# 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,

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.