I have a power supply, multimeter, and a copper wire with cross sectional area (A) A= 2.14*10^-3 m^2, Length L = 0.13 m and the resistivity of copper is p=1.7 × 10-8. The goal is to find the resistance of my copper wire, but I'm not sure what other instruments I need to determine the resistance, or how to setup the circuit to find out the resistance. After using the resistance equation, I found that the resistance should be about 1*10^-6 (ohms), but I cant get close to that answer.
How should the circuit be setup?
Electrical – measuring resistivity of copper wire
copperresistancewire
Related Solutions
The answer is
400+ mm^2 copper made up of 50+ wires per phase conductor x 3 conductors + 185 mm^2 copper made up from 30+ wires for the neutral conductor. See details below
If you are trying to implement this in practice and need to ask this question then your chances of dying if you try to use any answer you get is high.
If this is for an assignment then you are doing yourself a disservice by asking the question in this manner here rather than trying harder to find out for yourself by using the tools available.
Presumably you mean "If I have a 400V, 530 kW system, what conductor diameter do I need to carry the power 350 metres.
The answer depends on applicable regulations, load type, cable environment (buried, free air, ...) and more.
530 kW / 400V =~ 1,320A (not 132A mentioned elsewhere.)
This is the same as 3 x 230V phases at 530 kW / 230V / 3 phases =~ 550 A/phase.
This is a very serious level of power and amount of current by any measure.
You don't mention what sort of load you are using - if your load has a reactive component then your VA rating may be higher or much higher than your Watts and you may need larger cables again. There are many important and critical issues to answer when designing at this power level. If you do not use the services of an experienced professional then disaster is about certain.
Read this!:
This superb 70+ page document "Review of Power Cable Standard rating Methods" from a 2004 IEEE publication will provide you with a very solid grounding in what is required, and why.
Real world product - example only:
As a guide, the very largest cable on this useful page is possibly almost large enough for your load in some cases. The figures I use below may differ slightly from the table values. The results are close enough to provide a feel for what would be experienced.
Losses over 350m at 500A /phase would be about 5 kW/phase or 15 kW total. Or about 3% of input power.
Voltage drop per phase would be V=IR = 500A x 0.06 ohm = 10 Volt.
That's resistive drop. Note from the table that reactive resistance (reactance) is slightly higher than resistance for a cable this large, so that actual drop would be more like 15 Volts/phase.
The table below from here gives some sort of indication of what is required.
4.2 microns, not 4.2 microns squared. Area is not a unit of thickness. Divide your answer by pi, take the square root, and multiply by 2 to yield diameter
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Best Answer
As MKEITH said, copper has a temperature coefficient; its 0.4%, or 4,000 ppm, per degree Centigrade.
Copper, as with other materials has a thermal timeconstant (or, when inverted, a parameter named thermal-diffusivity).
For a cubic meter of copper, the time constant is 9,600 seconds (about 3 hours).
For 10cm cube (4" on a side), the time constant is 100X faster or 96 seconds.
For 1cm cube, the time constant is another 100X faster, or 0.96 seconds.
For 1mm cube, Tau is another 100X faster, or 0.0096 seconds.
Thus measurement has be made extremely quickly, or the temperature rise must be very small.