# What is the maximum 12/2 wire length from a 20A breaker to a baseboard heater?

What is the maximum distance from a 220V/20amp circuit breaker to an 2000 watt electric baseboard heater using 12/2? At what distance would it require 10/2?

Using the values from Chapter 9 Table 8 of the National Electrical Code, you can calculate the maximum circuit distance as follows.

# Find allowable voltage drop

The NEC recommends a maximum 3% voltage drop at the "farthest outlet of power, heating, and lighting loads, or combinations of such loads" (210.19(A) FPN No. 4). So you'll want to start by calculating the maximum voltage drop as follows:

`Max Voltage Drop = Voltage * 3%`
`Max Voltage Drop = 240V * 0.03`
`Max Voltage Drop = 7.2V`

# Calculate the maximum length

To calculate the maximum length of the circuit, you'll first have to know how much resistance the conductor being used has. Table 8 from Chapter 9 of the NEC, makes this simple. If you're using 12 AWG solid uncoated copper, you'll find that it has a resistance of 1.93 ohms per thousand feet.

For the following formula, you'll need to know the resistance per foot. This is easily calculated, by dividing the value above by 1000.

`1.93 ohms per 1000 feet / 1000 = 0.00193 ohms per foot`

Next you'll need to know the amount of current flowing along the conductor. Since you didn't specify this in the question, I'll use 20 amperes as an example.

## The formula

`Length = maximum voltage drop / ( 2 * Ohms per foot * current)`
`Length = 7.2 V / ( 2 * 0.00193 Ohms * 20 Amperes)`
`Length = 7.2 / ( 0.00386 * 20)`
`Length = 7.2 / 0.0772`
`Length = 93.264248704663212435233160621762 ft.`

# tl;dr

If you have a circuit with 20 amperes of current flowing along a 12 AWG uncoated solid copper conductor, and you want to keep the voltage drop below 7.2 V (3% or 240 V). You'll want to keep the length of the circuit (one way) less than 93' 4". Unless may calculations are wrong (which they sure could be).

### NOTE:

Since you didn't mention the actual current drawn by the heater. The above calculations are for example purposes only.

Using Ohm's law, you can figure out the current based on the wattage you've stated.

`I = P / V`
`I = 2000 watts / 240 volts` `I = 8.33333 amperes`

You can then plug this into the formula above.

`Length = 7.2 Volts / ( 2 * 0.00193 Ohms * 8.33333 Amperes )`
`Length = 7.2 / ( 0.00386 * 8.333333 )`
`Length = 7.2 / 0.0321666666666667`
`Length = 223.8341968911917 ft.`

• Why did you multiply the denominator by 2? Is it correct?
– user46734
Dec 17 '15 at 2:44
• @Leo Because while the distance between the breaker and the outlet is n, the distance the electricity has to travel is n x 2. Since the electricity has to flow to the outlet, and back again. Dec 17 '15 at 12:08

You don't mention the wattage or amperage which is needed to do the calculation, but normally you don't have to consider voltage drop until you're up around a 100' at 240v.

"220V" is actually 240V.

The 2000W heater draws 8.3 amps.

Our friendly neighborhood Southwire voltage drop calculator says the following about that with #12Cu wire:

Heaters ae extremely tolerant of voltage drop, so there is no trouble at all using 5% or even 10-20%. But even if we went 3%, the answer would still ge 250' or so.

## If you want to go bigger wire, don't use copper.

A lot of people think incrementally: if we’re at #12Cu, the next size up is #10Cu. That's fine if cost is no object.

However if your mind is on your money, you should transition to aluminum at this point. #10 copper costs about the same as #2 aluminum, which has 3 times the ampacity.

I would favor #6 aluminum, since it is the largest size that will play well with inexpensive "MAC Block Connectors" - after that you must splice with pricey Polaris. You would need to splice because most likely your heater either isn't rated for aluminum wire, or isn't rated for LARGE aluminum wire.

## Is this design correct?

But if you're worried about voltage drop for a couple of heaters, you have a design issue in the first place.

For instance one should not do many long runs to a single location. It's not even allowed for an outbuilding, and for a large house it's a great waste of wire. In either case you would run fat feeder from your supply to a subpanel.

Imagine you have four of these guys running, for 33A, and you have a nice fat 4/0 feeder. What does the voltage drop calc say now?

So even at 300' that would be 1% drop. And I'll bet you 4/0 Al is cheaper than you were prepared to spend on copper.

The nice thing about feeder is you have to calculate it for your maximum load, but most of the time you are drawing far less, so your voltage drop is much better than design figures,