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

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

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

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

Example
What is the current that a heating unit of 3.4 kW absorbs at 440 V?

I = 3400 / 400 · I = 7.7 A

Alternating 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

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 cosj = 0.89 and an efficiency of 90%?

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

 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)