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Power setting on hairdryers

  1. Sep 11, 2008 #1
    A hair dryer hooked up to a DC power supply has 2 power settings.

    One is 600 watts and the other is 1200 watts.

    How is a hairsdryer able to have 2 power settings when resistance is constant and the voltage is constant?

    Am I missing a valuable component inside a hairdryer that allows this to happen? What's going on?
  2. jcsd
  3. Sep 11, 2008 #2


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    The implementations I've seen are that the heating element is split into 2 parts.
    In one mode only one section is powered in the other both sections are powered.
    So the resistance in this case is not constant.
  4. Sep 11, 2008 #3
    so, with a higher resistance at constant voltage...the current must be lower. If current is lower at higher resistance, the power consumed is lower. so, at 600 watts setting, the hairdryer is connecting to the part that contains more resistance. IS THIS CORRECT?

    it's a bit awkward because you would think that the higher resistance would produce a higher power setting.
  5. Sep 12, 2008 #4


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    Or they're two resistors connected in parallel?
  6. Sep 12, 2008 #5


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    What hair dryers are hooked up to a DC voltage source? That looks like a typo to me. Beyond that, the other replies are correct.
  7. Sep 12, 2008 #6
    this is a hypothetical situation...of course, AC voltage at 60 hz and peak voltage of 170 V is standard in the usa...but just to understand how a hairdryer power setting is setup...DC current is used.
  8. Sep 12, 2008 #7
    i'm not sure if that is correct.....

    look at your standard hairdryer with 2 power settings. the setting at higher power usually creates more heat for your hair drying purposes. the 600 watt setting should not create more heat, and the 1200 watt setting creates more heat. TRY IT YOURSELF...TURN ON A HAIRDRYER WITH 2 POWER SETTINGS....WHAT'S GOING ON?????
  9. Sep 12, 2008 #8


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    Let's assume you have two resistors, both of resistance R (say, 24 ohms). On the low setting, only one of these resistors is connected:
    [tex]P=\frac{V^{2}}{R}=\frac{120^{2}}{24}=600 W[/tex]

    On the high setting, both of these resistors are connected in parallel, resulting in a total resistance of R/2 (in this example, 12 ohms). Consequently, the power consumed is:
    [tex]P=\frac{V^{2}}{R}=\frac{120^{2}}{12}=1200 W[/tex]

    It's because they are connected in parallel (and not series) that you can twice the resistance (if you just added up the numbers, ignoring their connectivity) and twice the power.
  10. Sep 12, 2008 #9
    oh...i haven't gotten to parallel and series resistors set ups.

    so, if resistors are set up in parallel, you don't add the resistance like you add capacitance...is that right???
  11. Sep 12, 2008 #10


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    :confused:How do you start connecting up capacitors before resistors?:confused:

    But yes, you add them the opposite of how you add capacitances:
  12. Sep 19, 2008 #11
    It could have a single element and an electronic (switched mode) power controller.

    I seem to remember one which also altered the motor speed as well as the heating power.
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