# Calculating actual power (wattage) created by a simple generator

1. May 13, 2014

### Steve S

Hi All,

I am thinking about building a simple electrical generator, 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 Watts

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

4.824 / 0.5 Ohm = 9.648 Watts

Similarily if I put a load resistance of 0.01 Ohm, i get a figure of 482Watts

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?

2. May 13, 2014

### lecandotnet

Hi Steve,

You asked for an output formula for generator, I don't have it all, but I have had what I could not find it by Google yet. I am building a similar wind power project, It is easier to buy one then we have nothing to say here.

3 factors excluding input: Strong magnet, best coil, and efficient configuratikn for maximum output from low speed with durability.

On coil, we known that device can not burn in in the short run, so ampere-turn must be met the standard table. Wind power using wire size 18 for lasting, a few using size 22 with reasons

From formula for coil:
L=E/I = (&mu;)*N^2*A/l

You have N, A, l, and (mu) from B = (mu)*I*N , current capital I not lower case l for former lenght.

From above, you and I needed more N and the shorter l lenght for higher L, and we have smaller diameter wire size where your neodymium magnet and configuration are at best for profit.

If you are disagree, please tell us why?

Want to see my works ?

http://www.lecan.net/Immigrants.html

3. May 13, 2014

### UltrafastPED

You will probably do better in the electrical engineering forum ...

4. May 15, 2014

### lecandotnet

Last edited: May 15, 2014
5. May 15, 2014

### Staff: Mentor

The easy one:
The maximum output electrical power is equal to the maximum input mechanical power times the generator's efficiency. Typically for a commercial generator, that's on the order of 95%. So 1 kW of mechanical input power yields 0.95 kW of electrical output power.

6. May 15, 2014

### lecandotnet

Last edited: May 15, 2014
7. May 16, 2014

### Steve S

Hi All,

thanks for your response so far.. I will try and clarify each of the points in turn

1) I might be wrong on this, but the formula you have given is to work out the inductance of a coil (in Henrys), which is used to define how a coil acts in a similar manner to a resistor - so this is the wrong equation to apply when trying to calculate generated current.

2) My working out for 4.8V is as follows (i'm assuming it is a perfect coil, and also not considering effects at the end of the coil):

Coil Length = 0.5m
Magnet Travel Speed = 2 m/s
T = 0.25s (time for magnet to travel coil length)

N = 200 turns
B = 0.6T (or 6000 Gauss)
Coil Diameter = 8cm --> A = Pi * 0.04^2
A = 0.01005

E = (-)200 * [ (0.6 * 0.01005) / 0.25 ]

E = 4.8V

3) I'm aware of the performance of commercial generators (i'm an electrical engineer), but I'm trying to understand how the figured and background theory work - as the concept i'm trying to come up with, would be very different to a typical commercial generator.

But yes I agree the output power must equal then input power, minus losses. So if my magnet weighed 1Kg and was travelling at 2.5m/s - then its maximum kinetic energy would be = 1/2 mv^2

K.E. = 2 Joules / sec (2 Watts) ?

4) thanks for the link - but the explanation on the Yahoo answers, highlights the point I made in my original post. Using the simple Ohms law relation V/R = I if V is fixed at 4.8V, and you use a very low resistance load - it would appear that you can get a very large wattage output - which clearly isn't the case for the small magenta i'm considering. (see point 3)

Thanks
Steve

8. May 16, 2014

### Pharrahnox

Sorry, I can't be of anymore help than to correct one of your formulas that you used in your original post: P = V2/R, not P = V/R. So at 4.824V and 1Ω, power would be 4.8242/1 = 23.271W.

And you said the induce current would be 4.824W, but it should be in amperes, not watts - just making sure you know the differences between the two.