power factor, voltage and current waveforms

Discussion in 'Homework Help' started by m_p, Mar 5, 2011.

  1. m_p

    Thread Starter New Member

    Nov 23, 2010
    6
    0
    hello,
    i need to draw the waveforms of the voltage and current in an inductive circuit which has a power factro of 0.866 (which give a phase angle of 30 degrees).

    i know that the current lag 90 degree behind the voltage, but i am not quite sure i understand how the power factor affect the waveforms.
    any help would be much appreciated.
    thanks
    mario
     
  2. Vahe

    Member

    Mar 3, 2011
    75
    9
    I think you should consider a series LR circuit. Apply a sinusoidal input to the series circuit in the form of
     v_{in}(t) = V_m \sin ( \omega t) .
    Then compute the current that is being drawn by the LR circuit from the voltage source. The input current should have the form
     i_{in}(t) = I_m \sin ( \omega t - \theta) .
    In other words, the current will lag the voltage which is what you expect in an inductive circuit. Power factor is simply  \cos \theta .

    Cheers,
    Vahe
     
    Last edited: Mar 5, 2011
  3. Vahe

    Member

    Mar 3, 2011
    75
    9
    If you compute the power factor of the series LR circuit, you should get
    <br />
PF = \frac{R}{\sqrt{(\omega L )^2 + R^2}}<br />
    where  \omega = 2 \pi f and  f is the frequency in Hz.

    In your case you are given the power factor, so you have to pick a frequency and a value for R and you can solve for the L that can give you the desired power factor. You may also be given the frequency (for example, 60 Hz) in that case, you will still have 1 equation and 2 unknown. So you choose the R and compute the L value, for example.
    There is no unique solution unless one of the elements are specified and a
    frequency is given.

    Cheers,
    Vahe
     
  4. m_p

    Thread Starter New Member

    Nov 23, 2010
    6
    0
    hi vahe, thanks for your reply.
    i think my tutor wants us to just draw a standard plot of voltage/current like the one below, but it also gave us the value of the power factor... i don't know why but i assume it must be shown somewhere on the graph
    [​IMG]
     
  5. Vahe

    Member

    Mar 3, 2011
    75
    9
    Given that graph, first mark the time difference between the plots. Call this time difference  t_d . Note that this time difference will be some fraction of the total period of the waveforms for the inductor current and voltage. If the waveforms have frequency  f , then the period in seconds will be  T = 1/f . Now we note that the full period  T corresponds to  2 \pi radians or 360°. Therefore  t_d corresponds to  2 \pi t_d / T radians or  (t_d/T)360°. This is exactly the angle  \theta and the cosine of this angle is the power factor.

    If you do this for the waveforms you show, you will find that  td = 0.25 T. In other words, the peaks have a difference of a quarter of a period. So the angle theta is one quarter of 360° or just 90°. The power factor here is 0.

    In your case you are asked to provide something that has power factor of 0.866; therefore,  \cos \theta = 0.866. This means that  \theta = \cos^{-1} 0.866 = \pi/6 radians or 30°. This means that your peak to peak difference has to be 30°/360° or 1/12 of the period. The actual time difference will depend on the frequency you choose but 30° will always correspond to the 1/12 of the period.

    Cheers,
    Vahe
     
    m_p likes this.
  6. m_p

    Thread Starter New Member

    Nov 23, 2010
    6
    0
    thanks Vahe, i understand it now.
     
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