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Finding the phase angle of current through a capacitor?

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  1. Apr 15, 2015 #1
    1. The problem statement, all variables and given/known data
    Calculate the current in the capacitor shown in the figure below if the voltage input is V(t) = 29cos(377t-30) V, C = 1 micro Farad

    2. Relevant equations
    i = C dv/dt

    3. The attempt at a solution

    I essentially differntiate V(t) and multiply by 1 * 10^-6. I then get a negative amplitude, so I add 180 degrees to the phase angle to get a positive amplitude,

    i = 1*10^-6*29*377 * (-sin(377t - 30) = 0.0109sin(377t + 150)

    What is the phase angle? The correct answer is 60 degrees. I'm confused about how to get this. I'm getting 150 degrees.
     
    Last edited: Apr 15, 2015
  2. jcsd
  3. Apr 15, 2015 #2

    Delta²

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    You should convert the current equation to cosine form using familira trigonometric identities.
     
  4. Apr 15, 2015 #3
    Does it always have to be converted to cosine?

    If I convert it to cosine, the phase angle changes from 150 to 240
     
  5. Apr 15, 2015 #4

    Delta²

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    Current and Voltage have to be both in sine or cosine (and with the same sign of amplitude) to be able to discuss about the phase angle.

    You ve made a mistake in the conversion, it is sin(x)=cos (x-90)=-cos (x+90).
     
  6. Apr 16, 2015 #5
    I have sin(x+150) in my above equation. I want to convert it to cosine
    EDIT: I seem to have some confusion.

    IN my book it says cos(x) = sin(x+90) (just as you said).

    This means if I want to convert to cosine from sin,

    cos(x+60) = sin(x+60+90)
     
  7. Apr 16, 2015 #6

    Delta²

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    You are correct now, so it will be cos (x+60)=sin(x+150), (x=377t), there you got a phase angle of 60, THOUGH when we talk about phase angle we usually mean the phase difference between voltage and current which is 90 in this case.
     
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