Suppose we have a circuit with a voltage source, a switch open and an inductor all in series. If we close the switch, the potential difference of the voltage source is instantaneously applied to the inductor. As the current starts to build up, induced voltage from the inductance opposes it. If the induced voltage (back-emf) is equal and opposite to the applied voltage, and the net voltage is zero, what drives the current then?
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1\$\begingroup\$ This will be interesting to read especially if someone can answer it assuming that the inductor’s internal resistance or losses is zero. \$\endgroup\$– Andy akaCommented Dec 10, 2019 at 19:29
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1\$\begingroup\$ Consider that the back EMF doesn't exist if there is no current. \$\endgroup\$– The PhotonCommented Dec 10, 2019 at 19:50
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2\$\begingroup\$ The source drives the current, by virtue of applying a voltage across the inductor. The current is initally zero, but it has a gradient, hence \$L\frac{di}{dt}\$ is generated to match the source voltage. The current continues with the same gradient as long as the source is connected. \$\endgroup\$– ChuCommented Dec 10, 2019 at 20:42
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1\$\begingroup\$ Nice question... IMO the induced voltage is absolutely equal only in the first moment after closing the switch so the net voltage and current are zero. Then it begins gradually decreasing... the net voltage and accordingly the current, gradually increase. So the inductor forms the increasing current by adjusting its opposing voltage. Indeed, it is difficult to imagine this without some internal resistance (just as it is difficult to imagine what happens at the first moment when a perfect voltage source is connected to a capacitor without any resistance). \$\endgroup\$– Circuit fantasistCommented Dec 10, 2019 at 22:03
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1\$\begingroup\$ ... So both inductor and capacitor can be considered as "dynamic opposing voltage sources" - the inductor fully opposes the input source in the beginning while the capacitor fully opposes it at the end. \$\endgroup\$– Circuit fantasistCommented Dec 10, 2019 at 22:23
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2 Answers
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The seemingly paradoxical image of "a current that is flowing despite nothing is driving it" stems from the misconception that you always need a force to move charges. You don't!
Just think about those old-fashioned cathode ray tubes, where you accelerate electrons inside a highly charged capacitor, and the electrons exit the capacitor through a small hole at the end with a certain velocity. Even in the absence of any electric field they will keep this velocity (current!) right until they hit the opposite phosphorescent side of the screen where they get decelerated abruptly and get their energy converted into visible light (and some X-ray!). This lack of resistance is all because these electrons move in vacuum and they move almost in the same direction, so they don't collide with each other.
This is very different from a solid state conductor, where electrons get scattered at the atom bodies, ie. the crystal structure of the solid, and thus constantly lose energy. This is similar to the molecules in a gas that will eventually lose any initially coordinated movement direction, and showing their undirected internal energy as temperature. In order to refuel the charge carriers in a solid resistive conductor (ie. non-superconductor), a persisting electric field is needed.
In an ideal inductor there is no resistance. Think of a coil shaped cathode ray tube as an approximate mental picture. Then what governs electron movement are the law of induction and Ampere's law:
$$\nabla \times E = -\frac{\partial B}{\partial t}$$ $$\nabla \times B = j$$
the latter of which is the simplified version for slowly varying fields (no EM waves).
So the only thing the laws of nature require for the existence of a current is a magnetic field with non-vanishing curl (second equation). If there is no magnetic field (i.e. in the beginning right after pulling a huge Frankenstein switch), there is certainly no current. Moreover, the shape of the current determines the shape of the magnetic field and vice versa. If the shape of the currents is given for all times (because we squeeze the charge carriers through the windings of a coil), the only thing that can vary about the magnetic field is its magnitude with time.
On the other hand, from the law of induction (first equation) we learn, that if the shape of the magnetic field is constant and only its magnitude changes with time, it will be the source of a rotational (closed) electric field of a given shape. In a coil the induction voltages along the individual windings (multiple rotations around an axis) add up to the total inductor voltage. Again, vice versa, any electric field with non-vanishing curl will give rise to a magnetic field changing in time.
Finally we can combine both equations into one by forming the curl of the law of induction and substituting Ampere's law into it:
$$\frac{\partial j}{\partial t}=-\nabla\times\nabla\times E$$
What we see here is the mathematical way of saying that a change of current in time is caused by (the second curl derivative of) a circular electric field. The remaining steps towards U=-L*dI/dt is just geometry applied to the specific inductor under consideration, and the boundary conditions it is subject to.
Conclusion: since there is no resistance, we don't need any voltage driving the current according to Ohm's law. Rather the current change rate is driven by the rotational electric field represented by the constant external voltage that is guided through the coil windings (as soon as the switch has been actuated). Most importantly the "net voltage" is not at all zero, but rather it is the external voltage that is defined by the given boundary conditions.
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1\$\begingroup\$ It seems the general property of some "inertness" can explain the phenomenon of "current without voltage" ("movement without force"). Indeed, there is always a need for initial voltage (force) as a kind of potential energy. But in such elements it is converted and accumulated in a form of some kinetic energy… and they can produce current without voltage... movement without force... \$\endgroup\$ Commented Dec 19, 2019 at 10:29
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2\$\begingroup\$ Yes, it is actually Newton's first law (inertia) that comes into play here. The reason why we are so familiar with intertia in everyday life, but we don't usually encounter it in electrical engineering, is because electrons are so tiny and sensitive to random disturbance. Everyone will agree that a train is very inert once it is rolling, but already for dust particles submerged in the air inertia doesn't play that much of a role (whereas viscous forces do). Macroscopically resistance shares some similarity with viscosity, but on the quantum mechanical level it is a little more complicated. \$\endgroup\$– oliverCommented Dec 19, 2019 at 11:00
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1\$\begingroup\$ Well, what do you call a "true current source"? Normally the notion of a technical source (voltage or current) also includes control, i.e. if a load is connected, the controller restores its setpoint from a reservoir behind the scenes. But if you leave control out of the discussion, then an inductor could be considered a current source (in the sense that it maintains the present current to some extent). Likewise a capacitor is a voltage source because it maintains the present voltage. But any inductor and capacitor is not ideal in that they have some resistance (+inductance +capacitance). \$\endgroup\$– oliverCommented Dec 21, 2019 at 11:54
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1\$\begingroup\$ This belief may be due to a capacitor requiring an open circuit to maintain its voltage (so nobody will notice its series resistance without load). Whereas an inductor requires a closed circuit to maintain its current. But as soon as you have a closed circuit, everyday experience tells us that there will be some resistance in this circuit, which then inevitably links the current to a voltage (and thus seemingly defying "current without voltage"). If superconductivity were a more everyday phenomenon, people would probably be more inclined to accepting an inductor as a current source. \$\endgroup\$– oliverCommented Dec 22, 2019 at 11:12
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1\$\begingroup\$ @Alex: as you can see, the current is not determined by the applied voltage but by the curl of the magnetic field. What you call back EMF is somehow similar to a ficticious force in mechanics: if you accelerate a steel ball on an inclined plane, the inertial force seems to balance the external force (gravity). In your logic, the ball should not move because the "net force" is zero. But the inertial force is a fictitious force that just expresses change of motion. Likewise, the back EMF is a "ficticious voltage", that just expresses the change of current due to the increasing magnetic field. \$\endgroup\$– oliverCommented Mar 25, 2020 at 8:24
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We don't generally refer to it as "back EMF" when we're driving the coil, that term is more popular when we stop driving it.
The voltage across the coil is proportional to the rate at which the amplitude of the magnetic field is changing. When we put a fixed DC voltage across a coil (and yes, it can be superconducting) the current (and thus the magnetic field) increase at a constant rate proportional to the voltage across it, which we control.
Back EMF generally refers to the effect of not driving it once the field is established. If you open that switch, you're no longer driving it and the magnetic field collapses at the speed of light. This attempts to continue driving a current through the coil in the same direction, and since there's no current path available, a large reverse voltage will be seen across the inductor until such time as current flows...which it will take the first opportunity to do without regard for the damage done. This is why you often see diodes across relay coils, the current spike flows through the diode rather than damaging the driving circuit.
answered Dec 10, 2019 at 21:38
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\$\begingroup\$ It does not attempt to drive current in the opposite direction. It seeks to maintain the current in the same direction. \$\endgroup\$– Andy akaCommented Dec 27, 2019 at 16:49
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\$\begingroup\$ Thanks, fixed it. I was thinking of reverse voltage. Good check. \$\endgroup\$ Commented Dec 30, 2019 at 15:28