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Express current in time and frequency domain (capacitor)

  1. Aug 18, 2013 #1
    1. The problem statement, all variables and given/known data


    Calculate the current in the capacitor shown if the voltage input is

    a. v1(t) = 10cos(377t - 30°)V
    b. v2(t) = 12sin(377t + 60°)V

    [Broken]


    Give answers in both time and frequency domain.



    2. Relevant equations

    I = dQ/dt = C * dV/dt


    3. The attempt at a solution

    I = dQ/dt = C * dV/dt

    I = 1ρF * (d/dt)(10 * cos(377t - 30°)v

    Took derivative:

    I(t) = 1μF * (-3770 * sin(377t - 30°)

    I(t) = (-3.77 x 10-3)(sin(377t - 30°))A


    If the above is correct, where can I start with converting this to frequency domain though? I know it's supposed to be something like in form e^t.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Aug 18, 2013 #2

    gneill

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    Staff: Mentor

    What you've done so far is fine. You've got the current expressed as a sine function, while the starting point was a voltage expressed as a cosine function. You may find that, sometimes, it is desirable to express the result using the same trig function as you began with (i.e., then both current and voltage time domain functions use the same trig function).

    For the frequency domain you'll be working with phasors. Have you covered them?
     
  4. Aug 19, 2013 #3
    So ω = 377 and the phasor current for that looks to be

    I = (-3.77 x 10-3)∠-30° ? Would that be it?
     
  5. Aug 19, 2013 #4

    gneill

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    Staff: Mentor

    Sure. But usually the magnitude of a phasor is a positive value. You can normalize the phasor by folding the negation into the angle.

    And, if you want to place the current phasor on the same reference axes as the voltage phasor (so that the "PHASE" part of phasor is actually useful), you should first convert the current's sine to cosine so the trig function matches that of the driving voltage. Then the phase angles will be referenced to the same starting point.

    So, the steps are:

    i(t) = -3.77 sin(ωt - 30°) mA
    i(t) = 3.77 sin(ωt - 30° + 180°) mA
    i(t) = 3.77 cos(ωt - 30° + 180° - 90°) mA

    So:

    i(t) = 3.77 cos(ωt + 60°) mA

    and your phasor is 3.77 mA ∠60

    I see that for part (b) the trig function in the time-domain voltage is sine, so you'll want to manipulate the current's expression to be likewise.

    Personally I find it easier to employ complex numbers to work with phasors and impedance in the frequency domain. Then essentially all the angle offsets are handled automatically by the complex arithmetic :wink:
     
  6. Aug 19, 2013 #5
    So is the second current:

    I = (4.524 x 10-3)(sin(377t + 150°)A ?

    Then the phasor

    I = 4.524 mA ∠ 150° ?
     
  7. Aug 19, 2013 #6

    gneill

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    Staff: Mentor

    Looks good!
     
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