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mikeames:
A power plant generator (alternator) can be treated as a voltage source with some internal impedance. From the power plant there is a distribution system that also contains internal impedance. Much of the distribution system is a distributed impedance.
Now greatly simplify this to two variable voltage DC sources each with an internal resistance of 10 ohm and feeding a single load of 50 ohms load.
The equations for this circuit are:
Iload = I1 + I2
Vload = Iload * Rload
Vload = V1 - I1*R1
Vload = V2 - I2*R2
If Vload is to remain constant, then a reduction in V2 requires V1 to be adjusted higher. This causes more of the load current to come from V1, and thus more energy is being supplied by V1 both from increased internal source voltage and the resultant increase in I1 current.
Obviously V1 can not be allowed to supply too much energy or the generator would be overheated. This is aside from the issue if there is enough steam energy to drive the generator.
If V2 is lowered to the point where V2 = Vload, then V1 is supplying all the energy. If Vload is 10 V and the resistances are as above, and V2 is 10 V, then V1 is is 12 V. If V1 is changed to 11 V, then what is the required V2 voltage to maintain Vload at 10 V?
With any such system there are stability problems.
Does this help?
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