# Back-Emf vs. Voltage

Discussion in 'Physics' started by shespuzzling, Jan 5, 2010.

1. ### shespuzzling Thread Starter Active Member

Aug 13, 2009
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This issue has been bugging me for a while. I can't seem to figure out the difference between voltages across different circuit elements, namely, why the voltage across a battery pushes current in a certain direction, while the voltage that appears across a resistor is more an indication of energy usage and doens't actively move current in any direction. However, back-emf induces a current in a direction counter to whatever inital direction the source was moving current in. I've attached some images of the two circuits I'll use to explain my thoughts.

For the first circuit with two resistors in series, they each display a voltage and the sum of the voltages equals the source voltage. If all 3 elements were replaced with black boxes and we only knew the voltage polarity across each element, then we'd have no way of telling which direction current would flow. However, when we know what the elements are, we know that current will flow clockwise from the positive end of the battery and through the two resistors. When the current flows, it is at whatever value is dictated by the voltage source/resistance values, and will stay at that value until the battery begins to die out. In other words, there is no back-emf in this case pushing current in the opposite direction.

In the second circuit with the inductor, however, the voltage polarity across the resistor and inductor are the same as in the first circuit with two resistors. BUT, in this case, the inductors voltage is technically a back-emf which PUSHES current in the counter-clockwise direction (in this case).

Why is there this discrepancy between a regular old voltage across a resistor that doesn't push current anywhere, a voltage across a battery that actively pushes in one direction, and the back-emf or voltage across an inductor that pushes in the OPPOSITE direction to current flow from the source? Are there just always these caveats when discussing voltage like "this one is a source voltage so it moves electrons" and "this one is a back-emf so it moves electrons in a certain direction according to Lenz's law" and "this one is a passive voltage so it only tells you how much energy it takes to move electrons from one end to another"?

Thanks in advance...I'd really love to get this problem sorted out.

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2. ### mik3 Senior Member

Feb 4, 2008
4,846
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By measuring the voltage only, you cannot tell if in which direction the current flows if the component is unknown. If you mesure both voltage and current you can tell whether its a source or if its is resistive, capacitive or inductive load.

3. ### shespuzzling Thread Starter Active Member

Aug 13, 2009
88
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Okay, thanks. So in the circuit with the inductor, is it safe to assume that while current is flowing in the clockwise direction from the source, current is ALSO flowing in the counterclockwise direction from the back-emf voltage across the inductor?

I guess a follow-up to this would be what exactly is the inductor doing to resist changes in current...is it actually pushing current in the opposite direction (thus attempting to neutralize any changes through itself)? How is the current ever able to reach a stable state? Why is the induced current from the back-emf never enough to fully resist changes in current through the inductor?

4. ### davebee Well-Known Member

Oct 22, 2008
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The inductor is concentrating a magnetic field to enable it to oppose changes in current.

The inductor isn't pushing current in the opposite direction; it is accumulating energy in the form of a magnetic field as current through it grows. When current through it starts to decrease, the magnetic field shrinks and returns its stored energy into causing more current to flow than just that due to the external voltage.

Think of sending current through the inductor like a person trying to push a heavy object. The object is hard to initially get moving, but once moving, it is hard to stop. The heavy object never actually causes any motion opposing your push; it just seems to resist your push as it slowly builds momentum in the direction of motion. And like the inductor, it tends to continue the motion even after you stop pushing.

5. ### shespuzzling Thread Starter Active Member

Aug 13, 2009
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Hmmm, okay, but how does the back-emf come into play then? I guess I'm trying to get a very literal idea of what exactly is happening inside the inductor both initially and then in its steady state. I thought that a changing magnetic field induces an emf, and if possible, a current as well and that the current will be in whatever direction opposes the change that produced it. Furthermore, since there is an induced current, it in turn induces a magnetic field which opposes the magnetic field originally induced in the coil.

I guess there are a couple of points I'm grappling with....first, the voltage that exists across an inductor seems to me to be fundamentally different than the voltage that exists across a resistor because the one across the inductor actively slows the current down and it takes some time before it reaches it's steady state, however across a resistor even though there is also a voltage, it does not oppose the flow of current, it actually determines how much current will flow from Ohm's law, and it all happens instantly.

If that's the case, why doesn't the voltage that appears across a resistor also slow down the flow of current (initially) until it is able to come to a steady state? That's why I assumed that a voltage across a resistor only signifies how much energy is lost per coulomb of charge.

Secondly, I know that a changing magnetic field induces an emf, and I know that inductors resist changes in current, but is it the induced current, the induced magnetic field, or the back-emf that actually does the "resisiting"?

Sorry if this isn't clear...I'm having some trouble getting everything straight.

6. ### davebee Well-Known Member

Oct 22, 2008
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I think you're right - where I'd said the changing magnetic field induces a current, actually the changing magnetic field induces an electric field. That would be the back-EMF. If the inductor were ideal, with zero resistance, then it would have no voltage across it due to resistance, so the only force on the charges in the inductor would be the back EMF.

And yes, time is part of the voltage change across the inductor but not part of the voltage formula across the resistor.

Suppose you assume that you have a battery, a wire and a resistor that has no inductance. Initially no current flows. Connect the wire to complete the circuit. Instantly the current V/R flows in the circuit. I guess it's because resistance is always right there in the resistor; there is nothing to build up in the sense of magnetic fields building up.

7. ### shespuzzling Thread Starter Active Member

Aug 13, 2009
88
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But if there is an induced emf, and it's a closed circuit, isn't there also an induced current? Similarly, that induced current will also induce a magnetic field.

So what exactly is causing the inductor to resist changes in current? Is it the back-emf, the induced current or the induced magnetic field (that opposes the one brought about by the source-current)? Assuming you have a circuit with a voltage source, an inductor and a switch, why is the current through an inductor 0 right after the swtich is closed? Doesn't current have to flow through the inductor in order to bring about a change in magnetic flux (because no back-emf would be induced if there was no current flowing through the inductor)?

8. ### rjenkins AAC Fanatic!

Nov 6, 2005
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There is a steady build-up of current, with the time dependant on the value of the inductor.

The nearest mechanical parallel is a flywheel. From stopped, a force will cause it to accelerate at some rate depending on it's mass and the force used.

Is it actually moving at the instant you start applying the force? Same question.
Energy is being stored as a magnetic field in one case, as inertia in the other.

9. ### davebee Well-Known Member

Oct 22, 2008
539
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The induced EMF is oriented in the direction to oppose the current that generated it, so there isn't really an induced current; instead, the current that already exists is increased or decreased in strength, whichever opposes the initial current change that started this all.

I can't really answer the question of why. There are plenty of formulas to describe how the inductor resists changes in current, and what the magnitude of that is, but I don't think they give a satisfactory answer of why it happens.

Like Robert just said, it's probably the same answer to the question of if you drop a rock, why doesn't it immediately slam down to the ground? Why is there a noticable period of acceleration? Is time itself basically equivalent to energy being exchanged between electric and magnetic fields, or between states of potential and kinetic energy?

10. ### shespuzzling Thread Starter Active Member

Aug 13, 2009
88
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Okay, I think I can get comfortable with the idea that it just takes time for the current to build up, and that's just the way it is, the same as the flywheel and rock analogy (I took mechanical physics a long time ago and don't remember much, but I'll take your word for it), but now I'm brought back to my initial question of why or if the voltages present across a resistor and an inductor are inherently different.

If you are assuming current to be positive charge moving, then I think it still holds that charge will move from a higher potential to a lower potential or, from the positive terminal to the negative terminal, and it will do this naturally. So, when a voltage exists across a resistor, it exists solely because some external source caused it to be, and thus positive current will flow into the positive terminal of the resistor and move from higher potential to lower potential. In fact, current HAS to flow from the positive end to the negative end, because a resistor isn't capable of pushing charge from lower potential to higher potential.

But, when an inductor is initially hooked up to a circuit with a voltage source, it's initial voltage is equal to the source voltage (assuming a single loop circuit) and has the same polarity as it would if it were a resistor instead. But this voltage is actually caused by a back-emf (because the inductor is really just wire and wouldn't have a voltage unless there was some other contributing factor, thanks davebee!) and it is supposed to oppose the change that brought it about. However, since it has the same polarity as a resistor, shouldn't that voltage behave the same way as the voltage across a resistor and want to push current in the same direction as the source voltage is pushing current?

That's where I'm confused about any opposition, it actually seems like the voltage across the inductor which starts out at it's maximum value is working WITH the source voltage that brought about the change in current, because that is the natural way for current to flow, from positive to negative. If the back-emf was to oppose any change in current, then the polarity would have to be reversed.

I really feel like I'm getting closer to enlightenment...thanks everybody for the help so far.

11. ### davebee Well-Known Member

Oct 22, 2008
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No, when an inductor is initially hooked up to a circuit with a voltage source, it's initial voltage is equal to the source voltage because it is due to the voltage source, not the back EMF (yet).

That initial voltage starts a current flowing. That change in current starts to build a magnetic field. That building magnetic field in turn induces an electric field (back EMF), oriented in the OPPOSITE direction as the initial electric field due to the initial voltage.

The sum of the initial voltage plus the opposite back EMF result in an initial very small total voltage, which explains why the initial current flow is very small through an inductor. The initial current is small, but its rate of change is maximum, so the back EMF is maximum right at the start.

As the current builds, its rate of change lessens, so the back EMF lessens, and gradually the current reaches the steady state determined by voltage and the resistance of the circuit.

12. ### shespuzzling Thread Starter Active Member

Aug 13, 2009
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Ah-ha!! (I think)

So, the equation V=L(di/dt) describes the total voltage, that is, the voltage due to the source and the back-emf. I don't know why I thought they were separate...I guess I just didn't think that a source could drop a voltage across an inductor which is essentially just condensed wire, so I assumed the entire voltage must be due to the back emf.

I guess the best way to understand something like this that you can't actually see is just understanding why V=L(di/dt) and then analyzing the equation for a single loop circuit with a resistor and an inductor and a voltage source...I didn't buy that before but

I'm still not entirely sure why/how a voltage can be dropped across the inductor in the first place due to the voltage source alone (because it's just wire!)...but I see that it does this based on the equation V(source)-IR-L(di/dt)=0.

13. ### davebee Well-Known Member

Oct 22, 2008
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No, V=L(di/dt) does not include the V=IR component. I'm sorry if I seemed to imply that; I'm just recalling my old physics classes as I write this stuff and I may mis-quote principles from time to time, plus, I'm just not as good at physics as some people. But I'm trying.

If you suppose the voltage source and the wire both have really low resistance, then the moment you connect the battery, a very large current would START to flow. But the instant it starts to flow, the di/dt will start producing a very large back-EMF across the inductor which "resists" the increasing current flow (but not in the sense of the resistor component).

As the current increases, its rate of change decreases, so the back EMF decreases, and the current approaches its steady state value based on I=V/R.

14. ### shespuzzling Thread Starter Active Member

Aug 13, 2009
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Is V=L(di/dt) the back emf...with no contributions from any other source other than the back-emf generated by a changing magnetic flux?

I don't see how the source could produce a voltage across an inductor anyway because it is really just coiled wire and if it weren't for the magnetic field, then current would flow through it as it does any other piece of wire with no change in voltage. This leads me to believe that L(di/dt) completely defines the back-emf.

If that is the case, the polarity in which this back-emf is displayed across the inductor confuses me. If the back-emf is no different than any other voltage displayed across an element, then I would think that it would push current form positive to negative through itself (this would in fact aid the current from the source and seems to go against nature). If, on the other hand, there is a difference between an emf and a voltage across a resistor (where the emf has something going on inside of it that doesn't allow current to take the shortest path from cathode to anode and travel in that direction through the element, so current will instead travel from cathode to anode without going through the element; and the voltage across a resistor is able to push current right through the resistor itself, so it's still travelling from + to - but in the overall more direct path) then I guess that would answer my question, that there is in fact a difference between voltage across a resistor and emf.

So, my logic tells me that the back-emf that is displayed across an inductor has the same kind of emf as the battery in that it has a difference in potential, but the shortest path for current to flow (through the inductor from + to -) isn't possible, so it instead pushes current from + to - through the circuit. The voltage across a resistor has no say in the matter and will always push current from + to - through itself.

Does this make any sense?

15. ### davebee Well-Known Member

Oct 22, 2008
539
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I think you're pretty close to describing things correctly, as far as I understand them.

Another way to look at the circuit would be to look at just the physical things going on, since the circuit is really just the same as any physical gadget; it's just harder to visualize because most of its parts are invisible.

Instead of thinking in terms of voltage and current and resistance, look at the physical parts that are interacting - charges and neutral molecules and electric fields and magnetic fields.

A battery has too many positive charges at one end and too many negative charges at the other end. Because of that, there is an electric field between the ends of the battery.

When you connect a wire to the battery, the electric fields pull charges from the neutral molecules in the ends of the wire. As each charge is pulled off a neutral molecule by the battery's electric field, a new electric field is generated between that charge and the formerly neutral wire molecule, and this charge-pulling/new electric field generation happens all down the length of the wire.

As each charge accelerates into motion, a magnetic field is generated around that charge. As the magnetic field increases in strength, it in turn generates a new electric field. That electric field is different than the one created by charges pulled off neutral wire molecules, and has a direction opposite from the original electric field that accelerated the charge in the first place. The force on a charge due to that field is called back electromotive force, or back EMF.

An inductor orients the magnetic fields and the generated electric fields from many different charges to be pointed in the same direction. Each charge still only generates its own magnetic field, but now is also affected by the magnetically generated electric fields of millions of its neighbors. Since the change in the motion of charges is (di/dt), and the amount of coiling including the effects of core material is described by L, then the total generated electric field (which causes back EMF) is determined by L(di/dt).

Voltage is the measure of the total force on a charge due to all electric fields, so since there would be no current without at least some minimal electric field from the battery, you have to say that the voltage across the inductor is the sum (with the correct polarity taken into account) of the voltage from the battery along with back EMF voltage.

I'm not sure whether this will help or just confuse things more, but it's an alternate way to look at the circuit that may be more technically correct in terms of just what causes back EMF.

16. ### shespuzzling Thread Starter Active Member

Aug 13, 2009
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That's actually pretty illuminating...sometimes it helps me to imagine things on the smallest scale possible to get a real understanding of what's going on. Thanks!

So would it be fair to say that voltage is independent of current direction, and that the direction of current (or, the direction of "push" by the voltage) can be determined only by analyzing the circuit and what kind of elements you are dealing with?

17. ### davebee Well-Known Member

Oct 22, 2008
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I don't think that would be correct. I think that in general, Ohms law will apply, meaning that the relationship V=IR will be true, including directionality of voltage and current.

If you're dealing with reactive components like inductors or capacitors, you will need to substitute reactance in place of resistance.

18. ### shespuzzling Thread Starter Active Member

Aug 13, 2009
88
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Wow, thanks for sticking with me on this one. Let me try again.

The electric field INSIDE the inductor is effectively neutralized for a split second after a swtich is closed, because the original field is in one direction, while the back-emf is in the opposite direction, and thus no current will flow inside, and this stops current from flowing throughout the entire circuit. Therefore, the inductor acts like an open circuit, and the voltage displayed across it is equal to that of the soruce. The overall push, however, is still due to the battery, and outside of the inductor current still wants to go from positive to negative, it just can't because of the back-emf. But as the magnetic flux starts changing less rapidly, the back-emf is reduced (but the emf from the source remains the same!) and thus current slowly creeps up to whatever it's constant value is.

If I got this right you've helped me work through a year's worth of agonizing over inductors....I've got my fingers crossed.

19. ### davebee Well-Known Member

Oct 22, 2008
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Yes! I think that sounds pretty much right.

The tricky part is how there can be such a strong back EMF even though actual current flow is practically zero, and the answer must be that even though the amount of current may be just about zero, the change in current is also an extremely powerful property of the circuit.

I've enjoyed working through this. I know the basic physics formulas explain circuits in raw math, but it's much more challenging to try to explain them in plain English!