please can any one help me how do i work out volts drop in a cable?

drop = A x cable lenght x 0.08 x 2

For AC, voltage drop per metre x cable length x current.

Multiply the answer by 2 for DC circuits.

Have a look at this tip

How_to_combat_voltage_drop.pdf

Your table looks like its wrong!!!

7/ 0.2 alarm cable is 8 ohms per meter

volts_drop.bmp

Your table looks like its wrong!!!

7/ 0.2 alarm cable is 8 ohms per meter

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)

Edited June 20, 200619 yr by georgahti

Your table looks like its wrong!!!

7/ 0.2 alarm cable is 8 ohms per meter

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..

Edited June 20, 200619 yr by georgahti

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