I understand thoroughly the power equations for balanced sinusoidal three-phase circuits. I haven't studied split-phase systems that much, but I think I understand how the power equations would be.
I saw this Westinghouse portable split-phase gasoline generator: wGEN6000. Here's the nameplate:
Figure 1. Nameplate of the Westinghouse wGEN6000 generator from the webpage.
Figure 2. Nameplate of the Westinghouse wGEN6000 generator from the actual generator.
To begin with, I think the "6000 W" rating is not correct; it should be 6000 VA (apparent power), not 6000 W (active power). You can see they used it as active power, since 25 A = (6000 VA)/(240 V) and 50 A = (6000 VA)/(120 V). I say it's not correct because in general the generator won't be able to deliver 6000 W continuously, only 6000 VA continuously. It could deliver the 6000 W continuously if the power factor of the load was 1. I suppose they list it as watts because the average homeowner doesn't know the difference and all they know is the watts of their appliances.
The second point, and why I'm really asking this question, is regarding the current. Why can we run more current at 120 V RMS (50 A RMS) than at 240 V RMS (25 A RMS)?
I know the answer "because for the same power, the higher the voltage, the lower the current: \$S = V \, I\$". That's true for transmission lines assuming the load draws constant active power, it's also true for induction motors since they act like constant-active-power loads, and it's also true in transformers and DC-DC converters because they are almost lossless two-port devices. But I don't think that explanation is valid here. To see why, let's consider how the windings of the stator (output) of the generator are:
Figure 3. Stator winding.
Correct me if the above diagram is wrong. I skipped the rotor winding, prime mover, etc. Now suppose as a first case that I connect a 240-V load that draws 50 A:
Figure 4. A 240-V load that draws 50 A connected to the two lines of the generator.
According to the webpage nameplate, the generator will burn, because it can only safely deliver 25 A (line current) at 240 V.
Now suppose as a second case that I connect a 120-V load that also draws 50 A:
Figure 5. A 120-V load that draws 50 A connected to the two lines of the generator.
According to the webpage nameplate, the generator won't burn, because it can safely deliver 50 A (line current) at 120 V.
So this doesn't make sense to me. We can clearly see that in both cases, 50 A RMS flow through the upper stator winding, so for me either the winding should burn in both cases or shouldn't burn in both cases.
Can someone explain me where I'm mistaken? Or if the nameplate is mistaken?
The only explanation I can see is that the manufacturer though that, for 240 V, the "25 A" is per phase (which is correct) but you must multiply it by two because there are two phases (which I think is wrong.)