# Inductor response to applied voltage

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1. Jun 16, 2015

### VantagePoint72

I'm having a bit of trouble wrapping my head around why inductors behave the way they do in certain circuits. Every physical explanation for why $V = -L\frac{dI}{dt}$ that I've seen explains how the voltage across the inductor is developed as the current changes: the changing current means a changing magnetic field, which induces an electric field along the wire. This electric field provides an EMF which, if the inductor has zero resistance, equals the potential difference that develops across the ends of the inductor.

However, in many—most, even—circuits of interest it is a (possibly time dependent) voltage that's applied across an inductor and current thus builds up according to the integral of the above equation. While I can see well enough how this works mathematically—the equation is valid whether $I$ or $V$ is the independent variable—I have a hard time turning the above argument around in order to understand physically what's happening. In terms of Maxwell's equation (and perhaps the Lorentz force law, if necessary) how can you derive the relation $\frac{dI}{dt} = -V/L$ when $V$ is applied across the inductor? To take the simplest example, consider an ideal inductor in series with a battery: the current through the inductor simply increases linearly as $I(t) = \frac{V}{L}t$ in the opposite direction as the applied voltage. How should I understand this behaviour from first principles?

2. Jun 16, 2015

### Staff: Mentor

You have a confounding negative sign that doesn't belong.

With an inductor, the voltage across it is related to its current as v = L. di/dt

You apply a voltage to it and the current slowly builds (providing you're starting from zero).. Compare this with a resistor where the current immediately jumps to its final value. These resistor and inductor currents are in the same direction.

3. Jun 16, 2015

### VantagePoint72

OK, well, my confusion was not over the direction of the current, though in any case I think we're just using different sign conventions for what directions are positive for the EMF and the current. You're using the passive sign convention, I was just using the same positive direction for both like EM books (including the one in front of me, Griffiths) often do.

That doesn't really have anything to do with my question though.

Last edited: Jun 16, 2015
4. Jun 17, 2015

### Svein

The $V_{L}=-L\frac{dI}{dt}$ equation does not have anything to do with the applied voltage, but the EMF generated across the inductor as a result of a change in the current through the inductor. Thus, the voltage that drives the current through the inductor is $V_{b}+V_{L}=V_{b}-L\frac{dI}{dt}$.

5. Jun 17, 2015

### VantagePoint72

$V_L$ absolutely has something to do with applied voltage—in the example in my OP, it's equal to the applied voltage. That must be true for Kirchoff's voltage law to be satisfied. Connecting an ideal inductor to a battery that supplies voltage $V_b$ causes a current whose growth satisfies $\frac{dI}{dt} = -V_b/L$ to develop through the inductor. To put my question another way: connecting an inductor to a voltage source causes current to flow through the inductor at just the right rate so as to produce an EMF that exactly equals the applied voltage. Appealing to the underlying facts of electromagnetism, why is that?

6. Jun 17, 2015

### VantagePoint72

Also: your expression is not correct. The total voltage across the inductor is simply $V_b$. Again, this is necessarily true by Kirchoff's law.

7. Jun 17, 2015

### Svein

I did not say that. I said "Thus, the voltage that drives the current through the inductor is..". Which means:
• If the current tries to rise too fast, VL increases and reduces the driving voltage, reducing the current.
• If the inductor is suddenly disconnected, there is no Vb to drive the current. This means that $\frac{dI}{dt}$ is suddenly large negative, so $V_{L}=-L\frac{dI}{dt}$ is suddenly large positive - a phenomena we have to deal with whenever we try to disconnect a relay or a solenoid.

8. Jun 17, 2015

### VantagePoint72

This "driving voltage" you're talking about does not sound like a very coherent concept, and is not present in any discussion about induction I've read before. There is the voltage across the inductor that you can actually measure with a voltmeter. This is $V_b$. There is the emf that opposes the increase of current, which is also $V_b$ by Kirchoff's law. $V_b$, the voltage across the inductor, is by definition the voltage that drives current through the inductor, just as it would be the voltage driving current through a resistor or any other circuit component. The quantity $V_{b}-L\frac{dI}{dt} = 0$, and as the current is steadily rising interpreting this as the "voltage that drives current through the inductor" does not make any sense. How could a current increase with time if it's being driven by a quantity equal to zero?

9. Jun 17, 2015

### Svein

Sorry, this is not so.

10. Jun 17, 2015

### VantagePoint72

It's either zero or $2V_b$ depending on the sign convention you're intending (which you've been inconsistent about) since, as I've said several times, the magnitudes of $V_L$ and $V_b$ are equal by Kirchoff's law. With the sign implied by your comment that, "If the current tries to rise too fast, $V_L$ increases," it is zero.

Thank you for your time, but I would prefer to have someone else try to answer. This talk of "driving voltage" is fundamentally incorrect.

11. Jun 17, 2015

### Svein

Kirchoff's law says something about the sum of currents. If you are singlehandedly trying to rewrite the laws of reactances, feel free to do so, but do not expect anybody to agree with you.

12. Jun 17, 2015

### VantagePoint72

Kirchoff's current law says that. Kirchoff's voltage law says that the sum of voltages over components in a loop is zero, and hence requires the magnitude of $V_L$ to equal the magnitude $V_b$ in this scenario. I am not rewriting anything; you are apparently unaware of, and mistaken about, some very basics ideas of circuit theory.

Last edited: Jun 17, 2015
13. Jun 18, 2015

### Tom_K

Physically what is happening is exactly the same as what happens when a force is applied to a mass to accelerate it. F = ma and VL = L di/dt

Electrical inductance is the same as physical or inertial mass and electrical current is the same as velocity, di/dt is acceleration.

If you understand one, you understand the other.

14. Jun 19, 2015

### tech99

May I try another approach. When you connect a battery to any circuit, there is initially a fast impulse as current starts to flow, and this contains high frequencies. Each electron which starts to move has to build a magnetic field, and this stops it from moving instantly - it exhibits mechanical inertia - so it cannot fully respond to a fast impulse. If there is iron near it, for instance, it has to build an even greater magnetic field. The energy it is storing away in the field is a consequence of its inductance, and anything which increases the magnetic field increases the inductance and also the inertia effect. As the initial impulse dies away, a steady current can flow and the magnetic field is by then fully built. As others have said, it is an exact parallel to accelerating a mass.

15. Jun 19, 2015

### Staff: Mentor

Edit: In an RL circuit there is no current impulse when a battery is connected. Consequently, there is no current impulse to "die away".

16. Jun 19, 2015

### Tom_K

There is a voltage spike across the inductor that does die away as current flows, so in that sense there is an impulse or transient response.

17. Jun 19, 2015

### Staff: Mentor

I wouldn't call that a spike, but it certainly is a voltage that dies away. The post I responded to I read as referring to a current impulse on switch-on, and there isn't one.

18. Jun 19, 2015

### sophiecentaur

This was in the OP and it is the crux of the problem / dilemma. When you stick zeros and infinities into the values in what are perceived as practical problems, you can get daft answers. Include a finite series resistance and you will get a no-nonsense answer.

19. Jun 19, 2015

### tech99

Yes, I agree, but there is a voltage impulse in the form of a step function.

20. Jun 19, 2015

### Staff: Mentor

An impulse is a short-duration pulse, a practical substitution for a delta function. A step is not an impulse; though if you have two steps, one up quickly followed by one down, you can create an impulse.

21. Jun 20, 2015

### VantagePoint72

This doesn't really make my question go away, it just complicates it (which is why I avoided adding such details in the OP). If you have a more realistic inductor and battery that each have resistance (or, equivalently, an RL circuit with ideal components) then at the instant the voltage is connected, the current is zero and so for a moment the problem is basically the same as the no resistance case. As current begins to flow through the inductor, it produces a back EMF and, for a moment, this back EMF is more or less equal to EMF of the battery. This we are assured by Kirchoff's voltage law. What I'm trying to get my head around is a more dynamical understanding of why this happens. I think there's been a lot of confusion about what I'm really asking (though there seems to be some argument that @tech99 's answer isn't correct, his/her approach to the question has been the closest to what I'm looking for), so let me give an analogy. It's not a perfect analogy, but it gets at what I'm trying to say:

Suppose you drop a weight from some height onto a spring and, neglecting air resistance and friction, ask how much the spring compresses. This is a simple calculation with conservation of energy. However, another (albeit more complicated) way to get the same answer is to show how the various forces act over time and work out the kinematics of the falling mass. For most purposes, just doing a conservation of energy calculation is the most sensible approach. However, sometimes going the long way is helpful for developing a better feel for what's going on. One may not doubt that, in the end, energy is conserved but can still wonder how exactly things conspire in a step-by-step manner so that energy winds up being conserved at all times.

For a brief moment after an RL circuit is switched on (or indefinitely after an ideal resistor is connected to an ideal battery), the voltage across it produced inductively is equal to the voltage being applied to it. Kirchoff's law, which, where it applies, is essentially just a statement of conservation of energy, tells us this must be so. I'm not questioning Kirchoff's law. I'm just trying to get a better understanding for how it is that inductors really do what they do in order to ensure it is satisfied. Faraday's law of induction tells you the voltage that will be produced by a particular change in current. That's how it's derived from Maxwell's equations. However, if I didn't know about Kirchoff's law, it might not be obvious to me the rate at which current accelerates through an inductor when a voltage is applied to it is just right so as to produce a back EMF of the same magnitude. What I'm missing is a physical understanding of how Faraday's law comes about "in reverse"—when the question is how the current behaves in response to an applied voltage, rather than the induced voltage by a particular current function. In terms fields and forces, what is happening with the charge carriers?

22. Jun 22, 2015

### sophiecentaur

It is 'near enough' equal but not Equal. That implies current can pass.
But the model is still inadequate because you are implying that a perfect step function has been applied. This is unrealistic; there will always be finite changes of applied voltage and that will be due to 'practical' values of impedance in the circuit. You can't pick and choose which bits you want to be ideal and which ones will be practical, if you want a realistic answer. The "moment" you refer to is highly relevant and the Impulse Response of the circuit determines now long this "moment' is applicable.
To apply Conservation Laws, you must apply them rigorously and I think that Kirchoff can be misleading when inductors are concerned. After all, the whole circuit must be included in your Inductor model when you take it to the level of extreme that your model is implying.
I can see, from the way you want to involve "charge carriers" in this, that you want to have a foot very firmly on Practicality. Hence, you have to include resistances (and hence, Time Constants) so your step function is no longer a step function

23. Jun 23, 2015

### VantagePoint72

You do not need imbalanced forces for current to able to pass. The ideal inductor has no resistance, forces are supposed to balanced. In any case, that has nothing to do with my question—my question is unchanged even if the forces are only nearly balanced in the less than ideal case.