DC Generator Question: How Can Loading Occur When Terminals are Open?

[TheAnalogKid83]

I've had this question going in my mind for probably half a year, but I just have not had time to think about it too closely, so maybe someone with experience can help me.

With a DC generator, it's torque is proportional to the current flowing through the input terminals. Basically the voltage applied creates current flow through the internal circuit of the motor, and then this current is converted by a constant to a torque, which is then converted through a friction and inertia transfer function to give you angular velocity of the motor shaft. The key point to my problem is that there is no open at the terminals, and the current has a path to flow.

Now with a generator, you basically have a reversed motor. The physical force is generating electrical engergy. So the turning of the shaft should generate a voltage on the terminals, the back EMF. If you short the terminals, you have maximum current flowing through the generator, and thus a maximum torque. The loading on the physical force should increase greatly when you short the terminals. The open terminals should have 0 torque, and the load should be purely mechanical (friction at steady state). Is this the case? When turning the generator that has open terminals, there still feels like the motor is generating negative feedback, and there still is measurable loading.



I've never had an electro mechanical course, so maybe I am sounding ignorant, but how can you have a loading on the physical force acting on a generator when the terminals of the generator are open and there is no current flow?

 

[anorlunda]

If you short the terminals, you have maximum current flowing through the generator, and thus a maximum torque.

That is your problem. It requires both voltage and current to produce power. If you short circuit a generator, it makes no external power. However, the $I^2R$ losses internally soar because of the high current.