# What is the theoretical limit to coil inductance over coil resistance?

## Main Question or Discussion Point

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?

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NoTime
Homework Helper
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.

The magnetic field for a given current is a constant.
The phrase "magnetic field" may refer to the strength of the "magnetic field" as well as the "extent" of the magnetic field. You may have 1 amp of current going through a meter of wire, or you may have 1 amps of current going to a thousand feet of wire. The current, however, is limited by the resistance of that wire and the rate change of current is limited by the inductance of that wire. The 1 amp going through a short wire will create a magnetic field with less extent than 1 amp going through a long wire, however, achieving those amps depend both on the applied voltage and the resistance of the wire.

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.
Higher voltage means greater ability to change the magnetic field. Rapid large changes in the magnetic field means greater induced voltage. Higher voltage in motors allows stronger magnets to turn faster, that is, they support greater changes magnetic flux in a given period. If the voltage is increased, applied power can still be reduced by reducing the current. But to produce a magnetic field with a large enough extent (i.e. magnetic flux), the current needs to be driven through a high inductance.

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?"

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NoTime
Homework Helper
The phrase "magnetic field" may refer to the strength of the "magnetic field" as well as the "extent" of the magnetic field. You may have 1 amp of current going through a meter of wire, or you may have 1 amps of current going to a thousand feet of wire. The current, however, is limited by the resistance of that wire and the rate change of current is limited by the inductance of that wire. The 1 amp going through a short wire will create a magnetic field with less extent than 1 amp going through a long wire, however, achieving those amps depend both on the applied voltage and the resistance of the wire.
I'm not interested in semantics. If you wish to refer to the extent of a magnetic field then you should properly qualify your statement.
Yes, E=IR always applies.

Higher voltage means greater ability to change the magnetic field. Rapid large changes in the magnetic field means greater induced voltage. Higher voltage in motors allows stronger magnets to turn faster, that is, they support greater changes magnetic flux in a given period. If the voltage is increased, applied power can still be reduced by reducing the current. But to produce a magnetic field with a large enough extent (i.e. magnetic flux), the current needs to be driven through a high inductance.
Your ideas here are incorrect. You only need to unplug the field winding of a DC motor in order to find this out (very dangerous by the way).

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?"
Other than limits in our ability to construct a physical device, there isn't one.
Edit: Relativistic and Quantum limits not considered here as they are well beyond what we can construct and not well defined.

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kmarinas86 said:
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?"
Other than limits in our ability to construct a physical device, there isn't one.
Edit: Relativistic and Quantum limits not considered here as they are well beyond what we can construct and not well defined.
What kind of motors tend to maximize this aspect of performance?

NoTime
Homework Helper
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.

I said:

kmarinas86 said:
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?"
You said:

Other than limits in our ability to construct a physical device, there isn't one.
Edit: Relativistic and Quantum limits not considered here as they are well beyond what we can construct and not well defined.
Why is there no other limit but our ability to build such a device? Surely our inability to build them depends on the decipherable. Generally, what are the reasons for not developing such devices? If in the future we improve our ability to convert power into magnetic flux, to what effect may this have on our ability to improve the efficiency? If some electric motor already in production is 95% efficient, should it be said that the electric power used by the motor cannot produce significantly bigger and quicker changes magnetic flux despite your claim there is no limit to that? Or did you misread my question?

NoTime
Homework Helper
If in the future we improve our ability to convert power into magnetic flux
This seems to be the root of your confusion.
No power is involved creating magnetic flux.
It is just a change of one potential form into another.

This seems to be the root of your confusion.
No power is involved creating magnetic flux.
It is just a change of one potential form into another.
But wouldn't power be involved when that magnetic flux is used to drive a magnet?

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.

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NoTime
Homework Helper
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.

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.
Right. I know that the energy stored in an inductor is proportional to the current squared, but it is also proportional to the inductance. Based on your answers earlier, you seem to agree there is no limit to the inductance to resistance ratio (save for quantum physics, relativity, and practical build considerations). But if that were true, then there would be no other limits as to the energy that can be stored in an inductor vs. power that goes into that inductor. Granted, energy and power are not the same thing. Yet, when I think of the capacity of battery, I think ampere hours. It would be possible to increase inductance and reduce current while using less power as well, while at the same time making the machine capable of creating a bigger potential. (Try the converse, imagine trying to build a motor with macroscopic force but with coils the size of the spider mite. It will hardly do anything no matter how much power you can put into it!)

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/techdata/brooks/brooks.html [Broken]

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The Electrician
Gold Member
Right. But if that were true, then there would be no other limits as to the energy that can be stored in an inductor vs. power that goes into that inductor.

Am I wrong?
There IS a limit "as to the energy that can be stored in an inductor vs. power that goes into that inductor." It's called the conservation of energy. Power is energy per unit time. If you wind an inductor with superconducting wire, then the energy in the magnetic field of that inductor is exactly the time integral of the power supplied to the inductor. The instantaneous power supplied to the inductor is simply the product of the instantaneous voltage applied and the instantaneous current that the applied voltage produces. You can't have more energy in the magnetic field than the accumulated sum (integral) of the power you supply.

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!)

NoTime
Homework Helper
Yet, when I think of the capacity of battery, I think ampere hours.
While amp hours is the capacity of a battery.
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.

While amp hours is the capacity of a battery.
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.

He says:

The Electrician said:
There IS a limit "as to the energy that can be stored in an inductor vs. power that goes into that inductor." It's called the conservation of energy. Power is energy per unit time.
Obviously power is energy per unit time, so energy divided by average power is time. So the longer the time that power must hold to develop the magnetic field in the coil, the higher the ratio of the coil's energy could be relative to the power sent to the coil! As he says:

The Electrician said:
You can't have more energy in the magnetic field than the accumulated sum (integral) of the power you supply.
The Electrician said:
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!
Say I have a capacity of 1 amp-hour stored at 9 volts. There are many possibilities to consider:

9 amps going through a 1-ohm coil for 1/9th of one hour. The power loss is (1)(9)(9) watts.
3 amps going through a 3-ohm coil for 1/3rd of one hour. The power loss is (3)(3)(3) watts.
1 amp going through a 9-ohm coil for one hour. The power loss is (9)(1)(1) watts.

The power going into the 1-ohm coil is (9V)(9 amps) which is the same as the power loss!
The power going into the 3-ohm coil is (9V)(3 amps) which is the same as the power loss!
The power going into the 9-ohm 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?
///

The Electrician
Gold Member
So the longer the time that power must hold to develop the magnetic field in the coil, the higher the ratio of the coil's energy could be relative to the power sent to the coil!
You use the unqualified word "power" to mean "instantaneous power"; I took it to mean "accumulated power" because that's what determines the energy in the field of an ideal inductor.

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 non-zero 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 non-zero 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.

9 amps going through a 1-ohm coil for 1/9th of one hour. The power loss is (1)(9)(9) watts.
3 amps going through a 3-ohm coil for 1/3rd of one hour. The power loss is (3)(3)(3) watts.
1 amp going through a 9-ohm coil for one hour. The power loss is (9)(1)(1) watts.

The power going into the 1-ohm coil is (9V)(9 amps) which is the same as the power loss!
The power going into the 3-ohm coil is (9V)(3 amps) which is the same as the power loss!
The power going into the 9-ohm coil is (9V)(1 amp) which is the same as the power loss!
This is just the sort of behavior I describe above.

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NoTime
Homework Helper
Say I have a capacity of 1 amp-hour stored at 9 volts. There are many possibilities to consider:

9 amps going through a 1-ohm coil for 1/9th of one hour. The power loss is (1)(9)(9) watts.
3 amps going through a 3-ohm coil for 1/3rd of one hour. The power loss is (3)(3)(3) watts.
1 amp going through a 9-ohm coil for one hour. The power loss is (9)(1)(1) watts.

The power going into the 1-ohm coil is (9V)(9 amps) which is the same as the power loss!
The power going into the 3-ohm coil is (9V)(3 amps) which is the same as the power loss!
The power going into the 9-ohm coil is (9V)(1 amp) which is the same as the power loss!

What? I don't get it.
Well, you are correct.
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.