# How to calculate quality factor for RLC circuit?

• zenterix
zenterix
Homework Statement
Consider the RLC circuit shown below. At ##t=0## the switch is closed.
Relevant Equations
a) For what values of ##R## will the initial sign of ##Q## be reversed at some later times?

b) Assuming the value of ##R## satisfies the condition in (a), find ##Q## and ##I## in terms of ##Q_0, R, L##, and ##C##.

c) If it takes 10 cycles for the energy in the circuit to decrease to ##e^{-1}## times its initial value, find the value of the resistance in terms of ##L## and ##C##.

$$\mathcal{E}=\oint\vec{E}\cdot d\vec{l}=\frac{Q}{C}+IR=-L\dot{I}\tag{1}$$

where ##I=\dot{Q}##.

After rearranging the expression we get

$$\ddot{Q}+\frac{R}{L}\dot{Q}+\frac{1}{LC}Q=0\tag{2}$$

$$\ddot{Q}+\gamma\dot{Q}+\omega_0^2Q=0\tag{3}$$

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

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

$$\gamma^2<4\omega_0^2\tag{4}$$

must be satisfied and we have oscillations.

(4) implies

$$R^2<4\frac{L}{C}\tag{5}$$

The solution to the differential equation is

$$Q(t)=e^{-\frac{\gamma t}{2}}(c_1\cos{\omega t}+c_2\sin{\omega t})\tag{6}$$

$$=e^{-\frac{\gamma t}{2}}A\cos{(\omega t-\phi)}\tag{7}$$

where

$$A=\sqrt{c_1^2+c_2^2}\tag{8}$$

$$\tan{\phi}=\frac{c_2}{c_1}\tag{9}$$

If we now impose the initial conditions

$$Q(0)=Q_0$$

$$\dot{Q}(0)=I(0)=I_0$$

then we find

$$Q(0)=A\cos{\phi}=0$$

$$\tan{\phi}=\frac{2LI_0+R}{2L\sqrt{\omega_0^2-\frac{\gamma^2}{r}}}\tag{10}$$

My question is about item (c).

What I learned about the quality factor is the following.

The quality factor is ##2\pi## times the number of oscillations for the energy to decrease by a factor of ##e^{-1}##.

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

Amplitude is proportional to ##e^{-\frac{\gamma t}{2}}##.

$$e^{-\frac{\gamma t}{2}}=e^{-1}\implies t=\frac{2}{\gamma}\tag{11}$$

and since the period is

$$T=\frac{2\pi}{\omega}=\frac{2\pi}{\sqrt{\omega_0^2-\frac{\gamma^2}{4}}}\tag{12}$$

then in a time period of ##\frac{2}{\gamma}## the number of periods (oscillations) is

$$\frac{2/\gamma}{2\pi/\sqrt{\omega_0-\frac{\gamma^2}{4}}}$$

$$=\frac{1}{\pi\gamma}\sqrt{\omega_0-\frac{\gamma^2}{4}}\tag{13}$$

Thus

$$Q=\frac{1}{\gamma}\sqrt{\omega_0^2-\frac{\gamma^2}{4}}\tag{14}$$

If I apply this last formula I get

$$Q=\pi\cdot 10=\frac{1}{\gamma}\sqrt{\omega_0^2-\frac{\gamma^2}{4}}\tag{14}$$

which, after subbing in for ##\omega_0## and ##\gamma## becomes

$$100\pi^2\frac{R^2}{L^2}=\frac{1}{LC}-\frac{R^2}{4L^2}\tag{15}$$

and we can solve for ##R^2## to obtain

$$R^2=\frac{4L}{C}\frac{1}{\sqrt{400\pi^2+1}}\tag{16}$$

$$R=\sqrt{\frac{L}{C}}\frac{2}{\sqrt{400\pi^2+1}}\tag{17}$$

The answer from MIT OCW is

$$R^2=\frac{L}{C}\frac{1}{\sqrt{400\pi^2}}\tag{18}$$

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

$$U_{sys}=\frac{Q^2}{2C}+\frac{LI^2}{2}\tag{19}$$

(19) by itself is constant.

The resistor dissipates energy at a rate of ##-RI^2##.

Thus, if the system starts with energy

$$U_{sys}=\frac{Q_0^2}{2C}+\frac{LI_0^2}{2}\tag{19}$$

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 ##e^{-1}## then we need to solve

$$e^{-1}\left (\frac{Q_0^2}{2C}+\frac{LI_0^2}{2}\right )=\int_0^t RI(t)^2 dt\tag{20}$$

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 ##e^{-1}##?

Last edited:
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.

zenterix said:
View attachment 348359

$$\mathcal{E}=\oint\vec{E}\cdot d\vec{l}=\frac{Q}{C}+IR=-L\dot{I}\tag{1}$$
Check all of the signs in this equation. Which way around the circuit did you define the closed loop integral of ##\vec E##? Think about the direction of ##\vec E## between the capacitor plates and the direction of ##\vec E## inside the resistor.

zenterix said:
where ##I=\dot{Q}##.
The circuit diagram defines the direction of positive current and defines ##Q## as the charge on the left plate of the capacitor. From these definitions does ##I = \dot Q## or does ##I = -\dot Q##?

zenterix said:
$$\ddot{Q}+\frac{R}{L}\dot{Q}+\frac{1}{LC}Q=0\tag{2}$$
This equation looks correct.

You can simplify the calculations for parts (b) and (c) by noting that the switch is closed at time ##t = 0##. So, what is the value of ##I_0##?

At ##t = 0##, 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?

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

$$\int_{capacitor}\vec{E}\cdot d\vec{l}=V_{C,before}-V_{C,after}=-\frac{Q}{C}$$

$$\int_{resistor}\vec{E}\cdot d\vec{l}=V_{R,before}-V_{R,after}=IR$$

However, we also have ##I=-\dot{Q}##.

$$-\frac{Q}{C}+IR=-L\dot{I}$$

$$-L\dot{I}-IR+\frac{Q}{C}=0$$

$$-\dot{I}-\frac{R}{L}I+\frac{Q}{LC}=0$$

$$\ddot{Q}+\frac{R}{L}\dot{Q}+\frac{Q}{C}=0$$

TSny said:
You can simplify the calculations for parts (b) and (c)
Indeed, ##I_0=0## simplifies the calculations.

In fact, after my OP I noticed this.

We find that

$$\tan{\phi}=\frac{R}{2L\omega}$$

$$Q(t)=e^{-\frac{\gamma t}{2}}Q_0(\cos{\omega t}+\frac{\gamma}{2\omega}\sin{\omega t})$$

$$I(t)=e^{-\frac{\gamma t}{2}}Q_0\left (-\frac{\omega_0^2}{\omega}\sin{\omega t}\right )$$

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

At ##t=0## 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 ##e^{-1}##?

zenterix said:
$$\int_{capacitor}\vec{E}\cdot d\vec{l}=V_{C,before}-V_{C,after}=-\frac{Q}{C}$$
$$\int_{resistor}\vec{E}\cdot d\vec{l}=V_{R,before}-V_{R,after}=IR$$
However, we also have ##I=-\dot{Q}##.

$$-\frac{Q}{C}+IR=-L\dot{I}$$
$$-L\dot{I}-IR+\frac{Q}{C}=0$$
$$-\dot{I}-\frac{R}{L}I+\frac{Q}{LC}=0$$
$$\ddot{Q}+\frac{R}{L}\dot{Q}+\frac{Q}{C}=0$$
Good.

zenterix said:
Indeed, ##I_0=0## simplifies the calculations.

In fact, after my OP I noticed this.

We find that
$$\tan{\phi}=\frac{R}{2L\omega}$$
$$Q(t)=e^{-\frac{\gamma t}{2}}Q_0(\cos{\omega t}+\frac{\gamma}{2\omega}\sin{\omega t})$$
$$I(t)=e^{-\frac{\gamma t}{2}}Q_0\left (-\frac{\omega_0^2}{\omega}\sin{\omega t}\right )$$
According to these equations, does ##I = -\dot Q##?

zenterix said:
At ##t=0## 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 ##e^{-1}##?
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.

TSny said:
According to these equations, does I=−Q˙?
I forgot to add a minus sign to the entire expression.

I simply computed ##\dot{Q}## but ##I## is ##-\dot{Q}##.
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 ##R\neq 0## then does equation (20) in the OP make sense?

If so then it gives the time for the energy to drop by ##e^{-1}##.

The number of oscillations is then just the period, which is ##2\pi/\omega## 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.

zenterix said:
I forgot to add a minus sign to the entire expression.

I simply computed ##\dot{Q}## but ##I## is ##-\dot{Q}##.
OK.
zenterix said:
Ok, so if ##R\neq 0## then does equation (20) in the OP make sense?
If so then it gives the time for the energy to drop by ##e^{-1}##.
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 ##1/e##.

zenterix said:
The number of oscillations is then just the period, which is ##2\pi/\omega## divided by this time.
The other way around: The number of oscillations is the time ##t## divided by the period.

zenterix

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