- #1

zenterix

- 657

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- 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##.

Using Faraday's law we have

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

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}##?

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