Voltage Across Inductor in AC

[Anonymous.Person]

I am trying to understand Inductor these days and I dont know if somebody has asked this question before on this forum so I am gonna ask here

First of all I know the basics about Inductor...I know How Inductor behaves when DC voltage is applied across it.
Also I know how Inductor behaves when AV voltage is applied through it...
I know that it is back emf that resist current flowing through inductor and I also know that back emf is produced when magnetic field across inductor changes due to change in current....
My question is that if back emf is opposing to source voltage(which it is I believe) then how come in AC circuits voltage across Inductor reaches to max value of that of source voltage? For Example If source voltage is 5Volt AC and emf producing across inductor is 3V lets say...then inductor's peak value should always remain 2volt AC...Isn't it?

[amspire]

I am trying to understand Inductor these days and I dont know if somebody has asked this question before on this forum so I am gonna ask here

First of all I know the basics about Inductor...I know How Inductor behaves when DC voltage is applied across it.
Also I know how Inductor behaves when AV voltage is applied through it...
I know that it is back emf that resist current flowing through inductor and I also know that back emf is produced when magnetic field across inductor changes due to change in current....
My question is that if back emf is opposing to source voltage(which it is I believe) then how come in AC circuits voltage across Inductor reaches to max value of that of source voltage? For Example If source voltage is 5Volt AC and emf producing across inductor is 3V lets say...then inductor's peak value should always remain 2volt AC...Isn't it?

No. An ideal inductor has zero ohms resistance So for AC, the applied voltage always has to equal the back emf otherwise infinite current would flow (ideally). In practice inductors do have some resistance, so the back emf has to be slightly less then the applied voltage so that the inductor magnetizing current can flow through the inductor's resistance.

The magnetizing current that flows in an inductor is exactly the current that will generate the correct magnetic field to generate the EMF that exactly equals the applied AC in an ideal inductor.

[Anonymous.Person]

So then in ideal Inductor if Supplied voltage is equal to the back emf then how would current flow? I mean instantaneous voltage across inductor would be zero then?
Edit: Also in real life inductor if back emf is slightly less then overall voltage drop across inductor should be very little.

[amspire]

So then in ideal Inductor if Supplied voltage is equal to the back emf then how would current flow? I mean instantaneous voltage across inductor would be zero then?
Edit: Also in real life inductor if back emf is slightly less then overall voltage drop across inductor should be very little.

In an ideal inductor the resistance is zero so current does not need a voltage drop to flow.

In a real inductor, the difference between the applied AC and the back EMF is exactly the right difference to allow the correct current to flow through the inductors resistance. In other words, if you found the difference between the Applied AC and the back EMF was 0.1V AC, then if you unwound the wire in the inductor into a straight line, and applied 0.1V AC, you would get the same amount of current as in the inductor.

[Anonymous.Person]

I think I getting your point about real inductor but I am still confused about Ideal Inductor...
even if resistance is ideally zero still AC source voltage is equal to back emf...how would current flow?
can you post a source AC voltage vs Inductor(Ideal) Voltage drop waveform?

[Zero999]

I think the difference between AC, steady state and step response is being confused.

Under steady state conditions, i.e. once the inductor has been left connected to the power supply for long enough for the current to stabilise, the inductance is unimportant. The current is simply proportional to the series resistance of the inductor. If it's an ideal inductor, the current will be infinite, which is obviously impossible.

When a voltage is suddenly applied, i.e. the inductor is subjected to a step change in voltage, the initial current is zero, then it slowly rises to the steady state value, limited only by the series resistance. In an ideal inductor, the current will linearly increase towards infinity, with the rate of change proportional to the applied voltage and inversely proportional to the inductance. In a real world inductor, the current will logarithmically increase towards the value limited by the series resistance.

When the current through the inductor is suddenly interrupted, the voltage across the inductor, at that instant will be infinite. In a non-ideal inductor, the series resistance actually makes no difference to the voltage. It's the parallel resistance which limits the voltage and the capacitance too. In real life, an arc will form across the switch contacts or the semiconductor driving it will undergo avalanche breakdown, which can be destructive, hence the need for a transient suppressor or fly-back diode.

In an AC circuit, the current through the inductor depends on its impedance, which is simply Z = 2pi*LF, in an idea conductor, where: Z = impedance in Ohms, L = inductance in henries and F = frequency in Hz. In an ideal inductor, the always current lags the voltage by 90^o. The current can be calculated using Ohm's law: I = V/Z. In a non-ideal inductor, the impedance is Z = ((2pi*LF)²+R²)^0.5.

[amspire]

I think I getting your point about real inductor but I am still confused about Ideal Inductor...
even if resistance is ideally zero still AC source voltage is equal to back emf...how would current flow?
can you post a source AC voltage vs Inductor(Ideal) Voltage drop waveform?

You are just not understanding what zero ohms means. It means any amount of current can flow with zero volts drop due to resistance.

A better question is in an ideal inductor, why is the current the correct amount?

As mentioned in an ideal inductor, if the current was less then the correct amount, the back emf would be less then the applied voltage and, as it has zero inductance, that would mean an infinite current. Rather then going to infinite current, the current would rise until the back emf was at least cancelling the applied AC voltage. So the current cannot be too low.

If the current was too high, the back emf would be higher then the applied voltage and so the current would start reversing. As soon it started dropping on the way to reversing, the back emf would drop until it was equal to the applied AC or lower.

So even with zero volts to make the current run, it cannot be too high or too low. The only possible value is the correct value.

The only current that can be stable in an ideal inductor is exactly the current that produces the back EMF that matches the applied voltage.

With all these things, this is one way to think about an inductor. There are always several different ways you can think about inductors and capacitors and they are all correct.

There are lots of properties of zero ohms that are totally weird. For example, if you have a zero ohm ring, and you start a DC current running in it, it runs forever and the ring behaves as a magnet. No voltage force is needed to keep the current flowing. We are used to requiring a voltage to make a current flow because we are used to seeing resistance in every conductor we use.

[amspire]

I think the difference between AC, steady state and step response is being confused.

Under steady state conditions, i.e. once the inductor has been left connected to the power supply for long enough for the current to stabilise, the inductance is unimportant. The current is simply proportional to the series resistance of the inductor. If it's an ideal inductor, the current will be infinite, which is obviously impossible.

We were only talking about AC which I took to mean a sine wave. DC and step changes were not mentioned.

You are totally right about the equations for inductor impedance, but we were talking more about what is happening in the inductor instead of what equations apply. The equations are a next step of course.

[Anonymous.Person]

Ok I am reaching to your point ....
I have a simple scenario now..If lets say 12volt AC is applied to a real Inductor(with minimum coil resistance) then because of back emf voltage across inductor would be less than 12voltAC source supply?

[amspire]

Ok I am reaching to your point ....
I have a simple scenario now..If lets say 12volt AC is applied to a real Inductor(with minimum coil resistance) then because of back emf voltage across inductor would be less than 12voltAC source supply?

A better way to think about it is this. Say you have an inductor with the 12V AC sinewave applied.

You measure resistance - 0.2 ohms.
You measure current - 3.0A AC.

That means the voltage across the resistor is3 x 0.2 = 0.6V.

So is the back EMF is 12V AC - 0.6V = 11.4V AC?

No, the EMF is in phase with the applied 12 V (close enough anyway), and the current is lagging the 12V AC voltage by 90 degrees and so the voltage across the resistance is lagging 90 degrees.

If you do the calculation (12V at 0deg) - (0.6V at -90 degrees), you actually get a back emf of root(12*12 - 0.6*0.6) = 11.98V AC back emf. The back EMF will be slightly out of phase with the applied 12V, and because it is out of phase, the difference provides the voltage to drive current through the resistance.

The above is a simplification. At this point, it is way easier to just use the equations that Hero999 mentioned. You can get all the correct answers from the equations.

You are probably going to ask Why isn't the back EMF in phase with the current producing it? Why is the current lagging by 90 degrees? It comes down the the fact that Faraday's law states that the EMF is proportional to the rate of change of the magnetic flux. If you have a coil and a magnet, the maximum voltage happens when you are moving the magnet the fastest.

Now the magnetic flux in the inductor is proportional to the current. When the current increasing the fastest? When it passed through zero heading positive. So the peak of the emf sinewave is when the current is at zero and going positive - the current is lagging the emf by 90 degrees.

[Anonymous.Person]

Many Thanks I think I have understand Inductor now!
Edit: Just last question...using your given scenario when I will measure voltage across inductor using AC Voltmeter how much it will show?

[Zero999]

Many Thanks I think I have understand Inductor now!
Edit: Just last question...using your given scenario when I will measure voltage across inductor using AC Voltmeter how much it will show?

What do you think?

[amspire]

12V AC

[Zero999]

12V AC

Dough, I was hoping he'd figure that out for himself!

[amspire]

12V AC

Dough, I was hoping he'd figure that out for himself!

Sometimes it is the really dumb sounding questions that are the most confusing.

Anonymous.Person, you can actually measure the back EMF in an inductor directly. How? Add another winding - you have just made a transformer. If the inductors had 100 turns and you added another 100 turns of wire, it has exactly the same magnetic flux passing through it so it will see exactly the same back EMF.

So what do you see on this second winding. I said for an ideal inductor, the back EMF has to exactly match the applied voltage. That means in our example but with an ideal inductor, the second winding will see 12V AC exactly in phase with the applied 12V AC.

In a real inductor, like the one I suggested with the 0.2 ohms resistance, you will get 11.98V AC out and the phase will be very slightly shifted. If you put a 11.98K ohm resistor across the second winding, 1mA will flow into the resistor and that current is in phase with the voltage. You will get the same 1mA current in phase with the applied voltage on the first winding - so the source is now providing real power. The magnetising current of the inductor that generates the back EMF doesn't change - it essentially stays the same as long as the input voltage stays the same.

[Anonymous.Person]

Thanks I have understand now!