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I'm seeking clarity on understanding what happens when a constant voltage is applied to an inductor. I gave this explanation in another thread and it was regarded as wholly misguided in the following exchange from this thread: Inductor Back EMF Direction Relative to Changing Current

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I want to understand the lesson that the corrector was trying to impart. The two perspectives detailed in the above exchange are listed below

  1. A constant voltage applied to the inductors terminals causes a di/dt transient when first applied, and then di/dt goes to 0 as the inductor reaches steady state.

Apparently this is not the right way to understand inductors. The correct way:

  1. A constant voltage applied to the inductors terminals causes a ramp in current (|di/dt| > 0) until the applied voltage can no longer sustain that current.

(1) and (2) agree that a constant voltage applied to the inductors terminals causes a nonzero di/dt, but (1) says that nonzero di/dt only occurs when the constant voltage is first applied (such as when a switch closes) which allows the electric field of the source voltage to reach the terminals of the inductor, creating a voltage across the inductor, which causes electrons to flow (nonzero di/dt) which causes a changing magnetic field d(phi)/dt by Amperes Law, and this d(phi)/dt produces an EMF by Faraday's law. Initially this induced EMF is equal to the applied (constant) voltage Vs. As the (di/dt) approaches 0, the induced emf is reduced from its initial value (Vs) to 0, so that the voltage across the inductor is equal to the source voltage (Vs - emf = Vs - 0 = Vs), which is constant. Now this constant voltage produces a constant current in the inductor and there is no 'ramp in current' anymore. Is there something incorrect in this analysis?

Does it seem that the corrector in the above exchange is simply getting at the idea that if the inductor has no resistance (a fictional concept), the current would asymptotically tend toward infinity, so the transient response of the inductor when a constant voltage is applied would never end as the current will never stop increasing so di/dt never goes to 0? And then we are talking about the ability of our source to provide infinite current, which is a complete distraction from the topic of inductor behavior in a practical sense. Seems to be an utterly misplaced and highly semantic correction about the difference between a real and ideal inductor, rather than a useful correction about the practical relationship between applied voltage, current, and induced emf in a inductor. Does this seem to be the purpose of the correction from others perspective? I want to be sure I'm not missing some piece of useful information. Thank you

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  • \$\begingroup\$ Seems that the second perspective assumes a very pure inductor having zero internal resistance, while the first perspective assumes a practical inductor that includes some resistance. The resistance of the first perspective determines the final steady-state current. \$\endgroup\$
    – glen_geek
    Dec 20, 2021 at 16:36
  • \$\begingroup\$ "Think of an -initially discharged- capacitor connected to a constant current source at t=0. There will ideally be a linear voltage ramp developed across the capacitor" Is this more intuitive? \$\endgroup\$
    – carloc
    Dec 20, 2021 at 18:13

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You have to forget common sense when dealing with ideal inductors.

The only rule which applies to ideal inductors is V = L di/dt. Or, if you prefer, di/dt = V/L. For any non-zero V, di/dt remains constant, and the current increases (or decreases) without limit.

Your visualization has gone wrong because you've ignored the fact that an ideal inductor has zero resistance. A real inductor will behave as you think, with the final current being determined by Ohms Law, and unrelated to the inductance.

The flip side of inductive behavior is what happens when you build up current in an inductor, then open the switch. The opening switch, in theory, causes an instantaneous change in current, as the current drops to zero, and this should produce a (theoretically) infinite voltage spike. In practice, this won't occur, since no switch has the ability to block an infinite voltage. However, you can put a small gap in parallel with the switch, and an arc will appear at the gap, protecting the switch. This is how spark plugs work.

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I'm seeking clarity on understanding what happens when a constant voltage is applied to an inductor.

If you are talking about inductor concepts then it is natural to consider ideal inductors first before moving on to a non-ideal inductor (with DC resistance). There's a good reason; the maths gets so much harder so, we learn inductor concepts in two stages; ideal and non-ideal then we add some more non-idealities.

So, if you just say "inductor" it is natural to default to the ideal-inductor.

But, if you are talking about a non-ideal inductor (one with some DC resistance) then, as soon as any ramping current begins, there is a gradual rising volt-drop across the series resistor that steals the applied voltage from being across the ideal inductor.

So initially, di/dt begins as if just an ideal inductor is connected to the applied source voltage. Then, with the passage of time, di/dt reduces in an exponential fashion and, eventually di/dt becomes zero. At this point there is a large DC current flow with all the applied voltage appearing across the resistor.

Context is needed

Because it was I who was involved in the conversation with the OP, its important to be fair and take into account the backdrop to the question and answer. There were many misconceptions arising that needed to be corrected such as these: -

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Given the detail of the misconceptions held (and their fundamental nature), it led me to understand that we have to talk about an ideal inductor rather than unpick the errors hitched to a model that was non-ideal.

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A pure inductor would get "charged" indefinitely reaching higher currents, as long as the voltage is applied.

This is how current is charged into superconducting cables.

For a resistive inductor, as the current rises more and more voltage will drop across the resistance and less across the inductance..As a result, the current saturates at The voltage divided by the resistance.

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