Amps through an extension cable

[johnh]

So, almost every electrical text has a table showing what gauge of wire is needed to send X amps Y feet.
Ex: 12 amps down a 25' cord requires 16 AWG wire.
Anything smaller could introduce unnecessary volage drop and pose a fire hazard, or so I've been told.

My question is this:

Let's say (for example) that I'm running 10 amps 150'. The book says I need 10AWG wire. It also says that I need 12AWG for 100' and 14AWG for 50'. Instead of 10AWG at 150', can I run 100' feet of 12AWG followed by 50' of 14AWG? Is this to code? Can anyone explain the physics here?

Assume SOOW or equivalent and appropriate rated connectors.

Thanks for the help.

 

[DuckJordan]

So, almost every electrical text has a table showing what gauge of wire is needed to send X amps Y feet.
Ex: 12 amps down a 25' cord requires 16 AWG wire.
Anything smaller could introduce unnecessary volage drop and pose a fire hazard, or so I've been told.

My question is this:

Let's say (for example) that I'm running 10 amps 150'. The book says I need 10AWG wire. It also says that I need 12AWG for 100' and 14AWG for 50'. Instead of 10AWG at 150', can I run 100' feet of 12AWG followed by 50' of 14AWG? Is this to code? Can anyone explain the physics here?

Assume SOOW or equivalent and appropriate rated connectors.

Thanks for the help.

Unfortunately no, Think of it like a water system, The reason you need 10AWG for 10 amps at 150' is just like how water is delivered, to get that much through your going to need a larger pipe, taking one pipe that's slightly smaller, and another one the rest of the way to 150' but also adds to equal the diameter doesn't produce the same flow does it?

This is why we can't have tapering connections to our devices. Although someone else can probably explain it a lot more clearly, the simple answer is no, you still need the 10 gauge cable. One thing thats good to note is most of the time you don't need the full power to wherever your going. Just because an outlet can deliver 15 amps doesn't mean that fogger or hazer is using every bit of those 15 amps.

 

[johnh]

Unfortunately no, Think of it like a water system, The reason you need 10AWG for 10 amps at 150' is just like how water is delivered, to get that much through your going to need a larger pipe, taking one pipe that's slightly smaller, and another one the rest of the way to 150' but also adds to equal the diameter doesn't produce the same flow does it?

This is why we can't have tapering connections to our devices. Although someone else can probably explain it a lot more clearly, the simple answer is no, you still need the 10 gauge cable. One thing thats good to note is most of the time you don't need the full power to wherever your going. Just because an outlet can deliver 15 amps doesn't mean that fogger or hazer is using every bit of those 15 amps.

Thanks DuckJordan. I agree that you don't always need all 15 amps, but I'm imagining needing to power a lot of equipment a long ways away (you need all of it, hence the 10 gauge cable). This brings up a follow-up question:

The electrician who installed the wall receptacle should have used a similar formula calculating the distance from the service panel to the receptacle, factored in the current rating for the circuit, and run wires of the appropriate size. On a 15-amp residential, most likely 12AWG. Wouldn't this create a problem if I plug my 150' 10AWG extension in and try to draw 10 amps through it?

 

[DuckJordan]

Re: Amps through and extension cable

Thanks DuckJordan. I agree that you don't always need all 15 amps, but I'm imagining needing to power a lot of equipment a long ways away (you need all of it, hence the 10 gauge cable). This brings up a follow-up question:

The electrician who installed the wall receptacle should have used a similar formula calculating the distance from the service panel to the receptacle, factored in the current rating for the circuit, and run wires of the appropriate size. On a 15-amp residential, most likely 12AWG. Wouldn't this create a problem if I plug my 150' 10AWG extension in and try to draw 10 amps through it?

Yes it would, but thats why you have a multi-meter right? You would first find out what your getting for voltage at your specific outlet, then you'd find out just what you have for resistance on that circuit. then you'd take Volts divided by resistance, Then you have the exact amps at your disposal.

 

[johnh]

Re: Amps through and extension cable

Yes it would, but thats why you have a multi-meter right? You would first find out what your getting for voltage at your specific outlet, then you'd find out just what you have for resistance on that circuit. then you'd take Volts divided by resistance, Then you have the exact amps at your disposal.

So, track current at the outlet rather than on the circuit, and design the system/wire accordingly. Makes sense. Thanks.

 

[Chris15]

Re: Amps through and extension cable

I think it may be useful to think of voltage drop as a budget...

If you fix the current draw and the amount of voltage drop you can tolerate, then you can arrive at a fixed maximum resistance of cable. If I were designing building wiring to Australian code, that allowable voltage drop is 5% from point of supply to outlet. I think you will find that the same 5% is probably the value used in determining maximum length of a cordset.

Like any other budget, it's worth noting that in some applications 5% will be too much voltage drop and in others you might even be able to afford to go to say 10%...

So for our purposes, let's consider that length x resistance per unit length = resistance.
We'll call that resistance per unit length y. Say we have 2 pieces of cable make up the total. l1y1 + l2y2 = R.
Let's say that the total allowable resistance is 150.
150' at y=1 (10AWG) = 150
100' at y = 1.5 (12AWG) = 150
50' at y=3 (14AWG) = 150.

So if you took 100' of 12AWG and 50' of 14AWG, then you'd get 1.5*100 + 3*50 which gives you a total of 300, double the "budget"...

Note that I've simplified this down and linearised some things that in reality aren't very linear at all. But for the sake of this, it's within safety margins...

Yes it would, but thats why you have a multi-meter right? You would first find out what your getting for voltage at your specific outlet, then you'd find out just what you have for resistance on that circuit. then you'd take Volts divided by resistance, Then you have the exact amps at your disposal.

Can you explain this a bit more please?

 

[MNicolai]

Re: Amps through and extension cable

So, track current at the outlet rather than on the circuit, and design the system/wire accordingly. Makes sense. Thanks.

I don't have much time to explain right now, but that is not correct. Current will the same at all points in the circuit. Through the hot wire at the receptacle, at the device, and on the neutral return wire at the receptacle. If the cable is 500' long with only one device connected to it, current will be the same at all points. However, there is a minute resistance in the wire. For every x feet of wire, there will be y ohms of resistance. For 5 miles of stranded 10AWG copper wire, there's roughly 52 ohms of additional resistance in the circuit. So speaking in hypotheticals, if you created a 5mi long, 10AWG dead-short across the hot wire (120v) and the neutral wire (0v), for every mile of wire, voltage would drop 24v. So 1mi away down the wire, you're at (120v-24v=96v).

The practical application of this is that the smaller diameter of the wire, the more resistance-per-foot it will have. So a 10AWG cable drops far fewer volts in addition to the device (aka "the load") connected at the end of it than a 20AWG cable would. Running a 16AWG extension cable a hundred feet is therefore not a very good idea because the voltage drop at the other end of it may be significant enough to cause your device to malfunction.

For example, a 15A load at the other end of a 300' 14AWG copper cable turns out to be roughly 28v. Connect a device to the end of that, and you're not plugging it into a 120v source, you're connecting it to a 92v source. Many electronics designed for 120v will cease to function at 92v. Meanwhile, if you connect the device directly at the receptacle, or only 15' away from the receptacle, the voltage drop across the cable is negligible and you're giving your device a source voltage of roughly 120v still, therefore it will function as normal.

Another part of the physics is that the exact voltage drop will be proportional to the load current. A 5A load on a 200' cable will create a smaller voltage drop than a 15A load will. I'll leave this part of the physics for someone else to explain.

 

[shiben]

Re: Amps through and extension cable

So, track current at the outlet rather than on the circuit, and design the system/wire accordingly. Makes sense. Thanks.

Track the voltage at the outlet. Knowing the available current, say 15A, you use V=IR to figure out that if you get 115V at the outlet, you have 7.7 Ohms of resistance in the circuit.

I think your getting screwed up on Voltage, Current, and maybe even Power. All are different measurements. Voltage is a measure of potential difference and Electromotive force. It is defined based on the Amp, with V=W/A, which is why you can find power (Watts = W) by doing V*A. For further monkeying, you can take the square root of Power times Ohms to get voltage. Current is measured in Amperes, which is actually a defined unit (although still a SI base unit), meaning that 1 Coulomb of electrons pass a specific point in 1 second when there is 1 Amp of current. Power is a measurement of energy conversion, technically speaking defined as Joules/seconds, which incidently lets you figure out joules used in your system based on known power, voltage and time the circuit is open. This sort of stuff is now moving into "Consult a professor" areas, mainly because a AP or college level physics instructor is better at this kind of thing than me. However, all this to say, Watts, Voltage, Amps, are not interchangeable in discussing this sort of thing. Voltage drop is a funcion of mainly friction, incidently (the electrons bump into more non-copper atoms in a smaller wire meaning they provide less electromotive force, thus giving us voltage drop). This is also why overloaded wires heat up. Basically, consult your tables, know your load (if your going to have 1 S4 750 on your 150' cable, the load (amps drawn) at 120V is 6.25 amps. Now this obviously does not factor in voltage drop, so youll need to figure that out too, which increases current. I would imagine this entire equation that results ends up being some sort of nasty calculus, thus why we have tables telling us how big to make conductors of given lengths for a given load and voltage situation so as to avoid doing the math ourselves. But if it says you need 10 AWG copper, then get that. Dont try and monkey with it becasue your load situation is "unique".

 

[Dionysus]

FIRST: Generally may I first state that the use of long 'extension cables', especially for large loads it to be avoided whenever possible.

You have to deal with two general issues here:
1) Voltage Drop
2) Overheating of cable (simplified statement)

Also remember that electrical code does state that 'extension cables' cannot be used save for 'temporary' uses. Even with 'temporary installations', it is to be avoided.

Also in general it is far better to run out to some sort of distribution for 'temporary' situations where the distance is great. Especially if you will be running multiple circuits to the location. There really are a lot of factors.

Also in general your statement of 16AWG wire is to be avoided for a circuit with a 15amp OC Device. You should use 14AWG minimum, however 'lamp cords' and some small extension cables for HOUSEHOLD USE are 16AWG.

Basically any cable as a particular amount of impedance per foot, along with other factors. De-rating cable involves a number of calculations based on number of conductors, cable type, identified vs neutral, conductor size, ambient temperature, etc. I'd rather got go into these too deeply, sufficed to say if you run into the situation that you need to know it, you should contact a professional with the appropriate experience.

In a house the wire size is indeed determined by the same factors, and the smallest size used for line-voltage is 14AWG. In a house it is rare to need to step up to 12AWG, but I've seen it happen. You see in any situation it is INTENDED that you WILL NOT be attaching a long extension cord into a general-purpose receptacle. Just like you cannot draw 15amps on every receptacle in a house, it is intended that noone would need to do so in a residential occupancy. There are some 'special use' receptacles, like the ones in your kitchen that are either 15amp "splits" or 20amp "GFCIs". This is because people tend to use large appliances in the kitchen. The rest of the house has 15amp circuits with usually around 8-12 receptacles on a circuit.
Again special purpose receptacles are provided for the stove and dryer, or in a workshop usually again.

The receptacles in a residential occupancy are also by code required to be a certain distance apart to avoid the use of any 'extension cables'. In commercial and industrial occupancies you simply put circuits where they are expected to be needed. If you say add something that requires power somewhere in these places, you are required to have an electrician place a receptacle there (if there is not one already).

So back on track here...
You are dealing with the impedance (that is resistance, and any inductance/capacitance) along that big run of cable. The system is most likely to fail at the 'weakest link'. So if you plug in a 150' 10AWG extension cable running a large load into a 15amp circuit run off of 14awg cable, you are most likely to reach failure where you plug-into the 15amp receptacle. This is where the most heat will build up.
KABOOM.

This is why generally when you are doing a show a long way from any power you run large conductors, dedicated from the service (or sub-panel) to the location.

Now I have no idea why you asked this question in the first place. If it is for a particular event that you are doing, or for general inquiry and learning. If for something specific, if you want to be further advised, please contact an electrician. If for educational purposes I'd be happy to, and so would any of the electrical engineers I'm sure, write out much better.

I realize this is much more a rant than anything else. So please, any questions or comments or soforth? need clarification? I tried to dumb everything down, and I am not an electrical engineer. An electrician yes, in Canada. Our code-books are quite similar however.

 

[Chris15]

I didn't really want to get too deep into Electrical theory but it looks like we need to recap a few basics...

IF you have a load like a lamp or a bar heater or a hot water urn at the end of your cable, the current will DECREASE as the available voltage decreases. Particularly with non ohmic lamps this will be decidedly non linear.

IF you have a device powered by a switchmode power supply, say anything with an electronic ballast, a PC, whatever, the current will INCREASE as the available voltage decreases.

You CANNOT calculate the resistance (let's leave complex impedance out of this for now please guys) by just metering the open circuit voltage at an outlet. V(drop) = IR. When I is practically zero in the case of a meter, V(drop) is negligble and you have NO information about the circuit. You are trying to solve for 2 variables with only one equation. It CANNOT work. You need a second measurement, in theoretical calculations that would be the short circuit current. In practice that will trip breakers so you would need to SAFELY measure both voltage and current under a load condition...

To answer the original question, no what is proposed is NOT to any known code and is not exactly safe or reliable either.

 

[museav]

Instead of 10AWG at 150', can I run 100' feet of 12AWG followed by 50' of 14AWG? Is this to code? Can anyone explain the physics here?

Simple answer, it's still a 150' total run and that is what matters. Take your example to the next level with three 50' long, 14AWG cords, making the entire 150' 14AWG, and you can start to pretty easily see how that concept falls apart. Then add to that the potential practical and code issues associated with any daisy-chaining of extension cords.

 

[derekleffew]

Article on effects of voltage drop on incandescent fixture intensity and color temperature by TV lighting veteran Bill Klages: Commemorating The PAR Can Fixture, by Bill Klages .