Proof of common sense when maths is lacking concept

[toneboy1]

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!

 

[Ratch]

toneboy1,

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).

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.

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.

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.

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?

You failed to specify what the problem is.

Ratch

 

[milesyoung]

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).

 

[toneboy1]

First of all, thanks for the reply;

toneboy1,



Irrelevant point, the length of time an appliance is not 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.

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'.

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.

Indeed.

You failed to specify what the problem is.

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?

 

[milesyoung]

There's your problem - less power does not mean less ohmic resistance.

 

[toneboy1]

There's your problem - less power does not mean less ohmic resistance. .

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?

 

[milesyoung]

Why would it have less resistance?

 

[toneboy1]

Why would it have less resistance?

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

 

[Ratch]

toneboy1,

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

Ratch

 

[toneboy1]

toneboy1,

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

Ratch

oh, ok, fairenough. What about a large motor to a small one?

 

[milesyoung]

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.

 

[toneboy1]

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.

I see what your saying, if you endulge me, so if I had two motors, both with the same gauge 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

 

[phinds]

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.

 

[toneboy1]

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.

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

 

[milesyoung]

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..

 

[toneboy1]

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.

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

 

[milesyoung]

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).

 

[toneboy1]

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).

I know, that's 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...

 

[milesyoung]

Hehe, yes - less power but also less current, which means less field strength and less torque. It's really not that simple.

 

[toneboy1]

Hehe, yes - less power but also less current, which means less field strength and less torque. It's really not that simple.

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.

 

[Ratch]

toneboy1,

I see what your saying,

I am not so sure about that.

...if you endulge me, so if I had two motors, both with the same gauge wire, one with more windings and larger, one with less and smaller, the smaller one would have more current and thus use more power?

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

 

[toneboy1]

toneboy1,

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

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?

 

[toneboy1]

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..

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

 

[milesyoung]

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).

 

[Ratch]

TB1,

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?

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

 

[AlephZero]

When a DC motor is rotating it acts like a generator, and creates a "back EMF", in other words a DC voltage that is opposite to the applied voltage. If you prevent the motor from turnng, the current through it is be given the Ohm's law. That is called the "stall current" for the motor and it can be much higher than the normal working current.

When the motor is turning, the back EMF (which is proportional to the RPM) reduces the current. So as the motor speed increases, the current decreases and the electrical power decreases.

The amount of mechanical work being done by the motor usually increases with the RPM, and the result is that the motor runs at the speed where the electrical power equals the mechanical power. Below that speed, the electrical power "wins" and speeds the motor up. Above that speed, the mechanical work "wins" and slows it down.

The same idea applies to AC motors, but it gets complicated trying to explain it in words rather than doing the math, and from your original question, I'm guessing you are not familiar with how to analyse AC electrical circuits yet.

 

[Runei]

The fan and the heater are both connected to an AC outlet. So we have to consider the AC system.

The heater would be made up of some sort of simple ohmic resistance that can handle a lot of power.

The power dissipated by the heater would then be (V_rms^2)/R.

The fan is a coil/resistance circuit. Possibly the resistances are not very big. But the coils may have a large inductance. Because of the large inductance, the circuit will not draw very much current because it is running in AC mode. Thus, the resistances are not dissipating very much power.

Heater - Small resistance - High power.
Fan - Large indutance leads to less AC current - Low power.

 

[Ratch]

milesyoung,

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).

Perhaps you know more about this than I do, but let me give you my take on what happens.

The average current is zero, just like it is when you apply an AC voltage across a plain resistor. DC resistance is the same as AC resistance, specifically the resistance of the motor coils and brush resistance. I don't know what effective DC resistance is. When a motor is running with no load, its voltage and current are way out of phase, and it is taking relatively little power from its voltage source. When a motor has a load, its impedance lessens because its reactance lessens, and its impedance becomes more resistive. This causes the motor to draw more power from the voltage source due to the phase change. During this time the resistance has not changed. It is only the reactance that has changed, which in turn changes the phase and increases power consumption.

Ratch

 

[milesyoung]

Ratch,

I can't argue with any of that. It seemed like at the start that there were some trouble with the basics, so I was trying to stick to the definition of resistance and avoid the details of AC/motors: A DC motor would appear as having a higher resistance at its terminals when running due to back-emf.

 

[yungman]

How can you say who use more power if there is no specification of the heater and fan? You can have a really big fan and a tiny heater!