Does a fan draw the same amount of power when using it at different speeds? Some of the people around me say so. But ${\rm power} = \frac {V^2}R$, and $V$ (the voltage) is always the same in the household supply. Then does the power differ for different resistances of the regulator or for different speeds? The highest speed should consume the most power due to having the lowest resistance, while the lowest speed will consume the least power due to having the highest resistance.
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$\begingroup$ If lower the resistance, higher the speed then more the power consumed.( since V is inverse to resistance) $\endgroup$– physics2000Commented Mar 30, 2018 at 4:17
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$\begingroup$ @physics2000 so isn't my understanding right? $\endgroup$– user190600Commented Mar 30, 2018 at 4:56
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$\begingroup$ yes! You are right. $\endgroup$– physics2000Commented Mar 30, 2018 at 5:08
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The formula for dissipated power
$$ P = \frac{V^2}{R} $$
is valid only for purely resistive elements, like straight piece of wire or a resistor. It is not valid for elements that manifest strong EM induction effects, like coil or a fan where there is an electromotor with coils of wire inside.
This is because the Ohm's law of proportionality of current to voltage
$$ I = V/R $$
is not valid for AC current in a coil. One must account for the self-induced emf force in the coil which decreases the current; the faster the fan spins, the greater the decreasing effect.
The answer to your question is that the faster the fan, the greater the transfer of energy from EM energy into air flow energy, but also the higher the impedance of the motor, so it is not so easy to say whether the fan draws more or less. It would require more detailed analysis, taking into account the construction of the motor. But if the construction was done as efficient as possible, I think the result would almost certainly be : the higher the speed, the greater the power draw.
answered Mar 30, 2018 at 12:59
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$\begingroup$ "A general result of EM induction on the coil is that large current may flow in the wires, but it dissipates much less energy than DC current would." by that do you mean that current in AC circuits are greater than DC circuits or something else? Please clarify $\endgroup$– user190600Commented Apr 10, 2018 at 1:26
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$\begingroup$ I expressed myself in a wrong way. I was thinking of the fact that when the effective current in the coil is $I$ and effective voltage $U$, average power transferred (not only dissipation into heat, but including the useful work) from power source to the appliance is $UI$ if the voltage and current are DC, but $UI.k$, which is less, if they are AC ($k$ is efficiency factor, depending on the phase shift of current from voltage). But this is not very useful in the context of the answer, so I removed that part. $\endgroup$ Commented Apr 10, 2018 at 10:43