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.
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.