Voltage drop is given by the following formula :

U : Voltage of the DC or AC system (V)

This is phase-phase voltage for 3-phase system; phase-neutral voltage for single-phase system.

Example :

- For western European countries a 3-phase circuit will usually have a voltage of 400 V, and single-phase 230V.

- In North America, a typical three-phase system voltage is 208 volts and single phase voltage is 120 volts.

NB: for DC voltage drop in photovoltaic system, the voltage of the system is U = Umpp of one panel x number of panels in a serie.

ΔU : voltage drop in Volt (V)

b : lenght cable factor, b=2 for single phase wiring, b=1 for three-phased wiring.

ρ1 : resistivity in ohm.mm2/m of the material conductor for a given temperature. At 20 celcius degree °C the resistivity value is 0.017 for copper and 0.0265 for aluminium.

Note that resistivity increases with temperature. Resistivity of copper reaches around 0.023 ohm.mm2/m at 100 °C and resistivity of aluminium reaches around 0.037 ohm.mm2/m at 100 °C.

Usually for voltage drop calculation according to electrical standards it is the resistivity at 100°C that is used (for example NF C15-100).

ρ1 = ρ0*(1+alpha(T1-T0)), here ρ0 = resistivity at 20°C (T0) and alpha = Temperature coefficient per degree C and T1 = temperature of the cable.

T1 : Temperature of the cable (default value = 100°C).

Note that from experience, a wire with a correct sizing should not have an external temperature over 50°C, but it can correspond to an internal temperature of the material around 100°C.

L : simple lenght of the cable (distance between the source and the appliance), in meters (m).

S : cross section of the cable in mm2

Cos φ : power factor, Cos φ = 1 for pure resistive load, Cos φ< 1 for inductive charge, (usually 0.8).

λ : reactance per lenght unit (default value 0.00008 ohm/m)

Sin φ : sinus (acos(cos φ)).

Ib : current in Ampere (A)

NB : For DC circuit, cos φ=1, so sin φ=0.

Voltage drop in percent :

ΔU(%) = 100 x ΔU/U0

Where :

ΔU : voltage drop in V

U0 : voltage between phase and neutral (example : 230 V in 3-phase 400 V system)

Energy losses in a cable is mainly due to resistive heating of the
cable.

It is given by the following formula :

E = a x R x Ib²

Where :

E : energy losses in wires,
Watt (W)

a : number of line
coefficient, a=1 for single line, a = 3 for 3-phase circuit.

R : resistance of one active
line

Ib : current in Ampere (A)

R is given by the next formula :

R = b x ρ1 x L / S

b : lenght cable factor, b=2
for single phase wiring, b=1 for three-phased wiring.

ρ1 : resistivity of the
material conductor, 0.017 for copper and 0.0265 for aluminium (temperature of the wire of 20°C) in ohm.mm2/m. Resistivity of copper reaches around 0.023 ohm.mm2/m at 100 °C and resistivity of aluminium reaches around 0.037 ohm.mm2/m at 100 °C.

L : simple lenght of the cable
(distance between the source and the appliance), in meters (m).

S : cross section of the cable
in mm2

NB : for direct current the energy losses in percent is equal to the
voltage drop in percent.

Chart
: Example of voltage drop losses according to wire cross
section for a PV system of 3 kWp with 50 m of solar DC string cable.

See also professional photovoltaic softwares that include voltage drop calculation

Many free photovoltaic softwares from inverter manufacturers include voltage drop calculation. Read more...