What is the theoretical limit to coil inductance over coil resistance?by kmarinas86 Tags: coil, inductance, limit, resistance, theoretical 

#1
Feb2408, 04:22 PM

P: 1,011

Is there a theoretical limit to the time constant of an electric motor?
I was think, if inductance is simply the conversion of current into the magnetic field, and if increase in resistance limits that current for a given power supply, that having a higher time constant would in effect allow for a stronger magnetic field for a given current. But what is the limit to this? Could I develop a series of optimized coils such that the magnetic field I produce with a given power supply can be of successively increasing magnetic fields? 



#2
Feb2408, 08:59 PM

Sci Advisor
HW Helper
P: 1,572

The magnetic field for a given current is a constant.
I suspect you are confusing voltage with current here. Power supplies are defined in terms of voltage. While a given power supply might have a maximum current, it may never end up supplying that maximum current Perhaps you could clarify your question. 



#3
Feb2608, 06:54 PM

P: 1,011

Let's say we systematically reduce applied power by decreasing current, yet increase inductance such that the smaller change of current need to produce that current can produce the same change of magnetic flux. Say this is done by reducing amps by a factor of two and increasing inductance by a factor of two. Resistance increases by a factor of two. The power is then reduced by a factor of two while the change of magnetic flux that can be produce corresponds to the voltage which was kept the same. If these steps were repeated over and over again, at what point do we say, "We can no longer decrease the current nor increase the inductance of our device such that we can produce the same magnetic field strength for less input power"? Keep in mind that even though ability to quickly switch the magnetic field on and off is dependent on the ratio of inductance to resistance (time constant), this time constant be predefined by setting the resistance in proportion to increased inductance, such that it not affects the maximum allowable change of magnetic flux supported by the given applied voltage. If you do not understand the above, may I ask this? "What is the maximum magnetic flux per second that can be used to turn a magnet, as limited by the input power?" 



#4
Feb2708, 10:50 AM

Sci Advisor
HW Helper
P: 1,572

What is the theoretical limit to coil inductance over coil resistance?Yes, E=IR always applies. Edit: Relativistic and Quantum limits not considered here as they are well beyond what we can construct and not well defined. 



#5
Feb2808, 11:51 AM

P: 1,011





#6
Feb2808, 01:36 PM

Sci Advisor
HW Helper
P: 1,572

None. It is not an aspect of motor performance.
Motor performance is rated as output_power / input_power. Performance will always be less than 1. Major elements affecting motor performance are eddy current losses and IR losses. Eddy current losses are minimized by the motor core construction and have nothing to do with coil winding. The only thing I can think of that might relate to what you seem to be asking is shaft speed. Even then it is not the only determinant for shaft speed. 



#7
Feb2808, 04:44 PM

P: 1,011

I said:




#8
Feb2808, 11:25 PM

Sci Advisor
HW Helper
P: 1,572

No power is involved creating magnetic flux. It is just a change of one potential form into another. 



#9
Feb2908, 12:53 AM

P: 1,011

Say I can create and destroy 1 billion teslas in 1 cubic kilometer of space every minute without consuming any power. What do you think I have done, say, to an automobile? That's overunity and supposed to be impossible. 



#10
Feb2908, 05:19 PM

Sci Advisor
HW Helper
P: 1,572

Perhaps I was less then clear.
In many ways you can think of a magnetic field as a battery. It stores potential energy. Think of a rock sitting on a hill. The rock has a potential energy of x. What hear you saying is that the potential energy value x is going to be different depending on how fast you pushed the rock up the hill. If it takes you a second or a month to push the rock up the hill x will have the same value. 



#11
Feb2908, 11:23 PM

P: 1,011

This program gave me certain results that suggested that the inductance to resistance ratio increases with inductance as well as the thickness of the wire used: http://www.datavoyage.com/coilmaestro/ Am I wrong? Brooks coil: http://home.san.rr.com/nessengr/tech...ks/brooks.html 



#12
Mar108, 01:01 AM

P: 703

You can have less, of course. Real copper wire has resistance and that resistance wastes some of the supplied power as heat. And as you apply a higher and higher current, you reach a point where the ratio of heat production versus increase in the magnetic field energy becomes unfavorable; the wire melts! But, understand that even with superconducting wire, the ratio of the energy in the magnetic field to the power supplied doesn't increase without limit. That's what energy is; it's the accumulated supplied power. Always, in every case. Whether you're considering energy in an electrostatic field, energy in a magnetic field, mechanical energy stored in a spring, etc. If you pump water uphill into a lake, as some utilities do to store energy for later use in generating electricity, the gravitational potential energy stored by the water up on the hill is never more than the accumulation of the power used to pump it up there. In the real world, it's always less because of inevitable losses. Nobody has ever found a way around this, despite what the free energy people say (and a lot of people have tried!) 



#13
Mar108, 09:12 PM

Sci Advisor
HW Helper
P: 1,572

You might thing of this as the number of rocks on the hill. It has nothing to do with potential. As for the rest I think you should read what The Electrician wrote. 



#14
Mar108, 11:37 PM

P: 1,011

He says: 9 amps going through a 1ohm coil for 1/9th of one hour. The power loss is (1)(9)(9) watts. 3 amps going through a 3ohm coil for 1/3rd of one hour. The power loss is (3)(3)(3) watts. 1 amp going through a 9ohm coil for one hour. The power loss is (9)(1)(1) watts. The power going into the 1ohm coil is (9V)(9 amps) which is the same as the power loss! The power going into the 3ohm coil is (9V)(3 amps) which is the same as the power loss! The power going into the 9ohm coil is (9V)(1 amp) which is the same as the power loss! What? I don't get it. /// (1)(9)(9) watts  (9V)(9 amps) = 0 watts (3)(3)(3) watts  (9V)(3 amps) = 0 watts (9)(1)(1) watts  (9V)(1 amp) = 0 watts Huh? /// 



#15
Mar208, 05:34 AM

P: 703

First, I'm going to say some things that assume the inductor is wound with superconducting wire. If you allow the power supplied to the inductor to vary, then you can always allow it to become zero for a microsecond or so after you have supplied a nonzero amount of power for a while, and then the ratio of energy stored in the magnetic field to the power being supplied will be infinite. If some energy has been stored in the magnetic field already, then you can adjust the power delivery so that the ratio of energy in the field to the power supplied is any value you choose. But, so what? This is of no particular significance. An inductor wound with superconducting wire can store energy in a magnetic field indefinitely. All you have to do is supply some power for a while and then short the coil. The current will continue to circulate, sustaining the field. In that case the energy in the field will be the time integral of the power supplied, which is the same as the average power times the length of time that average power was supplied. After the power is no longer being supplied, and the inductor is shorted, the ratio of the energy in the field to the power being supplied will be infinite. But, again, so what? By the way, if a superconducting inductor is being supplied with power from a constant voltage source, the current will increase indefinitely. If the voltage source cannot supply an indefinitely large current, then when the current from the supply becomes limited, power will no longer be supplied to the inductor. Now, the behavior for inductors wound with real, copper, wire at room temperature is quite different. When power is supplied from a constant voltage source to such an inductor, only some of it is stored in the magnetic field, and that only happens during the first few time constants. Ultimately, the current from a constant voltage supply will be limited by the resistance of the coil. When that happens, power will still be supplied by the source, but it will all be dissipated as heat; none of it will go into the magnetic field any more. This limiting current will determine the strength of the magnetic field, and the energy stored in it. The higher the power supplied to the resistance, the higher the current and the stronger the magnetic field; the lower the power supplied to the resistance, the lower the current and the weaker the magnetic field. So, this means that with a real inductor, if the power is supplied at a low rate the ratio of the energy stored in the field to the power will not be larger than if the power is large. With a real inductor, to reduce the power supplied, the current must be reduced; the current cannot be reduced without also reducing the applied voltage, and vice versa. If the voltage being applied to a real inductor to maintain the current (after several time constants) is reduced to zero, the energy stored in the magnetic field will be dissipated in the resistance in a few time constants. This is very different from what would happen with an ideal inductor. With an ideal inductor, wound with superconducting wire, assume that some power has been supplied for a while. The current will have ramped up to some nonzero value. The power being supplied can be reduced to zero without reducing the current to zero; all that is necessary is to reduce the applied voltage to zero. The current will continue to flow, maintaining the magnetic field and its energy. I think you may be confusing the behavior of real inductors with that of ideal inductors sometimes. 



#16
Mar208, 11:31 PM

Sci Advisor
HW Helper
P: 1,572

You do not get it. At t=0 (when you circuit is switched on) I=0. It is definitely not 9 amps, 3 amps or 1 amp. During the period when the coil is charging (or the magnetic field is being established) the current will be less than the value set by IR. Once the current reaches the value set by the equation E=IR the coil can no longer charge and the magnitude of the magnetic field has reached its maximum value for the particular circuit. Note that the converse is that when the circuit is switched off current will continue to flow at the value it had when the switch was opened. No battery required as the magnetic field on the coil provides the potential. Since R becomes very large with the switch open the E in E=IR becomes very large as well and will break down or arc any insulation to continue the current flow. This effect can be very unpleasant if you happen to be holding the coil leads when the switch is opened. For a large inductance where a significant amount of energy gets stored it can kill you, even though your original battery voltage might have been 1 volt. 


Register to reply 
Related Discussions  
Air Coil Inductance Formula  Electrical Engineering  1  
Self inductance of a solonoidal coil  Introductory Physics Homework  1  
Inductance of a coil in an AC LR circuit  Introductory Physics Homework  4  
Finding the Inductance of a coil  Introductory Physics Homework  2 