# Power stored in an Inductor

1. Mar 10, 2012

### *FaerieLight*

Is it valid to say that an inductor stores power? I know an inductor stores energy in its magnetic field. Power = dE/dt so is it right to say that power is stored in the magnetic field? Or is this completely wrong, and inductors don't absorb power at all?

Thanks.

2. Mar 11, 2012

### phyzguy

An inductor stores energy. The energy can be stored or removed with a large power for a short time or with a small power for a long time. Think of a lake with a river flowing into it. Does the lake store water or does it store water flow? It stores water - you can fill up the lake with a trickle of water flowing for a long time, or with a flood of water for a short time

3. Mar 11, 2012

### *FaerieLight*

Thank you for that. That makes sense. I was confused because I was trying to reconcile this with the fact that a transformer with the secondary coil not attached to any load acts like an inductor. I don't know if this is a valid assumption, first of all. If it is correct, and transformers are supposed to conserve power (so the power input to the first coil will be the same as the power output through the second coil), then power input to the first coil must be 0, as there is no output current through the second coil. But P = VI and there is a current flowing through and a voltage across the first coil, so what is going on?

4. Mar 11, 2012

### phyzguy

The reason why an inductor dissipates no power, even though there is a voltage and a current flowing, is that the voltage and the current are out of phase. If you plot the voltage and the current over a whole cycle, you will see that during part of the cycle power is flowing in to the inductor, and during part of the cycle power is flowing out of the inductor, and when I integrate the power over the whole cycle the total power is zero.

5. Mar 12, 2012

### *FaerieLight*

Right! Thanks for that as well. So in a transformer that does have a load connected to the secondary coil, does the primary coil still behave like an inductor? That is, are V and I out of phase in the coils of an ordinary transformer?

6. Mar 12, 2012

### phyzguy

Good question. I'm not sure without simulating it, but I think what happens is the following. If there is no load on the secondary, then the primary looks like a pure inductance, and the voltage and current are out of phase and no power flows into the primary, like I said earlier. If I then add a small resistive load to the secondary, there will be a phase shift between the voltage and current in the primary so that they are no longer exactly out of phase. Power will then be flowing into the primary, where it is transferred to the secondary and finally dissipated in the resistive load. A more detailed explanation would require a simulation. Have you ever used SPICE?

7. Mar 12, 2012

### kmarinas86

An LC circuit can store "power", but it is reactive power, and its magnitude can be completely arbitrary. It is based on stored energy and the rate of oscillation.

8. Mar 12, 2012

### jim hardy

An ideal transformer with open secondary would approximate an inifinite inductance
ie its primary would draw no magnetizing current.
and any load connected to the secondary could be represented as a load in parallel with primary .

A real transformer has imperfect core hence finite inductance so draws a small magnetizing current .

At any instant the energy stored in an inductor is $LI^{2}/2$
and phyzguy's explanation above is 'spot on' .

Does that help?

9. Mar 13, 2012

### *FaerieLight*

@phyzguy: No, I've never used SPICE, but I just looked it up and it looks like a really fun program to use! I will try it out when I get a spare moment.

@jim hardy, phyzguy and kmarinas86: Thank you for your answers! So if a transfomer with no secondary load acts like an inductor with infinite inductance, and draws very little current (ideally zero magnetizing current), then where does the power/energy produced by the power source (P=VI) go? Where in the circuit is the power dissipated, if it is at all? If the inductor has no current, then it is not magnetized, and so it does not store any energy in its magnetic field.

10. Mar 13, 2012

### jim hardy

If I is zero there's no power produced. V X zero is zero.

11. Mar 13, 2012

### maimonides

That´s strange.
I´ve always thought an ideal unloaded transformer behaves like an (ideal) inductance. (Why should the inductance of the primary disappear? It is here without a secondary, and nothing changes with an unloaded secondary).

With AC, its P = U*I*cos$\phi$.
Power is zero because the current is 90° phase shifted (relative to voltage)

The energy of the magnetic field goes back to the generator. It needs considerable currents to do so, which is a load on the grid. (and thats why low power factor (cos$\phi$) is a bad thing; you may even have to pay for it)

Last edited: Mar 13, 2012
12. Mar 13, 2012

### jim hardy

Okay, i exaggerated a little. More accurately i got lazy in my thinking and explaining.

An ideal inductor would be designed for some specific inductance. It would behave as you describe.

An ideal power transformer would be designed for zero loss
so it would have an infinitely permeable core , hence require zero magnetizing current
Confession - i took a shortcut in thinking -
that'd be infinite inductance, in addition to being ideal inductance.

{EDIT - Hey i did say infinite inductance - maybe i'm not a so bad guy after all.}

In an ideal inductor of less than infinite inductance , of course as you said, energy flows instant by instant back and forth between the source and the inductor 's magnetic field.

It is very informative for a beginner to graphically draw a sine and cosine wave representing voltage and current 90° out of phase, one above the other.
Then in each quadrant observe that their product, instantaneous power, alternates between positive and negative sign - reflecting that bidirectional exchange of energy.
Aha - i love when the math agrees with the word picture !

The resulting graph shows pictorially that over a complete cycle the net energy exchanged is zero.
And that's why capacitors and inductors can oppose flow of current without generating any heat, which is a very useful trait.

That exercise planted the concept in my mind , i hope it helps somebody else.

Do not be offended by my oversimplifications i do it to help those who struggle as i had to.
Also it is a useful tool of thought to exaggerate something to the ridiculous extreme, evaluate it there, then back up to reality.

Sorry for any confusion.

BTW - take a high quality toroid power transformer and operate it at say 1/4 rated voltage. Note how miniscule its magnetizing current becomes.

Last edited: Mar 13, 2012
13. Mar 14, 2012

### maimonides

I see.
Your ideal transformer has infinite permeability, while I assume finite permeability, but no saturation and no hysteresis ( => no magnetic losses).
This way it´s easier to get (mathematically) defined flux in the core (I wouldn´t know how to calculate flux for infinite permeability without additional assumptions) and a well defined flux gives a well defined secondary voltage.
I´ll have to think about the "magnetizing current" matter.

14. Mar 14, 2012

### *FaerieLight*

So...in an inductor of infinite inductance, no energy is transferred to the inductor, while in an inductor of finite inductance, energy is transferred back and forth between the inductor and the power source?

15. Mar 14, 2012

### maimonides

Infinite inductance means 1/$\infty$ current.
Field energy is proportional to I*$\mu$; so you get 1/$\infty$*$\infty$ and you´re lost without additional assumptions. (I think de l´Hospital will not help, since the infinities are "unrelated").
But with an appropriate assumption , yes, you are right.

That´s why I do not think that the infinite inductance model is a good way to describe a transformer.
So think of a normal inductor first. (with core). You don´t assume infinite $\mu$ when you work out Maxwell´s equations to get current in an inductor (and you don´t need to! You just assume infinite saturation limit and no hysteresis). So you get defined inductance, current and magnetic flux.
Then add the secondary (open at first). Voltage is ~d$\Phi$/dt. Then you load it and you can work out the current in the secondary and its effect on flux and primary current.

Afterthought (edit):
Energy goes back and forth in one period; so integrated over an integer number of periods there is no energy transfer. (No load case, of course)

Last edited: Mar 14, 2012
16. Mar 14, 2012

### jim hardy

OOps while i was typing, your post appeared. Thank you.

At first reading i take your 'normal inductor' approach as the first refinement a student should make to my infinite inductance thought experiment.

Taking limit of LI^2 brought me to $\infty$/$\infty$^2.

Of course there's no real infinite inductors, they're just a tool for thought.
Backing away from my logical extreme to reality gets one to exactly your word picture.

As a plodder, i years ago had to go through those intermediate thought steps, extrapolate to logical limit then back up to reality..
I try to remember what were my stumbling blocks and offer what worked.

old jim

17. Mar 14, 2012

### jim hardy

Yes, over any 360 degree period it'll be zero.

My point was high inductance is desirable , to minimize that magnetizing current.

Perhaps i'll adjust my terminology for this thought experiment.
Would you entertain this train of thought:
It's a lifelong process of refining our thoughts.

Aside - there was once a manufacturer of transformers who made a line of them with brand name "Ideal". Professor told us boys: "They're very good transformers, but ideal in name only."

18. Mar 14, 2012

### maimonides

No objections.
I haven´t got that magnetizing current matter clear in my head yet. I suppose it is an EE way of thinking (I come from the physics side). AFAIK historically transformers were grouped with motors and generators, and negligible reluctance (thats where the inifnite inductance comes from) may be a useful model there.
I´ll be offline for a couple of days.

19. Mar 14, 2012

### jim hardy

I learn from everybody i meet, am still cogitating on your statement

which said to me you come from a different background and thought process than i. I hope you expound a little when you get back.
I like to be able to get to a result by diverse paths - it both makes me more confident of my understanding and lets me converse in more diverse circles.

Thanks !

old jim

20. Mar 15, 2012

### jim hardy

Aha finally it 'clicked', and it's one of those little quirky fine points that require constant refinement of our ideas..

and,,

Okay here's why the inductance of an ideal transformer must be infinite, or as you say disappear.

In your mind remove the secondary for a while.
IF there is inductance there is magnetizing current even if core is lossless wrt hysteresis and eddy currents, just pure permeability.
It takes amp-turns to push along the B-H curve even it it's a straight line with no hysteresis loop.. or in my EE thinking they're wattless amp-turns.
So magnetizing current flows in amount V / (2∏ f L)

Now replace that secondary but leave it open.
Nothing changes.

and primary current increases by (Load on secondary X turns ratio)
because every secondary amp-turn requires a primary amp-turn , as we know and accept.

HOWEVER:

Ratio of currents Ipri / Isec is not inverse of turns ratio, Nsec / Npri
That's because Ipri includes that pesky magnetizing current , Imagnetizing = V / (2∏ f L) .
Ipri is:
Ipri = (Isec X Nsec/Npri) + Imagnetizing

so ratio of Ipri/Isec is

Ipri/Isec = [ ( Isec X Nsec/Npri) + Imagnetizing ] / Isec

or Ipri/Isec = Nsec/Npri + (Imagnetizing/Isec)

Now that's an inteersting observation - an unloaded transformer with zero secondary current has a current ratio of infinity regardless of its turns ratio.
LATE EDIT for CLARITY ...........
Sure, division by zero is meaningless.
But division by nearly zero is not.
A very lightly loaded transformer has a current ratio very different from its turns ratio.
That's because the magnetizing current is a reasonable fraction of the total current.
END EDIT ..............

SOOO --- If an "ideal" transformer is to have current ratio equal to inverse of turns ratio as we are taught,
it MUST have zero magnetizing current
meaning it must have infinite inductance
which requires infinite permeability or your zero reluctance.

And THAT's the fine distinction between an ideal inductor and an ideal transformer.
An ideal transformer requires an ideal inductor of infinite inductance.

Okay okay it's a minor point since magnetizing current is small enough to neglect for most practical applications
but if we are to be rigorous in our thought processes we must have it available at the back of our mind for those situations where it DOES become non-negligible.

Like this thread - we could sure use a hand, good friend Maimonides !

How many turns on a 50:5 CT over in electrical engineering
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