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I know that Inductor opposes an instant change in current direction. But it is used in AC circuits and in low pass filter circuits. It's a little bit confusing for me. If it resists instant change in current, then how it is possible that AC supply can pass through an inductor? To say clearly, I need to know how AC passes through an inductor and how it allows a low-frequency AC signals in filtering. Also, clearly explain what happens exactly in the inductor when AC passes through the inductor? Thanks, in Advance.

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Inductor current can never be a step, but of course it can change. The faster you want it to change, the more voltage you have to apply to the inductor.

One way to look at all this is that the impedance of a inductor goes up with frequency. This doesn't mean that all "AC" is blocked.

By putting a inductor in series with a signal, followed by a fixed impedance to ground, high frequencies are attenuated more than low frequency ones. Some level of slowly changing voltage is slow enough that the output voltage pretty much follows the input voltage.

You can do the reverse and put a resistor in series followed by a inductor to ground. Now the inductor shorts the low frequencies more than the high frequencies, so this is a high pass filter.

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  • \$\begingroup\$ Thanks for the answer. But I need to know what happens exactly in inductor for the positive half cycle and negative half cycle of AC and the inductor's reaction of the instance that Positive voltage crosses zero and entering into negative half cycle and vice versa. Hope I told my doubt clear! \$\endgroup\$ – Dhans Jul 10 '17 at 11:30
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    \$\begingroup\$ @Dhan: If you want to know exactly, then follow the inductor equation. \$\endgroup\$ – Olin Lathrop Jul 10 '17 at 12:13
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Also, clearly explain what happens exactly in inductor when AC passes through the inductor?

The clearest explanation I know of is the formula that relates voltage and current for an inductor. The formula is: -

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

Or put in words, if the rate of change of current (\$\dfrac{di}{dt}\$) is one amp per second then 1 volt is constantly developed across a 1 henry inductor. Using the same formula you can reverse things around and say that if you suddenly put 1 volt across a 1 henry inductor then you will get a rising/ramping current of 1 amp per second.

You can also relate this very same formula to an AC situation i.e. a situation where the driving signal is a sine wave. So, if the current through the inductor is a sine wave then the voltage across it must be the integral of a sine wave. This is a cosine wave: -

enter image description here

As you can see the voltage and current waveforms are separated by a phase angle of 90 degrees.

If you want a mechancial analogy of current rising linearly when you apply a fixed voltage across an inductor you could think in terms of a rubber band attached to a solid wall and driven by a DC motor at a constant speed: -

enter image description here

When the motor starts, the rubber band is unraveled and zero mechanical torque is needed when the motor is switched on but, as the rubber band acquires twists, the torque rises linearly with time.

The constant motor speed is the equivalent of applying a fixed DC voltage to the inductor and the torque rising is exactly the same as current rising. The solid wall is equivalent to 0 volts.

It's also equivalent in terms of energy storage; the rubber band gets longer as it forms twists and this stores energy: -

enter image description here

And this is exactly the same format as the inductor equation for energy (\$\dfrac{LI^2}{2}\$)

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  • \$\begingroup\$ Thanks for the mathematical form. From the image you attached, how inductor makes a 90 degree phase lag in current when apply AC? \$\endgroup\$ – Dhans Jul 10 '17 at 11:47
  • \$\begingroup\$ You get 90 degrees because that is the relationship between sine and cosine. V = L di/dt ensures that when di/dt is maximum (as it passes through zero because the slope is greatest) you get the peak in voltage. As it falls through zero the slope is maximum but negative hence you get a negative peak in voltage. \$\endgroup\$ – Andy aka Jul 10 '17 at 11:50
  • \$\begingroup\$ Sorry. I think I haven't asked clearly. You give a justification again in mathematical form. What I am asking is, the question is same written above, but I need an explanation of events that happens in inductor when passing AC which makes a lag in current. \$\endgroup\$ – Dhans Jul 10 '17 at 12:03
  • \$\begingroup\$ As I said, the clearest explanation of what happens with the voltage and current associated with an inductor is a mathematical one. Now I'm not one for resorting to math when there is a common form of words that can be used but, in the case of inductors and capacitors, I am certain that a mathematical explanation is the one that will always prove most useful. On this occasion, if you are looking for words I must disappoint you. \$\endgroup\$ – Andy aka Jul 10 '17 at 12:10
  • \$\begingroup\$ I'll use what you said: "how it is possible that AC supply can pass through an inductor?" - by simple examination of the formula you can see that current can pass through an inductor and that it does so inversely proportional to frequency. \$\endgroup\$ – Andy aka Jul 10 '17 at 12:14
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Here's a hand-waving answer with no maths.

An inductor is just a coil of wire. It may have a ferrite or iron core, but it doesn't have to.

A coil of wire is an electromagnet. Pass a current through it, and it generates a magnetic field. A DC current gives a steady magnetic field, while an AC current gives an alternating field.

If you apply a varying magnetic field to a coil of wire, you get a generator. Most practical generators move the coil while keeping the magnets fixed, because that's easier. But it works the other way round as well. The voltage you get out of a generator depends on how fast the magnetic field is changing. So if you spin a generator really fast, you get more voltage than if you spin it slowly.

Putting the two together, if you pass a DC current through an inductor, you get a steady magnetic field and nothing else happens.

If you pass AC through it, you generate an alternating magnetic field. But an alternating magnetic field turns it into a generator, which generates a voltage to oppose your current.

If it's low frequency AC, then the magnetic field is only changing slowly, so the voltage it produces is quite weak. But for high frequency AC, the field is changing faster, and the reverse voltage is stronger.

So inductors don't block DC, but they do block AC, and they block high frequencies more strongly than low frequencies.

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View an inductor as a box with two pieces-----one stores energy, and the other converts change of energy into an internally-generated voltage that happens to be oppositely wired.

These two boxes are in series.

If the energy attempts to abruptly change, piece #2 generates a huge opposing EMF.

I'm reading through a book recommended by "jonk" ----- "Matter and Interaction: electrical and magnetic interactions" with authors Chabay and Sherwood.

I'll read the chapter on inductors, and update my answer.


edit

deltaVsolenoid = emf_inductor - Rsolenoid * current

If Coil resistance (Rsolenoid) is very low, the deltaVsolenoid is almost the same as emf_inductor, almost cancelling.

Key: if sums were reversed, the current would grow to infinity.

from page 919 of the book, where the math included coulomb and non-coulomb electric fields that oppose.

For your use: Inductance = MU0 * Turns^2/Length * pi * Radius^2

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