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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:
ΔU= b( Rcosφ+Xsinφ ) I b U o ⋅100
ΔU = cable voltage drop, %
Uo = nominal line to neutral voltage, V
R = cable resistance, Ω
X = cable reactance, Ω
Ib = cable design current, A
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:
ΔU= b( R+jX ) I b U o ⋅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 2R (assuming the live and neutral circuits are of equal resistance). For parallel cables, R and X are the per line resultant values. For example, nph cables in parallel, with the resistance of one line conductor is Rc0ph then the R = Rc0ph/nph.
The above equation), has the relative voltage drop expressed as a percentage of Uo. Multiplying by Uo 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:
- the resistance of a single conductor, Ω/m
- the reaction of a single conductor, Ω/m
- the impedance of single conductor , Ω/m
- cable design current, A
- the nominal line to neutral/earth voltage (for single-phase a.c. or d.c.), V
- the nominal line to line voltage (for three-phase systems), V