Voltage Drop
Voltage Drop
(OP)
I have a question regarding voltage drop calculations. I'm sure it's painfully simple, but it's been troubling me for a little while.
The electric services where I work are predominantly 120/208VAC 3-phase. Apartments and single family homes get 120/208VAC "single phase" or network service, which is two hots and a neutral from the 3-phase distribution system.
When I need to do 3-phase voltage drop calculation, I turn to Chapter 9, Table 9 in the NEC and calculate voltage drop as
VD (line-neutral) =R x Circuit Length x I
To calculate line to line voltage drop, I multiply by 1.73, so VD = 1.73 x R x L x I
When I need to calculate voltage drop in 120/208VAC "single phase" circuit, i.e. two hots and a neutral or commonly two hots and no neutral I want to follow the explanatory text following 215.2(A) in the NEC Handbook. and calculate voltage drop as VD = 2 x L x R x I.
What is the right way to calculate line to line voltage drop for 208VAC? Do I need a factor of 1.73 or 2?
Thanks,
John
The electric services where I work are predominantly 120/208VAC 3-phase. Apartments and single family homes get 120/208VAC "single phase" or network service, which is two hots and a neutral from the 3-phase distribution system.
When I need to do 3-phase voltage drop calculation, I turn to Chapter 9, Table 9 in the NEC and calculate voltage drop as
VD (line-neutral) =R x Circuit Length x I
To calculate line to line voltage drop, I multiply by 1.73, so VD = 1.73 x R x L x I
When I need to calculate voltage drop in 120/208VAC "single phase" circuit, i.e. two hots and a neutral or commonly two hots and no neutral I want to follow the explanatory text following 215.2(A) in the NEC Handbook. and calculate voltage drop as VD = 2 x L x R x I.
What is the right way to calculate line to line voltage drop for 208VAC? Do I need a factor of 1.73 or 2?
Thanks,
John






RE: Voltage Drop
Use 2 for single phase L-L or L-N VD.
Be sure to use the effective 'Z' of the circuit from table 9 where you are indicating 'R' in your thread.
RE: Voltage Drop
I don't have an NEC in front of me, so I can't check Table 9.
"Theory is when you know all and nothing works. Practice is when all works and nobody knows why. In this case we have put together theory and practice: nothing works... and nobody knows why! (Albert Einstein)
RE: Voltage Drop
The BALANCED portion of the load (i.e. the smaller of the two 'hot' currents) will flow out and back through the hots, so VD = 2 * R * L * I applies.
The UNBALANCED portion (the difference between the two hot currents, if any) will flow back through the neutral, so you would use the other formula IF THE NEUTRAL is the same gauge wire as the hots.
RE: Voltage Drop
I'm still having trouble seeing where I draw the line. If I have three hots (120 degrees out of phase )and a neutral, I use VD=1.73*L*R*I. If I have only two hots (still 120 degrees out of phase) and a neutral, do I use the same formula? What if I drop the neutral?
Thanks,
John
RE: Voltage Drop
So in your last question John, you won't be 120 degrees out of phase since there is only one phase coming off the grid.
In larger power systems 3-phase is used and you will see the 1.73 factor.
Regards,
Jim
RE: Voltage Drop
This configuration is common in high density urban areas, but rare else where.
Thanks,
John