# Voltage Drop

Voltage drop is calculated in accordance with CENELEC technical report CLC/TR 50480 "Determination of cross-sectional area of conductors and selection of protective devices", dated February 2011.

The voltage drop (as a percentage) is given by:

$\Delta U=\frac{b\left(R\mathrm{cos}\phi +X\mathrm{sin}\phi \right){I}_{b}}{{U}_{o}}\cdot 100$

where

Δ*U* = cable voltage drop, %

*U _{o }*= nominal line to neutral voltage, V

*R = c*able resistance, Ω

*X*= cable reactance, Ω

*I*= cable design current, A

_{b}b = circuit factor (=2 for d.c. and single-phase, =1 for three-phase)

Note: within myCableEngineering, we use complex arithmetic and the above is evaluated as:

$\Delta U=\frac{b\left(R+jX\right){I}_{b}}{{U}_{o}}\cdot 100$

Note: *R* and *X* are per line conductor. For example, the resistance of a single-phase two core circuit (live and neutral) would be 2*R* (assuming the live and neutral circuits are of equal resistance). For parallel cables, *R* and *X* are the per line resultant values. For example, *n _{ph}* cables in parallel, with the resistance of one line conductor is

*R*then the

_{c0ph}*R*=

*R*/

_{c0ph}*n*.

_{ph}The above equation), has the relative voltage drop expressed as a percentage of Uo. Multiplying by *U _{o}* will give the actual voltage drop for d.c. and single-phase circuits and for three-phase circuits, this needs to be multiplied by √3.

Adjusting the CENELEC equation to take into account referring the voltage drop to line-line voltage for three-phase systems, setting R, X in ohms, and using complex forms gives:

ΔU, V/m | ΔU, % per m | |
---|---|---|

d.c. systems | ||

a.c. systems, single-phase | ||

a.c systems, three-phase | ||

- the resistance of a single conductor, Ω/m |