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While calculating total copper loss of two dc generators connected in parallel which is valid approach?

  1. calculate resultant resistor of two parallel connected generators then use
    ohmic loss formula with total load current.
  2. calculate two individual copper loss (current delivered by individual
    generator and its own resistor), then add up both copper losses.
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Just because there are two DC generators running in parallel, you can't assume that they are equally delivering the same power to a load. Therefore you have to treat them as individuals. Each generator will have its own feedback system and if one inherently wants to generate 100 mV than the other, it will supply the higher voltage and the other will "back off" on the basis that it thinks it is supplying too much voltage. Result is: 1 generator supplies nearly all the power to a given load.

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Use model for V/RPM for each generator and ESR from DC Ohms from each and compute conduction losses with power diode isolation to prevent shorting of each other or some impedance matching network between two sources and load.

Then apply KVL and Bode plots with generator inductance and load capacitance (battery and cap) with ESR included for both step load regulation transient and DC load regulation errors to compute current sharing.

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