# Calculating a simple generator's output power (wattage)

1. May 16, 2014

### Steve S

Hi All,

I am thinking about building a simple electrical generator (to use wave power), and I am trying to make sure I clearly understand the theory and expected results before starting the project. I have two basic questions:

1) How do I calculate the generators maximum current and wattage, if i know the induced emf?

2) How is this generators output power related / limited by the input mechanical power?

At the moment the concept is a simple renewable energy source which is the prime mover, acting to drive a magnet up and down a cylindrical coil, with N turns. So I believe this should be a simple classical problem.

I am clear that the voltage induced is calculated by Faraday's law - and I am comfortable with how this would be develop. Based on my initial setup of 200 turns, 6000gauss magnet, a cylinder of radius 4cm and a magnet travel speed of 0.25m/s i get a voltage of ε = 4.824V

I am conformtable with Ohms law, however, where i am confused is how I calculate the actual current and generators maximum wattage. Suppose for arguments sake I have a 1 Ohm load resistor, and neglect the impedance of the generator coil - I think that the induced current would be :

4.824V / 1 Ohm = 4.824 Amps
as P = V^2 / R
P = 23.27Watts

But if I reduced the loads resistance to say 0.5 Ohm, i get:

4.824V^2 / 0.5 Ohm = 46.54Watts

Similarily if I put a load resistance of 0.01 Ohm, I get a figure of 232.7Watts

So I am unclear as to how the generator can seemingly produce more power, by reducing the load resistance. Surely the actual power produced is limited by the amount of input energy coming in from the magnet?

Is there a way to calculate the maximum theoretical output and if so, can someone provide some guidance on how to link the input mechanical energy to the output power?

For reference I have also posted this in the Physics forum, and so far haven't had much success.

2. May 16, 2014

### Staff: Mentor

If you "try" to draw more output power than the mechanical input, then its rotation will slow down, producing a lower voltage so that output power will always be less than input. A generator becomes much harder to turn when you have it connected to low resistance load. There is no magic here.

You need to have wires thick enough to carry the maximum current without overheating.

3. May 16, 2014

### Steve S

Thanks for the response. I'm aware this would happen with a conventionally driven alternator, where the prime move is a gas turbine or diesel engine - but in the case i'm looking at the mechanical input power is essentially fixed i.e. the magnets physical action is caused by gravity, or a flow of water etc..

4. May 16, 2014

### donpacino

Here is a mechanical analogy. You are pushing a 10 lb sled with a certain amount of force and a certain speed. If I put another 50 lbs on the sled a few things can happen.
1.you dont change the input force. Therefore your speed decreases.
2. you increase your input force such that the speed stays the same.

So in the case of the generator, if you are applying a constant force to the generator, it will have to slow down when the load increases.

5. May 16, 2014

### Steve S

I understand your mechanical analogy - but the generation system I am describing is one that creates an impulse of power - it is not a traditional rotating magnet in a field (or vice versa).

In my scenario a magnet is dropping, due to gravity, through a coil - so there would be no effect that would cause the magnets speed to slow? Or does the EMF field created cause a magnetic physical force acting upwards on the magnet to slow its descent?

6. May 16, 2014

### donpacino

That is correct. If it were not, you would have broken the laws of physics as we know it.