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A {L}  m long, {S} mm2 cable with {p1Label}  conductors is carrying a current of {Ib} A. The supply voltage is {Vsupply} V. 

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The voltage drop is [[0]] V.

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Ib = {Ib} A

\n

S = {S} mm2

\n

L = {L} m

\n

p1 = {p1} Ωmm2/m 

\n

$\\Delta V= 2 (L)(Ib)((p1 * 0.8/S)+0.000048) $

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The maximum length this cable this cable can be to comply with 5% voltage drop limit is  [[0]]m.

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The max allowed voltage drop, $\\Delta Vmax$, is 5% of the supply, 0.05 of 230V = 11.5 V

\n

We must find the cable length, $Lmax$ that has this voltage drop.

\n

Rearrange $\\Delta Vmax= 2 (Lmax)(Ib)((p1 * 0.8/S)+0.000048) $ to get ...

\n

$Lmax= \\Delta Vmax/( 2(Ib)((p1 * 0.8/S)+0.000048) )$

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Ib = {Ib} A

\n

S = {S} mm2

\n

p1 = {p1} Ωmm2/m 

\n

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