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So it's my understanding that an ideal inductor with a constant voltage applied should have a constant ramp for the current through it (a triangle wave), but an inductor that's not ideal (with resistance) will have an exponential charging curve for it's current (like with an RC circuit's voltage). But is there a low resistance at which point the current would increase at a constant rate like an ideal inductor? If the RL time constant is larger or smaller than the time it takes for an ideal current source with constant current I to charge the inductor with energy equal to (1/2)LI^2, will the inductor basically act as an ideal inductor?

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    \$\begingroup\$ Why would it? For any non-zero resistance it will be RL. \$\endgroup\$
    – Eugene Sh.
    Jun 8, 2018 at 20:49
  • \$\begingroup\$ Ideal means ideal, R=0 (zero) for ideal inductor. Even if R is extremly small it still makes inductor non-ideal. \$\endgroup\$ Jun 8, 2018 at 20:50
  • \$\begingroup\$ When R equals zero, but for most non eBay inductors, when used to their specification or data sheet it will not be much of a divergence from the ideal slope. \$\endgroup\$
    – Andy aka
    Jun 8, 2018 at 20:51

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But is there a low resistance at which point the current would increase at a constant rate like an ideal inductor?

Sure that point is where the inductor is ideal meaning that the series resistance is zero.

So you could say that in the real world (where there are no ideal inductors only ones that have some series resistance, also assuming no tricks such as superconductivity at low temperatures) there is no inductor which behaves as an ideal inductor. And you would be correct.

The thing is, how significant is the difference caused by the series resistance. Let's assume that the current decreases 0.1% because of the series resistance then you already need a pretty accurate current measuring setup to be able to measure that change.

In electronics many things are not ideal. Even a 0 ohm resistor has a series inductance of a few nH. Does that affect the behavior of all circuits? No only some RF (high frequency) circuits. RF designers know this and take the that (parasitic) inductance into account. In non-RF circuits the small inductance of a resistor is insignificant.

Same with inductors. A designer using an inductor will take into account that it has some series resistance. If the series resistance is too high then maybe the designer selected the wrong inductor. A designer should choose the right inductor such that the series resistance become insignificant.

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    \$\begingroup\$ I am in the real world and I have inductors with zero resistance. They are pretty cool. \$\endgroup\$ Jun 8, 2018 at 21:03
  • \$\begingroup\$ I have inductors with zero resistance You must be kidding, perhaps you're not measuring them right. Or are you keeping them near absolute zero perhaps? \$\endgroup\$ Jun 8, 2018 at 21:05
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    \$\begingroup\$ Well, he said they are cool.... \$\endgroup\$
    – Eugene Sh.
    Jun 8, 2018 at 21:05
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    \$\begingroup\$ @Bimpelrekkie around 4.2K \$\endgroup\$ Jun 8, 2018 at 21:33
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An inductor connected to a DC voltage source will act 'as if it were ideal' as long as the voltage drop over the internal resistance is negligible compared to the DC voltage applied.
In other words: It's a matter of how much current flows through the inductor, or, how long the voltage has been or will be applied. So the lower the applied DC voltage, or the shorter it will be applied, the higher the internal resistance of the inductor can be while still acting as if it were 'ideal'.
Meaning: the current will look like a straight ramp within the current and time measurement accuracies.

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  • \$\begingroup\$ So what you're saying is that an inductor can act like an ideal one if the DC resistance of it is lower than the rest of the circuit? \$\endgroup\$
    – Tom
    Jun 10, 2018 at 0:28
  • \$\begingroup\$ No, I was only talking about the inductor. If the voltage drop over the internal resistance of the coil is much lower than the voltage applied over the coil, then it acts as if it were ideal. Other components don't matter, as long as you know the voltage over the coil. Of course this voltage depends on other components though... \$\endgroup\$
    – HarryH
    Jun 10, 2018 at 9:40
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Depends upon the Q (the Quality Factor) and the dampening of any ringing associated with parasitic capacitances. For non-ringing, pick the resistor as

Rdamp = sqrt( L / C)

Thus 1uH and 0.01uF produces sqrt(1u / 0.01u) = sqrt(100) == 10 ohms.

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You're correct in identifying the RL time constant as a useful parameter.

An inductor will appear more or less ideal if the time constant is much larger than characteristic times in the circuit. This means if you short-circuit it, the current will continue with little loss, until something else happens in the circuit. We think of many capacitors as ideal because they are good enough that when open-circuited, their voltage tends to remain fairly constant.

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