# Amperage Formula Needed

1. Sep 2, 2005

### GarageTinker

Hello All,
I have a personal "garage tinkering" project I'm working on which involves the production of magnetically induced electricity. I've been using Faraday's Law to figure the estimated voltage output of various coils under various conditions. However, I have been unable to find a/the formula/s that will allow me to calc the output amperage of the coils. I know it involves the resistance of the material, cross-section and overal length of the wire used in the coil, and I may well have seen it but looked right past it not recognizing it as what I needed. Would someone please let me know the formula I'm looking for? I'd appreciate any assistance in this. Thank you.

2. Sep 3, 2005

### zoobyshoe

The question is more complex than you think. A coil doesn't have an inherent output voltage or amperage. The output is going to be dependent on the strength of the magnetic field with which you induce a current in the coil, and the speed with which you change the strength of the magnetic field.

Roughly: the stronger the magnetic field you use to induce current, the more current you can get out of any given coil. The faster the motion, the higher the voltage.

If you have your coils wound on a ferrous core, and are inducing a current by passing a permanent magnet near the core, the strength of the magnetic field you induce in the core is also dependent merely on how close you get it to the ferrous core: the closer, the stronger. This is due to the inverse square law for the strength of magnetic fields: double the distance and the field strength drops to 1/4 the strength, not 1/2 the strength. Cut the distance in half and the field strength is four times stronger, not twice as strong. Inverse square formula: I = 1/r2 where I = intensity, and r = distance.

All these considerations are different from the fact that the amount of current delivered by a generator has to be limited to the amount of current the wire can withstand without over heating.

So, you really can't get the answer you want from only considering the number of turns and size of the wire of your coils.

In addition, amperage is generally reackoned from the standpoint of the load, not the generator. The load being the device you are using the generator to operate, be it a motor or a lamp or whatever. For any given load, energized at any given voltage, there will be an automatic amerage drawn given by Ohm's Law I = E/R. I = current (amperage) E = voltage, and R = resistance.

To figure this you need to measure the voltage your generator is delivering, and the resistance of the load. Divide the voltage by the resistence, and you have your amperage.

The less resistence, the more amperage the load will draw, untill, if the magnetic field of your generator is strong enough, you will overheat the wire in the generator and ruin the insulation. The resistence of the load limits the amperage "drawn" from the generator.

So, if your generator coil wire is too thin, and your magnet too powerful, you can burn the wire out by operating it without enough resistence in the load. This is why we have fuses: they are gaged to burn out before the wire does.

All this can be worked out on paper if you have the means to measure the strength of your magnetic field, which, as a garage tinkerer, you probably don't. It requires a special device.

For tinkering, though, rough rules of thumb help: if you don't seem to be getting enough amperage, get a stronger magnet, or get your magnet closer to the coil. If your voltage is low, spin the magnet faster, or wind more turns of wire onto the core. If the wire is overheating before you get enough amperage to operate the load, us a larger diameter wire.

I hope this helps. If not, maybe you can describe what you're trying to do in more detail.

3. Sep 4, 2005

### GarageTinker

Thank you very much for answering my question. "Volts / Resistance = Amps". Perhaps what I should have described is that I am trying to ascertain what the possible output (volts and amps) of a coil will be under different imposed circumstances; changes to the gauge and length of wire which the coil is made from, the speed of movement of the magnetic field and a couple of others. I own a DC Magnatometer which reads in Gauss units so measuring the field strengths of the magnets I'm using isn't a problem. Thanks again, I really do appreciate your help and your time.