Three-phase calculation confusion

[oldsparky52]

Yes, basically when one is at +104, and one is at -104, the voltage between them will be at 208.

Thank you .... again.

So if A phase is +104 to neutral, and then B Phase is -104 to neutral, at that same exact moment in time, would C phase be -16 volts to neutral (or +16 volts to neutral?)? If not 16 volts, what would it be?

Thanks for taking time for these simple questions. Now that I'm retired, I have more time to learn things I probably should have already known.

 

[oldsparky52]

Yes, basically when one is at +104, and one is at -104, the voltage between them will be at 208.

If at 90 degrees the A phase is 120V to neutral, what degree (or degrees since it happens twice in a cycle) would it be 104V to neutral? (I sure hope that question makes sense).

 

[Besoeker3]

Yes, basically when one is at +104, and one is at -104, the voltage between them will be at 208.

Surely only if the phase displacement between them is 180°

 

[david luchini]

Surely only if the phase displacement between them is 180°

Surely not...Are you feeling OK?

 

[david luchini]

If at 90 degrees the A phase is 120V to neutral, what degree (or degrees since it happens twice in a cycle) would it be 104V to neutral? (I sure hope that question makes sense).

At 60 degrees (it only happens once in a cycle.) At -120 degrees it will be at -104v to neutral.

 

[GoldDigger]

At 60 degrees (it only happens once in a cycle.) At -120 degrees it will be at -104v to neutral.

Actually, under the given conditions, at 60 degrees and 120 degrees it would be 104. At -120 degrees and -60 degrees it will be -104. There are two points (to either side of peak) in each half cycle where the lower voltage is present. Sine waves are like that.

 

[oldsparky52]

At 60 degrees (it only happens once in a cycle.) At -120 degrees it will be at -104v to neutral.

Thanks.

Interesting, 60 degrees is 67% of 90 degrees and 104V is 80% of 120V.

 

[david luchini]

Actually, under the given conditions, at 60 degrees and 120 degrees it would be 104. At -120 degrees and -60 degrees it will be -104. There are two points (to either side of peak) in each half cycle where the lower voltage is present. Sine waves are like that.

Yes, that's true. I still had the peak voltage in my mind.

 

[Besoeker3]

Surely not...Are you feeling OK?

Draw it out. The triangle.
One side 104 in one direction/phase. a second 104 on another phase. Now look at the difference shown by the third side of that triangle.

 

[david luchini]

Draw it out. The triangle.
One side 104 in one direction/phase. a second 104 on another phase. Now look at the difference shown by the third side of that triangle.

There is no triangle. He's asking about instantaneous values. There is only 104-(-104)=208.

 

[Besoeker3]

There is no triangle. He's asking about instantaneous values. There is only 104-(-104)=208.

Draw out how that would happen with three phase per the title of the thread......

 

[GoldDigger]

Draw out how that would happen with three phase per the title of the thread......

It works out just fine. The phasors are length 120, not 104. The projections onto the x-axis will be the instantaneous values of the voltage, as stated, and will add arithmetically, being parallel.
What is most confusing is that we are switching back and forth between using RMS amplitude and using peak amplitude, not getting into trouble since the ratios are constant.

 

[Besoeker3]

It works out just fine. The phasors are length 120, not 104. The projections onto the x-axis will be the instantaneous values of the voltage, as stated, and will add arithmetically, being parallel.

Three phases are not in parallel.
What am I missing?

 

[GoldDigger]

Three phases are not in parallel.
What am I missing?

The projections of each phasor onto the x-axis (or dot product with unit vector along x-axis) can be considered to represent the instantaneous voltage, a scalar.
You can either say that the two scalars add, period, or you can be somewhat imprecise and say that the two projections are parallel, both being along the x-axis.

 

[Besoeker3]

The projections of each phasor onto the x-axis (or dot product with unit vector along x-axis) can be considered to represent the instantaneous voltage, a scalar.
You can either say that the two scalars add, period, or you can be somewhat imprecise and say that the two projections are parallel, both being along the x-axis.

They peak 120deg apart.

 

[david luchini]

would C phase be -16 volts to neutral (or +16 volts to neutral?)? If not 16 volts, what would it be?

It would be 0 volts.

 

[david luchini]

Draw out how that would happen with three phase per the title of the thread......

He's asking about the instantaneous voltage between A and B. How would you draw that out in 3 phase?

 

[kwired]

This was the avatar of a once active member that is now deceased. Kind of shows you the voltage wave of each phase in relation to the others, slowed down of course, if you increased rotation to 3600 RPM the waves would be at 60Hz.

 

[Besoeker3]

He's asking about the instantaneous voltage between A and B. How would you draw that out in 3 phase?

Waveforms. Try it.

 

[kwired]

Still representation.

Note that 120 volt AC is RMS voltage or effective DC equivalent voltage and that actual peaks are about 170.

Look where V2 and V3 are at relative to same time V1 is at the + peak. They are both at same level on the - side.

Look where V2 and V3 are at relative to same time V1 is at zero crossing, they are same voltage but opposite polarity.

Look at any point along the time line and see where each phase is at that instant in time.