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#9204 04/21/02 10:01 AM
Joined: Apr 2002
Posts: 4
S
Junior Member
I ran across a chart that showed the maximum length of circuits. It claimed a maximum length of 50ft for a 120v, 20A circuit on 12 gauge wire.

I have a 120v, 20A circuit that is approx. 85 feet long, on 12/2 wire. It is used ONLY for a window AC unit that draws 5.6 amps.

Does the amperage draw of the appliance factor in, or should this circuit have been run with 10/2 regardless of the amperage draw of the circuit?

Or does this have anything to do with the amperage draw? THe chart mentioned voltage drop. Does this mean the end of the circuit may be receiving <120v ?. Is this a dangerous situation? What kind of problems could I expect if any?

Thanks.

Joined: Aug 2001
Posts: 599
N
Member
For a given length of a circuit, voltage drop increases with an increase in amperage. Your circuit has a 1.5% voltage drop which is perfectly acceptable. The NEC advises limiting voltage drop to 3% on branch circuits. Voltage drop, however, is not an enforceable requirement in the NEC. (Except for a few non residential applications)
PS: The chart you were looking at was probably baised on the circuit drawing a full 20A.

Joined: Apr 2002
Posts: 4
S
Junior Member
Many thanks.... A couple of more questions

Do you mind sharing your math (how you came up with 1.5% voltage drop)?

Are there any drawbacks to this situation?

Would you say that my AC unit is happy with the 98.5% voltage it is running on?

What if I *were* using this circuit for items that were pulling 17-18 amps (near capacity)?

Thanks again!

Joined: Aug 2001
Posts: 599
N
Member
My math was a voltage drop calculator program that you can download from www.mikeholt.com. If you want to do it the hard way.....
The formula for single phase DC voltage drop is VD= 2KIL/CM
VD= Voltage drop
K= Specific resistance for type of conductor @ 75 Deg. C. If I remember correctly it's 10.4 for copper and higher for aluminum. About 17.
I = Current in amps
CM= Circular mil area of conductor

Notice I said this formula is for DC circuits and we are dealing with AC. This formula is pretty close for most applications. AC has an added components to factor in like the type of conduit the wire is in, single phase or three phase and has a resistance known as impedance that DC does not have. It gets much more complicated so the above formula is generally as accurate as we ever need to get.

No draw back to your circuit at all.

Your unit is probably thrilled to death. Equipment like ac units are designed to operate at a fairly large range of voltage. Usually about + or - 10%.

If you were pulling near capacity of this circuit it would be advisable to increase the conductor size. Like I said voltage drop in this situation is not code mandatory. Only advised for proper operation. However, it is good engineering practice.

Joined: Jan 2002
Posts: 324
A
Member
I came out at 1.83vd using exact K for 12 awg. Approx K for copper is 12.9



[This message has been edited by arseegee (edited 04-21-2002).]

Joined: Aug 2001
Posts: 599
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Member
The Zoltek program gets 1.904 V dropped. The EDR calculator gets 1.9 V dropped. Both are factoring in AC resistance, the fact that there is no conduit and 100% power factor. Since the air conditioner is an inductive load it's probably running about a 85-90% power factor and the voltage drop would be slightly lower.

Joined: May 2001
Posts: 160
C
Member
I would think that the voltage drop is the same even though the power factor is less than one indicating an inductive load.The real part of the impedance (resistance)is the same.
Chris

Joined: Apr 2002
Posts: 4
S
Junior Member
Thanks again guys...


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