# Balanced Three Phase Star Load

1. Feb 21, 2014

### DunceKirchhoff

2. Feb 21, 2014

### DunceKirchhoff

Anyone??

3. Feb 21, 2014

### Staff: Mentor

It's a bit of a pain having to follow links to question bodies. Consider loading your pics to PF rather than some other image hosting site, and including them in the post body (via the "Go Advanced" editing frame where icons on the frame give you the ability).

The only issue I'd have with your results as shown is that the question asks you to use VAB as the reference phasor for the calculations, which should, I think, yield a "defined" angle for VAB of zero degrees rather than 30°. The other voltages would then similarly be shifted by -30° for part (a), and would carry through to the voltages calculated for part (b) and currents for part (c).

4. Feb 21, 2014

### DunceKirchhoff

Hi Neil

Sorry will do that in the future.

Yeah usually the ref is 0 degrees agreed.
But in this question the line voltage is not given but the phase voltage Vao.
So given that information I have tried to work it out.
Also the line voltage leads the phase voltage by 30 degrees.

Could be wrong

5. Feb 21, 2014

### DunceKirchhoff

I am wrong sorry, yes it has asked me to use vab as a reference.
So vab=0 degrees
Vbc -120 degrees
Vca 120 degrees
Whats the phase voltage?

6. Feb 21, 2014

### Staff: Mentor

For VAO', VBO' and VCO' you can just shift their normal values by the same -30°. Confirm by re-calculating VAB etc., from them to see that the resulting angles match what you got for part (a).

7. Feb 21, 2014

### DunceKirchhoff

So basically I have put vab, vbc and vca where vao, vbo and vco should be? And the phase angle is -30 instead of 30?

8. Feb 21, 2014

### Staff: Mentor

$V_{AO'} = 100\ ∠\ 0°$
$V_{BO'} = 100\ ∠ -120°$
$V_{CO'} = 100\ ∠ \ 120°$

which I believe is a "standard" set of angles for the lines of a star configuration. Now you want to have the voltages VAB, etc., come out having the desired set of angles where VAB has an angle of zero degrees. Those angles were determined in part (a).

Since all the various relative angles are fixed on a phasor diagram, if you had to shift VAB by -30° to make it the zero degree reference then ALL the angles will shift by the same amount. It's equivalent to rotating the whole phasor diagram by 30°. So if you were to subtract 30° from each of the above you should find that if you recalculate VAB etc. from VAO' VBO' and VCO', the angles will turn out to match the ones from part (a).

9. Feb 21, 2014

### DunceKirchhoff

Ok thanks I got you. Well explained.