What would the voltage drop be at the output of a 6/4 cable 50 feet long, with a 124V, 60 cycle, AC input, if 10 feet are laid out straight and 40 feet are coiled on a 16 inch diameter cord reel?
I think inductance enters the calculation, but don't know where to begin with this.

Thanks, JIM

 

With an air core, I wouldn't think you wouldn't get a huge amount of change from the wire. And I would think it would help keep the voltage from dropping as much. Alas, I would like to see the math on this one.

 

Hmmm!!! funny q?
If there is a load at the end then what is it
and if you were to give 124VAC to both ends and say perhaps you connected the return in reverse then I'd say BAAANG!!
No time for calculation. I guess!!!.

 

You need to know the RMS current in order to calculate the voltage drop.
If you know the wattage or VA rating of the load, divide it by the voltage to obtain the current.
AWG 4 copper wire has a resistance of 0.2533 Ohms per 1,000 feet, or 0.2533mOhms per foot.
AWG 6 copper wire has a resistance of 0.4023 Ohms per 1,000 feet, or 0.4023mOhms per foot.
If you have an extension cord 50 feet long, then you actually have 100 feet of wire that the current must travel through. So:
If the cord is AWG 4, it will have a total resistance of 2.533mOhms (0.002533 Ohms); so a 10A current will drop 253.3mV
If the cord is AWG 6, it will have a total resistance of 4.023mOhms (0.004023 Ohms); so a 10A current will drop 402.3mV.

The inductance of the coil is so low (roughly 70uH; figuring a 1" wide spool and about 9.3 turns on a 16" form) that it would have very little effect on the current at that frequency.

 

You need to know the RMS current in order to calculate the voltage drop.
If you know the wattage or VA rating of the load, divide it by the voltage to obtain the current.
AWG 4 copper wire has a resistance of 0.4023 Ohms per 1,000 feet, or 0.4023mOhms per foot.
AWG 6 copper wire has a resistance of 0.2533 Ohms per 1,000 feet, or 0.2533mOhms per foot.
If you have an extension cord 50 feet long, then you actually have 100 feet of wire that the current must travel through. So:
If the cord is AWG 4, it will have a total resistance of 4.023mOhms (0.004023 Ohms); so a 10A current will drop 402.3mV
If the cord is AWG 6, it will have a total resistance of 2.533mOhms (0.002533 Ohms); so a 10A current will drop 253.3mV.

The inductance of the coil is so low (roughly 70uH; figuring a 1" wide spool and about 9.3 turns on a 16" form) that it would have very little effect on the current at that frequency.

Sorry I didn't give enough or complete information up front.
The cable is four strands of AWG6 and is known as a 50A cord, but in reality it is two circuits of 124V each on it's own 50A circuit breaker.
The load is variable from 0 to 40A.
The cord is wound in a housing 16"x16" in a spiral fashion starting at the bottom and spiraling upward loop upon loop until the cord exits in the straight 10 feet to the plug.

What I want to know is what is the voltage drop from 124V at 30A.

The question is inspired by the fact that at times the supply voltage is marginal say 105V and I am concerned that with an additional voltage drop from the cord that I might wind up with an excessively low voltage that would be damaging to the equipment I am running.

If you think it is foolish of me to pose the question, say so and I will understand.

Thanks for your help so far, JIM

 

Ack - just realized I swapped the resistance values for AWG 6 and AWG 4.

OK, one 50 foot length of AWG 6 will have roughly 20.1mOhms (0.0201 Ohms) of resistance.
Coincidentally, that's the same resistance that two AWG 6 wires with a total length of 100 feet will have, or four 50-foot lengths (two for hot, two for neutral).

30A x 0.0201 Ohms = 0.603v (603mV) drop for both wires with a total of 30A current, or 15A per wire. If you had a 50A load, you would have a 1.005v drop.

Now if you are just talking one breaker's load (meaning a single hot line and a single neutral line) then the wire total resistance (2x50', or 1x100') will be 4.03mOhms, so with a 30A load, you will have a voltage drop of 1.206v. With a 50A load, you will drop 2.01v across the wire.

Does that make sense to you?

 

Ack - just realized I swapped the resistance values for AWG 6 and AWG 4.

OK, one 50 foot length of AWG 6 will have roughly 20.1mOhms (0.0201 Ohms) of resistance.
Coincidentally, that's the same resistance that two AWG 6 wires with a total length of 100 feet will have, or four 50-foot lengths (two for hot, two for neutral).

30A x 0.0201 Ohms = 0.603v (603mV) drop for both wires with a total of 30A current, or 15A per wire. If you had a 50A load, you would have a 1.005v drop.

Now if you are just talking one breaker's load (meaning a single hot line and a single neutral line) then the wire total resistance (2x50', or 1x100') will be 4.03mOhms, so with a 30A load, you will have a voltage drop of 1.206v. With a 50A load, you will drop 2.01v across the wire.

Does that make sense to you?

That all does make sence, but what do you think about the inductance?
Are you still of the opinion that it is negligible and therefore inconsequential?

JIM

 

Inductance can severely make a difference..... However, is it a cable or a conductor that you may be referring to?........

 

Also remember that as your voltage decreases, the amperage increases. So your 30 amp load at 120 volts is now 34.3 amps at 105 volts.

 

That all does make sense, but what do you think about the inductance?
Are you still of the opinion that it is negligible and therefore inconsequential?

JIM

If you have the four wires coiled together in a spool that's 16" in diameter for 40 feet, which is about 9.3 turns per wire, each wire will have an inductance of around 70uH. Inductance in series adds, inductance in parallel subtracts. So, the net result is that all four wires wind up having a net inductance of around 70uH.

The inductance of a motor will be so high in comparison (perhaps 9mH) that the inductance of the wiring represents about a 3% phase shift, but not necessarily a loss of power.

 

As you are winding the complete cable (containing both legs of a balanced circuit), the inductive effect of the coiling is zero.

The actual problem with coiling is heat build-up - a cable running at a current that causes little heating in free air can melt and even catch fire if several layers are left coiled on a drum, as the plastic insulation is also pretty good thermal insulation.

 

Thanks all for your insight. I now have sufficient information.

JIM