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A.C. Circuits - Graphs

  1. Aug 31, 2009 #1
    A.C. Circuits -- Graphs

    http://img136.imageshack.us/img136/5782/image64.jpg [Broken]
    In this graph, the curve of current is larger than that of voltage...but in some books the voltage curve is above the current curve (like the one below)...Which one's correct?
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Aug 31, 2009 #2
    Re: A.C. Circuits -- Graphs

    It's arbitrary. You are shown a graph with two curves of different units.
  4. Aug 31, 2009 #3
    Re: A.C. Circuits -- Graphs

    So basically it doesn't matter if you drew both curves of same height, right?
  5. Aug 31, 2009 #4
    Re: A.C. Circuits -- Graphs

    It doesn't matter, you can scale the y-axis to any size.
  6. Aug 31, 2009 #5
    Re: A.C. Circuits -- Graphs

    For capacitors, The ratio of voltage to current at a frequency f (where f = w/(2 pi) where w is the frequency in radians per second), is

    V/I = -j/wC (where the j indicates a 90 degree phase shift). For an inductance:

    V/I = +jwL

    So for any given C or L, the ratio V/I depends on frequency.
  7. Aug 31, 2009 #6
    Re: A.C. Circuits -- Graphs

    I'm sorry I don't understand what you are trying to explain..
  8. Aug 31, 2009 #7
    Re: A.C. Circuits -- Graphs

    Hi Usair-

    Suppose I have a capacitor C and I put a voltage V on it at frequency f.

    The current in it is given by I = (2 pi f) C V. But the phase of the current is shifted by 90 degrees. So we use the short-hand notoation j, so that

    I = j (2 pi f C) V

    This is because of the equation for charge on a capacitor is

    Q = C V

    Now suppose V = V0 sin(wt) where w = 2 pi f

    Then I = dQ/dt = C dV/dt = w C V0 cos(wt)

    So there is a 90 degree phase shift (betwen the sine and cosine), and the amplitude of I depends on frequency f, as well as on C and V0.

    I hope this helps

    Bob S
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