Question:

Reduce 580watts to 380watts?

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I have a heating element that is 580watts and my question is would it work if i put 2x 100watt light globes in series to drop the heating element resistance down to approx 380watts ?

Link to original question http://au.answers.yahoo.com/question/index;_ylt=AjmMM1xvRil7nO5F3P.gkCrg5gt.;_ylv=3?qid=20080620220307AAqnjbS

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  1. Only if you are very lucky.

    240v and 580w is 2.42 amps, resistance is 99 ohms.

    All things equal, at 380 watts and 99 ohms you need:

    P = E²/R

    380 = E²/99

    E = 194 volts

    I = 194/99 = 1.96 amps

    But heating elements and light bulbs do not obey Ohm's law because the resistance varies with temperature.

    But hoping the numbers are somewheres close to correct, you need to drop 240-194 colts or 46 volts, at about 2 amps, or 90 watts. R would be 46/2 = 23 ohms.

    Two 240 volt 100w light bulbs will have a resistance each of:

    P = E²/R

    100 = 240²/R

    R = 580 ohms, or 290 ohms for two of them.

    so, neglecting the nonlinearities due to temperature, the two bulbs are no where close. They total 290 ohms, you need 23 ohms. If you can find two 150 watt 120 volt bulbs, it will be closer, at 48 ohms.

    Be prepared to fiddle, start with the 2 US bulbs (120v) and go on from there.

    Realize that even if you find the perfect set of bulbs, your power dissipation will be 380 watts in the heater BUT the total power dissipated will be 480 watts, and you will be paying for the 480 watts in your electric bill. If you just bought a 380 watt heater, you would be paying for only 380 watts.


  2. I'm not sure if you bothered to read the previous answers mate, or didn't understand them. Using the lamps is a nasty kludge, and is not a good way to go. Their resistance changes dramatically with current.

    The (lamp) dimmer is the way to go. They are $8 at any building supply store. Resistive loads are very benign, just get a dimmer rated 3A or more, and you're done.

    You will have great controllability, and can make it as cool or hot as you like. Measure the voltage and current to verify power output if that is necessary. If you need to know the power, put a precision power resistor in series, say 0.1Ohm, and measure the voltage drop across it to accurately measure the current. Use a voltmeter to measure RMS voltage across heater, multiply together.

    I was suggesting the stove components because you didn't say you are down under; 240V dimmers are not very available in the US. In Australia, almost all the power is 240V 50Hz; they are easy to come by.

  3. As I read through the stuff and on your link alot of the answers seem more academic.

    What exactly are you trying to do with this element?

    First, I think you actually want to run those lights in parallel not series.

    If its just a heating element just run it on 120V- I forget all the formulas to prove it (its been 20 years) but it should give you about 1/2 the heat.

    Or get your self a rheostat, and if you are going to go through the trouble of cutting a rheostat in you might as put in a thermostat- that way it would be automatic- goes off when it gets to hot and comes on when it get to cool.

    on your link wings was probably most accurate. Try it on 120v. And the dimmer he is talking about would be a rheostat.

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