Chopped sine wave power factor

[Simon]

So I have my question:

A single phase full-wave AC voltage has a resistive load of R = 10Ω
and the input voltage Vs = 100 V (rms), 50 Hz. The delay angle of
thyristors T1 and T2 are equal and π/3.
Determine:
i. RMS output voltage V0
ii. Input power factor

So the one that has me stumped is the power factor. This is my uni course so as always I treat it like alternate reality and try to stick with the formula's and examples given in the module. But the example of calculating power factor has errors.

So all of what I find online talks about reactive loads but this is a resistor. Having calculated the output RMS voltage I can calculate the output RMS current and so have the power used. Power factor is the power that was used over the power that had to be generated regardless of what was used? so the instantaneous amps would still be the input voltage over the load resistance?

[T3sl4co1l]

Not all. There should be resources showing the generalized power factor too.  Reactive power is the linear / AC steady state case, but it works for any components orthogonal to the source (which, cosine is orthogonal to sine, so reactive power is an example of this).

That's probably not a great approach, as you need the Fourier series and Parseval's theorem to find the orthogonal (harmonic sine waves) components, and then to add them up.

If nothing else, you can take S = Vrms * Irms (input values), and P being the output power (assuming no losses elsewhere -- no voltage drop on thyristors -- if this is consistent with the material?), then PF = P/S.

Which, because Irms is the same (input and output) in this circuit, and the load is resistive so Irms = Vrms(out) / R, it's simply Vrms(in) / Vrms(out).

It's maybe not obvious that the second (ratio of RMSs) is equivalent to the first (doing component-wise analysis; though Parseval's theorem basically being RMS over the spectrum, which might be a pretty big hint), so a proof would be instructive.

Tim

[Simon]

Yes the answer i have given is RMS in over out as if I calculater the powers as the square of the voltage over the resistance I just end up with a ration of the two voltages.