FORMULAS
Required conductor cross-section for electric cables
| Given | Direct current | Single-phase alternating current | Three-phase alternating current | |
| Voltage drop, current | A = | 200 · L · I y · U° · U² |
200 · L · I · cosφ y · U° · U |
173 · L · I · cosφ y · U° · U |
| Voltage drop, power | A = | 200 · L · P y · U° · U² |
200 · L · P y · U° · U² |
100 · L · P y · U° · U² |
| Power consumption, current | A = | 200 · L · I² y · P° · P |
200 · L · I² y · P° · P |
300 · L · I² y · P° · P |
| Power consumption, current | A = | 200 · L · P y · P° · U² |
200 · L · P y · P° · U² · cos²φ |
100 · L · P y · P° · U² · cos²φ |
The formulas stated for alternating and three-phase current do not give any consideration to the inductive resistance. This resistance is a function of the distance of the individual conductors between one another. Direct current I = P / U · ŋ Example I = 3400 / 400 · I = 7.7 A Alternating current I = P / U · cosφ Example: I = 1900 / 230 · 0.77 · 0.79 = 13.6 A Three-phase current I = P / 1.73 · cosφ · ŋ · U Example: I = 22000 / 1.73 · 400 · 0.89 · 0.9 = 39.7 A
P
=
Power in W
U
=
Outer conductor voltage in V
I
=
Outer conductor current in A
cosφ
=
Phase shift
ŋ
=
Efficiency
P
=
Power in W
U
=
Voltage in V
I
=
Current in A
ŋ
=
Efficiency
What is the current that a heating unit of 3.4 kW absorbs at 440 V?
P
=
Power in W
U
=
Voltage in V
I
=
Current in A
cosφ
=
Phase shift
What is the current consumption of an alternating current motor of 1.9 kW at cosj = 0.77 and an efficiency of 79%?
The voltage 230V, 50 Hz.
What current does a three-phase motor of 22 kW consume at 400 V, 50 Hz, with cosj = 0.89 and an efficiency of 90%?
I
=
Power in Ampere
y
=
Conductivity (copper 56, aluminium 34)
L
=
Conductor length (single) in metres
P
=
Transmission power in watts
P°
=
Power loss in % of the transmission power
A
=
Conductor cross-section in mm²
U°
=
Voltage drop in % of the operating voltage
U
=
Operating voltage in Volts
cosφ
=
Power factor (usually assumed to be 0.8)
