Power loss in electrical installation :
10 kVA / 500 V / 20 A = I^2 R, R = 25 0hm, x=R= V/m
And, y=x V/m + c
Polar form : V/m sin phase angle(90 deg).
25 V/m, and V active 220 V, so the equation : V = 245 Volt
https://www.tokopedia.com/amp/find/elektronik-murah-jakarta
50 kVA generator was supplying load of 10 kVA induction motor. Suddenly the conductor got burnt after a year. Generator voltage 380 V active / with about NYM 4 x 4 mm2. The conductor should be : NYM 4 x 6 mm2 with total length = 25 m.
Voltage on conductor :
Label = 500 V / 3 cores or 4 cores
Actual voltage when electric motor has 660 V/25 A
50 kVA generator was supplying load of 10 kVA induction motor. Suddenly the conductor got burnt after a year. Generator voltage 380 V active / with about NYM 4 x 4 mm2. The conductor should be : NYM 4 x 6 mm2 with total length = 25 m.
Voltage on conductor :
Label = 500 V / 3 cores or 4 cores
Actual voltage when electric motor has 660 V/25 A
1. Third harmony : 660 V
2. Second harmony : 220 V cos phi x 3
3. Capacitance (ohm) on conductor : i ^2 t ( two poinrts ) equals 31x31 x 0'005 as coulomb divided by voltage. Put into ohm per meter to find cable length (6 mm2) from electric motor to electric panel as much as 5 to 7 m.
