RV extension cord max length

[Stevenfyeager]

For an SO 50 amp RV extension cord, do you know a maximum length before having significant voltage drop? I can use a voltage drop calculator like I usually use, but I know SO cords are rated differently because of the heavy insulation. A customer wants to use probably 80’. I’m not even sure how long of cords are available. Thanks

 

[mayanees]

For an SO 50 amp RV extension cord, do you know a maximum length before having significant voltage drop? I can use a voltage drop calculator like I usually use, but I know SO cords are rated differently because of the heavy insulation. A customer wants to use probably 80’. I’m not even sure how long of cords are available. Thanks

An old rule of thumb that I heard years ago was that VD is only a consideration if you exceed the length based on the magnitude of the voltage, and it tends to hold true. The max length for a 240 Volt circuit would be 240 feet before consideration should be given to voltage drop.

 

[LarryFine]

Keep in mind that voltage drop is based solely on current and circuit resistance.

The lower the system voltage, the more detrimental a given voltage drop becomes.

 

[petersonra]

For an SO 50 amp RV extension cord, do you know a maximum length before having significant voltage drop? I can use a voltage drop calculator like I usually use, but I know SO cords are rated differently because of the heavy insulation. A customer wants to use probably 80’. I’m not even sure how long of cords are available. Thanks

The insulation does not change the VD. That is determined by the size of the conductors inside the insulation. 10 AWG is 10 AWG regardless of what the insulation is. The thing that matters the most though is how much current you have flowing through those conductors. At zero current the VD is zero. The VD goes up as the current goes up, and it is not linear. It is related to the square of the current. So you might have 5 volts drop at ten Amps, but 20 Volts drop at 20 Amps.

 

[DrSparks]

An old rule of thumb that I heard years ago was that VD is only a consideration if you exceed the length based on the magnitude of the voltage, and it tends to hold true. The max length for a 240 Volt circuit would be 240 feet before consideration should be given to voltage drop.

I like that

Sent from my BE2028 using Tapatalk

 

[Todd0x1]

An old rule of thumb that I heard years ago was that VD is only a consideration if you exceed the length based on the magnitude of the voltage, and it tends to hold true. The max length for a 240 Volt circuit would be 240 feet before consideration should be given to voltage drop.

How does that work? 6ga 50 amps 240v = 162ft for 3%.

 

[Todd0x1]

For an SO 50 amp RV extension cord, do you know a maximum length before having significant voltage drop? I can use a voltage drop calculator like I usually use, but I know SO cords are rated differently because of the heavy insulation. A customer wants to use probably 80’. I’m not even sure how long of cords are available. Thanks

Depends on the load. Is the RV all electric heat? How many feet of branch circuit wire are there between the panel and receptacle? More to it than just cord length.

 

[hillbilly1]

Keep in mind, most RV loads are 120 volts, so using 240 volts would be assuming a reasonably balanced load.

 

[kwired]

Keep in mind, most RV loads are 120 volts, so using 240 volts would be assuming a reasonably balanced load.

True. The AC units are often the biggest load, they are often about same unit installed in a RV with only 30 amp 120V supply, there just happens to be two of them in a RV with a 50 amp supply.

 

[ggunn]

The VD goes up as the current goes up, and it is not linear. It is related to the square of the current. So you might have 5 volts drop at ten Amps, but 20 Volts drop at 20 Amps.

I don't think so. Power lost to resistance heating goes up as the square of the current (P=I^2R), but voltage drop is linear with respect to current (Vd=IR).

 

[Hv&Lv]

I don't think so. Power lost to resistance heating goes up as the square of the current (P=I^2R), but voltage drop is linear with respect to current (Vd=IR).

?

 

[ggunn]

?

View attachment 2559192

I don't know what that graph is showing but I do a lot of voltage drop calculations. For a 2 wire system Vd = IR = 2DIr where D is the distance in feet, I is the current in Amperes, and r is the wire resistance in ohms per 1000 feet. Ohm's Law.

 

[petersonra]

I don't think so. Power lost to resistance heating goes up as the square of the current (P=I^2R), but voltage drop is linear with respect to current (Vd=IR).

U r right

 

[Hv&Lv]

I don't know what that graph is showing but I do a lot of voltage drop calculations. For a 2 wire system Vd = IR = 2DIr where D is the distance in feet, I is the current in Amperes, and r is the wire resistance in ohms per 1000 feet. Ohm's Law.

I do too…

Trying to figure out what that graph is showing also..

 

[GeorgeB]

The VD goes up as the current goes up, and it is not linear. It is related to the square of the current. So you might have 5 volts drop at ten Amps, but 20 Volts drop at 20 Amps.

you're confused ... e=i*r, voltage drop equals amps current times ohms resistance; it is linear.

 

[ggunn]

I do too…

Trying to figure out what that graph is showing also..

It's not labeled well but the exponential character of the lines leads me to surmise that it has something to do with power. Of course that's just a guess.

 

[wwhitney]

The exponential character of the lines

I think you mean power law (y = x^a for a constant a), not exponential (y = a^x), an exponential curve would look different.

Anyway, if you look at a few points on the curve, it's clear that each curve is of the shape y = 1/x. So the graph is showing, for a given size and length of wire, the current that gives some fixed voltage drop.

Cheers, Wayne

 

[ggunn]

I think you mean power law (y = x^a for a constant a), not exponential (y = a^x), an exponential curve would look different.

Anyway, if you look at a few points on the curve, it's clear that each curve is of the shape y = 1/x. So the graph is showing, for a given size and length of wire, the current that gives some fixed voltage drop.

Cheers, Wayne

1/x = x^-1. Seems to me that it's exponential and the exponent is -1. But mathematical rigor is not my forte. ;^)

 

[winnie]

Voltage drop goes up as current goes up, and is _linear_ if resistance remains constant.

Power loss goes up as current goes up, and is proportional to the _square_ of current if resistance remains constant.

For practical voltage drop calculations we assume resistance is constant, however a detailed calculation of voltage drop includes the change in resistance when the conductor heats up. I'd expect voltage drop in an SO cord to be minutely different than voltage drop in THHN/conduit of the same wire gauge, just because of thermal conductivity differences. I also wouldn't worry about this difference for any cables in spaces where humans could walk.

For the OP, a 6 AWG cord used at 50A and 240V lets you go 162 feet with a 3% voltage drop, per the Southwire calc which assumes copper at 75C.

-Jon

 

[wwhitney]

1/x = x^-1. Seems to me that it's exponential and the exponent is -1. But mathematical rigor is not my forte. ;^)

As I mentioned, that's called a power law, where the variable x is in the base of the exponentiation (x², etc). An exponential law is where the variable x is in the exponent (e^x, etc.).

Cheers, Wayne