I think I am starting to understand various aspects of electrical generators, and I wanted to run this by the forum and have people correct my explanations, if needed. I understand there are many types of motors and this thought-experiment is just a permenant magnet near a coil. If anyone can verify this explanation, I'd be most grateful: According to Faraday's Law, Induced voltage is the negative of the change in magnetic flux linkage. Flux linkage is only related to the magnetic B-field linked within a conductor (such as a coil loop). The direction of the voltage/current induced (eddy currents) will be induced to create a magnetic field which opposes the change in the external field. So basically, if I have a magnet moving through a coil, as long as I move it at a specific speed through the loop, it will induce a specific voltage from the loop no matter what the load is. The difference that the load makes, however, is that it dictates how much mechanical energy you need to input into the generator in order to overcome the counteracting magnetic field created by the coil. This can be seen by looking at Ampere's Law, which shows that a magnetic field created by a loop is proportional to amount of current in the conductor. Ideally, mechanical power in = electrical power out. Let's say I have a generator connected to a 5Ohm load. I am able to move the permenant magnet at a certain speed to produce 40V. P = V^2 / R P = 320 Watts Electrical current = 8Amps In an ideal situation, I should be able to calculate how much force/torque I need to apply to the permenant magnet based on the power equivalency, the mass of the PM, the field strength of the PM, the geometry of the coil relative to the PM, and the velocity needed to get that specific rate-of-change of flux linkage. If I increase the resistance, I lower the required power: Load = 50 Ohms P = V^2/R P = 32 Watts Electrical current = 0.8 Amps It should be mechanically easier to move the PM around the coil since it is unable to create a large counteracting field due to the smaller coil current. What if I shunt the generator output? Load = 0 Ohms P = infinity I would have to apply an infinite amount of mechanical force/torque to produce this electrical output. I either wouldn't be able to move the magnet at all, or in even trying would produce such a large current through the non-ideal resistances of the wire that it would create a huge current and likely burn up the shunt. What if I connect a capacitor to the output? From a discharged state, the capacitor would look like a 0-ohm shunt (probably want to add some limiting series resistance due to the prior thought-experiment). As the capacitor charges the voltage accumulated on the capacitor causes the overall current to exponentially decay, requiring less mechanical power over time. Once the capacitor is charged, I can let go of the PM, and now that there is energy stored in the capacitor, it will want to discharge through the same coil, and everything works in reverse (the cap is now acting as a power source for an electric motor). If I wanted to keep the energy in the cap, I should probably build a full-wave rectifier circuit on the output of the generator and a diode to prevent the reverse currenting.