Single Phase Half-wave Purely Inductive Load

Anyone know the formular for the Mean voltage for a Single Phase Half-wave Purely Inductive Load? I can't find it anywhere and I need it for my homework. I'm not as much of a slacker as I sound.

The RMS value? [LATEX]\frac{1}{\sqrt{2}}[/LATEX] of the peak voltage, but it really doesn't matter if the load is inductive or not. Sure you are asking the right question?

Thanks for replying so quick. I can't beleive these forums actually work. Now I think about it it is not the right question, I've worked out Vmean as 0 which is right as the sin wave has an equal proportion in positive and negative over the cycle. But when I try and calculate current I get
Imean = Vmean/Z
Imean = 0L0°V/10L90°Ω
Imean = 0L-90°

As I'm trying to divide 0 by something I get zero. What I am doing wrong?

If you simply find the mean of any AC voltage (that has no DC offset) it will always be zero - which is useless, of course. You need to use the RMS of the peak voltage. Then you will get a non-zero number to calculate these other things.

In your (the OPs) original post you said it was a half-wave signal which to me means half-wave rectified. The average value (Vmean) is not zero. It's also not the RMS value.

Hint: The average value of a full-wave rectified sinewave is .637 of the peak (see this). I leave it as an exercise to the reader to calculate the value for a half-wave sinewave.

Another hint: The current through an inductor from a rectified sinewave is not AC, it is (pulsating) DC

Hi,

The average voltage over a full cycle is 0, but the average of the absolute value is 0.6366 approximately or 2/pi exactly and this often applies to the full non rectified wave too.
The average voltage over a full cycle of a half wave rectified signal is 1/pi exactly which is approximately 0.3183 .
The RMS voltage over a full wave is 1/sqrt(2) which is approximately 0.7071 .
The RMS voltage over a full cycle of a half wave rectified wave is 1/(2*sqrt(2)) which is approximately 0.35355 .