1. I don't understand the electrical torque in a generator. Isn't the counter torque related to the armature current in the generator?
  2. Why does electrical torque decrease during a fault in the lines that connect the generator to the infinite grid? Fault means more current in the armature winding and more current should mean more counter torque.
  3. Why would the rotor accelerate?
  • \$\begingroup\$ Have any context of what generator is accelerating like this? Is the generating regulated to counter the opposing magnetic drag that is produced? A generators goal usually is a certain frequency, if the drag increases, it will be forced to accelerate to maintain this frequency... but context is needed to be sure we’re talking about the same thing. \$\endgroup\$ Dec 17, 2019 at 11:30
  • \$\begingroup\$ If there are multiple generators on the same line, this too will have an impact. More context is needed to the setup being discussed. \$\endgroup\$ Dec 17, 2019 at 11:36
  • \$\begingroup\$ I'm talking about synchronous generators in which there is a mechanical torque produced by prime mover connected to the rotor through the shaft and the counter torque produced by armature load current.Mechanical torque equals to the electrical torque during steady state but when there is a fault,electrical torque decreases and mechanical torque causes rotor to accelerate.What I don't understand is why this happens in case of faults. \$\endgroup\$
    – engyhya
    Dec 17, 2019 at 11:38
  • \$\begingroup\$ Does your fault take any real power the prime mover should provide? \$\endgroup\$
    – Jeroen3
    Dec 17, 2019 at 12:53
  • \$\begingroup\$ Well looks like it doesn't but how does the magnetic field seperate the real power current and fault current ? So it can produce an electrical counter torque.Both the load current and fault current flows through the armature winding. \$\endgroup\$
    – engyhya
    Dec 17, 2019 at 13:03

2 Answers 2


Sorry if I disturb with elementaries, but synchronous generator runs ahead the grid a certain phase angle to be able to generate some power into the grid.

If for some reason the load suddenly drops off or decreases substantially but the motor which rotates the generator still keeps up the same torque the generator gets more speed and it can drift so much more ahead the grid that the grid doesn't any more brake the generator. That happens easily if the generator is used near its limits and runs nearly 90 degrees ahead the grid. If the generator happens to slip more than 90 degrees ahead the grid getting more ahead happens automatically - like sliding downwards on a slippery surface. The generator drops out of sync and hopefully gets quite soon disconnected by the protection devices.


The electrical power out of the machine is P = ((VG x VR)/X)sin(d) where VG is the generator voltage, and VR is receiving bus voltage. X is the reactance between the two. d is the angle difference between the two voltages. We use per unit typically when solving, so, at normal conditions both voltages are near 1.0. When a fault occurs in the nearby system VR is depressed, so the P that can be electrically moved out if the machine is reduced proportionately. So, the excess power being put into the machine (mechanical torque) goes into accelerating the rotor.

I would recommend Kimbarks 3 volumes as they are classics and volume one is the introduction you need to this topic.

Edward Kimbark Power System Stability, Volumes I, II, III, 3 Volume Set

regards, russ


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