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Topic Title: voltage drop
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Created On: 23 September 2011 08:57 PM
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 23 September 2011 08:57 PM
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jonah71

Posts: 41
Joined: 05 March 2006

A circuit is wired in 2.5mm pvc twin and earth cable and is 25 metres long, the current is 18 amps and the resistance is 0.37 ohms
Volt drop = I X R = V
18 x 0.37 = 6.66 volts
Where does the 0.37 come from ?
Volt drop for 2.5 is 18 mV/a/m
So i would work it out at 18 x 18 x 25 = 8.1
1000
Am I missing something ?
 24 September 2011 12:27 AM
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eswnl

Posts: 144
Joined: 29 November 2008

Can you repost to say:

What is the question, what variables are given and what is it you are asked to calculate?.
 24 September 2011 11:25 AM
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Avatar for OMS.
OMS

Posts: 19900
Joined: 23 March 2004

The "volt drop" in mV/A/m is 18 at 70 C.

The cable has probably been measured at 20C - ie room temp

So:

Cable resistance at 70C = 18mV/A/m x 25m = 0.45 Ohm

Correction for the resistance measurement temperature from 70C to say 20C would be about 0.8

So - 0.45 x 0.8 = 0.36 Ohm (close enough to the stated 0.35 ohm)

Poor question really, because with 18A flowing through a 2.5mm2 cable it would be at approx 40C

Does that help at all

Regards

OMS

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