it's exam time again, RMS voltage of chopped sine and power factor

[bdunham7]

so the power factor is indeed the real power that goes through the resistor over the apparent power which is what the grid was ready to supply but did not get used. using the voltage I got albeit wrong the power factor is therefor correct. But apparently the voltage is right and the power factor wrong.

Several of us appear to be trying to tell you the same thing although perhaps I've been too brief.  I don't know about this learning module or what the teacher is trying to get across with a problem like this, but frankly learning a 'formula' for calculating the PF of a phase-controlled resistive load is not something I would take the time to do--or pay for.  I'm pretty sure the answers are 1) 89.7V and 2) 0.897 and although The Electrician's simplification also gets you there, I much prefer to let that 0.897 'fall out' of the full calculation. 

The problem with trying to reduce PF to concepts that you can intuitively understand, like "what the grid was ready to supply but did not get used", is that they're just not quite right.  The most common basic understanding of PF starts with an ideal AC voltage source that can source or sink any amount of current, the the PF is a characteristic of a load to which this voltage is applied.  Saying that reactive (or 'recirculating') current did not get used is not really all that helpful and could be very misleading. 

PF as a concept--and the math--can be applied to any system including DC, but it was devised as a way of understanding AC power systems, so it isn't typically referred to as 'power factor' outside of that context.  The basic idea can be useful in other contexts--like an audio amplifier driving a complex multi-driver speaker system.  The math is a lot harder there and I've never seen an explicit PF calculation by audio engineers.

Anyway, for a source voltage V and load current I, V_rms= (∫V² dt)^1/2 and I_rms = (∫I² dt)^1/2 and both are always positive numbers--so apparent power is always positive.  Real power W_avg is simply ∫ VI dt and can be a positive or negative number.  In the case of purely resistive load, I=V/R, so W simply becomes V_rms²/R or I_rms²R.  This is where that 0.897 'falls out'. Apologies for my poor math notation skills and any silly errors I may have made.

PF = W_avg/(V_rms * I_rms).  If all three values are provided or obtained by measurement, the calculation is straightforward.  If one of the values is missing and cannot be obtained by using some constraint in the problem (like a resistive load) then you know that you can't calculate PF.  If the math is hopelessly complex, then at least you still understand conceptually what the PF is even if you can't calculate it.  For an example, take a half-wave circuit consisting of a heating element driven by an AC source through an ideal rectifier.  Even if you don't have a formula and only have basic algebra skills, you should be able to figure it out.

[Simon]

I'm trying to simplify it to the level that the tutor is to make a comparison and figure out who is right about what. Welcome to modern UK university studying! The tutor is frankly incompetent! they seem to have some knowledge somewhere but also seem to be often writing abut topics they are not familiar with, have just read up about and are now regurgitating the information incorrectly whilst producing near illiterate words to fill a page. Honestly a copy writer could do a better job!