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12V DC Motor Wind Turbine Answered

Hi im trying to build a wind turbine.

The purpose of the wind turbine is to determine the maximum amount of power it can produce with different loads. It needs to produce at least two watts of power during testing, I tried using a potentiometer but it started smoking. I need to record voltage, current and power.

I needed to know how to set this up, and if I can use resistors instead of a potentiometer.

I have

power supply

dmm

resistors

blade

12V DC

Wires

Discussions

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Jack A Lopez

17 days ago

One of the nice things about resistors is the linear relationship between voltage and current, namely,

V=R*I

also called, Ohm's law.

Combine Ohm's law with P=I*V, the usual expression for electric power, and I get some nice expressions for power dissipated by a resistor, namely,


P = I*V = I^2*R = V^2/R

So, the power dissipated by a resistor is a predictable quantity.

I can estimate, in advance how much power a resistor is going to dissipate in various circumstances.

Like for example, if I have a 50 ohm resistor, with 10 volts across it, how much power is it dissipating? Answer: P = (10 V)^2/(50 ohm) =100/50 = 2.0 watts

So for your generator toy,I would just get a handful of 100 ohm resistors, and arrange some number of them in parallel, to give you some loads smaller than 100 ohms: e.g. 2 in parallel for 50 ohms, 3 in parallel for 33 ohms, 4 in parallel for 25 ohms.

Also the power handling ability does not decrease when I change my load in this way, because the load is being shared across a plurality of resistors.

In contrast, if my load is a potentiometer, then the power handing ability does decrease as I decrease its resistance towards zero, because all the power is being dissipated in the fraction of the resistance in the pot, between the wiper and the other terminal.

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Jack A LopezJack A Lopez

Reply 17 days ago

I forgot to mention this yesterday, but regarding the power rating for these 100 ohm resistors, or whatever you use, the power rating for any one of them should be greater than the max power any one, alone, would experience.

I am guessing about 2 watts or 5 watts for each 100 ohm resistor, since the voltage needed to make a 100 ohm resistor dissipate that much power is around 14 volts or 22 volts, respectively.

Since your generator is, nominally, some sort of 12 volt DC motor, I am guessing it will not produce much more voltage than 12 volts or so, especially if it is turning at less than its usual speed. By that I mean, whatever speed it turns, when driven as a motor.

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Downunder35m

18 days ago

You can use what you like as a load ;)
Ok, jokes aside, there better options of course.
I prefer simple light bulbs as they won't care about getting hot and won't go up in smoke.
As you are on 12V anything from a car or 12V halogen system will do.
One break light would offer you a load of 5 and 20W for testing.
Adding them in series increases the resistance while connecting them in parallel lowers the resistance and through that the power in Watt.

Bonus with light bulbs is that you also have a visual output - the brighter the light the higher the voltage.
If you need more control over the load then I suggest to use Nichrome or Kanthal wire.
You can measure the resistance of the wire with the multimeter or just calculate the required lenght based on what is printed on the label.