Refresh My Memory
Refresh My Memory
(OP)
I have 2 15 kva loads connected to a 3 phase Y generator 220
phase to phase voltage. Load connected A-B and A-C. The load current is 15 kva/(0.22 kva 1.73 = 39 amps. If the load was connected Phase to neutral the neutral current would be
39 x 1.73 = 64 amps. However in this case the load is connected phase to phase with phase A assuming the common position. I
want to know the current in Phase A. Please refresh my memory as to the proper calculation.(Should be simple I admit)
Thanks
phase to phase voltage. Load connected A-B and A-C. The load current is 15 kva/(0.22 kva 1.73 = 39 amps. If the load was connected Phase to neutral the neutral current would be
39 x 1.73 = 64 amps. However in this case the load is connected phase to phase with phase A assuming the common position. I
want to know the current in Phase A. Please refresh my memory as to the proper calculation.(Should be simple I admit)
Thanks






RE: Refresh My Memory
15 KVA @ 127 Volts = 118 Amps
15 KVA @ 220 Volts = 68 Amps
A phase 68 Amps x 1.73 = 118 Amps
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Refresh My Memory
RE: Refresh My Memory
RE: Refresh My Memory
When equal loads are connected from two phases to the neutral, the three currents (A phase, B phase and neutral) are equal.
Consider three equal loads each connected from a phase to neutral.
In the absence of harmonics the neutral current will be zero. If you remove one load, the remaining current will be equal but opposite in polarity.
For this reason the Canadian Code does not allow any reduction in the size of a neutral conductor serving two phases and requires that the neutral be counted as a current carrying conductor for the purpose of de-rating.
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Refresh My Memory