brother
Senior Member
Had a guy tell me that the load had to calculated with the 3 phase formula for the voltage drop, even though it is only 2 hots and 1 ground for the load/circuit. About 125 feet. This comes from a 3 phase panel 120/208. However as stated the load is a 2 hots 208 circuit.
3850 watts
208 volts
16 amps.
Is what is on the name plate.
Don't have my book handy, but If recall correctly the 3 phase formula, is VD =1.73 X12.9 X 16 X 125FEET/10380 (#10 copper)= 4.3volts
Since its only 2 hots, and not the 3 hots,(3 phase) wouldn't this need to be calculated using the single phase formula? VD= 2 x K X I X D/CMIL Basically 2 x 12.9 x 16 x 125feet/10380 =4.97 volts
It would be a 20amp circuit, so #10 copper is being used regardless. but looking at the name plate 3850watts/208=18.5amps if you use his 3 phase formula, 3850/208 x 1.73=10.699 amps
His 10.69 amps is no where near the name plate amps of the manufacturer. So isn't the use of the single phase formula the correct one to use in this?
Its late so forgive my brain spasm, haven't had to do these load calcs in a while.
3850 watts
208 volts
16 amps.
Is what is on the name plate.
Don't have my book handy, but If recall correctly the 3 phase formula, is VD =1.73 X12.9 X 16 X 125FEET/10380 (#10 copper)= 4.3volts
Since its only 2 hots, and not the 3 hots,(3 phase) wouldn't this need to be calculated using the single phase formula? VD= 2 x K X I X D/CMIL Basically 2 x 12.9 x 16 x 125feet/10380 =4.97 volts
It would be a 20amp circuit, so #10 copper is being used regardless. but looking at the name plate 3850watts/208=18.5amps if you use his 3 phase formula, 3850/208 x 1.73=10.699 amps
His 10.69 amps is no where near the name plate amps of the manufacturer. So isn't the use of the single phase formula the correct one to use in this?
Its late so forgive my brain spasm, haven't had to do these load calcs in a while.