Formulas

Formulas

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
= Power loss in % of the transmission power
A = Conductor cross-section in mm²
= 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

P = Power in W
U = Voltage in V
I = Current in A
ŋ = Effectivity

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

I = 3400 / 400 · I = 7.7 A

Alternativ ström

Alternate current

I = P / U · cosφ

P = Power in W
U = Voltage in V
I = Current in A
cosφ = Phase shift

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

Trefasström

Three-phase current

I = P / 1.73  · cosφ  · ŋ · U

P = Power in W
U = Outer conductor voltage in V
I = Outer conductor current in A
cosφ = Phase shift
ŋ = Effectivity

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