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In a RL circuit, a battery, coil, and a resistor are connected together in series. In this case after closing the switch, from KVL, we have ε-iR-Ldi/dt=0, but in an ideal situation where R=0, ε= Ldi/dt.

I must conclude that di/dt=ε/L=const, so the current builds up to infinity. However, what I understand is that in such a case I should expect i=0 since an inductor works as if it was a battery of the same magnitude as the source, |ε|, but which happened to be of an opposite polarity.

Another question is this: let's assume that di/dt is constant as it is the case apparently, then why does this rate decrease in value when R≠0?

ocrdu
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Jack
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    What do you mean "works as if it was a battery of the same magnitude as the source"? That's not right at all; a very oversimplified approximation if anything. – Hearth Nov 27 '22 at 18:20
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    Besides, even if it was, that gives you an ideal wire loop with zero net voltage across it. That's an underdefined system, and has infinitely many solutions with any amount of current flowing through it. So it's not that the current should be zero; the current could be anything. – Hearth Nov 27 '22 at 18:21
  • I meant to say an induced emf is an emf after all so it is basically a source which in this case opposing the source polarity – Jack Nov 27 '22 at 18:22
  • @Jack yes, with infinitely many steady state solutions as explained by Hearth. An inductor in parallel to the source, however, has only one steady state solution: the current being infinite. – tobalt Nov 27 '22 at 18:23
  • You don't need to worry about the induced EMF when you're forcing a voltage across it, though. Putting an ideal voltage source in parallel with an ideal inductor forces the inductor's terminals to have a fixed voltage difference. – Hearth Nov 27 '22 at 18:24
  • @Jack what should I do when someone answers my question? – Andy aka Dec 28 '22 at 10:34
  • @tobalt the question I meant to ask was Why a current passes at all let a lone be infinite. if the induced emf in the inductor pushes against the battery with the same strength for ever, then how a current can pass through? – Jack Jan 24 '23 at 21:46
  • @Jack it doesn't. Its back emf drops over time. – tobalt Jan 24 '23 at 22:22

2 Answers2

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What you should understand is that an inductor opposes instantaneous changes in current. The induced emf in the coil is given by Lenz's law. But if the inductor is directly connected to a voltage source, then there is enough time for the inductor current to ramp up. If there is no resistance in the wire and the inductor is connected to a voltage source, the current through the inductor keeps on rising until it gets damaged. But if you keep a resistance in the circuit, the current will be limited by \$\epsilon/R\$.

Jonathan_the_seagull
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Why the current rate of change is constant in an ideal, zero resistance, coil connected to a battery?

For an ideal inductor: -

$$V = L\cdot\dfrac{di}{dt}$$

This is the fundamental inductor equation and, it means that if the applied voltage is constant and, the inductance doesn't vary (it can of course) then \$di/dt = V/L\$.

Why this rate decreases in value when R≠0?

When current is low, the volt drop due to \$R_{SERIES}\$ is small and the inductor voltage is pretty much the applied voltage.

But, as current rises, the \$R_{SERIES}\$ volt drop causes the voltage across the inductor to fall hence, the rate of change of current rising falls as current rises. Current eventually settles at a constant value of \$V/R\$.

Andy aka
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