I was just looking for some insight on this problem. It is not a homework problem, my boss just asked me to figure it out.(adsbygoogle = window.adsbygoogle || []).push({});

Now, there is 120V supply, with 600 feet of cable (copper) running to a load, and 600 ft neutral coming back to the supply. Ignoring the resistance of the load, he asked what was the voltage at the load due to the voltage drop in the cables with a current of 12A.

Now, he wanted to know the values for #12 wire, and #8 wire. Below is what I came to...

VD = 1.732*k*Q*I*D/CM (equation from... h t t p : / / ecmweb.com/mag/electric_code_calculations_17/)

VD: Voltage drop

k: Direct current constant (12.9 for copper)

Q: Alternating current adjustment factor (skin effect), I thought it doesn't apply here...

I: Current (I think, it said "The load in amperes at 100%, not 125% for motors or continuous loads." I just put the 12A here)

D: Distance in ft

CM: Cable in Circ-Mils

#12 Wire

VD = 1.732*k*Q*I*D/CM

= 1.732*12.9 * 12A * 600ft / 6529.8CM

= 24.636V

Now, that is what I thought the voltage drop in the 600ft cable to the load was. Now I am confused because at first I thought that since there is also 600ft of cable back to the source, the voltage across the load would be 70.727V seeing as 120V - 24.636V (cable to load) - 24.636V (cable back to source) = 70.727V across the load.

Then I thought again, well... in 3 phase, if perfectly balanced, there is no current in the neutral lead back to the source. So the voltage across the load would be around 95V.

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For #8 Wire

VD = 1.732*k*Q*I*D/CM

= 1.732*12.9 * 12A * 600ft / 16509CM

= 9.744V

Now then the same situation as for the #12 wire, if I am doing any of this correctly... should be something like 120V - 9.744V *2 = 100.511V across the load or... 120V - 9.744V = 110.255V.

Any help or requests for further understanding so I could respond to get more help would be greatly appreciated. Thanks in advance.

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# Voltage Drop (Three phase)

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