# AC Inductor phase shift

1. Apr 20, 2012

### fonz

This is quite a frustrating problem for me so hopefully somebody can describe it in such a way that it settles in my mind.

It is the concept of voltage and current being out of phase. Inductors are frequently described as responding to changes in current but does this really make sense?

The big problem for me is how the inductor is reacting to changing current on an alternating voltage supply. It is the voltage that is changing so how is it not responding to the changing voltage?

Trying to visualise what is happening when a voltage is applied across an inductor. Initially when the voltage is applied what will the current be? My own intuition would suggest that since initially there is no magnetic field to oppose the change there will be a large current inrush to build up the magnetic field correct? But how is this current 90 degrees out of phase with voltage?

It is difficult for me to explain what it is I cannot understand but hopefully by providing an explanation for these two questions I will begin to understand.

Thanks
Dan

2. Apr 20, 2012

### FOIWATER

I think I can explain it to you as I thought about it when it did not make sense to me.

You will have to read each thing I write to make sense of it, and if something doesn't make sense you will have to learn more about each thing.

Once you connect AC voltage across an inductor, it draws a specific amount of current, based on resistance. The current through the inductor creates a magnetic field around the coil. As you said, the voltage is what is constantly changing. But because the voltage is constantly changing, the current is constantly changing (in magnitude) and so the magnetic field is constantly changing around the inductor as well.

Think about it, as the voltage changes the current changes, (it is increasing and decreasing) so to the magnetic field is growing (expanding away from the inductor, and then collapsing onto the coil as the current (which creates the magnetic field) is decreasing in value.

Now as the field collapses, it CROSSES THE COIL. now the conditions for induction exist. we have a voltage induced back into the coil each and every time the current value is on the negative half cycle of the ac waveform. The voltage that is induced, OPPOSES the source voltage (see lenz' law) so, you now have a LOWER value of voltage applied.

This is the nature of inductive reactance, inductors have both resistance (due to the resistance of the wire composing the coil) and reactance. You can see, the reactive part is simply due to the voltage that the inductor creates when the current changes. It is created in such a direction that it opposes the source voltage, and as so it seems to drop voltage. But it does not it simply stores energy in the form of a magnetic field and every so often (60 times per second) returns the energy to the circuit.

Now think about what you commonly hear people say, inductors opposes current changes. How can this be so, is your question. Do you now see? this can be so because as current passes through a coil, and if the current is sinusoidal, it creates a magnetic field that is also sinusoidal. Which means the field expands and collapses on the coil depending on frequency (which is why inductive reactance depends on frequency X = 2pifL) And each time it collapses on it, it induces an opposing voltage into the coil that appears to impede the source voltage.

The current is out of phases ninty degrees (and so too, is the field) because it physically takes time to build the magnetic field around the conductor, and it stores the energy there. This does not happen at the same rate that electrons travel through a circuit.

Hope this helps,

PS: This accounts for why inductors appear as shorts in DC. Imagine, if the current is not changing, the field is now static (not changing as well) If it doesn't change, then the field can not cut the inductors and induce a voltage into it that opposes source voltage. IE, at 0 frequency inductive reactance is zero as well. but as soon as you open the circuit, or stop providing voltage across the coil, the field WILL collapse.. the voltage that is induced will cause a current to back feed in the circuit. This is the purpose of connected a diode reverse biased across coils in automotive circuits containing computers (many applications). It is called a free-wheeling diode, or flyback diode and allows the current produced by the coil to return to the other end of the coil rather than feeding through the circuit and damaging other equipment.

Last edited: Apr 20, 2012
3. Apr 20, 2012

### Averagesupernova

You have it backwards. When voltage is first applied to an inductor the current will be nearly zero. Then it steadily ramps up to max. In order for there to be a 90 degree phase shift it has to be this way.

4. Apr 20, 2012

### FOIWATER

is the reason the current is initially zero because this is the time the field is expanding? obviously not, because the magnetic flux is in phase with the current, and is out of phase with the voltage lagging by ninty degrees.

fonz, the physical reasoning behind lagging / leading currents in inductors/capacitors respectively (especially including resistors) has been a sort of hot topic on a few threads here, the math becomes especially important when describing WHY they are out of phase. My previous post just describes the action of an inductor in an ac circuit, which isn't really what you asked but may give you some understanding that you didn't have prior to reading it not sure.

5. Apr 20, 2012

### DragonPetter

Here is one way to think of it mathematically. I don't know if this is of any use or already obvious to you. It can at least show you that this magic number of 90 is a result of differentiation

given the inductance relation:
$V(t) = L\frac{di(t)}{dt}$

let $V(t) = sin(t)$

then rearrange:

V(t)dt = Ldi(t)

integrate:

$\int{sin(t)}dt = L\int{di(t)}$

$cos(t) = Li(t)$

$i(t) = \frac{cos(t)}{L}$

Now, knowing that sine and cosine are 90 degrees out of phase, we know $cos(t) = sin(t-90°)$

so we can say
$i(t) = \frac{sin(t-90°)}{L}$ which shows a phase shift of -90 degrees from the voltage function, so the current is lagging behind the voltage.

Last edited: Apr 20, 2012
6. Apr 20, 2012

### psparky

7. Apr 20, 2012

### psparky

Really cool explanation without using math.

Bravo.

8. Apr 20, 2012

### Antiphon

As an exercise for the advanced student I pose a little brain teaser.

1) All real wires exhibit some inductance even if very tiny.
2) The differentiation carried out above with sines and cosines is obviously correct and the phase is always 90 degrees regardless of the tiny nature of the inductance.
3) therefore you can NEVER have voltage and current in phase!

9. Apr 20, 2012

### Fuxue Jin

A piece of wire consist of three component, DCR, inductance L and capacitance C.

With DC current, frequency is zero, V and I follows V = IR, always in phase.

With AC current, when frequency is low, L dominate, when frequency is very high, C dominate.

Phase shift between V and I depends both on L and C for that specific frequency.

10. Apr 20, 2012

### Antiphon

Ok- so far so good. Then by all means, let us ensure that the inductance dominates any capacitance by setting the frequency F = 1 femto-Hertz. Clearly then at this frequency we will have AC voltage that takes 31.6 million years to complete a cycle with the voltage and current around 7 million years apart!

11. Apr 21, 2012

### fonz

Ok thank you for contributing so far, I've had a think and I'm starting to get to grips with this however I seem to have come across a bit of a paradox

If an unmagnetised inductor is connected to a voltage source I would say that the initial current is zero because the magnetic field of the inductor induces an emf to counter the source voltage. This would mean the voltage drop initially is equal to the source voltage. As the source voltage decreases the inductor generates an emf to sustain it's magnetic field and current increases?
How can an emf be induced initially to counter the source if there is no magnetic field?

12. Apr 21, 2012

### FOIWATER

Do you know this for a capacitor? would it be possible to PM it to me?

13. Apr 21, 2012

### Antiphon

Your thinking is confused in that the voltage applied to the inductor never changes.

If you connect a 1 Volt source to a 1 Henry inductor the voltage will be 1 Volt forever and the current I as a function of time t will be I=t.

The magnetic field will grow linearly in time as well.

14. Apr 22, 2012

### fonz

The trouble I am having is that the argument for why when full voltage is applied at t=0 the current is zero is that the magnetic field in the inductor induces an emf to counter the applied source voltage.

Initially where does the magnetic field come from to generate this back emf?

By Faraday's law, induced emf is a result of current flow but no current is flowing?

My only explanation of this is that when the voltage is applied, the inductor offers virtually no resistance so maximum current tries to flow. The inductor reacts to this by absorbing this current into it's magnetic field. This changing magnetic field produces the counter emf to prevent current from flowing.

Can somebody please confirm this explanation?

15. Apr 22, 2012

### Antiphon

There is no "back" emf.

When the voltage is applied to the inductor, that is the EMF.

For a 1 volt source connected to any inductor, the EMF is 1 Volt.

16. Apr 22, 2012

### psparky

The entire explanation lies right here.

L*di/d(t)=i(t)

It covers everything you said above.....

17. Apr 22, 2012

### psparky

C*dv/d(t)=i(t)

I find derivation to be most satisfying.

18. Apr 22, 2012

### fonz

Thanks!

19. Apr 22, 2012

### psparky

An inductor's reactance of JωL fills in the rest nicely as you now realize after reading all these wonderful threads.

20. Apr 22, 2012

### Fuxue Jin

Don't we need to separate the initial state and steady state for the original question?

For steady state, the voltage and current simply follow the rule V = L di/dt. If the voltage is sine wave AC, then it will be 90 phase shift for a PURE ideal inductor.

For initial state, no matter how quick the voltage ramps up from zero to whatever the level is, it STILL takes time, even though it is very short. AND the current will be starting from ZERO.