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Voltage Drop

Last updated on 2020-02-27 4 mins. to read

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:

$Δ U = \frac{b (R \cos φ + X \sin φ) I_{b}}{U_{o}} \cdot 100$

where
        ΔU         = cable voltage drop, %
        U_o= nominal line to neutral voltage, V
        R           = cable resistance, Ω
        X           = cable reactance, Ω
        I_b           = 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:

$Δ U = \frac{b (R + j X) I_{b}}{U_{o}} \cdot 100$

Note: R and X are per line conductor. For example, the resistance of a single-phase two core circuit (live and neutral) would be 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, n_ph cables in parallel, with the resistance of one line conductor is R_c0ph then the 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		$<math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mn>100</mn><mi>Δ</mi><mi>U</mi></mrow><msub><mi>U</mi><mi>o</mi></msub></mfrac></math>$
a.c. systems, single-phase		$<math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mn>100</mn><mi>Δ</mi><mi>U</mi></mrow><msub><mi>U</mi><mi>o</mi></msub></mfrac></math>$
a.c systems, three-phase		$<math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mn>100</mn><mi>Δ</mi><mi>U</mi></mrow><msub><mi>U</mi><mi>n</mi></msub></mfrac></math>$
- 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

BS 7671 Voltage Drop Tables

BS7671 "Requirements for Electrical Installations", the IET Wiring Regulations, Appendix 4 voltage drop tables, values are given in mV/A/m (or effectively mΩ). However, these tables are related to the line to neutral voltage for single-phase circuits and line to line voltage for three-phase circuits.

To convert the three-phase table values so that they relate to the input resistance (or reactance) required by CENELEC 50480 it is necessary to divide the BS 7671 three-phase values by √3 (the square root of three). No adjustment is required for single phase values. Given values of voltage drop for three-phase balanced systems are related to the line voltage.

Note: the values in the tables are given at the maximum conductor operating temperature. myCableEngineering uses circuit theory (see ) to calculate impedance and voltage drops are calculated using CENELEC 50480. Our testing shows that the difference in resistance calculation by circuit theory and that given in BS 7671 is small.

Subtopics

BS 7671 Voltage Drop

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