How to measure the power of a load with imaginary part ?

Discussion in 'General Electronics Chat' started by shengwuei, Jun 7, 2010.

  1. shengwuei

    Thread Starter Member

    Aug 22, 2008
    17
    0
    Hi,

    I have a load with impedance 10-20j.
    When the load is driven by a sine wave, the peak-to-peak voltage measured on load is about 100V(RMS value 100/2/1.414=35.4V). Can I take the real part of the load(10 Ohm) and estimate the power applied on the load as the following ?

    35.4V x 35.4V / 10 = 25W

    What confused me is, since the impedance is far from matched between my driving circuit and the load, the waveform measured on load is distorted a lot compare to a sine wave, is it still valid to take the peak-to-peak value of waveform on load to calculate power applied on the load ?

    If not, what's the correct way to do it ?
    Or could I use a power meter to measure the power ?

    Thanks.
     
  2. timrobbins

    Active Member

    Aug 29, 2009
    318
    16
    The loss in watts is the rms voltage across the effective resistance - ie. V*V/R. If the load can be represented as 10 ohm in parallel with a reactance, then the effective resistance is 10 ohm. If the load is 10 ohm in series with a reactance, then you need to convert to a parallel form. You need to measure rms voltage, and not use pk value coupled with ideal sinewave conversion factor.

    Ciao, Tim
     
    shengwuei likes this.
  3. shengwuei

    Thread Starter Member

    Aug 22, 2008
    17
    0
    Hi Ciao,

    Thanks for the reply.

    One more question, does higher Vpp(or Vrms) value measured on load always mean higher power on load even if the load has imaginary part with it ?

    Thanks again.
     
  4. Bychon

    Member

    Mar 12, 2010
    469
    41
    The real voltage causes power in the real load.

    That's the end of the list.
     
Loading...