Understanding 120V branch loads in 100A 240V system
OK, given I've got a 240V 100A main box. That means I could, in theory, supply 100A @ 240V to a circuit (not referring to code, just theory). I could instead do 100A 120V on one side and 100A 120V on the other side, yes? So, my house could use 200A at 120V and be ok (again, theory, not code). HOWEVER, that would only be true if I balanced load over the two 120V feeds, right.
Say I've got a box like the (bad) diagram below, given that boxes alternate breakers the odd #s are on one leg, the evens on the other, yes? (00 is main breaker) If I loaded all 200A up on the odd breakers I'd be screwed, right? That would be seen by the main breaker as 200A and it would trip. But if I put 100A on odd, and 100A on even, that would be balanced at the main breaker, be seen no more than 100A per leg, and I'd be fine. As I'm assessing my box, I want to make sure I understand how this lays out. The Box Code:
 
50/50 each on that even odd 115Vs CBs .....240Vs total is 100 amps on the entire panel

Yes. you are correct. I don't know what the previous post means.

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Each leg may draw 100 amps but it's not considered a 200 amp load. 100 amps @ 240 volts. Each individual feeder wire is rated to carry a 100 amp load. It's a 100 amp service. 
The key to understanding this is the same as understanding a multiwire branch circuit.
Scenario 1: Svc Leg 1: 100 amps Svc Leg 2: 100 amps Svc Neutral: 0 amps This is a perfectly balanced load. Your "200 amp@120v" scenario. Scenario 2: Svc Leg 1: 0 amps Svc Leg 2: 100 amps Svc Neutral: 100 amps This is a perfectly unbalanced scenario. Scenario 3: Svc Leg 1: 50 amps Svc Leg 2: 100 amps Svc Neutral: 50 amps You notice the neutral carries only the difference. In this way, no conductor ever carries more than 100 amps. 
OK, I guess I'm still somewhat confused (by the breaker box and the MWBC :)
Say I've got a magical 100A 120V lightbulb (that's what a 12,000 watt bulb. :) Anyway, I can plug that on on Leg 1 without a problem, right? I'm drawing 100A on leg 1. I can unplug it and plug it in on leg 2, I'm still only drawing 100A @ 120V. But can I plug 2 of them in? One on leg 1 and one on leg 2? In that case, I'm consuming 200A @ 120V, no? Yes, I agree, that's 100A @ 240V, but that's because 100A on leg 1 and 100A on leg 2. If I can do this, I assume, however, that I could _not_ plug the two bulbs in both on leg 1 (or leg 2) because I'd then be drawing 200A off a single leg (and 0A on the other leg), and each leg can only supply 100A. Do I have it? 
Yes that's correct. The key though is that when two if them are plugged into different legs the svc neutral carries 0 amps. It is exactly as if the lamps were in series on a 240v outlet.

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Very informative thread, and hopefully someone can answer my questions.
1.) If a balanced load constitutes each leg using the same current and the neutral equaling zero, how would a 120v load on each leg, even if the same amperage, be balanced, since 120v circuits uses a neutral? Is it because the phases are 180 degrees apart and cancel at the transformer? 2.) Also, how does the current of a 240v load on one leg return to the transformer? Same leg? Other leg? Much thanks 
1. Yes the two legs of the 120/240 volt service are 180 degrees out of phase. So you can say that for one leg the current "comes" on that leg and "returns" on the neutral and for the other leg the current "comes" on the neutral and "returns" on the leg. When current "comes" and "returns" on the same wire at the same time, you get cancellation.
2. The current serving a 240 volt load "comes" on one leg and "returns" on the other leg. You can also think of a system with 120 volt loads on both legs as having a 240 volt load equal to the smaller of the 120 volt loads plus an additional 120 volt load on the leg that had the larger of the two original loads. Balanced means here that there is no "additional 120 volt load" and as a result no current is flowing in the neutral at that moment. Some electrical engineering teachers have you pretend it is direct current for a moment and they use plus and minus to show how currents add and cancel. >>> so my house (with its 100 amp service) use 200 amps at 120 volts and still be OK Yes you can have 120 volt lights and appliances totalling 200 amps. all on at the same time depending on what circuits they are plugged into, specifically half on each leg. From the point of view of the pole transformer and the point of view of the panel, no more than 100 amps are going through any one of the service conductors. 
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Anyway most people understand what is meant by saying 180 degrees out of phase when referring to residential voltages. I wouldn't get in a big argument over it. I'll put up a diagram of the 240 volt wave signature first. The point to look at here is I have a common crossing point of the neutral with both waves red and yellow so the 2 waves are shifted 180 degrees maintaining a single phase voltage of opposite magnitude. If I were to put the waves 'out of phase' by some certain number of degrees there would never be a common crossing where I could connect both waves to the neutral and get 0 volts. The next two diagrams will show the multiwire branch circuits relationship to the utility transformer and graphically show the cancellation of current in the shared neutral. The last diagram is showing why the NEC requires that multiwire branch circuits must pigtail the neutral to a receptacle in order to keep the neutral from being opened when a receptacle is removed on a multiwire branch circuit serving mutiple receptacles. You can see if the neutral is opened the circuit becomes a 240 volt circuit with the loads in series to each other and current begins using the the 2 legs of the service to complete the circuit to the transformer. It is also why the NEC requires one grounded leg or neutral under one screw at the service neutral bar so that you don't inadvertantly disconnect the neutral of a multiwire branch circuit. 
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240v x 100A = 24 kW = the steady state power limit.
The max power you could pull for a half cycle or so might be 600 kW, or more, before the breakers trip. This is the instantaneous power limit. 24,000W/120V = 200A but the CBs would trip to protect the wires. So the steady state current limit = ~100A. 
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In addition, just made the connections from the replies, that since 240v draw from each 120v, and each 120v is 180 degrees apart, and also since each supply leg is also a return leg (both are supply and load side), therefore, there is cancellation on each side of a 240v load. Quote:
Thanks for all replies. 
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