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An AC-circuit

  1. Nov 23, 2008 #1
    1. The problem statement, all variables and given/known data
    Hi all.

    Please take a look at the attachted circuit.

    I have found the following expression for the complex current in the resistor (the hat indicates that it is complex):

    \widehat_{I (t)} = \frac{{\omega ^2 LC {\widehat{U_0} }}}{{\omega ^2 LRC + i\omega L - R}}\exp ( - i\omega t),
    where U_0 is the amplitude of U(t), and the hat indicates that it is complex.

    Now I wish to find the real current, and I want to write it as:
    I(t) = I_0\cos(\omega t + \phi).

    But how do I do this? I have spent like 2 hours trying, but I don't know how to rewrite the complex amplitude of the current to have a phase.

    Thanks in advance.


    Attached Files:

  2. jcsd
  3. Nov 23, 2008 #2


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    Hi Niles! :smile:

    (have an omega: ω and a phi: φ :wink:)

    The standard way would be to write the left bit as A + iB (multiply top-and-bottom by the complex conjugate of the bottom), and then

    Re(A + iB)eiωt = Re(A + iB)(cosωt + isinωt)

    = Acosωt - Bsinωt,

    which maybe you know how to rewrite as Ccos(ωt + φ) ? :wink:
  4. Nov 23, 2008 #3
    Ok, so you are telling me to multiply the first term on the RHS with the complex conjugate and then take the square root. Thus I get the amplitude of the current, and now we have to multiply that with exp(iφ).

    But this gives me: ... * exp(-i(ωt-φ))? Don't we want a plus instead?

    And thanks for replying!
  5. Nov 23, 2008 #4


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    Why are you taking the square root? :confused:

    (And you did multiply top and bottom, didn't you? To get a complex over a real?)

    Anyway, you don't multiply the amplitude with exp(iφ) …

    you can't find the amplitude (or the phase) until you've got the whole function …

    in other words, until after you multiply. :smile:
  6. Nov 23, 2008 #5
    Ok, so what I do is:

    1) Multiply the first term on the RHS with the complex conjugate of the denominator.

    2) I find the phase, and write the first term as: z = |z|*exp(iφ).

    3) Find the real part, and all ...

    4) Done? If yes, then I will go to bed.

    EDIT: Can you confirm me in that -0.1 * 2*Pi radians is equal to 2*Pi - 0.1 * 2*Pi radians?
  7. Nov 23, 2008 #6


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    I'm not convinced that you've got it.

    You're given (K + iL)/(M + iN) times eiωt

    that's (K + iL)(M - iN)/(M + iN)(M - iN) times eiωt

    = (K + iL)(M - iN)/(M2 + N2) times eiωt

    the bottom is real, so the whole of the left can now easily be put in the form A + iB …

    go on from there.
    Yes, adding multiples of 2π to the phase makes no difference … e2πi = 1

    … so if your result comes out as more than 2π, just subtract 2π (or 4π or … ) :smile:
  8. Nov 23, 2008 #7
    And then I write z = A + iB as z = |z|e, right?
  9. Nov 23, 2008 #8


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    ooh, I misunderstood what you wrote earlier …

    even though it was clear :rolleyes:
    Yes, that's right … then the real current will be the real part of |z|*exp(-iωt + iφ), = |z|*cos(-ωt + φ)

    To answer a previous point, cos(x) = cos(-x), so the minus sign doesn't matter (though it will mess up the sign of φ). :smile:

    Ignore some of my previous comments. :redface:
  10. Nov 24, 2008 #9
    Re: sorry!

    Nono, they were all good. Thanks for helping me - I see it much clearer now!

  11. Nov 24, 2008 #10
    I have found the phase φ to be -0.63, which is equivalent of 5.65. Does this mean that I am allowed to write:

    I(t) = Re [|z|*exp(-iωt - 0.63i)] = Re [|z|*exp(-i(ωt + 0.63)] = |z|*cos(ωt + 0.63),

    where Re [..] denotes the real part of the argument "...".
    Last edited: Nov 24, 2008
  12. Nov 24, 2008 #11


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    Yup! :biggrin:
  13. Nov 24, 2008 #12
    Great! Thanks!

    Just one last thing: Is it correct that sometimes the expressions for the current are very nasty? I mean, they are sometimes quite long :uhh:
  14. Nov 24, 2008 #13


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    oooh … dunno! :redface:

    i never did current 'n' stuff …

    i just know about the maths bit! :smile:
  15. Nov 24, 2008 #14
    Ahh, I see.

    But thanks anyway!
  16. Nov 24, 2008 #15
    Hah, yes.

    Consider a serial RCL circuit, that is, a voltage source connected to a condensator, a loop and a resistor. Let's say the Resistor is 200 ohm, the loop is 1 millihenry and the condensator is 20 nanofahrad. If you apply a simple voltage, not even timevariable, of 10 volt to this system, the current will look like this:


    (this assumes i(0)=0 and i'(0)=10000)

    If, instead, you apply an AC source, the voltage function (again assuming i(0)=0 but i'(0)=2*10^9) of which being v(t)=10*sin(20000*t), the resulting current function looks like this:

    i(t) = (1/20)*e^(-100000*t)*sin(200000*t)+B*sin(200000*t)+A*cos(200000*t).

    Explaining why this is so takes about two full pages.
    Last edited by a moderator: Nov 24, 2008
  17. Nov 25, 2008 #16
    Oh, I forgot to substitute the variables. A=1/85 and B=4/85
  18. Nov 25, 2008 #17
    I am familiar with Kirchhoff's laws (I guess that's what you are referring to?), and I used them to find my expression.

    But I guess I should stop fearing long expressions :smile:
  19. Nov 25, 2008 #18
    Nah, it's



    L=1 milliHenry, R=200 ohm and C=20 nanofahrad.
    Applying 10 volt (DC) to this system gets you

    0.001*i'' + 200*i' + 5*10^7*i = 0.

    Solving this equation:

    0.001*y^2 + 200*y + 50000000 = 0

    gives you two imaginary numbers,

    y1=100000 + j*200000
    y2=100000 - j*200000

    which means that

    i(t) = e^(-100000*t)*( A*sin(200000*t) + B*cos(200000*t) ).

    (This is the general solution to the "ordinary differential equation" shown above.)

    To find A and B, you need to know some starting conditions. We know i(0)=0 because the loop has "infinite resistance*" when you apply the initial potential.

    Knowing that i(0)=0 means you can rewrite, so:

    i(t) = e^(-100000*t)*( A*sin(200000*t) + B*cos(200000*t) )
    i(0) = e^(0)*(A*sin(0) + B*cos(0))
    0 = A*sin(0)+B*cos(0)
    0 = B*1

    so you can simplify the initial expression for the current to

    i(t) = e^(-100000*t)*A*sin(200000*t)

    from elsewhere (and I honestly can't remember) we know that i'(0)=10000, so A can be isolated and determined. I don't want to write out di/dt here because it is a damn long function, but suffice to say, it turns out A = 0.05 so

    i(t) = 0.05*sin(200000*t)*e^(-100000*t)

    All in all, current determination becomes complicated once you have a lot of loops and capacitors, but not so difficult that you should feel discouraged - there are a lot of shortcuts and neat tools at your disposal if you ever start in on the subject in a serious manner.

    *not true
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