# How would you explain self-induced voltage in an inductor?

1. Jul 16, 2015

### Moshe1010

Consider the following:
You have a circuit with an inductor and a battery. You have an open switch, that was open for gazillion years. The inductor is "uncharged", meaning has no stored magnetic field. Then, you close the switch. At time t=0+, right at the moment you closed the switch, the current at the circuit is zero due to the resistance of the inductor. This happens due to a voltage/emf change. I get the idea that Faraday's Law relates the change in current and change in magnetic flux with time and I also understand the graphs/time constants etc`.

These are the questions:
1. If there is no current at t=0+, how does the inductor make a change in voltage to resist the current? where this voltage is generated from?

2. As far as I understand, current is not even flowing out of the battery (at all) at t=0+, then again, what creates the change in potential energy in the inductors, that creates the change in magnetic flux which is affected by change in current if no current was flowing from the battery? To remind you, the inductor is "uncharged", no current was flowing in the inductor for gazillion years, so there is no magnetic field to create energy/voltage.

3. How is it different than connecting a resistor to a circuit with a battery (just a resistor and a battery, without anything else). At t=0+, right when you close the switch, current starts to flow from the battery to the resistor. This is not the case with an inductor. Would you say the battery has some sort of mental decision when to release a current based on the connected properties in the circuit? How does the battery know what is connected or not along the way?

Assume these are ideal batteries, wires,/coils/inductor (no resistance) and everything else.

Thanks for the help.

Last edited: Jul 16, 2015
2. Jul 16, 2015

### Staff: Mentor

The voltage is generated from the battery. It doesn't really have anything to do with the inductor. The inductor simply relates the voltage to the current, but doesn't generate either.

For your question 3, the difference between the resistor and the inductor is that they enforce a different relationship between current and voltage.

3. Jul 16, 2015

### Moshe1010

I order to oppose change in current at t=0, the inductor generates negative voltage. If you close the switch for a long period of time then remove the battery, the inductor would generate (for a limited time), the same current the battery was generating since it doesn't like a sudden change in current and wants to keep things as they are. Thus, I'm not sure what do you mean by "the inductor doesn't generate either".

4. Jul 16, 2015

### Staff: Mentor

No, the voltage is entirely determined by the battery. If you use a 9 V battery then the voltage is 9 V regardless of L. If you use a 12 V battery then the voltage is 12 V regardless of L. There is no voltage generated by the inductor. The voltage is entirely generated by the battery.

The resistor enforces the relationship $v=i\,R$ the inductor enforces the relationship $v=L \,di/dt$

Last edited: Jul 16, 2015
5. Jul 16, 2015

### Staff: Mentor

The way that I think of passive elements is not in terms of "generating" any voltage or current, but in terms of enforcing a relationship between the voltage and the current. If you just have an inductor then you cannot determine either $v$ or $i$. You need something else to generate either $v$ or $i$, and then the inductor determines the other.

6. Jul 16, 2015

### cabraham

Inductor does NOT generate "negative voltage". Inductor has no energy conversion going on inside. It is not a battery or generator. If a switch was closed connecting a battery and series resistor to a parallel inductor-resistor combo, the inductor would energize, and settle to a steady current of battery voltage divided by series resistor, i.e. Vbatt/R1. This is inductor current IL. Switch opens disconnecting Vbatt and R1, leaving the energized inductor in parallel with R2.

The current in L, IL, continues in the resistor R2. A voltage develops across R2, i.e. V2 = R2*IL. This voltage is NOT "generated" by the inductor at all. The inductor current is made up of electrons in the conduction band, the high energy level. When the electrons from the inductor arrive at the resistor, collisions occur between electrons and lattice ions. As a result some electrons lose energy and fall down to the valence band, a lower energy state than conduction band. To conserve energy, photons are emitted. The heat we feel around a wire carrying current is due to this action.

The voltage developed across R2 is a result of electrons colliding with lattice ions, and dropping from conduction to valence band. Since current is the motion of charged particles, when this motion stops, these particles build up an electric field. By definition, the line integral of this field is the developed voltage across the resistor. It's not generated by the inductor. An inductor cannot generate voltage. Inductance is the property of maintaining the present value of current. In the process voltages are developed, but not generated by inductors. Have I helped?

Claude

7. Jul 16, 2015

### DarioC

Moshe, looks like we have a failure to communicate here, but it isn't yours.
http://whites.sdsmt.edu/classes/ee322/class_notes/322Lecture4.pdf
That is a pretty good link for showing inductive kickback (which is what you are talking about) from a quick changing input current on an inductor.
Current has to flow first and the rate of change is the kicker (pun intended).
DC

test 1234

8. Jul 17, 2015

### Moshe1010

Thanks for the link but my question was different. I asked what happened at t=0+, when no current is flowing through the inductor, so no magnetic field is generated (it is but it doesn't happen because of the current flow, but due to the potential difference by connecting a source with some voltage output).

What you would call EMF then? EMF = VL = -N(dB/dt)

they both depend on current. While the first is net current and the other is change in current over time. At t=0+, you have di/dt, but not I. However, di/dt = change in current over change in time. It means that you have current final and initial. So let's say your initial is zero, and your final is some number, where is this number coming from? since at t=0+, di/dt is not zero, then it has some value. What is it(assume random number of voltage as a source)? or if the change if from 1--->2 or from 9999999 to 10000000 is still the same change - who is generating final current and who is generating the initial current, and how? your di/dt at t=0+ is not zero - this is the question basically.

I understand all the theory around inductors (on the basic, "ideal" inductor/battery/capacitor level - or I think I am at least). I get the idea that voltage potential would induce change in magnetic field, which would "create" an EMF (voltage) that will drive/resist current flow.

but my question is before it's building up/stores magnetic field - means it doesn't have any "energy" stored at t=0+ (just when you close the open switch). The inductor never has current flow thourhout it = nothing is stored = was never "energized".

and electors (charge) create electric fields which create voltage/potential. You are just saying that Faraday's law is false:
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/farlaw.html

this is again, not true for t=0+. At t=0+ the resistor doesn't play any role in voltage drop if an inductor is connected as well due to the current being 0. No current, no voltage drop.
This is a different story. Yes, voltages are developed based on the source. I may use the "wrong" terminology by saying "generated", I did not mean, at any stage, that the inductor would generate a greater voltage than the source (battery/capacitor).

9. Jul 17, 2015

### cabraham

Actually, in boost, buck-boost, and isolated flyback converters, the inductor voltage developed can exceed the supply value. But this developed voltage is due to either charge separation as in a capacitor, or lattice collisions as in a resistor. The inductor does not "generate" voltage, less than, or more than the supply.

Nothing personal but you don't know enough to be lecturing anybody. See my replies above in red.
Claude

10. Jul 17, 2015

### Moshe1010

Thanks Claude. Your ego has very big potential. You are still trying to explain something that hasn't been my direct question. P.S I'm not trying to lecture anybody. If you do, please pay attention to other's questions rather than explaining unrelated information (and arguing that basic laws in physics are "not true" since they are based on "observations".

11. Jul 17, 2015

### cabraham

I made it clear that Faraday's law in true beyond doubt. Observation was brought up to illustrate that the relation observed and recorded was between flux and its rate of change, and voltage. I then illustrated the sequence of events. When we state that v(t) = -N(d(phi)/dt), that establishes a relation among these quantities. But some have interpreted Faraday's law as to suggest that the changing flux "generates" the voltage, which I and others have pointed out is not the case. Does the flux generate the voltage? Does voltage generate flux? Again, the law must be interpreted and applied in light of boundary conditions.

An ac voltage source connected across a transformer primary will result in a flux determined by Faraday's law. phi = (1/N)*integral (v dt). Here, the flux is determined by the voltage and its rate of change and the primary turns. The voltage at the secondary is determined by the flux, its rate of change, and the secondary turns. The flux can determine the voltage and vice-versa. That was my point.

Regarding your question at t = 0 and 0+, here is my answer to what I believe you are asking, feel free to clarify if needed. A switch is closed and a battery is connected across a resistor plus inductor in series, an R-L circuit. Inductor starts with zero current. Is that correct?

At t=0, there are no electrons in the conduction band, but plenty in the valence band of the conducting materials. When the electric field from the battery travels through the wires, the valence electrons incur a force per F=qE, Lorentz force. In an insulator, the electrons are tightly held and it takes a very high voltage/electric field to move electrons from valence to conduction. But in a conductor it is much easier, less energy is required. As electrons move up in energy level into conduction, the electrons move around the loop migrating among atoms. We now have current in the loop. As this current circulates, electrons moving through the resistor incur lattice collisions and ionization. A layer of charges develops on each side of the resistor. The line integral of the E field associated is the voltage drop across the resistor.

A small forward voltage drop occurs across the inductor as well due to its winding resistance. If the inductor is wound with SC (super-conductors), then no voltage drop (dc) occurs. Eventually steady state is reached and the entire battery voltage (or almost that if L has resistance) appears across the resistor. Little to no dc voltage appears across the inductor.

At t=0(+) the current is 0 because the initial condition was I=0, and it takes E field force to move valence electrons up into the conduction band. This takes a little time. In the resistor, Ohm's law V=IR does hold in steady state, but during the very short time when the E field is moving through the loop, we must look at this problem using fields instead of circuits. Circuit theory does not explain how charges move due to forces, whereas field theory does explain that. An inductor tends to maintain its present current value.

In order to change an inductor's current, one must either move conduction electrons down to valence, or valence electrons up to conduction. An E field is generally needed for that. Without an E field the current stays the same, zero or otherwise. When a SC inductor is energized, then shorted in a perfect SC loop, the current sustains indefinitely. As conduction electrons move around the SC loop, no lattice collisions occur and no electrons drop down out of conduction. Thus the current value remains constant. The voltage is zero. Inductors cannot generate voltage, period.

If the loop has resistance, then the lattice collisions ionize atoms and form a potential barrier due to charge accumulation. This is how the voltage in the resistor develops. Not at all generated by the inductor. When electrons stop circulating and ionize atoms their charge develops a local E field and a voltage is developed. Inductor is blind as to what is going on.

I hope I've helped, feel free to ask for clarification. Thanks.

Claude

12. Jul 17, 2015

### Moshe1010

These are two different circuits.
First circuit = Inductor + battery
Second circuit = Resistor + battery
At t=0+ (right when the switch was closed), the first circuit doesn't have current. The second does. The first circuit doesn't even have current going out of the battery - at all - it means no charge was flowing out of the battery to the inductor. The second has net current and current going out of the battery. The question is "simple": Why? How does the battery know what is connected along the way (wire)? Does it have some sort of mental decision? I'm trying to understand it physically, not mathematically (and I understand that physical phenomena are explained by math in physics, but I try to have a mental picture of the entire situation and fail to do that).

I think I get it, maybe in a very flimsy way. Imagine a circuit with just an inductor and a capacitor (I chose capacitor to eliminate all the discussion on chemical energy in a typical battery). The capacitor is fully charged with two plates, one is + = higher potential, the other is - = lower potential. Connect the capacitor to an inductor, at t=0+, you get no current. This is due to to electrons in the capacitor creating E-field on the (-) side of the inductor and protons creating E-field on the (+) side of the inductor. This creates a potential difference, which induces magnetic flux that develops emf and oppose the current that is trying to go through the inductor at t=0+. On the surface, is this correct?

13. Jul 17, 2015

### Staff: Mentor

You can just as easily say that the current depends on the voltage. In a circuit where the voltage is set from something else, then it makes more sense to say that the current depends on the voltage. In a circuit where the current is set from something else, then it makes more sense to say that the voltage depends on the current. The general statement is simply that passive elements enforce the specified relationship between voltage and current. That remains valid for all passive elements regardless of the circuit.

In a circuit (such as this one) where the voltage across the inductor is externally supplied, then the inductor responds with the current which enforces $v=L \,di/dt$. In a different circuit where the current through the inductor is externally supplied, then the inductor responds with the voltage which enforces $v=L \,di/dt$.

From the voltage imposed by the battery.

$di/dt=v/L$ at all times, including at t=0+.

Last edited: Jul 17, 2015
14. Jul 17, 2015

### DarioC

Ah saved from a boring afternoon, post failure to get rid of the turbulence in my pressure cloud chamber.

Say we take a long, large gauge wire that is hooked to one terminal of a large, charged capacitor via a switch. Is the wire an inductor? U betcha yes.

So what happens when we flip the switch on?

I have a little problem with your concept of t 0+. That part is actually pretty simple. No voltage on capacitor end of wire. Voltage on capacitor end of wire. Sorry no mysterious in-between other state allowed. But the 12 volts say, is not the kickback voltage that I spoke of earlier. What exactly happens when you connect the switch is that the voltage tries to distribute itself down the wire to bring the entire wire up to 12 volts. Remember there is nothing connected to the other end of this wire.

You might want to think of the wire as being like a huge transport tunnel filled with BB sized electrons (ping pong balls might be good too). The numbers in even a cross section only of a 1 mm diameter wire are astounding.

The "bucket full" of electrons that the voltage shoves into the end of the wire do not fly down the wire at the speed of light (sorry about that). What they do is bump, with their charge fields, against the fields of the conduction electrons in the (metal) wire and there is a wave of displacement that does travel down the wire at the speed of light (more or less). If there is a closed circuit at the other end of the wire, an equal quantity of other electrons located there will pop out into the return circuit.

Without any connection this displacement wave will hit the end of the wire and turn around and bounce all the way back. It can be used to make a very narrow pulse generator. Now I am getting into a shaky area for me, but I suspect a coil inductor (simply?) behaves in a magnified way, with the individual turns increasing the effect.

Again, the power is either on or off, and the current rise time is likely very, very fast in the first nano, pico, femto,whatever, second. This is current into the coil, not necessarily thru the coil.

I can't find my notes, but I once calculated that an individual electron in a 1 meter long 1mm wire might take as long as 23 hours to get to the other end with a current of one ampere flowing. It is in a discussion somewhere here on the forum. The numbers of atoms and therefore the number of conduction electrons in even a small conductor are almost beyond comprehension when you start doing the calculations.

Oh, the bumping of electrons from the valence band to conduction band is not required as there are already electrons in the conduction bands of metal atoms. That is why they are conductors. Simi-conductors, yes, copper no.

Bottom line, you don't get something from nothing. Something has to move/change first).

DC

Last edited: Jul 17, 2015