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

With this calculator you can calculate voltage drops for single-phase and three-phase AC circuits calculated from the rated current. You will also calculate wire length, wire diameter, wire cross-sectional area, voltage or current.

## Voltage drop calculator from current 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 the circuit where 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 acceptable 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 receiving circuits, from a meter to any receiver, according to N-SEP-E-002, should not exceed 3%, and from the meter to a connector 0.5%, with power transmitted up to 100 kVA and 1 % at 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 exceeding 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:
$$\Delta U_\%=\frac{200 \cdot I_n \cdot L \cdot \cos \phi}{\sigma \cdot S \cdot U_{n}}$$

for three-phase circuits:
$$\Delta U_\%=\frac{\sqrt{3} \cdot 100 \cdot I_n \cdot L \cdot \cos \phi}{\sigma \cdot S \cdot U_{n}}$$

where:
ΔU% – voltage drop [%],
L – wire length [m],
In - rated current [A],
Un - rated voltage [V],
S – cross-sectional area of the line veins [mm²],
d - wire diameter,
σ – conductivity of the conductor [m/Ωmm²],
cosφ – phase shift factor,

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:
$$d = \sqrt {\frac{800 \cdot I_n \cdot L \cdot \cos \phi}{\sigma \cdot \Delta U_\% \cdot U_{n} \cdot \pi}}$$

Rated current:
$$I_n = \frac{\sigma \cdot \Delta U_\% \cdot U_{n} \cdot S}{200 \cdot L \cdot \cos \phi}$$

Wire length:
$$L = \frac{\sigma \cdot \Delta U_\% \cdot U_{n} \cdot S}{200 \cdot I_n \cdot \cos \phi}$$

Rated voltage:
$$U_{n}=\sqrt {\frac{200 \cdot I_n \cdot L \cdot \cos \phi}{\sigma \cdot \Delta U_\% \cdot S}}$$

Cross-sectional area:
$$S=\frac{200 \cdot I_n \cdot L \cdot \cos \phi}{\sigma \cdot \Delta U_\% \cdot U_{n}}$$

After transformations for three-phase circuits:

Wire diameter:
$$d = \sqrt {\frac{\sqrt{3} \cdot 400 \cdot I_n \cdot L \cdot \cos \phi}{\sigma \cdot \Delta U_\% \cdot U_{n} \cdot \pi}}$$

Rated current:
$$I_n = \frac{\sigma \cdot \Delta U_\% \cdot U_{n} \cdot S}{\sqrt{3} \cdot 100 \cdot L \cdot \cos \phi}$$

Wire length:
$$L = \frac{\sigma \cdot \Delta U_\% \cdot U_{n} \cdot S}{\sqrt{3} \cdot 100 \cdot I_n \cdot \cos \phi}$$

Rated voltage:
$$U_{n}=\sqrt {\frac{\sqrt{3} \cdot 100 \cdot I_n \cdot L \cdot \cos \phi}{\sigma \cdot \Delta U_\% \cdot S}}$$

Cross-sectional area:
$$S=\frac{\sqrt{3} \cdot 100 \cdot I_n \cdot L \cdot \cos \phi}{\sigma \cdot \Delta U_\% \cdot U_{n}}$$

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