Ask Me Help Desk - Electromagnetic Induction

Hey,

I have got a little experiment going just out of the sake of curiosity and I am running into some problems.

So I want to store some energy in a capacitor (10F 2.5V, huge I know) using electromagnetic Induction. I started with a setup with around 70 coils and I would shove a magnet though the coils and back really fast trying to induce as much voltage as possible. I have those lead wires run through a full wave rectifier and then through my bread board to my capacitor, but the voltage reading out of my capacitor was almost not changed at all, I then took some data right out of the rectifier instead of from the capacitor and I was generating around 4V. Discouraged I realized that I may have to increase the amount of current that is going to the Super cap so I upped the 70 coils to 1500 coils and changed the wire to really thin copper wire (it was like copper hairs). I used the same magnet and the same capacitor to up my voltage and to increase my current. Then I took some more data from the capacitor and the voltage coming out was still the same.

I thought that upping the amount of coils would charge my capacitor faster(or at all), and it didn't. Should I even be looking at the voltage output of the capacitor? Is there something about electromagnetic induction that I am forgetting? Are there missing parts to my circuit that I need to make my capacitor store energy?

Thanks,
Race

First, I think there's a bit of confusion here between voltage and current. Moving the magnet through the coil will induce a voltage (not current), whose magnitude depends on the nature of the changing magnetic flux that the coils are subjected to as you move the magnet. I suggest taking that output, connect it to your rectifier and then measure the output voltage (without the capacitor) to see whether it's working as intended.

Second - you don't mention what value of resistor you are using, which should be in series with the voltage source (your rectifier) and the capacitor. You can take a stab at what value to use from the fact that using really thin wire means that you want to limit the current to perhaps 100 mA max. So if your voltage source puts out, say, 5 volts, R should be around 50 ohms or greater.

The value of the resistor R times the value of the capacitor C gives the time constant for the circuit, which is a rough way of measuring how quickly the capacitor will charge to about 63% of its final voltage. For C=2.5 ahd R = 50 this is 1250 seconds! So trying to power it up with a voltage source with a frequency much faster than that will not work well at all.

Finally, because the voltage wave form coming out of the rectifer is not constant, the capacitor wlll tend to discharge whenever the supply voltage is below the voltage of the capacitor. Again - you need to supply a steadier voltage and/or reduce the time constant of the circuit.

I would strongly encourage you to use a much, much smaller value of C. Not only might you get this to actually work, but it's much safer as well.

Wow thank you so much!

I had no idea about the resistor thing because in my head I would use the resistance of the wire as my R in V=IR. So awesome. Thanks.

I thought my current would increase because I had added more coils to my circuit increase my V value in V=IR and keeping my R value the same would increase I. Is my thinking incorrect?

Thanks

Quote:

Originally Posted by racee

I had no idea about the resistor thing because in my head I would use the resistance of the wire as my R in V=IR. So awesome. Thanks.

You're welcome. If you only relied on the reistance of the wire then if you could ever get the capacitor to charge it would discharge with so much curren as to vaporize the wire!

Quote:

Originally Posted by racee

I thought my current would increase because I had added more coils to my circuit increase my V value in V=IR and keeping my R value the same would increase I. Is my thinking incorrect?

The voltage on the resistor is reduced as the capacitor charges, and hence the current flow in the circuit changes as well, even if the generator is putting out constant voltage. So it's clearer to think in terms of the voltage output of the generator, not its current.