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Finding values for an RLC circuit graphically

  1. Mar 6, 2016 #1
    1. The problem statement, all variables and given/known data
    Determine ##q(t=0)##, ##i(t=0)##, the phase difference ##\phi##, angular frequency ##w## and the time constant ##\tau## from the graph of the capacitor waveform below:(pulse voltage source)

    Capacitor_Voltage_Waveform.jpg

    For this circuit we have ##q(t)=Ae^{-\frac{t}{2\tau}}\cos(\omega t+\phi)## therefore
    ##v(t)=\frac{A}{C}e^{-\frac{t}{2\tau}}\cos(\omega t+\phi)##

    2. Relevant equations
    ##i=\frac{dq}{dt}##
    3. The attempt at a solution
    For ##q(t=0)## I get ##q(t=0)=A\cos(\phi)##, for ##i(t)=\frac{dq}{dt}=-A(\frac{t}{2\tau}e^{-\frac{t}{2\tau}}\cos(\omega t+\phi)+\omega\sin(\omega t +\phi)e^{-\frac{t}{2\tau}})## therefore
    ##i(t=0)=-A\omega\sin(\phi)##

    For the phase difference since at ##t=0## the function is starting at 0 and we have a cosine the phase is ##\phi=-\frac{\pi}{2}##

    To find ##\omega## I used the fact that a fourth of the period corresponds to ##\pi/2## so taking a time difference that satisfies this ##t_1=0## and ##t_2=0.533\mu_s## we have ##\omega t_2-\omega t_1=\omega(t_2-t_1)=\pi/2\Longrightarrow \omega=\frac{\pi}{2t_2}## (Since ##t_1=0##),
    This gave ##\omega=2.9rad/\mu_s##

    Now I'm stuck trying to calculate ##\tau##, I wanted to sub into my voltage formula ##t=2\tau## then I would have ##v(t)=\frac{A}{C}e^{-1}\cos(2\omega \tau-\pi/2)## since I wanted to look graphically for the time where the amplitude became ##\approx \frac{1}{e}## of the original but I'm not sure what to do with the ##\cos## term.
     
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  3. Mar 6, 2016 #2

    Simon Bridge

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    Hint: What happens to the size of each successive maxima?
     
  4. Mar 6, 2016 #3
    They decrease (Is the amount it decreases by related to ##\tau##?). Or am I just supposed to take two values of ##t## and ##v## and then plug these into the formula and divide one by the other to solve for ##\tau##?
     
  5. Mar 6, 2016 #4

    Simon Bridge

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    You can use your knowledge of v(t) for many different times to solve for the key variables, yes.
    It does help to exploit particular times though .. ferinstance, part of your issue was what to do with the cosine part right? well:

    At each peak, what does ##\cos(\omega t + \phi) = ##

    ... if ##t_1## and ##t_2## are two peaks - what is ##v(t_1)## and ##v(t_2)## ?
    (write it out without plugging values in...)
     
  6. Mar 6, 2016 #5
    Okay I wrote out the two terms and obtained this formula ##\tau=\frac{(t_1-t_2)}{2\ln\frac{v_2}{v_1}}## which using voltages and time values for two maximums gave me ##\tau\approxeq 2.3\mu_s##.

    If ##\tau=\frac{L}{R}## and ##C=1nF## how would I go about finding what ##L## and ##R## are equal to? One method I've thought of is to first solve for ##A## by picking ##t=\tau## then I can arrive at the equation ##A=\frac{CVe^{1/2}}{\cos(w\tau+\phi)}## and then substituting ##\tau=\frac{R}{L}## into my equation for ##V## and picking ##t## at a maximum I end up with ##\frac{R}{L}=-\frac{2\ln(VC/A)}{t}##. This would give me 2 equations with two variables but I'm curious as to if there exists an easier method.

    Edit: This ended up not working since with ##t=\tau##, ##V\approxeq -0.1##, and then I get ##A=-54.137## which makes my ln function undefined.
     
    Last edited: Mar 6, 2016
  7. Mar 6, 2016 #6

    Simon Bridge

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    You wrote: ##v(t)=\frac{A}{C}e^{-t/2\tau}\cos(\omega t + \phi)## ... at this stage you know everything except A - which you can get off one point: pick one where v is a maxima then you don't have to deal with the cosine terms.

    To find L and R you need two equations with L and R in them as unknowns... which is what you tried.
    The equations have to be independent.
     
    Last edited: Mar 6, 2016
  8. Mar 6, 2016 #7
    Ah so because I would've ended up with two equations that were both of the same form ##R/L=\mbox{constant}## it would've been useless?
     
  9. Mar 6, 2016 #8

    Simon Bridge

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    The way you were thinking yes: your second equation was ##R/L = \text{a number}##. That "a number" must be ##\tau## because ##\tau = R/L##
    In your problem statement you are not actually asked to find L or R ... do you know the impedence in the circuit?
     
  10. Mar 6, 2016 #9
    The total impedance of the circuit would be ##Z=R+j(\omega l-\frac{1}{\omega c})=\sqrt{R^2+(\omega l-\frac{1}{\omega c})^2}e^{j\theta}## where ##\theta=\tan^{-1}(\frac{\omega l-\frac{1}{\omega c}}{R})##, then with ##\tilde{v}=\frac{A}{C}e^{-\frac{t}{2\tau}+j(wt+\phi)}## and using ##\tilde{i}=\frac{\tilde{v}}{Z}## so $$\tilde{i}=\frac{Ae^{-\frac{t}{2\tau}+j(wt+\phi-\theta)}}{C\sqrt{R^2+(\omega l-\frac{1}{\omega c})}}$$ since ##i=\frac{v}{Z}## $$i=\frac{Ae^{-\frac{t}{2\tau}}\cos(wt+\phi-\theta)}{C\sqrt{R^2+(\omega l-\frac{1}{\omega c})}}$$
    Which I could set equal to ##dq/dt##... Is this correct or did I go overboard?
     
  11. Mar 6, 2016 #10

    Simon Bridge

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    I suspect that won't work. Note: you have i(t) already in post #1.

    Let's make sure I understand what is happening...
    I take it this is a series RLC circuit with no source so that ##\ddot q + \frac{R}{L}\dot q + \frac{1}{LC}q = 0##
    The plot is for v(t) taken across one of the components... with v(0) = some initial value before a switch is thrown.
    Looks like capacitor-voltage since ##q(t)/C = v_C(t)##

    If you can work out the role of the 1/LC parameter in the v(t) plot, then you can get L.
    I gotta go for a bit.
     
  12. Mar 6, 2016 #11
    Yep everything you said is correct except that the circuit is undergoing an applied pulse. Unfortunately I'm not sure what to set ##\ddot q+\frac{R}{L}\dot q+\frac{1}{LC}q## equal to since it's pulsating and not a sinusoidal or constant voltage source. I suppose I will try setting the RHS equal to a function of time.
     
  13. Mar 6, 2016 #12

    Simon Bridge

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    Yep - the RHS is v(t) ... the driving voltage. In this case that is the voltage profile of the pulses.
    You will need to make sure you use ##v_C## for the voltage across the capacitor.
    square pulses would be defined by top-hat functions.
     
  14. Mar 6, 2016 #13

    gneill

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    It looks like a transient response curve, so there's probably a signal generator set up to give it a kick at regular intervals, then the response is followed for some time and allowed to die out before the next kick. You're only concerned with the transient solution.

    You can model it as a voltage source with a current limiting resistor and switch that gets some steady-state current flowing through the inductor. Something like:

    upload_2016-3-6_23-56-6.png

    The capacitor charge and voltage will be zero initially, since at steady state the potential across the inductor will be zero. Then at "t = 0" the switch opens and the current path is isolated to the RLC circuit. The capacitor charge and voltage will then follow a decaying sinwave (or a suitably shifted cosine as you've done).
     
  15. Mar 6, 2016 #14
    A top-hat function would be one like if f is defined on 0 to 2: ##f(t)=1## if ##0\leq t\leq 1##, ##f(t)=0## if ##1<t\leq 2##?

    Sorry I'm having some trouble understanding this. So we're looking at the steady state of the circuit before the switch is opened, in the steady state the capacitor will become an open circuit and a total current ##i=\frac{V_s}{r}## will be passing through the inductor. Now when the switch is opened is the capacitor fully charged so it essentially acts as the voltage source and begins to discharge?
     
  16. Mar 7, 2016 #15

    gneill

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    No, the capacitor starts out with zero charge. As I stated, the potential difference across the inductor at steady state is zero. So the resistor and capacitor have zero potential across them. In the instant before the switch opens the inductor is carrying a current and the capacitor is uncharged.
     
  17. Mar 7, 2016 #16
    I'm confused as to how the capacitor charge will follow a decaying sine waveform if it's starting uncharged. I feel like I'm failing to see something obvious here.
     
  18. Mar 7, 2016 #17

    gneill

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    Take a close look at the curve in the image that you presented. The capacitor voltage starts at zero and doesn't hit a maximum until a quarter cycle later.

    It's the inductor that supplies the initial energy to the circuit when the switch opens. The capacitor first begins to charge when the inductor current is forced to flow in the RLC loop.
     
  19. Mar 7, 2016 #18
    Thanks this makes sense now, so then I should just use kirchoff's law on the loop to obtain ##L\frac{di}{dt}-\frac{q}{c}-iR=0## which after changing to one variable ##q## and multiplying by -1 becomes (I'm kinda concerned about this step actually since I just read http://web.mit.edu/sahughes/www/8.022/lec09.pdf and sometimes they use ##i=+\frac{dq}{dt}## and other times ##i=-\frac{dq}{dt}##) $$\frac{q}{c}+\dot{q}R-\ddot{q}L=0$$ which is solvable by substituting ##q=e^{rt}##, The only troubling thing is that there will be multiple cases for this solution.
     
  20. Mar 7, 2016 #19

    gneill

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    What values were you given for the circuit components? From what I can see in the thread you're told that the capacitor is 1 nF, is that correct?

    It may not be necessary to solve the differential equation if you already have the form of the solution. The parameters for the equation can be obtained by suitable measurements on the decay curve.

    upload_2016-3-7_2-0-19.png

    The angular frequency can be found by counting off the some number of full cycles and reading off the time. I've indicated the location where ten full periods just completes. Read the time (I've provided a "ruler" with a fine scale) and calculate ω. For the damping constant (attenuation factor ##\alpha##), pick a convenient location in the middle of the curve where the envelope just "kisses" a peak and use the time and envelope value there to find ##\alpha##. The envelope follows an exponential decay,
    $$E(t) = E_o e^{-\alpha t}$$
    By inspection it looks like ##E_o## is 1 V, but you could always measure a pair of values at a different peaks and slog through the math to fit the curve.

    You should be able to tell the initial capacitor charge by inspection since you can read the initial capacitor voltage from the curve.
     
  21. Mar 7, 2016 #20

    Simon Bridge

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    Thank you @gneill - I had just noticed discrepancies, hence post #10 ... I'll leave you to it to avoid confusion.
     
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