Calculating Vd of 120/208 V circuit

[unsaint33]

when calculating the max voltage drop of 3% for feeder (or branch circuit) for a 120/208V, do I want the Vd to be 3% of 120V or 208V? And does that change for the 5% of the total circuit?

 

[petersonra]

when calculating the max voltage drop of 3% for feeder (or branch circuit) for a 120/208V, do I want the Vd to be 3% of 120V or 208V? And does that change for the 5% of the total circuit?

if the feeder has the same current on all three legs, the VD will be the same L-L as L-N.

 

[ptonsparky]

Do you have a reduced neutral?

 

[unsaint33]

if the feeder has the same current on all three legs, the VD will be the same L-L as L-N.

So, in order to select the feeder conductors big enough to limit the Vd under 3%, should my 3% Vd be 3.6V or 6.24V? (3.6V is 3% of 120V & 6.24V is 3% of 208V)

 

[petersonra]

So, in order to select the feeder conductors big enough to limit the Vd under 3%, should my 3% Vd be 3.6V or 6.24V? (3.6V is 3% of 120V & 6.24V is 3% of 208V)

If you have 3% VD on the L-N side, that would be 116.4 V. That would make the L-L voltage 201.6 V, or a 3% VD.

if the feeder has the same current on all three legs, the VD will be the same L-L as L-N.

I should have said the VD in % is the same. Obviously the voltage drop in volts is not the same.

 

[iceworm]

So, in order to select the feeder conductors big enough to limit the Vd under 3%, should my 3% Vd be 3.6V or 6.24V? (3.6V is 3% of 120V & 6.24V is 3% of 208V)

They are all telling you straight.
3% Voltage drop on each 120V leg gets you 3% Voltage drop on the phase to phase 208V

It is a vector (trigonometry) problem. Vphase-neutral X sqrt(3) = Vphase-phase
120V X 1.732 = 207.8V

Works the same for the Voltage Drops
3% X 120V = 3.6V
3.6V X 1.732 = 6.24V

.03 X 207.8V = 6.23V

The only difference we are seeing is the round off error.

The attached sketches show the angles.

 

[ggunn]

They are all telling you straight.
3% Voltage drop on each 120V leg gets you 3% Voltage drop on the phase to phase 208V

It is a vector (trigonometry) problem. Vphase-neutral X sqrt(3) = Vphase-phase
120V X 1.732 = 207.8V

Works the same for the Voltage Drops
3% X 120V = 3.6V
3.6V X 1.732 = 6.24V

.03 X 207.8V = 6.23V

The only difference we are seeing is the round off error.

The attached sketches show the angles.

I deal with balanced lines (3 phase PV inverters), so for me the answer is simple; it's just the one way drop on an individual conductor @120V. There's no current on the neutral, so no voltage drop.