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Proof of common sense when maths is lacking concept

  1. Nov 18, 2012 #1
    Ok, (an embarassing question) well I know an electric fan is relatively low wattage compared to say an electric kettle (while both on for the same amount of time). But also that smaller resistance draws more current than larger resistance, and by this logic (v2/R or I*R) the smaller resistance uses more power.
    To reitterate the electric fan uses less power than the kettle (or water heater, air con etc.) so what is this model failing to take into consideration? Power factor or something?

    Thanks!
     
  2. jcsd
  3. Nov 18, 2012 #2
    toneboy1,

    Irrelevant point, the length of time an appliance is on has nothing to do the rate electrical energy is consummed. In other words, a 45 W bulb will consume energy at half the rate of a 90 W bulb no matter how long it is on.

    The correct term for power at a constant voltage is (I^2)*R, not I*R. Yes, according to (V^2)/R, a smaller restance will consume more power.

    You failed to specify what the problem is.

    Ratch
     
    Last edited: Nov 18, 2012
  4. Nov 18, 2012 #3
    Not quite sure what you mean, so I'm just taking a shot here - you'll have to forgive me if I've misunderstood the question.

    Less resistance gives you more current for the same voltage. The instantaneous power delivered to the circuit is current x voltage. The resistance of the electric fan at its terminals will be higher than that of the heater (if it draws less current for the same voltage you apply to the heater).
     
  5. Nov 18, 2012 #4
    First of all, thanks for the reply;
    I understand that, what I was trying to make clear that one appliance uses more than the other, I imagined someone saying 'but you leave a fan on and a kettle turns off after boiled'.

    Indeed.

    Ok, well to the fan uses less power yet, has less 'resistance' (being for the most part an inductive motor) I assume than say an electric water heater, so how is this seeming contradiction wrong?
     
  6. Nov 18, 2012 #5
    There's your problem - less power does not mean less ohmic resistance.
     
  7. Nov 18, 2012 #6
    On the contrary, it means more ohmic resistance, but what I'm saying is that say a fan would have LESS resistance than a heater but still use LESS power, or am I wrong, is a fan actually manifold more resistive?
     
  8. Nov 18, 2012 #7
    Why would it have less resistance?
     
  9. Nov 18, 2012 #8
    I just thought it would because its just coils of copper and copper isn't that resistive. (not sure how many meters but it would seem like if I had a huge motor with more resistance than it would be less power to run than a small one, which is counter intuitive)

    I would have thought, for an idea of proportionality a fan would be like 100 ohm fan and a heater 3k ohm (again, just for a relation, no idea what they would actually be)
    Thanks
     
  10. Nov 18, 2012 #9
    toneboy1,

    A small fan has more resistance than a heavy duty heaater, plain and simple.

    Ratch
     
  11. Nov 18, 2012 #10
    oh, ok, fairenough. What about a large motor to a small one?
     
  12. Nov 18, 2012 #11
    For a large motor you'll typically also use more material for the windings, so you can't really assume some kind of proportionality between size and winding resistance.

    Regardless of what circuits you're working with, if one draws more current on average than another for the same voltage, it will have less equivalent resistance.
     
  13. Nov 18, 2012 #12
    I see what your saying, if you endulge me, so if I had two motors, both with the same guage wire, one with more windings and larger, one with less and smaller, the smaller one would have more current and thus use more power?

    Thanks again
     
  14. Nov 18, 2012 #13

    phinds

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    Gold Member

    I think some of the confusion here is the distinction between resistance and impedance. Resistance is a coverall term for DC circuits but is meaningless for AC circuits, which use the term impedance which takes into account non-DC effects.

    That is, a coil of wire can have REALLY low "resistance" as measured by a DC ohmmeter, but when you run an AC current through it, the inductance causes a high impedance.
     
  15. Nov 18, 2012 #14
    Quite right, but power only uses the ohmic resistance 'real' part of impedance doesn't it? So for my last question the larger motor would still have more real impedance. (just trying to illustrate how I can clarify my counter-intuitive confustion)
    Cheers
     
  16. Nov 18, 2012 #15
    If we're just talking about motors in the sense that we model them as a series connection of a resistor and an inductor, then yes - the small motor will draw more current for the same voltage and thus use more power.

    Edit: And of course as phinds pointed out, I assume we're strictly talking DC here.

    Edit: You're not actually going to use this argument for anything important, right? There's a lot more to motors..
     
  17. Nov 18, 2012 #16
    that seems like it's violating concervation of energy (I know it mustn't be) but you know what I mean, the more you get (like more torque or whatever) the less power, for that matter it seems like I could just put a resistor on the end of the terminal and make it use even less power (though that would probably slow the fan down). Sorry this question must be so irritating but I hope you can see why I'm confused
     
  18. Nov 18, 2012 #17
    You are oversimplifying motors. Less windings also mean less torque - typically (there are a lot more factors to consider here but for the sake of argument).
     
  19. Nov 18, 2012 #18
    I know, thats what I said, so you make a huge motor, with heaps of windings (and torque) and yet is uses less power than a small one....
     
  20. Nov 18, 2012 #19
    Hehe, yes - less power but also less current, which means less field strength and less torque. It's really not that simple.
     
  21. Nov 18, 2012 #20
    M'mm, I'll accept that. I'm going to get my multimeter tomorrow and check what my fan and 500W distiller are, I really can't believe a fan would be so resistive.

    Thanks for your help.
     
  22. Nov 18, 2012 #21
    toneboy1,

    I am not so sure about that.

    Are the current and voltage in phase? If you applied AC line voltage to a single coil of high inductance, you could have a tremendous amount of current, and yet dissipate a very small amount of power. That is because the voltage and current are out of phase with each other. It will take a large amount of current to build up energy in the magnetic field of the coil during a quarter cycle, but that energy will be given back to the circuit in the next quarter cycle, and repeat the process in the reverse direction. Therefore, it will be large current and small power dissipation.

    You are trying to confuse AC power distribution with DC power. It doesn't work that way. You need to know the phase ramifications of AC before you can calculate AC power.

    Only the resistance part of the impedance dissipates power. The more the voltage and current are in phase, the more the resistance dominates the impedance.

    Ratch
     
  23. Nov 18, 2012 #22
    YES that's more the answer I've been looking for! THANK YOU, I knew a bloody fan wasn't that resistive.
    Out of curiosity, what about a DC motor, given that there is no phase change (or voltage lag) would one use more power than its ac counterpart?
     
  24. Nov 18, 2012 #23
    I wasn't just talking DC, yeah I know there is a lot more to motors but every analogy I use seems to make the question harder to explain, I knew there was just a simple answer, I just ended up using motors to make the point.

    EDIT: DC probably would have simplified things though :P
     
  25. Nov 18, 2012 #24
    I was just about to write that you were probably going to come back tomorrow angry that you've measured the resistance using an ohmmeter and you were right about the resistance etc :)

    Just to make my point clear - I've only been talking about DC current and ohmic resistance. If you measure the average current into your circuit for some voltage, that relationship will give you its effective DC resistance - it won't have much to do with the winding resistance you'll measure (which will be very low).
     
  26. Nov 18, 2012 #25
    TB1,

    A DC motor has a commutator that switches the current direction in the rotor. That causes magnetic fields to form and collapse, thereby storing and releasing energy. This in turn causes a phase change in the voltage and current. Its behavior is complicated. See the neat animation at this link http://en.wikipedia.org/wiki/DC_motor .

    Ratch
     
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