Discuss Problem: Voltage Drop...? in the Periodic Inspection Reporting & Certification area at ElectriciansForums.net

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lukas86

I was asked to solve this and I was wondering if I was going about solving it in the proper manner so wanted some info and hoping I could get some help.

There is a voltage of 120V, and resistance of 16 amps at the start of the ground connected to the breaker. The wire goes into the ground and goes for 600ft. The wire is also 1 ott in size, which I'm thinking is around 10mm-15mm in diameter. I'm not sure what exactly that is in mm or anything like that, I still have to look that up. What is the voltage drop over the length of the wire?

Thanks,
Luke
 
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Luke, who asked you this question?

Resistance of 16 AMPS!:confused:
ground connected to the breaker:confused:

If the cable is 600ft long, volt drop will probably be high but still :confused:

Is it a wind up?
 
Jimmus

I think you will find that this gent is one of our former colonial cousins(USA) with 120v and ground being mentioned.
 
Hey,

Canada ya. The high current mentioned is apparently right, but the voltage drop shouldn't be 6.7V, it should be closer to 1V.
 
I'm not too sure, I don't believe so no. I have done a few calculations, with the current being 16 Amps, to get a voltage drop of 6.7 volts, the resistance would have to be 0.4188 ohms. But in this case, the resistance of the cable is approximately 0.0294 ohms, and with this resistance, the voltage drop should only be 0.470 so I don't know. Maybe I'll grab an electrician to come down with me and show me where exactly the measurements were taken from. Check all the connections again.

Also, the 4/0 awg wire, has 19 strands of 12 gauge wire. Now the resistance compared to a 4/0 wire of solid copper will be less or more? I'm thinking less and there's very little different, but would just like to know as well.

Thanks
 
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Yeah im confused too are you saying voltage =120volts Current =16Amps length of cable = 600ft = 600*12*25.4 =182880mm = 182.88m resistance of cable =pl / a where p=resistance of Copper l=length of run a=CSA of conductor and using V=IR? What was that about ground? why not just type the whole question as it was posed. All the UK tables are metric and hence the resistance of Cu vairable that we use is based upon this scale. i.e. R of 1 metre of Cu is different to 1ft of Cu even if CSA the same. Ignoring reactance the resistance equation should work but you would have to use your own tables for the specific resistance of copper per unit length. Stranded wire has higher resistance than solid as lower usable CSA. Also ask whomever posed this question whether the borg effect is being taken into account as this will make the resistance futile
 
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