What gauge wire/cord do I need for these outdoor lamps?

I have two 700 watt outdoor halogen lamps, each connected to a splitter, then back to a power source. Given the distances illustrated below, what gauge cord should I use? migrated from electronics.stackexchange.comSep 15 '14 at 21:21

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To determine what gauge wire you need, you need to know two things: How much current is needed, and what current capacity (ampacity) corresponds to different wire gauges.

Ampacity

You can find ampacity charts online (for example). This chart also shows resistance. There's another factor you must consider, which is voltage drop. Long lengths of wire will have an associated resistance (because copper is not a superconductor), so you will need to take into account what that resistance is (perhaps even using a larger wire to accommodate it if necessary). This resistance means that the load won't receive the full voltage supplied at the other end of the cord; this is also known as the "voltage drop".

Ohm's Law

So, let's first calculate what current will be needed. Ohm's law allows you to determine voltage (volts), current (amperes), resistance (ohms), or power (watts) by knowing any two values. The splitter will connect both lamps in parallel, giving each 120 volts. You know that they are 700 watts, so you can find current by using the formula I = P/E.

I = 700 / 120 = 5.83 A

The ampacity chart shows that you would need at least 14 AWG (for power transmission) for the connections between your splitter and the lamps.

The current from the splitter to the source would of course be double (11.7 A), so you would need at least 10 AWG wire (11 AWG is uncommon, so I picked the next larger size).

The Lamp Runs

In the run of 14 AWG wire, you would have an additional resistance of 2 * 2.525Ω/1000 or 0.51Ω. (Remember the length of wire is actually double; one for line and one for neutral.) You can calculate the voltage drop of the wire by treating it like a circuit where the lamp is one resistor and the wire is another, then use Ohm's law to determine the voltage on both resistors. The lamp's resistance is (R = E^2 / P):

R = 120^2 / 700 = 20.5Ω

The total resistance (lamp plus wire) is:

20.5 + 0.51 = 21Ω

Now the total current with 120V applied is (I = E/R):

I = 120 / 21 = 5.7 A

This is less than the original current (5.83 A) because with more resistance (the long wire), less current can flow. You can also determine the voltage drop and how much power the wire itself is dissipating (E = R * I, P = R * I^2): 2.9 volts dropped, 16.6 watts dissipated. (This isn't a lot (less than 3% of the total voltage) so you could just use 14 AWG for these runs. "Upgrading" to thicker wire would present slightly less resistance, but the benefit would not outweigh the added cost of more expensive wire.)

The Home Run

Doing the math on the "home run" depends on what was selected for the runs to each lamp. We'll assume that you stuck with the 14 AWG for now, so the total current is 5.7 * 2 = 11.4 A. (We can treat the splitter and downstream wires and lamps as a load of 1368 watts, or a resistance of 10.5Ω.)

Total resistance using 10 AWG wire (and remembering to double the run length) for the home run:

10.5 + 0.2 = 10.7Ω

The total current would be:

120 / 10.7 = 11.2 A

Why does the current seem to be lowering each time we calculate things? Because the wire resistance limits how much current can flow, just like a resistor in a circuit. Two 700 watt lamps with superconducting cables would actually pull 5.83 A each, or 11.7 A in total. With the extra length of wires creating resistance, the whole configuration pulls 0.5 A less.

Because 10 AWG is rated for up to 15 A power transmission, it is sufficient.

If you were to use a cheap 16 AWG extension cord for the home run, rated for only 3.7 A, you would run into trouble.

• That chart above is VERY misleading. Can you explain what they mean by "chassis wiring" and "power transmission"? Folks should never go by made up charts like that. The NEC is the place to find accurate information regarding this. Also, just having the conductor size is not enough. You need to factor in the insulation type. – Speedy Petey Sep 16 '14 at 0:09
• @Speedy A lot of ampacity charts show values like the one I linked. "Chassis wiring" is meant for wiring in air, while "power transmission" is for wiring in bundles (ref), as you would find in a typical extension cord. Insulation type doesn't change how much current a given wire can carry directly, but matters because a wire carrying high currents at high temperature can burn or melt insulation that isn't rated for it. – JYelton Sep 16 '14 at 1:04
• Well, then a lot of ampacity charts are wrong for this application. You need to stick with AC building wiring and not some electronics or other site. Like I said, the NEC is the only chart you need. Also, the insulation absolutely DOES absolutely matter. Have you even seen a real ampacity chart form the NEC? The insulation temperature and voltage rating determine what column you use to figure ampacity. Also, your terms are not standard AC building wiring terms. Again, if you are going to give advice on this type of wiring do not confuse things with erroneous terms and charts. – Speedy Petey Sep 16 '14 at 3:27
• @Speedy I didn't say insulation does not matter, I said it doesn't change how much current a given wire can carry. The more current a wire carries, the higher the temperature will be, and thus you need to have an appropriate type of insulation. Different charts provide "safe" ampacities based on ambient temperature and insulation material. Yes, I have seen (and worked with) NEC information. What terms do you find erroneous or not standard? – JYelton Sep 16 '14 at 3:33
• It's great that you've seen and worked with NEC information. You need to stick to that when giving advice on AC building wiring. You need to separate your field and expertise from this type of advice since there is very little correlation. Terms like chassis wiring and power distribution, in the way you use them, are not common terms in this context and only serve to confuse. Also, I'll reiterate about the charts and links you provided, they are simply NOT relevant to this conversation. – Speedy Petey Sep 16 '14 at 11:06

Here is the ampacity chart from the NEC.

Table 400.5(A)

The thing to take into account here is voltage drop. 200' is a LONG way to go for this kind of load. Personally I would not go with less than 12ga cords. Remembering that the VD may be excessive at the end. A #10ga cord to the "splitter" would be the best bet.

• The ampacities given by NEC are less conservative than the table I used. For 10 AWG copper wire with a 60°C rating, the value given is 30 A (compared to 15 A). There are many factors involved, but I would hesitate to use a 10 AWG extension cord for a 30 A load. Also, I calculated the voltage drop for a 100' length of 14 AWG extension cord (200' total) to be < 3V, which I wouldn't necessarily deem excessive. – JYelton Sep 16 '14 at 1:12
• (I should have added that the voltage drop was at the ~5.7A load of the lamp. It would be more with a higher wattage load.) – JYelton Sep 16 '14 at 1:29
• Thank you, between your answer and JYelton's, I've got my info. – raffian Sep 16 '14 at 13:14

Using 14AWG for 100ft is 0.25R and two 100ft feeders of 16WG is 0.40R

Neglecting surge startup, if each hot lamp is 700W at 120V , R=21 Ohms thus two lamps ~10 Ohms

Losses are proportional to R (P=I^2Rtot) and 5% distribution losses are acceptable . The 14awg leg is 0.25/10 or 2.5% loss and the AWG16 loss is 0.4/21=2% , adds to <5% for acceptable loss .