Proof of common sense when maths is lacking conceptby toneboy1 Tags: current, electric, model, power factor, resitance 

#19
Nov1812, 10:09 AM

P: 427

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




#20
Nov1812, 10:13 AM

P: 174

Thanks for your help. 



#21
Nov1812, 10:13 AM

P: 315

toneboy1,
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 



#22
Nov1812, 10:16 AM

P: 174

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? 



#23
Nov1812, 10:20 AM

P: 174

EDIT: DC probably would have simplified things though :P 



#24
Nov1812, 10:27 AM

P: 427

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). 



#25
Nov1812, 10:31 AM

P: 315

TB1,
Ratch 



#26
Nov1812, 10:42 AM

Engineering
Sci Advisor
HW Helper
Thanks
P: 6,347

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. 



#27
Nov1812, 10:47 AM

P: 134

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. 



#28
Nov1812, 11:11 AM

P: 315

milesyoung,
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 



#29
Nov1812, 12:38 PM

P: 427

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 backemf. 



#30
Nov1812, 12:56 PM

P: 3,843

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!!!




#31
Nov1812, 01:00 PM

P: 427





#32
Nov1812, 01:02 PM

P: 315

yungman,
See post #9 of this thread. Ratch 



#33
Nov1812, 01:40 PM

PF Gold
P: 1,370

Electric Kettle: rated power: 1500 watts line voltage: 120 derived current: p/v=i > 1500/120=12.5 amps derived resistance: v/i=r > 120/12.5=9.6 ohms Electric Fan: rated power: 100 watts line voltage: 120 derived current: p/v=i > 100/120=.833 amps derived resistance: v/i=r > 120/.833=144 ohms I guess I don't understand where common sense comes into play with this problem. There are only equations. hmmm.... (scratches head, runs aerodynamic analogy through skull...) Ah ha! I think I understand what you are asking. Interesting. If mythical car A has an aerodynamic drag resistance of 144 units, it would make sense that it would consume more power than mythical car B with an aerodynamic drag resistance of 9.6 units. (I'm ignoring back EMF and inductive resistance here, as I'm obviously interpreting your question differently than others.) hmmm.. Where do we go from here? The mythical cars will both have to have engines. I'll give them both the same engine, rated at 120 mythical units, which I will call volts. And what does this tell us so far? It tells us how fast the cars will go! Volts/Ohms = Amps. Amps are the mythical unit for velocity that the cars will travel at. (This is somewhat of a bad analogy, as real Amps are a measure of a certain number of electrons passing a certain point per second. Kind of like the number of cars driving through an intersection per second. But it kind of implies motion, and I'll stick with it. Let's just pretend that each car is negatively charged with an excess of 6.241 × 10^{18} electrons(or 1 Coulomb), and each cars length is equal to 1 coulomb per 0.003 miles, and see what happens) Car A: (aka, the fan) volts: 120 ohms: 144 amps: .833 (egads. now I understand the PFEE heavyweights aversion to analogy) length: 13 feet velocity: 9 mph power: (Force, aka volts)*Velocity = 1076 oomphs(new mythical name of electroaerodynamic power unit) Car B: (aka, the kettle) volts: 120 ohms: 9.6 amps: 12.5 (why is length based on charge?) length: 198 feet velocity: 135 mph power: 16,200 oomphs So what does this all mean? They've both got the same motive force. But the car with the least resistance is using the most power. scratches head again. hmmm... It still doesn't make sense, but at least the maths worked out: 16,200 oomphs / 1076 oomphs = 15.06 1500 watts / 100 watts = 15  ok to delete. I enjoy nothing more than Sunday morning mental aerobics. 



#34
Nov1912, 02:30 AM

P: 174

Some good points were made and I did understand everything said and why, except for maybe you cheeto, but it was a hilarious departure nevertheless.
So to summerise, the more resistance of a load on a motor, the more it increases the current, which makes it use more power. So if it's an ac motor it will change the phase of the voltage or current, (which I dont know) so the peaks will be closer and I*V will be bigger, or if its DC then the back emf will decrease and it will use more power. COOL. Exactly the sort of explanation I was looking for. thanks EDIT: If any of you (possibly clever engineers) could, I have an algebraic nightmare (a few of them) in this thread: http://www.physicsforums.com/showthr...=1#post4165324 I'd REALLY appreciate some help! EDIT: What the hell, I have a Fourier question too for what it's worth: http://www.physicsforums.com/showthr...36#post4165336 



#35
Nov1912, 02:21 PM

P: 532

Hello Toneboy  it may be possible that if you just measure the Resistance of the motor it has less resistance then the heater  but STILL uses less power  why? Because a motor when running  is not a resistor. The structure of the motor generates a magnetic field, and as the motor rotates, it generates "back EMF"  and still a little more dynamic than just impedance  for example a DC motor generated Back EMF  but steady state DC in an inductor does not have back EMF.
Since now the motor is generating EMF ( simply put Voltage) that opposes the voltage being applied  this reduces the current. 



#36
Nov1912, 03:32 PM

P: 315

toneboy1,
Ratch 


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