Physical/conceptual reason for inductor voltage step response

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 2K views
halleff
Messages
4
Reaction score
1
TL;DR
I have both an intuitive and mathematical understanding of the capacitor's voltage response to a current step input but I don't physically understand its dual, which is the inductor current response to a voltage step.
I'm trying to understand the physical reason why when you drive an ideal inductor (no series resistance) with an ideal voltage step input (no series resistance), e.g. some Vin(t) = V0u(t), the output current will be a linear ramp. I can see how to derive this from the inductor equation, v = L di/dt. I know some background like Faraday's law which the inductor equation comes from but it's still not intuitively clear to me.

I think what I'm looking for is an explanation analogous to this explanation of the relationship between the voltage across a capacitor which is being driven by a step current input (see attached image):

1. For time t < 0, the current source is I = 0. Assume the initial voltage on the capacitor is 0. Assume an LTI capacitor, so that Q = CV.
2. At time t = 0, the current source starts providing I = I0 (and will keep providing I = I0 for all t >= 0).
3. By definition of current (dq/dt), the current source is gradually increasing the stored charge difference on the capacitor. Since we're assuming an LTI capacitor, the voltage across the capacitor is directly proportional to the stored charge difference.
4. Since the current and thus rate of change of charge is constant for t >= 0, the voltage across the capacitor will increase linearly. As time goes to infinity, the voltage across the capacitor will go to infinity.

This explanation works for me. When I try to come up with an analogous physical explanation for the voltage source/inductor case which is mathematically equivalent (see attached image), I don't get very far:

1. For time t < 0, the voltage source is V = 0. Assume the initial current in the inductor is 0. Assume an LTI inductor, so that Φ = Li. [Note: I assume this means the magnetic flux passing just through the inductor itself, where we approximate the field outside the inductor as 0?]
2. At time t = 0, the voltage source starts providing V = V0 (and will keep providing V = V0 for all t >= 0).

From there I'm not really sure where to go. I'm used to thinking about an emf being induced "because of" a change in magnetic flux, but this seems to be a case of a change in magnetic flux being "caused by" a change in voltage. I guess they should be the same but it's not clear to me why that would be. Even if the change in voltage at time t = 0 will cause a change in magnetic flux that induces a current, the voltage is constant thereafter, so why would the magnetic flux apparently keep changing and not become constant, thus leading the current to become constant?
 

Attachments

  • Screen Shot 2020-06-22 at 9.48.17 PM.png
    Screen Shot 2020-06-22 at 9.48.17 PM.png
    10.4 KB · Views: 309
Last edited:
  • Like
Likes   Reactions: sophiecentaur
Engineering news on Phys.org
The voltage drop in the coil must match the applied voltage. The only way to have a voltage drop across an ideal coil is a change in its magnetic field (as a change in magnetic field is directly linked to a voltage across its windings). The magnetic field is proportional to the current, so current goes up over time as well. You can interpret this as "the current increase goes up until it is sufficient to balance the applied voltage".
 
The inductor current is proportional to the flux it supports.
For a fixed voltage the inductor current and the flux rise linearly.
The creation of a magnetic field requires energy input.
For an inductor; E = ½·L·I²
For a capacitor; E = ½·C·V²

The voltage induced in an inductor is proportional to the rate of change of flux.
Likewise, the rate of change of current or flux is proportional to the applied voltage.
 
For me - they are completely symmetrical ? Cap has an E field and Inductor an M field.

A difference in charge make an E field, so it is proportional to voltage.

A moving charge makes an M field, so it is proportional to current

Both inductor and capacitor store energy, and this energy can not "instantly" change, it takes TIME.
 
  • Like
Likes   Reactions: anorlunda
mfb said:
You can interpret this as "the current increase goes up until it is sufficient to balance the applied voltage".
I think this gets round the 'cause and effect' vs the Maths, problem. And it's very succinct.