energy is converted to heat in the first complete cycle? How much energy is converted to heat in the complete train of oscillations?
Relevant Equations
a = R/2L, w_0 = 1/sqrt(LC), w_d = sqrt(a^2 - w_0^2), i(t) = e^(-at)[A_1 cos(w_d t) + A_2sin(w_d t)]
I tried to start by assuming that we need to integrate something over 1 period (2pi). Therefore, we need i(t)^2 R integrated over something. From there, I recognized that this is an underdamped model since R/2L < 1/sqrt(LC). I believe that i(t) can be represented by i(t) = e^(-at)[A_1 cos(w_d t) + A_2sin(w_d t)]. However, I am stuck on finding A1 and A2, and even if I found it, I don't know if this is the correct approach. I also tried letting 1 period = (1/2pi*(w_0))^-1 and working with that, but I can't see if it works.

I feel like there is supposed to be a much simpler approach, but I am not sure what it could be.


(Answer is given to be 6.9e-4 J, but I don't get process)
 
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testobesto said:
I feel like there is supposed to be a much simpler approach
I can think of a simpler approach for part (b)/ (c) ...
(How much energy is present in the circuit at ##t=0## :rolleyes:)

But for part (a) some work is unavoidable. Fortunately you already have an expression for the solution. Notice that in the plot in the link all responses start with ##i(0)=0##. With ##A_1 = 0##, what do you get for the fraction ##\int_0^T/\int_0^\infty## (where ##T = ##1 period) ?

##\ ##
 
Did you learn about RMS values of voltage and current? (this is for the OP)
 
nasu said:
Did you learn about RMS values of voltage and current? (this is for the OP)
Yes.
 
BvU said:
:welcome:


I can think of a simpler approach for part (b)/ (c) ...
(How much energy is present in the circuit at ##t=0## :rolleyes:)

But for part (a) some work is unavoidable. Fortunately you already have an expression for the solution. Notice that in the plot in the link all responses start with ##i(0)=0##. With ##A_1 = 0##, what do you get for the fraction ##\int_0^T/\int_0^\infty## (where ##T = ##1 period) ?

##\ ##
At t=0, the capacitor holds all charge, 8e-4 C. I take it that since A_1 = 0, we just evaluate for A_2? Or use phase shifts for a new identity? I also immediately recognized that for (c), the answer is the same at 8e-4 C, but only because I know that all charge can only leave through the resistor as t->inf
 
As you observed this is a highly resonant circuit (Q=10K !). As such, a good approximation for the peak inductor current can be found using conservation of energy at the 1st quarter cycle. Likewise, the resonant frequency can be approximated by assuming R=0. Although I doubt that you need to know the period since you are integrating over 1 period, ω will drop out of the result.

PS: The approximate solution to simple resonance like this has a nice simple form that is worth memorizing. You're likely to see it many more times.
 
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Would there be some function that gives the charge in the inductor and capacitor as a function of omega or time, then we evaluate that function at t=0 and t = 1 period, and take it so that the charge on the capacitor at 0 minus the charge on both the inductor and capacitor at 1 period would equal the charge that was dissipated on the resistor, due to the charges on the RHS and LHS on each side needing to agree? I have that as R = 0, the resonant frequency = 1/√LC = 31622.8 Hz
 
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