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# Voltage drop from power, single-phase and three-phase current, cable length, cross-section or diameter

With this calculator you will calculate the voltage drops for single-phase and three-phase AC circuits calculated from active power. You will also calculate cable length, conductor diameter, conductor cross-sectional area, phase or line voltage or active power.

## Voltage drop calculator for single-phase and three-phase circuits - information

Voltage drop - voltage reduction in electricity, that is, the difference in electric potential between two points in a circuit in which an electric current flows.

It is also a concept which in the energy sector can mean:
- reduction of the electric voltage between the beginning and the end of the supply line,
- voltage reduction below the rated voltage for a given power network.

Relative voltage drop is the ratio of voltage drop to rated voltage. The permissible voltage drop at rated load on the transmission line from the transformer to the electricity consumer must be less than 5% of the rated voltage.

Electric energy receivers, to ensure their correct operation, should be supplied with voltage close to the rated voltage. This sometimes requires the use of cables with a cross-section greater than the current carrying capacity. The permissible voltage drop in non-industrial electrical installations in load circuits, from the meter to any receiver, according to N-SEP-E-002, should not exceed 3%, and from the meter to the connector 0.5%, with power transmitted up to 100 kVA and 1% for a power greater than 100 kVA and less than 250 kVA.

For circuits made with cables, multi-core or single-core conductors with a conductor cross-section not greater than 50 mm² Cu (copper) and 70 mm² Al (aluminum), the reactances of these conductors are ignored.

Assuming the above assumption, the voltage drops are calculated from the dependence:

for single-phase circuits:
$$\Delta U_\%=\frac{200 \cdot P \cdot L}{\sigma \cdot S \cdot U^2_{nf}}$$

for three-phase circuits:
$$\Delta U_\%=\frac{100 \cdot P \cdot L}{\sigma \cdot S \cdot U^2_{n}}$$

where:
ΔU% – voltage drop [V],
P – active power [W],
L – wire length [m],
σ – conductivity of the conductor [m/Ωmm²],
Unf - phase voltage [V],
Un - line voltage [V],
S – cross-sectional area of the line veins [mm²],
d - wire diameter,

Given the diameter of a conductor, the cross-sectional area of a conductor can be calculated using the formula:

$$S = \frac{\pi \cdot d^2}{4}$$

where:
S – conductor cross-sectional area,
d – conductor diameter,

Conductivity (specific conductivity, specific electrical conductivity) is a physical quantity that characterizes the electrical conductivity of a material.

After transformations for single-phase circuits:

Wire diameter:
$$d = \sqrt {\frac{800 \cdot P \cdot L}{\sigma \cdot \Delta U_\% \cdot U^2_{nf} \cdot \pi}}$$

Active power:
$$P = \frac{\sigma \cdot \Delta U_\% \cdot U^2_{nf} \cdot S}{200 \cdot L}$$

Wire length:
$$L = \frac{\sigma \cdot \Delta U_\% \cdot U^2_{nf} \cdot S}{200 \cdot P}$$

Phase voltage:
$$U_{nf}=\sqrt {\frac{200 \cdot P \cdot L}{\sigma \cdot \Delta U_\% \cdot S}}$$

Cross-sectional area:
$$S=\frac{200 \cdot P \cdot L}{\sigma \cdot \Delta U_\% \cdot U^2_{nf}}$$

After transformations for three-phase circuits:

Wire diameter:
$$d = \sqrt {\frac{400 \cdot P \cdot L}{\sigma \cdot \Delta U_\% \cdot U^2_{n} \cdot \pi}}$$

Active power:
$$P = \frac{\sigma \cdot \Delta U_\% \cdot U^2_{n} \cdot S}{100 \cdot L}$$

Wire length:
$$L = \frac{\sigma \cdot \Delta U_\% \cdot U^2_{n} \cdot S}{100 \cdot P}$$

Line voltage:
$$U_{n}=\sqrt {\frac{100 \cdot P \cdot L}{\sigma \cdot \Delta U_\% \cdot S}}$$

Cross-sectional area:
$$S=\frac{100 \cdot P \cdot L}{\sigma \cdot \Delta U_\% \cdot U^2_{n}}$$

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