Funny how I was pondering about this question for a long part of the day yesterday, and today I was at stackoverflow and in the box with the featured questions from other SE sites this very related question appeared: Will a 3 phase induction generator fight its own rotation?. So, I decided to ask my related question here.

I know that a generator will fight its own rotation, but what I am wondering is this: suppose I put far more mechanical energy into a generator than its electrical load; what is going to happen then? Picture this scenario: we have a 1KW generator hooked up to a 100W lamp and a watt-meter. The generator has a crank. If I put about 110W of mechanical energy on the crank, the watt-meter will tell me that the generator supplies my lamp with about 100W of energy. (Assuming 10% loss.)

But now suppose that I put a lot more power into the crank; Suppose I go close to the 1KW mark. (Without reaching it, or exceeding it, so as not to burn my generator.) Where will all the excess energy go?

And one more thing: If the answer is by any chance that the generator will only resist the crank by 110W of mechanical energy, so putting 1KW of energy will be impossible, and I will just spin the crank so fast that I will either reach my own limit of how fast I can spin, or the generator will disintegrate due to centrifugal force, then please answer this to me: Is it possible to build a generator that will provide as much resistance as necessary to the mechanical force driving it so that the generator will consume only as much mechanical energy as required by its electrical load?

Let me tell you the reason why I am asking so that you can have more context to take into consideration, if needed, to answer the question: I was wondering about the possibility of storing energy in a mechanical arrangement such as a coil. So, on windy days, when my windmill power generator has surplus energy to give, I would be redirecting the surplus energy to a motor which winds a pretty huge coil, while on windless days I would allow the coil to slowly unwind, driving a generator. (There is, of course, a gear between the coil and the generator, so that the coil unwinds very slowly, while the generator turns fast.) Now, the coil will have a certain amount of mechanical energy stored in it, and it will want to always apply the same amount of torque to the generator, regardless of how much energy my household requires. So, I was wondering if there is anything that can be done to prevent the coil from needlessly unwinding, other than applying a break on it. (A solution which sounds very bad to me.) The ideal would of course be if the generator by itself could resist the unwinding of the coil in such a way that the tendency of the coil to turn minus the resistance of the generator would at all times equal the exact necessary force to turn the generator to produce the exact amount of energy required by its load. But I do not know whether this is electrically possible.

UPDATE Okay, it turns out that the answer to the first question in bold is what the sentence immediately following it anticipates, (the one which says "And one more thing: If the answer is by any chance that...") so the real question then becomes the second question in bold. Unfortunately, this was left unanswered. I was thinking about it, and I came up with something that according to my simplistic understanding of electricity and electromagnetism might (just might) constitute a proof that it should be possible to achieve what I am looking for, so let me explain it, and you tell me if it is correct.

Suppose that I hook up not only a generator on the coil, but also a separate motor. (Which I will have to do anyway in order to be able to wind the coil when my windmill has excess energy.) Now, suppose that the coil has so much mechanical energy stored in it that it wants to exert so much torque to my generator that it would be capable of producing 10KW while I only need 110W. So, suppose that I redirect the excess 9890W to the motor, which counters the torque of the coil. Wouldn't then this entire system reach an equilibrium? Wouldn't the coil end up unwinding at a rate which would correspond to the number of watts required by the lamp, plus about 10% of that for losses of the generator, plus about 10% of the 10KW for losses of the motor? So, wouldn't I then be losing about 1.1KW of mechanical energy stored in the coil instead of the whole 10KW?

And another question: if the answer to this is yes, then is it possible to have a single device which acts both as a generator and as a motor and achieves this behavior without having to have the generator and the motor as separate devices?

  • 1
    \$\begingroup\$ To answer your question (printed in bold): Every generator basically only "uses" the amount of mechanical energy it delivers on its output (+ some % mechanical and electrical losses). In theory, you should even observe this with a small DC motor: It should be much harder to turn at a high rate when the terminals are shorted. \$\endgroup\$
    – 0x6d64
    Jan 10, 2012 at 11:10
  • \$\begingroup\$ "Now, the coil will have a certain amount of mechanical energy stored in it, and it will want to always move my generator by that amount" - It would be better to say that it will always apply the same amount of torque to the generator - energy isn't an "over-time" quantity, and it probably won't actually apply a constant power. \$\endgroup\$
    – Random832
    Jan 10, 2012 at 14:43
  • \$\begingroup\$ @Random832 yes, of course you are right. I corrected my question. \$\endgroup\$
    – Mike Nakis
    Jan 10, 2012 at 14:48
  • \$\begingroup\$ Ideally, I would think you want your generator to turn at a constant speed - so you could either have a mechanical arrangement (either purely mechanical or reactive to changes in the output voltage) to make the coil unwind at a constant speed, or to have the coil unwind less on lesser loads, have some sort of automatic transmission to change the gearing ratio. \$\endgroup\$
    – Random832
    Jan 10, 2012 at 15:45
  • 1
    \$\begingroup\$ As a comment from a generator owner and "was heading down the off-grid road" before the power company got more reasonable in the 2009 recession - typical real efficiency for AC generator sets is closer to 50-75% than 90% \$\endgroup\$
    – Ecnerwal
    Dec 24, 2016 at 17:19

4 Answers 4


If you have a 100W electrical load and you drive 100W plus efficiency losses, say 110W, into the generator, things will be in a state of equilibrium, with 100W being converted from mechanical input power into electricity, and the other 10W of mechanical input power being eaten up by losses.

Now suddenly put 1kW of mechanical power into the machine; at that instant, before the rotational speed can change, the 100W electrical load will continue to present the same mechanical load to the prime mover. Things will not be in equilibrium, and the machine's rotational speed will accelerate. Depending on circumstances, this may or may not increase the electrical load. Certainly the generated voltage will go up, and any simple resistive load will therefor absorb more power, but maybe you have some regulation such that the load continues to draw exactly 100W.

So assume the load continues to draw exactly 100W. Where does the extra 900W of mechanical power go then? The machine's speed must increase until the losses equal the mechanical driving power; so it ends up turning extremely fast, the increased power going into increases in friction in the bearings, windage loss due to the rotating parts, eddy currents in the magnetics (and doubtless a couple of other things I forget at the moment), none of which are desirable.

You would find that, without exceeding the machine's electrical rating, you would quickly exceed its mechanical ratings, i.e., probably long before you got to 1000W, the rotation speed would be several times the suggested speed, and catastrophic failure would likely result. Note you can do this with no electrical load on the generator at all.

  • \$\begingroup\$ Exactly what happened to the windmill generator at the last house I owned, while the prior owners lived there. There are still pieces of windmill scattered all through the woods... \$\endgroup\$ Aug 17, 2014 at 4:35

The power fed into the generator equals the power taken from the generator. The latter includes heat or mechanical energy (e.g. vibration).

So any additional power is converted to heat or mechanical energy which may lead to destruction of the generator.

But consider, that it is not so easy to put more energy into the generator, because if there is not enough electrical load, the mechanical resistance (torque) will drop. So if you really want to put more energy into the generator you have to increasing RPM further and further, which increases friction (--> heat) and mechanical energy (vibration) in the generator; both will eventually lead to destruction of the generator.

  • \$\begingroup\$ It's worth noting that there are some mechanical systems such as wind-up springs which will, if allowed, put more power into a low-torque load than a high-torque one. Assuming only sliding or bearing friction (no fluid friction), the maximum power would be delivered at a revolution rate where the torque that contributed toward useful current would equal the frictional torque; if the motor is allowed to spin faster than than, more power will be mechanically into heat and the spring will wind down faster, but less electrical power will be generated. \$\endgroup\$
    – supercat
    Jan 10, 2012 at 15:50
  • \$\begingroup\$ @supercat that's precisely what I want to avoid. That's why I want to counter the movement of the spring, so that I consume only as much mechanical energy as I need for my electrical load, plus losses. \$\endgroup\$
    – Mike Nakis
    Jan 11, 2012 at 18:18
  • \$\begingroup\$ @MikeNakis: I would suggest that you'll probably need two switching converters: a boost converter to feed a variable voltage from the motor to a storage cap whose output must be allowed to vary widely, and a second supply (probably buck) to convert that to a usable supply voltage for your main circuitry. The use of two supplies will be necessary because the motor is apt to have some inertia; if your demand falls off substantially, you'll still have to convert the motor's kinetic energy to electricity and store it. The simplest approach (assuming 100mA generator torque) would probably be to... \$\endgroup\$
    – supercat
    Jan 11, 2012 at 22:34
  • \$\begingroup\$ ...operate in a discontinuous mode, where each time the output cap starts to get low, you start letting the motor freewheel until it gets up to a good speed, and then harvest energy from it, sitting at that speed, until you've got enough, whereupon you then try to harvest all remaining energy and leave the motor shorted through the initial inductor once the voltage has fallen to nearly nothing. Smoother operation might be desirable, but would probably require a microcontroller. \$\endgroup\$
    – supercat
    Jan 11, 2012 at 22:40

The generators I have worked on were all driven by a constant RPM motor. If the motor speed were to change, so would the generator frequency. Voltage and current were controlled by an excitation current which fed rotor coils of the generator. The generator outputs on the stator windings, eliminating the need for slip rings, or brushes. An electronic feedback circuit monitored/controlled the excitation current to keep the voltage steady.

  • \$\begingroup\$ Wait, I am not sure I understand. If the motor was a constant RPM motor, then why was there any need to monitor and control anything? Or was the entire assembly working at constant RPM precisely because of the monitoring and the controlling? \$\endgroup\$
    – Mike Nakis
    Jan 11, 2012 at 19:59
  • \$\begingroup\$ The RPM is fixed, mechanically. As load demands change the excitation current is varied to regulate the output voltage. This is accomplished by monitoring the output voltage in a feedback loop that adjusts the excitation current. \$\endgroup\$
    – SteveR
    Jan 11, 2012 at 20:25

I think we are avoiding the simplest method of all for storing electrical power while creating an additional drain on the system & then converting it back to mechanical power.

Step 1: Connect a power A/B switch on your building and wire to the fuse box.

Step 2: At the 120 out from the generation connect a 3 way plug, run one into your A/B power switch you just installed .

Step 3: Connect a high output car battery charger Jumper to the second outlet and run that into a bank of semi-truck battery's for storage:

Step 4: Run a (+)/(-) line from your battery pack to ON/OFF switch then to a adjustable switch to allow speed control of a DC motor such as a car starter set to gears to rewind your spring.

Step 5: Run a (+)/(-) line from your battery pack to DC to AC converter to supply electricity to the the structure in case of generator failure. Depending on your converter selection you could power a lot of stuff or just maintain power to your fridge, TV and a light or two for an extended period of time.

I have seen this done with survival cabins in the woods without the spring in step 3.

  • \$\begingroup\$ ...until the battery bank is fully charged, at which point the batteries can't act as a buffer any more and excess power again gets wasted. \$\endgroup\$ Aug 17, 2014 at 4:41

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