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I'm trying to physically understand the responses of this pair of (dual) circuits: a step current source driving a capacitor and a step voltage source driving an inductor.

schematic

simulate this circuit – Schematic created using CircuitLab

The capacitor output voltage is a ramp and the inductor output current is a ramp, as expected. I understand fine how to mathematically derive these responses, what I'm trying to do is physically explain them.

What I have for the capacitor is that if we assume zero initial charge on the capacitor, then its initial voltage is zero because there is no charge difference. At time t = 0 the current source steps to some \$I_0\$. This means that, by definition of current, the source is now transferring charge onto the plates of the capacitor such that a charge separation develops on the plates over time. Assuming an LTI capacitor such that \$ q = CV \$, the voltage ramps up since the charge difference on its plates is being increased at a constant rate.

I can't quite articulate an equivalent description of the inductor circuit. I have a basic understanding of the phyiscal origin/operation of an inductor but I'm struggling to put it together here. My understanding is that, assuming an LTI inductor, it is defined by the relationship \$ \phi = Li \$, where \$ \phi \$ is the total flux linked through in the inductor. I also know that by Faraday's law, \$ v(t) = -\frac{d\phi}{dt} \$, so we can differentiate the inductor relation to get \$ v(t) = L \frac{di(t)}{dt} \$ (dropped the minus sign). From here I can mathematically derive current as a function of voltage, but why is this constant (step) voltage source apparently causing a constant change in magnetic flux that causes the finite change in current?

When I think of a change in magnetic flux through a current loop that induces an EMF and a current, I think of something like this: B I don't see how the voltage source creating a constant voltage across the inductor's terminals does this.

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  • \$\begingroup\$ "why is this constant (step) current source apparently causing a constant change in magnetic flux ", because of Faraday's law, which you had just quoted. \$\endgroup\$ – The Photon Jun 23 at 3:36
  • \$\begingroup\$ @ThePhoton I think my confusion might be that in words (from wiki), Faraday's law says: "The electromotive force around a closed path is equal to the negative of the time rate of change of the magnetic flux enclosed by the path". So the emf induced in the current loop should be due to change in flux enclosed by the current loop, like in the last picture. But an inductor has almost all of its magnetic field inside of itself (which is why it can be modeled as lumped). So does the emf approximately all drop across the inductor because the inductor is what experiences most of the mag field change? \$\endgroup\$ – knzy Jun 23 at 4:06
  • \$\begingroup\$ Yes, I think. For one thing, the turns of the coil are mutually coupled, which increases the inductive effect. For another there may be a material in the center of the coil that further concentrates the flux. \$\endgroup\$ – The Photon Jun 23 at 4:15
  • \$\begingroup\$ In my imagination, inductors have nothing to do with magnetic flux - instead they give electrons lots of inertia. When you start pushing the electrons, they don't start moving as soon as you push them, they gradually speed up. \$\endgroup\$ – user253751 Jun 23 at 12:20
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but why is this constant (step) voltage source apparently causing a constant change in magnetic flux that causes the finite change in current?

You seem to be changing the way you express things when you say this so I'm going to clear this up first. You appear to be equating: -

$$\boxed{\text{a constant change in magnetic flux WITH a finite change in current}}$$

In truth, any change in flux is also a change in current. The two are inseparable; one follows the other. I just wanted to clear that up. You probably meant what I re-phrased anyway!

From a comment: -

So does the emf approximately all drop across the inductor because the inductor is what experiences most of the mag field change?

That is correct and what confuses some people is that if the back emf exactly matches the forward applied voltage, then how come there can be any current. Maybe that's what is confusing you. If that is the case, it can be answered but needs a little thought. Basically it boils down to this: -

$$\boxed{\text{back-emf tells us nothing about the current that flows in an inductor}}$$

enter image description here

One more thing, you say this: -

My understanding is that, assuming an LTI inductor, it is defined by the relationship ϕ=Li

This is not quite true. This is correct: -

$$\Phi = \dfrac{L}{N}\cdot i$$

In other words, inductance per turn = \$\Phi\cdot i\$.

Hope this helps.

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  • \$\begingroup\$ Thanks a lot for your detailed response. The circuit diagram you shared is interesting- if I understand it right, it's just showing how the element we call an inductor is meant to capture all the back emf that appears in the current loop and turn it into a terminal voltage? \$\endgroup\$ – knzy Jun 23 at 20:03
  • \$\begingroup\$ It's a just an inductor but I could have drawn it as a zero ohm resistor; the point is that the back emf appears in series with the "inductor" and it equals the applied voltage no matter what the voltage is or how quickly it is changing. Therefore, across the "inductor" can only be 0 volts and that means we have 0 volts (or DC) divided by 0 ohms (the inductor's impedance to DC is of course zero) and that means we can't define the current by trying to say that the back emf opposes the flow of current. Inevitably we accept that di/dt is applied voltage divided by inductance. \$\endgroup\$ – Andy aka Jun 23 at 20:24

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