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I am installing a 240 volt 40 amp electric cooktop. Will 12/3 NM cable suffice for this?
Secondly, I know I need a double-pole breaker because 240 volts is delivered using two out-of-phase 120 volt circuits. My question is, is a double-pole breaker labeled "20 amps" really two 20-amp circuits that together can supply 40 amps @ 240 volts? Or do I need a double-pole breaker labeled "40 amps"?
First off you'll notice the text "Column C to be used in all cases except as otherwise permitted in Note 3.)", in the title of table 220.55. This makes it simple. You have 2 appliances, so follow that over in the table, and you'll see 11 in Column C. So there you go, you can just use 11 kW. Done.
11,000 W / 240 V = 45.8333 A
So you'll need a 50 ampere breaker, and wire sized appropriately for the load.
Note 3 says:
- Over 1 3⁄4 kW through 8 3⁄4 kW. In lieu of the method provided in Column C, it shall be permissible to add the nameplate ratings of all household cooking appliances rated more than 1 3⁄4 kW but not more than 8 3⁄4 kW and multiply the sum by the demand factors specified in Column A or Column B for the given number of appliances. Where the rating of cooking appliances falls under both Column A and Column B, the demand factors for each column shall be applied to the appliances for that column, and the results added together.
Perfect, so instead of just using the value from column C you can do math. Let's step through it.
...it shall be permissible to add the nameplate ratings of all household cooking appliances rated more than 1 3⁄4 kW but not more than 8 3⁄4 kW...
8.4 kW + 5.0 kW = 13.4 kW
...and multiply the sum by the demand factors specified in Column A or Column B for the given number of appliances...
Let's check the table again... You have 2 appliances, both between 3 1/2 and 8 3/4 kW. So You'll look at column B, and find 65%.
13.4 kW * 65% = 8.71 kW
8710 W / 240 V = 36.2916 A
So using this method you can use a 40 ampere breaker, and appropriately sized wire. However, keep in mind that if you change the equipment, you'll have to do the calculation again. So while you can use this value, you may have to upgrade the circuit later if you change equipment.
I'm not exactly sure how note 4 comes into play, but I think it can be used if this is the only equipment on the branch circuit. Just for fun, let's run through that one too.
- Branch-Circuit Load. It shall be permissible to calculate the branch-circuit load for one range in accordance with Table 220.55. The branch-circuit load for one wall-mounted oven or one counter-mounted cooking unit shall be the nameplate rating of the appliance. The branch-circuit load for a counter-mounted cooking unit and not more than two wall-mounted ovens, all supplied from a single branch circuit and located in the same room, shall be calculated by adding the nameplate rating of the individual appliances and treating this total as equivalent to one range.
You're only concerned with the second half of this note, since you have one counter-mounted cooking unit, and one wall-mounted oven, all supplied by a single branch-circuit, and located in the same room. So you can add the nameplate values, and treat it as a single range.
8.4 kW + 5.0 kW = 13.4 kW
So you can treat the units as a single 13.4 kW range. Check the column C again, this time for a single range. You'll find a value of 8 kW. But wait... The column header says "(Not over 12 kW Rating)". Your range is 13.4 kW. That's bigger than 12 kW. Now you'll have to check note 1
- Over 12 kW through 27 kW ranges all of same rating. For ranges individually rated more than 12 kW but not more than 27 kW, the maximum demand in Column C shall be increased 5 percent for each additional kilowatt of rating or major fraction thereof by which the rating of individual ranges exceeds 12 kW.
That's easy enough.
13.4 kW - 8 kW = 5.4 kW
Since .4 is not a "major fraction", you can just use 5 kW. So you'll have to add 5% 5 times.
5% * 5 = 0.25
8000 W * 0.25 = 2000 W
8000 W + 2000 W = 10,000 W
That means you'll have to use 10 kW as your demand.
10,000 W / 240 V = 41.666 A
Which means you can use a 50 ampere breaker, and appropriately size wire.
It's referring to the electrical service supplying the house. Most homes in the US are supplied by a single phase service, which is often described as a 120/240 Volt system. If your house is supplied by a three phase system, it could be a 120/208 Volt system.
In both these cases you'll be able to install the equipment you have.
NEMA 14 devices are four wire grounding devices. Which means they'll require two ungrounded "hot" conductors (X,Y), one grounded "neutral" conductor (W), and one grounding conductor (G). When installed on a single phase 120/240 volt circuit, NEMA 14 devices can supply 120 and/or 240 volt loads. A NEMA 14-30 device is designed to supply 30 ampere equipment.
The circuit will require a 30 ampere double pole breaker, and four at least 10 AWG copper conductors. You'll connect one ungrounded "hot" conductor to each of the X and Y terminals. The grounded "neutral" conductor goes to the W terminal. And the grounding conductor connects to the G terminal.
Best Answer
Not even remotely close. You need 8 gauge copper or 6 gauge aluminum wire.
A 20 amp dual pole breaker supplies 20 amps at 240 volts (line to line), or 20 amps at 120 volts twice (line to neutral and line to neutral). You need a 40 amp at 240 volt breaker.
...and these two questions in combination strongly suggest that you get professional help before you cause yourself an expensive problem. Like the price of a new house expensive.