# Inductor questions (generate a voltage opposing the source voltage?)

• I
• annamal
In summary, the self-inductance of a circuit loop can create a back EMF, but it is usually very small and only comes into play when the change in current is significant, such as when opening or closing a switch. The inductance of a circuit loop is typically much smaller than that of an inductor due to the smaller number of turns and the larger surface area. Inductance can also create a forward EMF, similar to the inertia effect in a mechanical system, when the switch opens and the current begins to fall.f

#### annamal

1) Why does there have to be an inductor to generate a voltage opposing the source voltage? Doesn't the circuit create back voltage without an inductor?

2) In ##U = \frac{1}{2}LI^2 + \frac{1}{2}\frac{Q^2}{C}##
##E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2## where ##x(t) = Acos(wt + \phi)##
following this pattern where ##k = \frac{1}{C}## and x = Q, ##q(t) = q_0cos(wt+\phi)##, m = L, v = I; could x= I and k = L so that ##I(t) = I_0cos(wt + \phi)##?

1) Yes any circuit loop as a closed loop of current has some self inductance and creates some back EMF. But this self inductance is usually very small and we choose to ignore it. It comes into play only when the other factor of the back EMF, ##\frac{dI}{dt}## is very large, like when opening or closing a switch of the loop.
2) Yes you could have the I(t) you proposing but the analogy isn't perfect now because current is the first derivative of charge as velocity is the first derivative of position.

malawi_glenn
1) Yes any circuit loop as a closed loop of current has some self inductance and creates some back EMF. But this self inductance is usually very small and we choose to ignore it. It comes into play only when the other factor of the back EMF, ##\frac{dI}{dt}## is very large, like when opening or closing a switch of the loop.
2) Yes you could have the I(t) you proposing but the analogy isn't perfect now because current is the first derivative of charge as velocity is the first derivative of position.
Ok, 2) makes sense, but I am wondering more about 1)
Doesn't the back emf in the inductor only come into play when the dI/dt is large like when opening or closing the switch of the circuit (like in the self-inductance of the circuit)?

Are you familiar with the basic differential equations relating voltages and currents for inductors and capacitors?
$$v(t) = L \frac{d i(t)}{dt}$$
$$i(t) = C \frac{d v(t)}{dt}$$

Delta2
Doesn't the back emf in the inductor only come into play when the dI/dt is large like when opening or closing the switch of the circuit (like in the self-inductance of the circuit)?
It might be largest in those cases, but it may still "come into play" in other cases if it's significant in comparison.

Ok, 2) makes sense, but I am wondering more about 1)
Doesn't the back emf in the inductor only come into play when the dI/dt is large like when opening or closing the switch of the circuit (like in the self-inductance of the circuit)?
No because the L of a typical inductor is like 100-1000 (would dare to say up to 10000) times larger than the L of a typical circuit Loop. Remember that the back emf is ##L\frac{dI}{dt}## so if L is 1000 times smaller, you need 1000 times larger ##\frac{dI}{dt}## to generate the same back emf.
On why the L of a circuit loop is much smaller? Mainly because it is only one turn of wire. If we do the math we see that the L in inductors goes up as ##N^2## where N the number of turns. The larger surface area of a circuit loop , is neutralized by the fact that the B field varies spatially much more in this surface area (so that it has a smaller average value, compare with that in the interior of an inductor), while in an inductor the cross area is small but the B field is much more intense there.

malawi_glenn and annamal
1) Yes any circuit loop as a closed loop of current has some self inductance and creates some back EMF. But this self inductance is usually very small and we choose to ignore it. It comes into play only when the other factor of the back EMF, ##\frac{dI}{dt}## is very large, like when opening or closing a switch of the loop.
2) Yes you could have the I(t) you proposing but the analogy isn't perfect now because current is the first derivative of charge as velocity is the first derivative of position.
Incidentally, inductance creates a forward EMF when opening a switch. It acts in an analogous way to mass in a mechanical system, having inertia and resisting change.

Sorry what do you mean by "forward" EMF, haven't heard this term before.

When the switch opens, the current in the circuit begins to fall, but the inductor now creates a voltage which tries to keep it going. This is the inertia effect I mentioned. The inductor might create a spark across the switch contacts and this is a current flowing the same way as the battery current. It is a forward EMF not a back EMF.
The EMF across the inductor is - L dI/dt, where dI/dt is now a negative slope. So EMF = L dI/dt which is a forward voltage.

Delta2
When the switch opens, the current in the circuit begins to fall, but the inductor now creates a voltage which tries to keep it going. This is the inertia effect I mentioned. The inductor might create a spark across the switch contacts and this is a current flowing the same way as the battery current. It is a forward EMF not a back EMF.
The EMF across the inductor is - L dI/dt, where dI/dt is now a negative slope. So EMF = L dI/dt which is a forward voltage.
I think you are attempting to open a can of worms here, regarding what is the voltage and the back emf/forward emf in the ends of an inductor. If you want create another thread and I ll post there

My simple answer to you would be that the inductor voltage is always back EMF in the sense that it resists the change in current (it might resist the increase or the decrease in current). We don't call it forward EMF if it resists the decrease in current and back EMF if it resists the increase in current. But as I said create another thread if you want to discuss this in more detail.

rude man and cabraham
The EMF across the inductor is - L dI/dt, where dI/dt is now a negative slope. So EMF = L dI/dt which is a forward voltage.
No, the EMF is still ##-L\frac{dI}{dt}##. I agree that it is a positive quantity and that ##\frac{dI}{dt}## is negative. This voltage is "forward" in the sense that it's in the same direction as the EMF.

Ok, I get it, we call it forward EMF if it augments source's EMF, or back EMF if it opposes source's EMF. There are cases where there aren't any sources in a closed loop with inductor's and/or capacitors(a typical LC circuit with initial charge at capacitor) , how we should call it then?

Sigh... I personally find language one of the very worst ways to deal with these polarity questions. It's confusing, either to you or me. People make assumptions and describe in one sentence what may require a paragraph. Honestly, if you know the underlying physics, then context is everything and you just assume that the polarity the other guy is talking about is the correct one. That's how it works in practice, I think.

I very much prefer schematics with the current and voltage polarities defined with arrows and such. Random assignment is fine with me, although I do prefer the passive sign convention. Just show me what you mean. Then you write equations with the correct signs. Not only will I know what you mean, I'll know that you agree with me, and that it won't change later when the context of the question shifts.

The most common mistake EEs make in circuit analysis is messing up the polarities because they weren't careful and explicit.

Dale, sophiecentaur and berkeman
Ok, I get it, we call it forward EMF if it augments source's EMF, or back EMF if it opposes source's EMF. There are cases where there aren't any sources in a closed loop with inductor's and/or capacitors(a typical LC circuit with initial charge at capacitor) , how we should call it then?
In this case the circuit is resonant so the polarities of the various EMFs vary with time.

There are no polarity problems, and indeed a current has no direction, because it's a scalar quantity. What makes circuit theory complicated with the signs of the various involved quantities is that they are integral laws, and that's, why there are various "directions" involved, which lead to the signs of these various quantities.

For AC circuit theory involving only "passive elements" like resistors, capacitors, and coils, you just deal with Faraday's Law in integral form taking the line integral along each loop of your circuit and charge conservation at each node.

I'm not sure, which circuit you are really discussing here. From #1 it seems to be a capacitor and a coil in series connected to an AC voltage ##V(t)=V_0 \cos(\omega t)##.

In the above diagram the arrow indicates the orientation of the line integral in Faraday's Law,
$$\int_{\partial A} \mathrm{d} \vec{r} \cdot \vec{E}=-\mathrm{d}_t \int_A \mathrm{d}^2 \vec{f} \cdot \vec{B}.$$
This line integral gives, taking into account the chosen "polarities" of the voltage source and the capacitor,
$$\int_{\partial A} \mathrm{d} \vec{r} \cdot \vec{E}=-V(t)+C Q.$$
The magnetic flux has to be calculated with the surface-normal vector going into the plane and in this direction also is the magnetic field. So you have
$$-\mathrm{d}_t \int_A \mathrm{d}^2 \vec{f} \cdot \vec{B}=-L \dot{i}.$$
Together with ##i=\dot{Q}## you get
$$-V(t)+C Q=-L \ddot{Q}$$
or
$$L \ddot{Q} + C Q=V(t)=V_0 \cos(\omega t).$$
It's of course a bit artificial, assuming that there's no resistance (at least the coil in reality has some resistance, which you can put in series with the ideal coil which adds a term ##R i=R \dot{Q}## on the left-hand side of the equation).

Dale and Delta2
When the switch opens, the current in the circuit begins to fall, but the inductor now creates a voltage which tries to keep it going. This is the inertia effect I mentioned. The inductor might create a spark across the switch contacts and this is a current flowing the same way as the battery current. It is a forward EMF not a back EMF.
The EMF across the inductor is - L dI/dt, where dI/dt is now a negative slope. So EMF = L dI/dt which is a forward voltage.
Inductor cannot "create an emf". It has no internal source of chemical reaction like batteries do, nor any moving parts to generate.
When a circuit with an inductor is closed, the current is made up of charges moving in the conduction band, the highest energy level. When the switch opens, the charges in the conduction band continue to move. If the circuit is open, charges accumulate at the switch contacts, & this results in a voltage across said switch. If the faces of the switch contacts have a capacitance C, then dV/dt = I/C. If the the switch is double throw, where the battery is disconnected & circuit gets connected to resistor R, then V across R = I*R. The inductor acts as a current source, the voltage is determined by the impedance of the loaf.
Inductor does not generate voltage, that voltage varies depending on resistance. The voltage results from conduction band charges colliding with resistors lattice ions.

weirdoguy, hutchphd and DaveE
Inductor cannot "create an emf". It has no internal source of chemical reaction like batteries do, nor any moving parts to generate.
When a circuit with an inductor is closed, the current is made up of charges moving in the conduction band, the highest energy level. When the switch opens, the charges in the conduction band continue to move. If the circuit is open, charges accumulate at the switch contacts, & this results in a voltage across said switch. If the faces of the switch contacts have a capacitance C, then dV/dt = I/C. If the the switch is double throw, where the battery is disconnected & circuit gets connected to resistor R, then V across R = I*R. The inductor acts as a current source, the voltage is determined by the impedance of the loaf.
Inductor does not generate voltage, that voltage varies depending on resistance. The voltage results from conduction band charges colliding with resistors lattice ions.
If an inductor (e.g. solenoid) sees a time-varying externally generated B field within it it will generate an emf . Think an open secondary winding of a 2-coil transformer.
There is an E field in each winding such that per winding ## emf = - d\phi/dt = \oint \bf E \cdot \bf dl ## around the winding. If the winding has zero resistance an equal but opposite electrostatic field ## E_s ## must also exist since the net E field in the wire is zero.
That in turn mandates an external electrostatic field around the outside of the entire solenoid since the sum of winding voltages = the solenoid voltage. The line integral of this E field is the voltage across the solenoid. The circulation of the elecrostatic voltages inside & outside the solenoid = 0 by Kirchhoff.

vanhees71 and Delta2
It's just the EMF due to the (rapidly) changing magnetic field.

Ok, I get it, we call it forward EMF if it augments source's EMF, or back EMF if it opposes source's EMF. There are cases where there aren't any sources in a closed loop with inductor's and/or capacitors(a typical LC circuit with initial charge at capacitor) , how we should call it then?
You'd do far better just to call it an Induced emf and leave it to the Maths to sort out the sign for you. Back emf and 'deceleration' are terms to be used for preliminary or final arm waving chat about a situation and not when the actual result counts. Of course, Lenz's law always applies but that tells you the sign of the emf.

Last edited:
vanhees71, DaveE and hutchphd
@annamal
Just backing up a bit to question 2, you are making an analogy between an electrical and a mechanical circuit.
But as @Delta2 mentioned these analogies are not quite exactly perfect.

See https://en.wikipedia.org/wiki/Mecha...of force, torque is made analogous to voltage.

The one you made would be of the type of an impedance analogy,
But others can be used, such as the mobility analogy, which is a dual of the impedance model.
The impedance analogy classifies the variables as effort( F, V) and flow(Velocity, I ).
The mobility analogy is a dual of the impedance analogy, with have them switched with effort (F, I) and flow( Velocity, V ).
The variables come from the power conjugates - Force times Velocity and Voltage time Current, as you gleamed, except you used the energy of the system as the comparison, which is more Hamilton like, and
mentioned in the article.

One can extend the analogies of electrical and mechanical systems to other systems ( between each one if so desired ) - electrical, mechanical, fluid, thermal, ..., and extend it further with the duals of of systems.

Quite an interesting subject if you really want to get into it.
they do note some of the pitfalls.

sophiecentaur
Lenz's law always applies
Yup. That's how real world EEs figure it out. The voltage polarity is always in the direction that acts to minimize the change in the magnetic field. It's really simple, in practice.

sophiecentaur and vanhees71
Concerning the signs you can also simply get back to the roots, i.e., Faraday's Law, which is one of the fundamental Maxwell equations. The signs come out always correct when you strictly follow the conventions of Stokes's theorem: The direction of the surface-normal vector determines the direction of its boundary curve due to the right-hand rule.

hutchphd and sophiecentaur
Instead of many words let's do a calculation. Suppose we have a battery, a coil and a resistor in series and now you turn a switch, which takes off the battery and short-circuits the coil+resistor. The equation for the current then is (using Faraday's Law)
$$L \dot{i} + R i=0$$
with the initial condition ##i(0)=i_0=U_0/R##. The solution obviously is
$$i(t)=i_0 \exp(-R t/L).$$
You can also calculate the total energy dissipated in the resistor. The power is
$$P=R i^2 = R i_0^2 \exp(-2R t/L)$$
and thus the total energy dissipated
$$E=\int_0^{\infty} R i^2 = \frac{L}{2} i_0^2.$$

malawi_glenn, DaveE and sophiecentaur
Instead of many words let's do a calculation. Suppose we have a battery, a coil and a resistor in series and now you turn a switch, which takes off the battery and short-circuits the coil+resistor. The equation for the current then is (using Faraday's Law)
$$L \dot{i} + R i=0$$
with the initial condition ##i(0)=i_0=U_0/R##. The solution obviously is
$$i(t)=i_0 \exp(-R t/L).$$
You can also calculate the total energy dissipated in the resistor. The power is
$$P=R i^2 = R i_0^2 \exp(-2R t/L)$$
and thus the total energy dissipated
$$E=\int_0^{\infty} R i^2 = \frac{L}{2} i_0^2.$$
The problem tha many people have with that succinct presentation is the fact that it's a statement of relationships and says nothing about 'what made it happen'. We are so familiar with that approach that we don't tend to ask the question about the 'flow of causality' (I know that sounds dodgy but I think it gets the meaning across) The system works and we take that approach when dealing with Science and usually get the right answer but I sympathise with the feeling that it's not totally satisfactory.

In a talk about belief in God, I heard A C Grayling describe how we (humans) look for an agency to explain every experience. Science manages to sidestep this and delivers the goods without pages and pages of words and examples. That's unsettling for students.

Science is about objectively observable and quantifiable facts. The only language we have for that is math, and it's a very concise language. The underlying theory (in this case Maxwell's electromagnetism) provides a clear, causal picture of what's going on here. There's no need for religion to understand the results of the natural sciences.

sophiecentaur
Inductor cannot "create an emf". It has no internal source of chemical reaction like batteries do, nor any moving parts to generate.
Of course an inductor creates EMF. Saying an inductor creates EMF is equally as valid as saying a generator creates EMF. Mechanical motion and electrochemical reactions are not a necessary requirement to say that something generates EMF. Inductors simply use electromagnetic potential energy stored in the field as their source of energy instead of the chemical potential energy of a battery or whatever type of energy you define a generator to use (kinetic? electromagnetic?).

The inductor acts as a current source, the voltage is determined by the impedance of the loaf.
Current sources also generate EMF. The distinction between a voltage source and a current source is that the former attempts to provide a steady voltage to the circuit (by letting current fluctuate), while the latter attempts to provide a steady current to the circuit (by letting voltage fluctuate). Both provide EMF, otherwise you'd have no voltage or current in the circuit at all.

Note that voltage (electric potential difference) and EMF are not the same thing and are often confused for each other. An inductor generates an EMF, but the amount of EMF generated (the voltage) is determined by both the inductor and the rest of the circuit.

An electron falling from conduction band down to valence band changes net charge on that atom(s). If atom is neutral, acquiring an additional electron makes atom negative net charge of -1. If atom had a net charge of +1, acquiring electron makes it neutral.

This change in net charge results in an E field change. Voltage across resistor is integral of E dot dl. The resistor voltage changed as a result of resistor atoms in the lattice acquiring new electrons, or shedding electrons, resulting in change of E field.
Resistors do not work this way, nor does resistance in general*. As an example, replace the resistor with a very thin copper wire so that most of the resistance of the circuit is in this thin wire (some resistors are actually made out of thin coils of wire). Copper, as with other metals, doesn't have separate valence and conduction bands. Electrons may change states and energy levels, but they will not be bound to single atoms and no changing of charge happens.

*Partial exceptions include devices like LED's.

hutchphd, DaveE, artis and 2 others
Science is about objectively observable and quantifiable facts. The only language we have for that is math, and it's a very concise language. The underlying theory (in this case Maxwell's electromagnetism) provides a clear, causal picture of what's going on here. There's no need for religion to understand the results of the natural sciences.
Absolutely.
But , first time in, that’s hard for many people.

vanhees71
Absolutely.
But , first time in, that’s hard for many people.
Michael Faraday knew little mathematics yet he noticed the electromagnetic nature of light and provided the basis of Maxwell's Equations.

sophiecentaur and Delta2
Instead of many words let's do a calculation. Suppose we have a battery, a coil and a resistor in series and now you turn a switch, which takes off the battery and short-circuits the coil+resistor. The equation for the current then is (using Faraday's Law)
$$L \dot{i} + R i=0$$
with the initial condition ##i(0)=i_0=U_0/R##. The solution obviously is
$$i(t)=i_0 \exp(-R t/L).$$
You can also calculate the total energy dissipated in the resistor. The power is
$$P=R i^2 = R i_0^2 \exp(-2R t/L)$$
and thus the total energy dissipated
$$E=\int_0^{\infty} R i^2 = \frac{L}{2} i_0^2.$$
I think there is a problem with the proposed circuit, because when the switch is operated, the current falls almost instantly to zero, creating very large voltage, and all the energy of the magnetic field is dissipated in a spark across the switch contacts..

malawi_glenn and Delta2
I think there is a problem with the proposed circuit, because when the switch is operated, the current falls almost instantly to zero, creating very large voltage, and all the energy of the magnetic field is dissipated in a spark across the switch contacts..
It depends. @vanhees71 models it as an RL circuit where the switch is not open but it is closed and bypasses the originally present EMF source.

You have another circuit in mind where the switch is open (after again we have removed the original EMF source) and this circuit should be modeled as an RLC circuit where the capacitor C is between the switch contacts and has very small capacitance so that is charged very fast to a big voltage because all the energy of the magnetic field of the coil E, goes into energy in the capacitor ##E=\frac{1}{2}CV^2##, C is very small, V must be very big for some finite E (the energy originally stored at the magnetic field of the coil, before we open the switch).

vanhees71
It seems to me that you have achieved something very important. You have realized that many of the 'simple' EE questions you are likely to be asked are actually very complicated and seem to imply paradoxes. You have also appreciated that these apparent paradoxes can be rationalised by acknowledging the need to add extra 'components' or factors into the model or sometimes to wait for transitional effects to have died down. This is good!
Thanks, not that often when I hear positive feedback from you, I guess I've passed the test.

Science is about objectively observable and quantifiable facts. The only language we have for that is math, and it's a very concise language. The underlying theory (in this case Maxwell's electromagnetism) provides a clear, causal picture of what's going on here. There's no need for religion to understand the results of the natural sciences.
Sure. But some visual mind models and intuition does come in handy to better understand the maths. In fact for me it;'s always better to first get a "picture" of how something works in my mind and then it is far easier to learn the math side. I think that is because not all people are equally capable of high levels of abstraction. Some need to see the "connection" in the abstract math to the actual process within the circuit for example.

I just visualize the B field as a real (even though invisible) field around the coil and then I intuitively think, now energy must go somewhere it can't just stop existing, so it goes to current as that is the only way it can go, but current has no where to go, so it needs to develop a circuit through an arc, so the energy from the B field goes to the E field across the open switch contact capacitance. Boom problem solved.

Once I develop this mental model then the dry math makes sense. The problem with math alone is as I said - it's "dry", it;s basically just a language of abstraction that aims to represent some real physical phenomena.
Sure some folks have a natural "feel" for such language and for them they can learn it even without ever understanding what the symbols represent in real life physics. We call such folks theoretical mathematicians.

My former math teacher could solve all of Maxwell's equations and much more but she doesn't know a single thing about how current behaves in actual circuits.

Delta2
It depends. @vanhees71 models it as an RL circuit where the switch is not open but it is closed and bypasses the originally present EMF source.

You have another circuit in mind where the switch is open (after again we have removed the original EMF source) and this circuit should be modeled as an RLC circuit where the capacitor C is between the switch contacts and has very small capacitance so that is charged very fast to a big voltage because all the energy of the magnetic field of the coil E, goes into energy in the capacitor ##E=\frac{1}{2}CV^2##, C is very small, V must be very big for some finite E (the energy originally stored at the magnetic field of the coil, before we open the switch).
I think the devil is in the detail here. vanhees seems to describe a change over switch, going from battery to resistor. This will not work, as the inductor will instantly give up its energy during the changeover. Maybe someone can provide a circuit which will work, or maybe not.

I think the devil is in the detail here. vanhees seems to describe a change over switch, going from battery to resistor. This will not work, as the inductor will instantly give up its energy during the changeover. Maybe someone can provide a circuit which will work, or maybe not.
Switches are nearly always assumed to be ideal (instantaneous) in thought experiments like this. IRL, you would add a bit more engineering, like snubbers, clamp diodes, transient suppressors, make-before-break types, etc.

That really wasn't his point. It's a diversion from the basic physics here. You could use a step function for the voltage source without any switches, which is also a magical creation.

This will not work, as the inductor will instantly give up its energy during the changeover
This is not sure, I think if the changeover is very very fast the capacitor formed by the switch contacts will not have time to charge, so not much energy transferred from the inductor to the capacitor.