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, power	A =	200 · L · P y · P° · U²	200 · L · P y · P° · U² · cos²φ	100 · L · P y · P° · U² · cos²φ

I	=	Power in Amps
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)

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

Direct current

I = P / U · ŋ

Exempel
Vad är strömmen som en värmeanordning på 3.4 kW absorberar vid 440V?

I = 3400 / 400 · I = 7.7 A

Alternate current

I = P / U · cosφ

Example:
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.?

I = 1900 / 230 · 0.77 · 0.79 = 13.6 A

Three-phase current

I = P / 1.73 · cosφ · ŋ · U

Example:
What current does a three-phase motor of 22 kW consume at 400 V, 50 Hz, with cosφ = 0.89 and an efficiency of 90%?

I = 22000 / 1.73 · 400 · 0.89 · 0.9 = 39.7 A