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I'm wondering if the following statement is fully correct:

Voltage causes current through a closed circuit, but through an inductor it is the change in current that causes a voltage.

Obviously there is no current without voltage. In a simple DC circuit there's no doubt that voltage causes the current to flow.

However in AC circuit with an inductor, the voltage drop across the inductor is proportional to the rate of change of current. So we can have 0 voltage + peak current and vice versa.

But I wouldn't say that current causes voltage in this case, because there would be no current change without a voltage source connected to the circuit!

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    $\begingroup$ The implicit "causality" in the assertion is just plain wrong. $\endgroup$
    – nabla
    Nov 29 '15 at 18:32
  • $\begingroup$ So there is no cause-effect relationship between voltage and current - is that your point? $\endgroup$ Nov 29 '15 at 18:51
  • $\begingroup$ It is a mutual effect, happening at the same time. Mathematically this is modeled as $V(t) = L\cdot I'(t)$, and "=" is a symmetric relationship. $V$ doesn't come before $I$, and viceversa. $\endgroup$
    – nabla
    Nov 29 '15 at 19:01
  • $\begingroup$ @gerd that's what I was thinking - it's best to just accept the mathematical formula for this, and forget about cause and effect relationship. Under the hood there's a lot of quantum mechanics going on, and QM itself often doesn't work according to cause and effect rules - I think you could agree with that. $\endgroup$ Nov 29 '15 at 19:06
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    $\begingroup$ "Obviously there is no current without voltage." - this is false. $\endgroup$ Nov 29 '15 at 22:54
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The magic here is Jefimenko's equation (of causity). It is charge, and moving charges that produce and respond to a field.

Let's suppose you have a voltage field. These can exist, but if the charge is fixed, then little current flows. This is what a resistor does. On the other hand, if the field can cause a displacement of charge, then a current will flow until a counter-field exists to stop it.

Note for example, no current flows across an inductor (transformer) or capacitor. The energy is transferred, and creates a new current on the other side of these devices.

But there are situations where voltage does not cause current, sometimes the charge and field must build high enough for the charge to discharge through a spark, or bolt-of-lightning.

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In an inductor you are also using the voltage, and hence current flow, to "pump up" the magnetic field and when the voltage is removed or reversed the field collapses generating a reverse voltage and current.

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If you have a metal of some resistivity, it needs an electric field inside of it to sustain a current. If there is an electric field in it, the line integral $\vec{E}\cdot d\vec{\ell}$ (which is the electromotive force) along a segment is nonzero. Hence there's a voltage somewhere. If there weren't, there could be no current.

Likewise, one of Maxwell's equations, the Faraday equation, tells you that $\oint\vec{E}\cdot d\vec{\ell}$ is nonzero around a loop of an inductor containing a time varying magnetic field (and the magnetic field depends on the current, so it is nonzero around a loop of an inductor with a time varying current). So across the inductor, there is a voltage when the current is changing.

So in those senses, the statement is true!

You say:

However in AC circuit with an inductor, the voltage drop across the inductor is proportional to the rate of change of current. So we can have 0 voltage + peak current and vice versa.

but you don't say zero voltage across where. You probably are visualizing zero voltage across an alternating current source. That's fine, but there's still a voltage in the circuit across the inductor. (You really have to measure voltage across things, because what you really care about is electromotive force)

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  • $\begingroup$ If my circuit is made solely of AC source and inductor, then the voltage across the inductor is exactly equal (with opposite sign) to the source voltage. That's why I didn't mention it. $\endgroup$ Nov 29 '15 at 19:01
  • $\begingroup$ @user4205580 There are lots of little nuances and this is the reason people usually talk about LR or LC or RC circuits. If you just have an "L" circuit, the current will be as high as possible until the "R" of the wire itself, the battery, or the inductor comes into play, and you have to treat things as a three component circuit. If you really did have zero voltage across $L$ and zero voltage across the battery (and you measure these things in a lab, so think of a real circuit here), there would be no current! $\endgroup$
    – user12029
    Nov 29 '15 at 19:16
  • $\begingroup$ I think that Kirchoff's voltage law is always true, even if the resistance in my circuit is zero. It means voltage at source + voltage drop across the inductor = 0. $\endgroup$ Nov 29 '15 at 19:24
  • $\begingroup$ But I get your point that there is no such thing as zero resistance. $\endgroup$ Nov 29 '15 at 19:27
  • $\begingroup$ @user4205580 and if the voltage drop across the inductor and the voltage drop across the source are both zero, in a real circuit you build, then the current is zero. $\endgroup$
    – user12029
    Nov 29 '15 at 19:53

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