- #1
zenterix
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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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