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Optimum resistance for heat

  1. Mar 6, 2007 #1
    I'm a mechanical engineer getting confused with electrical engineering...
    I've made an electrical generator by spinning the shaft of a DC motor by connecting it to my bicycle.

    I've tested it with a multimeter, and on 1 kOhm, 2kOhm, 3kOhm, and 4kOhm resistors I'm producing about 70 Volts. Current is varying accordingly with V=IR. Therefore power P=IV is increasing as resistance decreases.

    My question is how do I find the optimum resistance to produce the most electric power? By this calculation, power would increase as resistance decreases. However there has to be some limit to this, as I was only putting in about 200 W into the bicycle. According to the equations a 10 ohm resistor would lead to 7 amps and 490 W, but obviously I'm not producing more power than I'm inputting.

    Can someone please help me approach this problem? What is the optimum resistance for optimum power?
     
  2. jcsd
  3. Mar 6, 2007 #2
    When the source resistance is equal to the load resistance, that's when maximum power transfer occurs.
     
  4. Mar 6, 2007 #3
    The source is a DC motor, rated at 24 V and 16 amps. Therefore the resistance is 1.5 ohms.
    But this is the motor specs, there are no specs for spinning the shaft and turning it into a generator. We are producing 70 V, so a resistance of 1.5 ohms result in 3267 W. This is not possible as I'm inputting roughly 100-250 W.

    Thanks.
     
  5. Mar 6, 2007 #4
    You can't go by the 24 volt/16 amp rating. This means that the motor is designed to run at 24 volts and draw 16 amps when under its maximum rated load. When running idle the motor will draw much less current. Keep reducing the resistance until the calculated power starts to drop. By calculated power I mean calculate it from measurements of voltage and current. What happens is the voltage will drop faster than the current increases. This is when the load impedance is equal to the source impedance as waht has mentioned. You will need some serious high wattage resistors to do this. I would recommend using sealed beam headlights instead. I've used these for testing power supplies before. You can wire them in various combinations to get different loads.
     
  6. Mar 6, 2007 #5

    dlgoff

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    "I would recommend using sealed beam headlights instead."

    yep. And I've even used 110v bulbs. You can get lots of different combinations by using different wattage ones.
     
  7. Mar 6, 2007 #6
    Those are the maximum rating for the motor, can't use that. You say max voltage rating is 24V, yet you measure 70V. The actual source resistance is something else entirely, and will depend on the rotation (rpm) of the generator.

    Note, Different rotational speeds will probably result in different source resistances.
     
  8. Mar 7, 2007 #7
    At the end of the day P=IV. V will be determined by the speed you are spinning your motor/generator at so you need to set the gearing to the motor such that it will deliver the required voltage for the load you intend to use, or 70V in your set-up. I is therefore the input power minus losses divided by the output voltage. If you put a big load on the end you will find it harder to turn the generator, thats whats happening with the resistors, the bigger the load the harder it is to turn the generator at x rpm to get 70V, but you probably don't notice it as it is well within your power input capabilities.
    The 16A rating is the rating of the motor windings. So what ever load you place on the end of the generator you should not pull more than 16A. The Voltage rating of the motor will be much higher than 24V, can't think how to calculate it though at the moment.
    As an example if I start my 1964 MG on a really cold day and then switch on all the electrical bits the car stalls at idle because the load on the generator is putting load onto the engine.
     
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