Column C
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
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.
Note 4
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.
You are correct that the first two options are allowed by Code (citing from the 2014 NEC here) -- the relevant passage is 210.11(C)(3) along with its Exception:
(3) Bathroom Branch Circuits. In addition to the number of branch circuits
required by other parts of this section, at least one 120-volt, 20-ampere
branch circuit shall be provided to supply a bathroom receptacle outlet(s).
Such circuits shall have no other outlets.
Exception: Where the 20-ampere circuit supplies a single bathroom, outlets for
other equipment within the same bathroom shall be permitted to be supplied in
accordance with 210.23(A)(1) and (A)(2).
However, this passage only applies to receptacles unless you take advantage of the terms of the exception; putting the bathroom light on a shared lighting circuit is Code-legal as long as the receptacles are on a dedicated circuit.
For the heaters, you can either put them on a per-bathroom circuit as long as they satisfy 210.23(A)(2) by not using up more than 50% of the circuit's ampacity:
(2) Utilization Equipment Fastened in Place. The total rating of
utilization equipment fastened in place, other than luminaires, shall not
exceed 50 percent of the branch-circuit ampere rating where lighting units,
cord-and-plug-connected utilization equipment not fastened in place, or
both, are also supplied.
or you can have them on a shared lighting circuit, if you decide to have a separate circuit for bathroom receptacles. Unfortunately, putting them on the same circuit as the bathroom receptacles when that circuit is shared between bathrooms is prohibited by the Exception to 210.23(A):
Exception: The small-appliance branch circuits, laundry branch circuits, and
bathroom branch circuits required in a dwelling unit(s) by 210.11(C)(1), (C)
(2), and (C)(3) shall supply only the receptacle outlets specified in that
section.
Best Answer
No, every hardwired appliance does not need to be on its own circuit. But...
Provision power for heaters with a 125% derate
You need to study the unit's specs carefully, and note the amps or VA drawn by it (note a resistive electric heating element will have VA identical to Watts, but a fan motor may knock that off a little). Then, multiply that figure by 125% (8A -> 10A). The +125% number is how much power you must provison (allocate) on that circuit.
So for instance if you have two heaters that are 8.1 amps each. Those derate to 10.125 amps. Two of them would be 20.25 amps, which is too much for a 20A circuit. Most heater manufacturers understand this intimately, and will size their heaters to be just under a threshold number, i.e. not 8.1A :)
Also beware of heaters whose spec rating is at 125V instead of 120V. Due to Ohm's Law, a reduction in voltage causes an equal reduction in amperage. So if you have an 8A heater @125V, then you have to multiply by (120/125) to get the 120V amperage.
An interesting idea
Oftentimes, people speccing power for bathrooms and kitchens will be thinking only about the Code book (which is complicated enough), and forget that the main user of a bathroom is people who do a lot of "Beauty stuff" in the bathroom with heat appliances, often several. For instance Code allows one single 120V/20A circuit to serve all receptacles in all bathrooms, and you know the men who specced that didn't have a wife and teenage daughter! That's because Code is written by the National Fire Prevention Association not by the National Hair Care Association.
Well, look at the rules.
So, if the heater provisions out to <=10 amps, I would be inclined to run a 4-wire circuit for each heater. Then, continue the circuit to a 2-gang box containing two GFCI receps, wired in multi-wire branch circuit fashion - sharing the neutral, and one hot per recep. Now, you can plug a curler and hair dryer in at the same time! (you'd be kissing 20A if the heater was on also, but the user could turn off the heater while using the appliances - they'll be making plenty of heat of their own.) So for about $30 extra, you've made the beauty enthusiasts in the house very happy.