How to calculate quality factor for RLC circuit?

  • #1
zenterix
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Homework Statement
Consider the RLC circuit shown below. At the switch is closed.
Relevant Equations
a) For what values of will the initial sign of be reversed at some later times?

b) Assuming the value of satisfies the condition in (a), find and in terms of , and .

c) If it takes 10 cycles for the energy in the circuit to decrease to times its initial value, find the value of the resistance in terms of and .
1721093401537.png


Using Faraday's law we have



where .

After rearranging the expression we get





If the system is critically- or over-damped then becomes zero at most once.

Therefore, for the sign of to be reversed at multiple times, we need to have an under-damped system and so the condition



must be satisfied and we have oscillations.

(4) implies



The solution to the differential equation is





where





If we now impose the initial conditions





then we find





My question is about item (c).

What I learned about the quality factor is the following.

The quality factor is times the number of oscillations for the energy to decrease by a factor of .

It is also times the number of oscillations for the amplitude to decrease by this same factor.

Amplitude is proportional to .



and since the period is



then in a time period of the number of periods (oscillations) is





Thus



If I apply this last formula I get



which, after subbing in for and becomes



and we can solve for to obtain





The answer from MIT OCW is



which is slightly different.

So, the first thing I want to tackle is why I didn't get (18).

But what I really want to know is the following.

The potential energy of the system is



(19) by itself is constant.

The resistor dissipates energy at a rate of .

Thus, if the system starts with energy



then it simply loses energy through the resistor over time.

If we want to know the time it takes for this energy to drop by then we need to solve



is this correct?

This latter calculation seems very complicated, but does its solution give us the time it takes for the energy to drop by a factor of ?
 
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  • #2
FYI, this problem is from MIT OCW's "Vibrations and Waves Problem Solving" course.

It is posed as a problem and the answer is available, but the solution not so much (broken latex).

It is problem 2 here.
 
  • #3
zenterix said:
View attachment 348359

Using Faraday's law we have

Check all of the signs in this equation. Which way around the circuit did you define the closed loop integral of ? Think about the direction of between the capacitor plates and the direction of inside the resistor.

zenterix said:
where .
The circuit diagram defines the direction of positive current and defines as the charge on the left plate of the capacitor. From these definitions does or does ?

zenterix said:
This equation looks correct.

You can simplify the calculations for parts (b) and (c) by noting that the switch is closed at time . So, what is the value of ?

At , where is all of the electrical energy in the circuit? After one cycle, where is all of the remaining energy? After 10 cycles, where is the energy?
 
  • #4
TSny said:
Check all of the signs in this equation.
You are correct about there being a mistake in the signs. However, by luck the resulting differential equation is correct.

Here are the corrected calculations





However, we also have .

Thus, Faraday's law should be









TSny said:
You can simplify the calculations for parts (b) and (c)
Indeed, simplifies the calculations.

In fact, after my OP I noticed this.

We find that







TSny said:
At t=0, where is all of the electrical energy in the circuit?

At the energy is all in the capacitor.

Without the resistor, this energy would be converted to magnetic field energy in the inductor and then back again to electric field energy in the capacitor in every cycle.

The total energy is always constant, however.

With the resistor present, energy is dissipated over time so that the total energy is decreasing in time.
TSny said:
After 10 cycles, where is the energy?
After each cycle, the energy is back in the capacitor.

My doubt is about the quality factor.

For an LC circuit, where the total energy is constant, what does it mean to calculate an energy decrease by a factor of ?
 
  • #5
zenterix said:


However, we also have .

Thus, Faraday's law should be




Good.

zenterix said:
Indeed, simplifies the calculations.

In fact, after my OP I noticed this.

We find that


According to these equations, does ?

zenterix said:
At the energy is all in the capacitor.

Without the resistor, this energy would be converted to magnetic field energy in the inductor and then back again to electric field energy in the capacitor in every cycle.

The total energy is always constant, however.

With the resistor present, energy is dissipated over time so that the total energy is decreasing in time.

After each cycle, the energy is back in the capacitor.
Yes. So, it should be easy to write an expression for the total energy at the end of any integer number of cycles. This should help with part (c).

zenterix said:
My doubt is about the quality factor.

For an LC circuit, where the total energy is constant, what does it mean to calculate an energy decrease by a factor of ?
Yes, if then the total electrical energy remains constant. So it doesn't make sense to ask how long it will take for the energy to decrease by . In this case the quality factor is infinite.
 
  • #6
TSny said:
According to these equations, does I=−Q˙?
I forgot to add a minus sign to the entire expression.

I simply computed but is .
TSny said:
Yes, if R=0 then the total electrical energy remains constant. So it doesn't make sense to ask how long it will take for the energy to decrease by e−1. In this case the quality factor is infinite.
Ok, so if then does equation (20) in the OP make sense?

If so then it gives the time for the energy to drop by .

The number of oscillations is then just the period, which is divided by this time.

I know we were given this number of oscillations, but I am trying to reason about the concepts to make sure my understanding is correct here.
 
  • #7
zenterix said:
I forgot to add a minus sign to the entire expression.

I simply computed but is .
OK.
zenterix said:
Ok, so if then does equation (20) in the OP make sense?
If so then it gives the time for the energy to drop by .
The integral in (20) gives the energy lost to heat. So, the left side should be the difference in the initial and final energies. If this correction is made, then the equation could be used, in principle, to find the time for the energy to drop by a factor of .

zenterix said:
The number of oscillations is then just the period, which is divided by this time.
The other way around: The number of oscillations is the time divided by the period.
 
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