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My question is the following: We have two 220V 20A generators connected in parallel to total power 8.8 kw 220 v 40 A load is 5.5 ohms I'm interested how each generator sees this load since every generator gives half the power. I put this question because one generator can not see this load like 5.5 ohms because it gives half power to not the full power I would like to have somebody explain it a little bit better because it makes me confused .

simplified question

my question is that since every generator in the parallel circuit gives half power in the ideal case it would mean that the generator 1 that gives 220v 20A that is half the power can not "see" the resistance of 5.5 ohms because if "seen" this resistance he had to give complete current in that case we have two problem generator 1 burned because it is not made for such a current problem two if generator 2 gives the same current then the law on energy conservation is not worth what is absolutely not possible thanks

The last attempt to explain what I asked is very simple

we have two 220v 20A generators and a load of 5.5 ohms if we connect this resistance to just one generator it exceeds its power because it is 220v and 20 A 4.4 kw and the current at that resistor is 40A which would mean that it exceeds the maximum power of one generator ok if we add another generator then together have enough power to supply load it if we know that every generator gives half power ie 220v and 20A it is impossible to "see "a resistance of 5.5 ohms

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    \$\begingroup\$ Are you sure that each generator is seeing half the load? Have you checked the current in each leg of the connection? That may tell you a bit more about what is really going on. \$\endgroup\$ – F. Bloggs Sep 29 '17 at 11:50
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    \$\begingroup\$ @loltor: Please edit your question again and put in capital letters and punctuation (commas, full-stops and question marks) where required to make proper English sentences. It is impossible to know where one "sentence" ends and another begins. It is very poor writing. \$\endgroup\$ – Transistor Sep 29 '17 at 19:10
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My question is that since every generator in the parallel circuit gives half power in the ideal case it would mean that the generator 1 that gives 220v 20A that is half the power can not "see" the resistance of 5.5 ohms because if "seen" this resistance he had to give complete current [edit: I think a full-stop was required at this point.]

enter image description here

Figure 1. Two equine traction units on a common load.

With reference to Figure 1:

  • Each of the parallel horses (generators) gives half of the power.
  • If one horse stops (generator going off line) then the other horse takes up the full load (the plough).
  • If one horse (generator) is not strong enough the system will slow down (generator frequency or voltage or both decreases).
  • If one horse is strong enough the system may continue but if the maximum pull (power) of the drawlines (fuses and cables) is exceeded they will break (blow / trip) and the plough (load) will be disconnected.

Note that the horses will never be exactly evenly matched but this doesn't matter. They both contribute to the effort of pulling the load and the fraction of load borne by each will be proportional to the effort of each.

in [sic] that case we have two problem [:] generator 1 burned because it is not made for such a current problem

No. The circuit breaker or fuse will trip to protect the generator.

two [sic] if generator 2 gives the same current then the law on energy conservation is not worth what is absolutely not possible

Sorry, I don't know what this group of words means.


Note that paralleling AC generators requires proper synchronisation control equipment. You can not just wire them up together.

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  • \$\begingroup\$ +1 for nice horses. I'd add if one horse pulls harder than the other, the load on the other horse decreases and it can speed up till the load on the harness matches its strength, and vice-versa. \$\endgroup\$ – Trevor_G Sep 29 '17 at 19:13
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When you parallel two AC generators you have to make sure they are completely synchronous. Both the voltage and the frequency must match so that there is zero current between the generators.

In the real world this is impossible to do. There will always be a small difference, and with the impedance of a generator this will result in large currents.

To prevent this the generator voltage regulator will reduce the voltage by a few percent if the reactive current rises. This "out-of-phase" current must be the unwanted current going to the other generator because it's not perfectly matched up, since we assume all loads are nice 1.0 power factor.

Both generators lower their voltage until a balance has been found. You can move this balance around by changing the percentage each will drop.

This is called droop, or VAR sharing, also perfectly explained in this question.

This is assuming perfect prime mover conditions.

I'm not that into CircuitLab, but this is basic model.
See what happens when you put one source off by 1 degree. Or by 1 Volt.
Then add Droop. (with the series resistor)

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Thanks for the answer but I did not ask that my question is unrelated to circular currents \$\endgroup\$ – lol tor Sep 29 '17 at 11:32
  • \$\begingroup\$ @loltor Then perhaps you should tell me what you'd like to know. Each generator does not only see the load. They also see each other. And that is the only other way for current to flow. \$\endgroup\$ – Jeroen3 Sep 29 '17 at 13:07
  • \$\begingroup\$ my question is that since every generator in the parallel circuit gives half power in the ideal case it would mean that the generator 1 that gives 220v 20A that is half the power can not "see" the resistance of 5.5 ohms because if "seen" this resistance he had to give complete current in that case we have two problem generator 1 burned because it is not made for such a current problem two if generator 2 gives the same current then the law on energy conservation is not worth what is absolutely not possible . \$\endgroup\$ – lol tor Sep 29 '17 at 15:47
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    \$\begingroup\$ @loltor I still have no clue what you're saying. One generator might very well be capable of driving your load. In fact, it will be probably able to perform 3 times nominal for a short time, unless it's a real cheap one. Protection electronics should prevent overheating in this case. I suggest you research some more generator fundamentals. I don't get the part of energy conservation in your comment. \$\endgroup\$ – Jeroen3 Sep 29 '17 at 18:17
  • \$\begingroup\$ The last attempt to explain what I asked is very simple we have two 220v 20A generators and a load of 5.5 ohms if we connect this resistance to just one generator it exceeds its power because it is 220v and 20 A 4.4 kw and the current at that resistor is 40A which would mean that it exceeds the maximum power of one generator ok if we add another generator then together have enough power to supply load it if we know that every generator gives half power ie 220v and 20A it is impossible to "see "a resistance of 5.5 ohms \$\endgroup\$ – lol tor Sep 29 '17 at 19:04
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Given 2 generators, one generator will produce more power than the other generator. The powerful generator will try to pull ahead (phase advancement) of the lower generator and will be required to provide more power. Thus automatic sharing of load, tho not at exactly 50%.

Now, for controlling that frequency.

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  • \$\begingroup\$ my question is that since every generator in the parallel circuit gives half power in the ideal case it would mean that the generator 1 that gives 220v 20A that is half the power can not "see" the resistance of 5.5 ohms because if "seen" this resistance he had to give complete current in that case we have two problem generator 1 burned because it is not made for such a current problem two if generator 2 gives the same current then the law on energy conservation is not worth what is absolutely not possible \$\endgroup\$ – lol tor Sep 29 '17 at 15:48
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I think there is a language AND theoretical barrier here -- Try to apply your failure mode to the following thought experiment:

  1. Same 5.5 Ohm load
  2. Single generator, rated at 220V, 40A

From the above, you would only need one generator. Problem solved, but not your question.

Now. Visualize the generator construction, and split everything in half. Visualize two 20A 220V generators that come from these two virtual halves. They are both in circuit.

That visualization is identical to wiring any number of generators in parallel, as long as they are synchronized, and the interconnecting cable is able to handle the currents.

There is no difference between wiring 10 or 100 or 13 identical generators in parallel and running synchronously, than having one generator of 10 or 100 or 13 times their capacity.

A 40A generator is basically the electro-magnetic components of 2 20A generators, wired in parallel (for example, instead of using AWG #10 enamel wire, it is using a gauge of twice the metal cross-section, or, a twisted pair of #10 wire). The magnetics are also doubled, and the everything is in parallel on one rotor, which is the same as two rotors in sync.

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  • \$\begingroup\$ Consumer generators aren't often designed to run connected in parallel, and the procedure that starts with two generators with their engines stopped , and ends with two generators phase locked with their engines running is not obvious to me, perhaps you could explain it. \$\endgroup\$ – james Jan 27 at 22:07
  • \$\begingroup\$ @james: Start the two generators off load. Adjust the voltages until they're the same. Wire a lamp across the contact that you close to connect the two generators. Adjust the frequency of one until the lamp "beats" very slowly (< 1 Hz). When the lamp goes out the generators are in phase and you can close the switch to connect them together. See youtube.com/watch?v=RGPCIypib5Q. \$\endgroup\$ – Transistor Jan 27 at 22:50
  • \$\begingroup\$ Sorry that was a rhetorical question - I was hoping you might add that information to your answer, although I'm not sure how I would adjust the frequency of a consumer generator. \$\endgroup\$ – james Jan 27 at 22:52

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