a1steve Posted June 20, 2006 Share Posted June 20, 2006 please can any one help me how do i work out volts drop in a cable? Link to comment Share on other sites More sharing options...
Guest Posted June 20, 2006 Share Posted June 20, 2006 drop = A x cable lenght x 0.08 x 2 Link to comment Share on other sites More sharing options...
Guest Posted June 20, 2006 Share Posted June 20, 2006 For AC, voltage drop per metre x cable length x current. Multiply the answer by 2 for DC circuits. Link to comment Share on other sites More sharing options...
SystemQ Posted June 20, 2006 Share Posted June 20, 2006 Have a look at this tip How_to_combat_voltage_drop.pdf System Q Ltd. Link to comment Share on other sites More sharing options...
Guest Posted June 20, 2006 Share Posted June 20, 2006 Your table looks like its wrong!!! 7/ 0.2 alarm cable is 8 ohms per meter volts_drop.bmp Link to comment Share on other sites More sharing options...
SystemQ Posted June 20, 2006 Share Posted June 20, 2006 Your table looks like its wrong!!!7/ 0.2 alarm cable is 8 ohms per meter System Q Ltd. Link to comment Share on other sites More sharing options...
Guest Posted June 20, 2006 Share Posted June 20, 2006 resistivity of copper cable is 0.0178 m ohms per one metre when A of the cable is 1mm2 formula to count voltage drop is simply: (Rcore) x (l x 2) x I where: Rcore = resistivity of the cable per one metre l = length of the cable 2 = 2 I = current ( U/R ) or (V/R there at the island) Link to comment Share on other sites More sharing options...
SystemQ Posted June 20, 2006 Share Posted June 20, 2006 Your table looks like its wrong!!!7/ 0.2 alarm cable is 8 ohms per meter System Q Ltd. Link to comment Share on other sites More sharing options...
Guest Posted June 20, 2006 Share Posted June 20, 2006 1mm2 solid --> 0,00178 ohms / m x 200 m = 0,356 ohms 0,356ohms x 0,1A = 0,0356 volts For AC, voltage drop per metre x cable length x current.Multiply the answer by 2 for DC circuits. announced - efficient -AC voltage is comparable with DC voltage. actual peak to peak voltage in 230VAC circuit is thou square2 x U --> 325,27V but that is irrelevant when calculating voltage drop in the cable.. efficient AC voltage was determined so that it generates as much heat as equivalent DC voltage when terminated to one ohm pure R.. Link to comment Share on other sites More sharing options...
Guest Posted June 20, 2006 Share Posted June 20, 2006 These are measure readings from our workshop not a book.The orginal cable tested was either CQR or Ventcroft I'm not sure which. Cqr state 92 Ohms per KM at 20 Celsius You Link to comment Share on other sites More sharing options...
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