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Proof of common sense when maths is lacking concept 
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#37
Nov1912, 05:21 PM

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It would be a great idea if this problem could be sorted out for Resistive Loads first. When we've got that all straightened out then we could move on to inductive and rotating loads. These are much more complicated and certainly not to be leaped into without getting the basics first.
For resistive loads, all you need to thing of is 1) P = IV : the Power relationship and 2) R = V/I : the way we define Resistance Combine these two together and you get two more useful formulae: 3) P = V^{2}/R and 4) P = I^{2}R Those four formulae will give you the answer to all possible questions about resistance, power and the way it can be supplied. 3) shows you that the power delivered (and used) goes down as the resistance is increased. 4) shows how the power lost in a supply wire, for instance, will increase as the resistance of the wire increases (that's if you keep the current the same) 


#38
Nov1912, 07:25 PM

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This is because the current in a fast spinning fan is largely not determined by the windings' resistance. As others have pointed out, the electrical model for a fan is not simply a fixed resistance and nothing else. It's that "something else" that is the key to explaining why it is that motors use more power the harder they work. 


#39
Nov1912, 10:01 PM

P: 174

Thanks Windadct. Rach: "The power will be greater because the voltage and current are in phase longer." good way to put it, but I didn't necessarily mean stall the motor, just slow it down.
"The "back EMF" is due to the collapsing magnetic field" Ah, right, ofcourse, thanks. As for the thread in the math section, I did, but I from what I've seen the people here stand a better chance of allowing me to figure it out. Thanks sophiecentaur, I think I've got the purely resistive relationships down and as far as I'm aware the phase relationship of an ac motor too. The only possible area of lacking I feel might be possible is the way the EMF reduces the current in a DC motor. That said I'm still satisfied. 


#40
Nov1912, 11:14 PM

P: 315

TB1,
Ratch 


#41
Nov1912, 11:25 PM

P: 174

Thanks, well I had no reason to think they wouldn't be the same, but you're saying they're not equal? That shouldn't be, are you sure? For argument's sake if I just had the LHS and wanted to reduce it to the form of the RHS would I still have to do the conjugate times numerator and denominator? (surely not) I was just looking for the simplest way to solve the LHS without expanding into 25 term polynomials and having to factorise cubics. I've used wolfram on the net and I have matlab and I think maple (which I've never used) but I don't know what I'd use to show me how to solve anything like that. 


#42
Nov1912, 11:57 PM

P: 315

TB1,
Ratch 


#43
Nov2012, 12:05 AM

P: 174

Thanks for pointing out that they're not equal, I could have wasted even more time.
I'll have to give maple a go when I get a chance. To varify. To clarify, these two steps were in someone's working out, and I didn't know how they got from LHS to RHS, it wasn't a question of proove they're equal or not (if there's been a misunderstanding). I still dont see how I'd factorise the denominator to get it like ()^2 + ()^2 from the denominator of your (1). To do this it doesn't matter if there are j's in the denominator. 


#44
Nov2012, 12:19 AM

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TB1,
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#45
Nov2012, 12:39 AM

P: 174

I don't see why it's necessary to eliminate all the complex numbers on the denominator or how it could be factorised to that, given the highest exponent of the solution saught will be 4, not 8. I could factorise the two term part by completing the square, but as for the 3 term polynomial squared, I wouldn't know where to start...Thanks for your help anyway. 


#46
Nov2012, 05:06 AM

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Reading this thread reminds me how similar electric motors and radio antennae behave (and even loudspeakers). The upfront resistance is usually just not relevant to their performance  a dipole is an 'open circuit' but it can look like a useful 70Ω at the right frequency. Electrical resistance turns up where you'd least expect it but Energy Conservation requires it. The Back EMF is only a 'way of looking at it', I think. There must be an alternative way of looking at it involving self inductance of a rotating machine. 


#47
Nov2012, 10:59 AM

P: 315

TB1,
Ratch 


#48
Nov2012, 11:47 AM

P: 315

TB1,
Just to follow up on your previous question about that math expression. If you look at equation (6) of the attachment, you can find an alternative form of that expression. If fact, there are many different forms. Once you have the denominator rationalized, you can use partial fraction expansions. Then the number of different expressions is only limited by your imagination. Ratch 


#49
Nov2012, 12:50 PM

P: 526

I'm going to throw in my 2 cents here because I think there's a small misconception with regards to selfinductance and its significance.
Let's consider a single winding of an AC motor which we'll model as a resistor in series with an inductor. It's true that a changing current through the winding will produce an opposing voltage in proportion to the selfinductance of the winding, but this isn't what limits the current in the circuit as the angular velocity of the rotor increases. Be it a DC or an AC motor, as the motor speeds up, the current is limited by the voltage produced by the increasing rate of change of magnetic flux through the winding, due to the increasing angular velocity of the rotor field. It's this component of induced voltage that's usually termed 'backEMF' in motors. 


#50
Nov2012, 02:29 PM

P: 315

milesyoung,
Ratch 


#51
Nov2012, 02:55 PM

P: 526

Give it a second readthrough  in the simplest case there are two causes of changing magnetic flux in the winding. One is a change in current in the circuit  the other is a change in the angular position of the rotor. The latter is what limits the current.
Edit: And just clarify further, I'm talking about two different components of opposing voltage  that's my whole point. 


#52
Nov2012, 03:15 PM

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milesyoung,
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#53
Nov2012, 03:23 PM

P: 526

How else would you explain an increasing opposing voltage in a DC motor? For a constant current the inductive voltage drop associated with the selfinductance is clearly not present. 


#54
Nov2012, 05:08 PM

P: 315

milesyoung,
Ratch 


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