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

  • #1
testobesto
4
0
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)
 
Physics news on Phys.org
  • #2
:welcome:

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) ?

##\ ##
 
  • #3
Did you learn about RMS values of voltage and current? (this is for the OP)
 
  • #4
nasu said:
Did you learn about RMS values of voltage and current? (this is for the OP)
Yes.
 
  • #5
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
 
  • #6
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.
 
Last edited:
  • #7
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
 

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

What is power dissipation in an RLC series circuit?

Power dissipation in an RLC series circuit refers to the amount of electrical energy converted into heat over one cycle of alternating current (AC). It is primarily influenced by the resistance in the circuit, as inductors and capacitors do not dissipate energy but rather store it temporarily.

How is power dissipated calculated in an RLC series circuit?

The power dissipated in an RLC series circuit can be calculated using the formula P = I²R, where P is the power dissipated, I is the RMS (root mean square) current flowing through the circuit, and R is the resistance. This formula indicates that power dissipation is directly proportional to the square of the current and the resistance.

What role do reactance and impedance play in power dissipation?

Reactance and impedance affect how much current flows through the circuit, thus influencing power dissipation. In an RLC circuit, the total impedance (Z) is a combination of resistance (R) and reactance (X), and it determines the RMS current. The relationship between voltage, current, and impedance is given by Ohm's law (V = IZ), where V is the voltage across the circuit.

Does the phase angle affect power dissipation in an RLC circuit?

Yes, the phase angle between the voltage and current in an RLC circuit affects power dissipation. The real power (or active power) dissipated is given by P = VIcos(φ), where φ is the phase angle. The cosine of the phase angle, known as the power factor, indicates how effectively the current is being converted into useful work. A lower power factor means more reactive power and less real power is being used.

What is the significance of the quality factor (Q) in power dissipation?

The quality factor (Q) of an RLC circuit is a measure of its resonant behavior and energy losses. A higher Q indicates lower energy loss relative to the energy stored in the circuit, meaning less power is dissipated as heat. Conversely, a lower Q signifies higher losses, leading to greater power dissipation. The Q factor is defined as Q = (Resonant Frequency) / (Bandwidth), and it influences the circuit's efficiency.

Similar threads

Replies
8
Views
845
Replies
6
Views
575
Replies
14
Views
3K
Replies
8
Views
2K
Replies
3
Views
2K
Replies
6
Views
7K
Replies
5
Views
4K
Back
Top