Resistance and capacitance in LCR circuits

[physicsStudent00]

Homework Statement

The output voltage, V_(out) (t) ‡ q(t) /C, effectively measures the charge on the capacitor as a function of time, where q(t) is an exponentially damped simple-harmonic oscillation:
a) Given that L = 0.05 mH, the square-wave frequency is 25kHz, and the ringing decays to 40% of its peak within half a square-wave cycle what is R?
b)If there are 17 cycles of the LCR ringing each half-cycle of the square-wave, what is C?

Homework Equations

file attached is the equations

The Attempt at a Solution

a)
the frequency : f=25*10^3
period = 1/f = 1/25*10^3 -0.00004
Period = 2L/R = 0.00004,
therefore R=2(25*10^-3)/(0.00004)=2.5

b) not sure how to incorporate the information of the 17 cycles into the first given equation. i think you have to use the first equation to find omega, then use omega to solve for omega_0 to then find C. also not sure if C is constant during the dampening of the LCR circuit.

 

[Charles Link]

Your first part is incorrect. You first need to find ## \tau ## where ## e^{-t/\tau}=.40 ## for ## t=T/2 ## where ## T ## is the period of the square wave. (The square wave is like a single step function applied to the circuit, but this step function gets applied repetitively so that the result can be observed much easier on an oscilloscope).

 

[physicsStudent00]

so we assume that Cos(ωt)=1? why do we assume that at half way through a cycle, isn't that assuming that t=0?

but going with that e^(t/tau)
⇒tau=0.00002*ln(e)/ln(0.4)=2L/R
∴R=2(0.05*10^-3)/tau=4.58

dumb question but what is ω and ω_0 for this equasion, what is it a measure of.

then to move into part b) how do i incorporate the 17 cycles given that it dampens exponentially, would i need to know what the end result is then use that to solve for ω, but if its assumed that cos(ωt)=1 like in part a how does that work.

sorry if I'm asking a lot i just really don't understand this content and my textbook isn't very helpful.

 

[Charles Link]

To give you some introductory information, the series RLC circuit has ## L \, \frac{d^2 Q}{dt^2}+R \, \frac{dQ}{dt}+\frac{Q}{C}=V(t) ##. With a square wave input, ## V(t)=V_o ## is constant (## V_o ##) for a half cycle. (Note that current ## I=\frac{dQ}{dt} ##). This differential equation has as its solution the form they gave you. That is basically the homogeneous solution. (There is also a ## sin(\omega t) ## in the homogeneous solution, but that is an extra detail that isn't needed to solve this. The solution to the inhomogeneous equation is simply ## Q(t)=CV_o ##, so the homogeneous solution is the only part that is oscillating. This occurs because of the resonance of the LC combination, and the frequency ## \omega ## a which it occurs is changed slightly by the resistor. With ## R=0 ## , ## \omega=\omega_o=\frac{1}{\sqrt{LC}} ##, where ## \omega=2 \pi f ##. For ## R \neq 0 ##, ## \omega=2 \pi f ## is given by the formula they give you. ## \\ ## I think your ## R=4.58 \, \Omega ## is correct. (I checked your arithmetic very quickly).## \\ ## For the next part, they essentially tell you what ## \omega ## is. ## T_1=(T/2)/17 ## with ## f=\frac{1}{T_1} ## and ## \omega=2 \pi f ##. From that and their formulas, you should b able to compute ## C ##. ## C ## is the capacitance of the capacitor, and yes, it is constant.

 

[physicsStudent00]

To give you some introductory information, the series RLC circuit has ## L \, \frac{d^2 Q}{dt^2}+R \, \frac{dQ}{dt}+\frac{Q}{C}=V(t) ##. With a square wave input, ## V(t)=V_o ## is constant (## V_o ##) for a half cycle. (Note that current ## I=\frac{dQ}{dt} ##). This differential equation has as its solution the form they gave you. That is basically the homogeneous solution. (There is also a ## sin(\omega t) ## in the homogeneous solution, but that is an extra detail that isn't needed to solve this. The solution to the inhomogeneous equation is simply ## Q(t)=CV_o ##, so the homogeneous solution is the only part that is oscillating. This occurs because of the resonance of the LC combination, and the frequency ## \omega ## a which it occurs is changed slightly by the resistor. With ## R=0 ## , ## \omega=\omega_o=\frac{1}{\sqrt{LC}} ##, where ## \omega=2 \pi f ##. For ## R \neq 0 ##, ## \omega=2 \pi f ## is given by the formula they give you. ## \\ ## I think your ## R=4.58 \, \Omega ## is correct. (I checked your arithmetic very quickly).## \\ ## For the next part, they essentially tell you what ## \omega ## is. ## T_1=(T/2)/17 ## with ## f=\frac{1}{T_1} ## and ## \omega=2 \pi f ##. From that and their formulas, you should b able to compute ## C ##. ## C ## is the capacitance of the capacitor, and yes, it is constant.

so for the new period shouldn't it be T_1=T*17, as what you are taking is that if in the same time part a) it now rotated 17 times instead of just once?
because following that and introducing it into the equation
ω=√(1/LC)-(R^2/4L)
C=27.66*10^-9

 

[Charles Link]

so for the new period shouldn't it be T_1=T*17, as what you are taking is that if in the same time part a) it now rotated 17 times instead of just once?

This is a high frequency oscillation, basically ## f=34 f_o ## where ## f_o ## is the frequency of the square wave. But to compute it using the period ## T_1 ## , I believe I did it correctly. (## T_1 ## is very short).

 

[physicsStudent00]

This is a high frequency oscillation, basically ## f=34 f_o ## where ## f_o ## is the frequency of the square wave. But to compute it using the period ## T_1 ## , I believe I did it correctly. (## T_1 ## is very short).

didn't realize that thank you very much
so by looking at it, it therefore makes sense that since the new period would be incredibly small shouldn't capacitance therefore be very small as the oscillations would have decreased be a large amount
following that and introducing T_1 into the equation
ω=√(1/LC)-(R^2/4L)
it solves to give
C=27.66*10^-9

thank you so much for all the help

 

[Charles Link]

did realize that thank you very much
so by looking at it, it therefore makes sense that since the new period would be incredibly small shouldn't capacitance therefore be very small as the oscillations would have decreased be a large amount
following that and introducing T_1 into the equation
ω=√(1/LC)-(R^2/4L)
it solves to give
C=27.66*10^-9

thank you so much for all the help

Just one more item: Calling ## f_o ## the frequency of the square wave is a poor choice (on my part)=I should have called it ## f_{sq} ## so it doesn't get confused with ## \omega_o=2 \pi f_o ##. This ## f_o ## is of course completely different from ## f_{sq} ##.

 

[physicsStudent00]

Just one more item: Calling ## f_o ## the frequency of the square wave is a poor choice (on my part)=I should have called it ## f_{sq} ## so it doesn't get confused with ## \omega_o=2 \pi f_o ##. This ## f_o ## is of course completely different from ## f_{sq} ##.

don't worry i know what you meant, so in general as the period is very small, frequency will be very large. that affects the dampening through making it dampen at a greater rate?

 

[Charles Link]

don't worry i know what you meant, so in general as the period is very small, frequency will be very large. that affects the dampening through making it dampen at a greater rate?

The damping is caused by the factor ## e^{-t/\tau} ##. The ## cos(\omega t) ## is an oscillation, as seen on an oscilloscope, that would be sinusoidal without the ## e^{-t/ \tau} ## term. With the ## e^{-t/ \tau} ## term, it becomes a damped sinusoid=the amplitude of the sinusoid gets exponentially smaller with time. The shorter ## \tau ## is, the quicker it gets damped. ## \\ ## ## \tau=2L/R ##. If ## R=0, ## ## \tau=+\infty ## so that ## t/ \tau=0 ## and there would be no damping.

 

[Charles Link]

And one additional item is worth mentioning here, in regards to the voltage signal that you observe on an oscilloscope. The oscilloscope trace is made by sweeping an electron beam horizontally in time across a TV (phosphorescent) screen at a selected speed that you can adjust. The voltage signal is the input to the oscilloscope that will cause a deflection of the electron beam in the vertical direction. The result is that the oscilloscope gives you a graph on the screen of voltage vs. time.