How Do I Work Out Volts Drop

[SystemQ]

Cqr state 92 Ohms per KM at 20 Celsius

You

[Guest]

Cqr state 92 Ohms per KM at 20 Celsius

You

[Guest]

It is the resistance of a pair that will give you the voltage drop.

no a single core give the same results using my method

compare the tables 100ma over 100M (7 / 0.2) there within a few % of each other

Your paired method must work by having twice the resistance but half the cable length?

Whether we would get the same results through out the table?? Don

[Truss and France]

Love it!

[SystemQ]

no a single core give the same results using my method

compare the tables 100ma over 100M (7 / 0.2) there within a few % of each other

Your paired method must work by having twice the resistance but half the cable length?

Whether we would get the same results through out the table?? Don

[Guest]

0,00178 * 1000 / ( 0,25 x 0,25 x pi ) = 9,065 ohms ( ! one core - end to end ! )

0,535 ohm difference most propably is caused by meter leads and two contact surfaces and meter inaccuracy..

edit: plus i expected CQR to be 0,5mm core diameter

and i seem to have decimal mistake somewhere - which propably is in copper resistance per metre @ 1mm2...

[esp-protocol]

Try this posted previously.

[Guest]

2LRI = Volt Drop

Where:

L is the length of cable in meters

R is the resistance per meter of the cable (see table above) in ohms

I is the maximum current in Amps

(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)

i notice some similarities...

drop = A x cable lenght x 0.08 x 2

but only now i got this.. was wondering for a long time from where did you get that 0.08 and A (was expecting I for amperes..)