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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:

Nameplate from the website

Figure 1. Nameplate of the Westinghouse wGEN6000 generator from the webpage.

Nameplate of the Westinghouse wGEN6000 generator

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:

Stator winding

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:

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:

120-V load that 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.)

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  • \$\begingroup\$ I suspect that the generator is capable of delivering 6000W as claimed, if the power factor is reasonably good, but the limiting factor may be mechanical rather than electrical. Similarly, the output wiring is presumably rated to 50A although the motor can’t deliver enough power to maintain speed if loaded above 6000W. So the placard could be correct, but that doesn’t guarantee that it is. \$\endgroup\$
    – Frog
    Jul 22, 2022 at 6:51
  • \$\begingroup\$ So, are you saying that we could safely draw 50 A continuously if we connect a 240-V load? \$\endgroup\$
    – alejnavab
    Jul 22, 2022 at 6:55
  • \$\begingroup\$ Unlikely, the motor would be unable to maintain speed and the voltage (and frequency, obviously) would drop. \$\endgroup\$
    – Frog
    Jul 22, 2022 at 6:57
  • \$\begingroup\$ But then, for the same reason, we can't safely draw 50 A continuously if we connect a 120-V load, right? Yet the nameplate says we can. If we can, please tell me where my reasoning is mistaken. Thanks. \$\endgroup\$
    – alejnavab
    Jul 22, 2022 at 6:59
  • \$\begingroup\$ The generator consists of several parts and you must operate within the limits of each part, i.e. current must not exceed 50A, voltage must not exceed 250V, power must not exceed 6000W and presumably VA must not exceed some limit (a little over 6000VA) \$\endgroup\$
    – Frog
    Jul 22, 2022 at 9:58

4 Answers 4

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It has a gasoline engine rated to give you 6000 VA.

So even if the generator and wiring etc could provide you 50A at 240V, it would require a gasoline engine that can output 12000 VA, and it does not have that.

But it can give you 50A at 120V which is 6000 VA, and it can give you 25A at 240V which is also 6000VA.

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  • \$\begingroup\$ Now this makes sense. Thanks. I'll wait about a day to see if someone else answers it. \$\endgroup\$
    – alejnavab
    Jul 22, 2022 at 7:04
  • \$\begingroup\$ So, for the same reason you explain, I shouldn't connect two 120-V loads that draw 50 A each, right? \$\endgroup\$
    – alejnavab
    Jul 22, 2022 at 7:09
  • \$\begingroup\$ That is correct. Two 120V load that draw 50A equal to one 240V load at 50A or one 120V load at 100A. \$\endgroup\$
    – Justme
    Jul 22, 2022 at 7:32
  • \$\begingroup\$ Thanks again. Just one last question, and sorry for this. This three-phase generator according to its nameplate is rated for 25 kVA, 220 V (L-L) and 127 V (L-N), and 114 A @ 220 V or 197 A @ 127 V. Does that mean that: 1) if I connect only one 220-V single-phase load that consumes 25 kVA, the generator can safely supply I = S/V = (25 kVA)/(220 V) = 113.63 A ≈ 114 A on two lines only; 2) if I connect a only one 127-V single-phase load that consumes 25 kVA, the generator can safely supply I = S/V = (25 kVA)/(127 V) = 196.85 A ≈ 197 A on one line only; ... \$\endgroup\$
    – alejnavab
    Jul 24, 2022 at 2:10
  • \$\begingroup\$ ... and 3) if I connect only three 220-V single-phase load (or one 220-V L-L three-phase load) that consumes 25 kVA in total, the generator can safely suply I_L = S_3ϕ/(√3 * V_LL) = (25 kVA)/(√3 * 220 V) = 65.61 A on the three lines. Right? \$\endgroup\$
    – alejnavab
    Jul 24, 2022 at 2:35
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I would expect the generator to provide two opposite-phase 120 V 25A outputs, for a total of 50 A at 120 V.

The total current available at 120 V would be 50 A, but a single 120 V load could not draw more than 25 A.

Something like this, where the transformer secondary represents the generator output windings.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ The 120 VAC outlets are probably standard NEMA 5-20 120V/20A receptacles which should be fused for 20A each, although there are 120V outlets for 30 A, (and 50 A, but rather rare). And none are rated 25 A. \$\endgroup\$
    – PStechPaul
    May 16 at 3:20
  • \$\begingroup\$ The specs show two 5-20R GFCI receptacles and one L14-30R. \$\endgroup\$
    – PStechPaul
    May 16 at 4:42
  • \$\begingroup\$ If that's what the manufacturer meant, then it's misleading, at least for an electrical engineer. In the diagram you show, there's no 50 A anywhere. There's only 25 A of line current and 0 A in the neutral wire. The generator is delivering 25 A, not 50 A total. It's like saying a three-phase generator supplying 5 A of line current is therefore supplying a total of 15 A, which simply doesn't make sense. Thanks. \$\endgroup\$
    – alejnavab
    May 27 at 21:07
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Why can we run more current at 120 V RMS (50 A RMS) than at 240 V RMS (25 A RMS)?

That's because the two 120 V - 25 A windings, that together source 50 A, can source only 25 A when connected in series for 240 V.

Here's how 120V - 50A or 240V - 25 A is drawn from the generator.

enter image description here

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  • \$\begingroup\$ If that's what the manufacturer meant, then it's misleading, at least for an electrical engineer. In the diagram you show, there's no 50 A anywhere. There's only 25 A of line current and 0 A in the neutral wire. The generator is delivering 25 A, not 50 A total. It's like saying a three-phase generator supplying 5 A of line current is therefore supplying a total of 15 A, which simply doesn't make sense. Thanks. \$\endgroup\$
    – alejnavab
    May 27 at 21:06
  • \$\begingroup\$ The two 120 V receptacles may be loaded up to 25 A each. The 240 V receptacle may also be loaded up to 25 A. The total of the three loads is not to exceed 6000 VA. You're right - the neutral current will be the difference of the two line currents. \$\endgroup\$
    – vu2nan
    May 28 at 11:01
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One overlooked detail is the generator manufacture may have a throw over switch for 120 or 240 volts for 120 the two separate windings supply 25 amps each in phase! with the switch on 240 the windings are in series and still putting out 25 amps. Your diagrams do not show two, separate windings and possible throw over in phase out of phase switch.

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  • 1
    \$\begingroup\$ Yes it may have, but if you click on the product link, look at the pictures or read the manual, you will see it does not. The manual even includes a wiring diagram. \$\endgroup\$
    – Justme
    May 16 at 12:44
  • 1
    \$\begingroup\$ "the generator ... may have a throw over switch for 120 or 240 V [operation]” Welcome to EE SE. It would be possible to do that but a look at the control panel and the circuit diagram in the operator's manual shows that this generator doesn't have such a switch. 120 V operation is 25 A per leg for a total of 50 A. It doesn't seem possible to pull 50 A on a single line. \$\endgroup\$
    – Graham Nye
    May 16 at 12:45

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