In a hypothetical circuit with only an inductor and a DC voltage source (no resistance), why is the voltage across the inductor the same as the source voltage? I get that the charges coming from the battery need to lose their energy, so do so over the inductor, but surely when the voltage across the battery and inductor are equal, there is no net voltage acting on the current, and so no change in current and so no back emf? Thanks :)
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$\begingroup$ I don't know what "no net voltage acting on the current" means. $\endgroup$– Alfred CentauriCommented Dec 16, 2016 at 13:56
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$\begingroup$ well if the voltage source is acting on the current in one direction and the back emf acts in the opposite direction by the same amount, don't the voltages cancel out? $\endgroup$– user294388Commented Dec 16, 2016 at 19:16
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$\begingroup$ Voltage doesn't act on current; 1V across a resistor gives a different current than 1V across a capacitor, or 1V across and inductor, or 1V across a diode, etc. etc. Circuit elements are defined by the relationship between the voltage across and current through. $\endgroup$– Alfred CentauriCommented Dec 16, 2016 at 22:42
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$\begingroup$ Because the inductor and the battery are in parallel? $\endgroup$– Stack Exchange Supports IsraelCommented Dec 18, 2016 at 9:07
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$\begingroup$ @AlfredCentauri thanks I think I'm starting to get it, so am I wrong to picture voltage as force that accelerates current (for a battery at least). My thinking had been the the battery would be 'accelerating' the current and the inductor 'decelerating' it, but this is wrong? $\endgroup$– user294388Commented Dec 20, 2016 at 9:53
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3 Answers
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An inductor is basically a passive element that could store the electrical energy in it's magnetic field. Ideally the inductor has no resistance even though it could offer a reactance to time varying currents. Now, in your hypothetical circuit, we can divide the situation into three stages:
When you switch on the current, the current through the inductor rises from zero to a particular value. So, there is a change in the current through the inductor, which in turn, by its property of self-inductance will try to oppose that change by creating a back emf which appears across the inductor. Suppose the inductance of the coil is $L$. Then the applied emf
$$V=L\frac{di}{dt}=-V_{ind}$$
where $V_{ind}$ is the voltage induced in the coil. The induced emf is utilized to oppose the change in the current.
After some time, the current in the circuit becomes stable. Then the energy stored in the magnetic field of the inductor is completely utilized and is zero so that the induced emf across the inductor vanishes. This means
$$V_{ind}=0\implies i=\textrm{constant}$$
as expected. However, one could not expect $V=0$. It's because the emf applied is completely utilized for the flow of current through the circuit. If the circuit has no resistance, then a maximum current flows through the circuit. The inductor then behaves like a connecting wire. In such a case, one should not expect any voltage across the inductor.
The last case is when you turn off the current. The current through the inductor falls from a non-zero value to zero. The inductor opposes this change in the circuit by converting the magnetic energy stored in the inductor into electrical energy to support the current flow. So there is a delay we can observe for the current to become zero.
You can expect a voltage drop to appear across the inductor only when there is a change in the current in the circuit. When the current becomes stable, there is no voltage drop across the inductor. As a matter of conservation of energy, one may write, in general, at any instant $t$, the applied emf (assuming zero resistance in the circuit):
$$V=L\frac{di}{dt}$$
If the circuit has an effective resistance $R$, then the above equation becomes
$$\underbrace{V}_{\textrm{applied emf}}=\underbrace{L\frac{di}{dt}}_{\textrm{voltage across inductor}}+\underbrace{iR}_{\textrm{voltage across resistor}}$$
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$\begingroup$ "there is an instantaneous change in the current through the inductor" - this is false; the inductor current must be continuous. Also, the sign in your first equation is incorrect; it should be $V=L\frac{di}{dt}$ $\endgroup$ Commented Dec 16, 2016 at 15:22
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$\begingroup$ thanks! although my textbook says that the current will keep increasing linearly from solving V=L(di/dt) and not settle, which I find pretty counterintuitive :/ (sorry I don't get latex!) $\endgroup$ Commented Dec 16, 2016 at 19:19
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$\begingroup$ @Alfred Centauri, What do you mean by the sign is wrong? The $V$ in the context is the applied emf and the voltage induced in the inductor appears across the inductor and will be opposite to the applied emf, by Lenz's law $\endgroup$– UKHCommented Dec 17, 2016 at 6:28
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$\begingroup$ there is an instantaneous change in the current through the inductor- Of course there is. Because on switching on the current, the current rises from o to a particular value. At tat time interval, there is a change. I dont understand why my answer is downvoted $\endgroup$– UKHCommented Dec 17, 2016 at 6:29
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$\begingroup$ "The V in the context is the applied emf" - correct and thus your sign is wrong. The correct equation is $V=L\frac{di}{dt}=-\mathcal{E}$ where $V$ is the voltage across the inductor (this can be measured) and $\mathcal{E}$ is the emf generated by the changing flux threading the inductor. If the link I provided in my comment doesn't convince you, try this one $\endgroup$ Commented Dec 17, 2016 at 12:15
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An inductor is, under the simplest interpretation, just a coil of wire. The only time it has an effect on the circuit different than just a wire element is when there is changing current flow through it. If you connect a DC voltage source to an inductor, you will not have any changing current flow. Therefore, you are basically short circuiting your battery. In this case, we can no longer approximate the wire (including the wire of the inductor) as having zero resistance.
The voltage of the charged particles flowing through the circuit will drop uniformly along the length of the circuit (assuming uniform wire composition). You would obviously have a very high amount of current since this is essentially a short circuit (but since there is no change, the inductor doesn't generate any magnetic fields, back EMF, etc).
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The reason why the voltage across the source and the voltage across the load are the same is because of Kirchhoff’s voltage law. It doesn’t actually matter what type of source and what type of load it is. The voltages across the elements of any two-element circuit must always be equal.
