# The voltage across an inductor isn't -Ldi/dt

Discussion in 'Electronic Design' started by [email protected], Mar 4, 2007.

1. ### Guest

I always thought the voltage across an inductor was -Ldi/dt. Yet it
cannot be, because for a conductor with zero resistance and carrying a
finite current, the electric field inside it must be 0 meaning there
is no voltage across it's ends. When we talk about a voltage existing
across an inductor, what we really mean is that the voltage across the
terminals of the voltmeter is -LdI/dt.

Take a look at this link for a full explanation:-

http://ocw.mit.edu/NR/rdonlyres/Phy...DB-AB80-49E0-BFAB-BCE00159F00E/0/lecsup41.pdf

2. ### Guest

That's nice. Now could you take a few minutes out of your busy
schedule to learn that "it's" means IT IS?

3. ### BobGuest

That *was* a dead giveaway, wasn't it?

Next, he's going to explain how to use sheep's bladders to prevent
earthquakes.

:-}

Bob

4. ### Boris MoharGuest

You can do that?

5. ### Phil AllisonGuest

<

** The date on the top of that text is a worry ....

....... Phil

6. ### John LarkinGuest

Nope, it's E = + L di/dt using conventional notation.
The electric field is nonzero because there's a changing magnetic
field, which creates potential. The magnetic field is changing because
of... dI/dT !
Same thing, if you have a voltmeter. Same thing if you don't.
You might do that.

John

7. ### Guest

Yeah, 2002 was a bad year for me too. Lost my job and stubbed my big
toe. Nail fell right off.

8. ### Phil AllisonGuest

** Google Groper half wit alert !!

"Phil Allison"

** Yawn .....

What a tedious moron.

...... Phil

9. ### Guest

There is no electric field inside a conductor of zero resistance
carrying a finite current. If there was, the current would be
infinite. Therefore there is no electric field inside an inductance of
zero resistance carrying a finite current. The line integral of E
across the terminals of the voltmeter - its measured voltage - equals
the rate of change of magnetic flux through the circuit consisting of
the inductor, the leads to the voltmeter and the terminals of the
voltmeter.
This is not obvious to me. With the voltmeter and leads connected to
the inductor, only V across the voltmeter contributes to the total
E.dl around the loop, since E is zero inside the leads and the
inductor. So the voltage across the voltmeter equals the total E.dl.
Take the leads away and E.dl across the voltmeter no longer equals the
total E.dl around the loop because E is no longer zero where the leads
were.
What is your impression of the paper if you have read it?
At least you have made an attempt to answer/correct my points. Thanks.

10. ### Jim ThompsonGuest

You need to read up on it. You are thoroughly confused. I hope you
are not a student ;-)

...Jim Thompson

11. ### John LarkinGuest

In a volume of space that is subject to a changing magnetic field,
there *is* a corresponding electric field. If there's a conductor in
that space, the electric field is present inside the conductor, too.
That's why you can have voltage across the ends of a transformer
winding.

No. Place a coil or loop in a time-varying magnetic field. Connect its
ends to a twisted pair or a coaxial cable and run that cable a mile
away, far from the magnetic field zone, and use the voltmeter there.
Same voltage.

Of course, if the voltmeter and its leads are in the field, you'll
have to take the additional potentials into account. In the case of a
voltmeter connected to the secondary of a transformer, the voltmeter
isn't exposed to much field, but it still measures the winding
voltage.
As noted, you can move the voltmeter far away, out of the field, and
accurately measure the non-zero coil or loop voltage. And a 100-turn
coil pumps move voltage into the meter than a 5-turn coil, or a bigger
loop more than a small one, so there must be voltage within the coil.

There *is* an electric field inside a conductor which is immersed
inside a changing magnetic field. "Ohm's law" is trashed by
electrodynamics. The electric field is like a gravitational field...
it permeates everything. If that were not true, every generator and
transformer and motor on the planet would instantly stop working.

In Hertz's seminal experiments, he detected em fields by using a
single metal wire that *almost* made a full ring, leaving just a tiny
gap. When exposed to a sufficiently strong em field, he observed
sparks jumping across the small gap. No voltmeter needed.

Classic electromagnetics.

John

12. ### Jim ThompsonGuest

Read it more carefully, you still are confused.

(Physics 8.02 is MIT FRESHMAN physics, 2nd semester ;-)

...Jim Thompson

13. ### Guest

If the current is finite and resistance is 0, then the charge inside
the conductor rearranges itself to create an electric field that
cancels the applied electric field. The net electric field is
therefore zero. Since the electric field is the measured force on a
unit positive charge placed there and an infinitesimally small i.e. 0
force is required to move it because the reistance is zero, then the
electric field has to be zero.
I think the answer is that Faraday's law is true for any passive
region that modifies the electric field, and the electric field in
Faraday's law is the total electric field in the passive region and
not just the induced electric field. So that the electric field
through the terminals of the voltmeter is that due to an
infinitesimally small current that generates a voltage across it's
internal resistance.
For a finite I , E.dl through 0R leads is still zero, from definition
of what an electric field is.
Suppose a conductor requires 0.01V applied across it to push 1A
through it. If it is now wound into an inductor and pushes 1A through
1R, the voltage through the 1R will be 1V, and through the inductance
still 0.01V. An electron in the inductance will still require the same
force to move it for the same current, whether that force is from a
static electric field or a changing magnetic field. If the force is
the same, the electric field there is the same.
The electric field inside a conductor depends on the resistance of the
conductor. But for a conductor of 0 resistance carrying a finte
current it is zero. Concentrate on the definition of an electric
field.
Yes. And the electric field in the gap was several orders of magnitude
greater than that in the metal wire.

Well, he makes the point that E.dl through an inductor is zero, yet
you have difficulty understanding why. The paper was written by a
distinguished professor at MIT.
Once again, thanks for attempting to discuss my points.

14. ### Jim ThompsonGuest

[snip]

You're mixing up DC and AC components with complete abandon.

...Jim Thompson

15. ### John LarkinGuest

It's not. Just quit being silly. If that were so, nothing would work.

Since the electric field is the measured force on a
OK, OK, I give up. Believe whatever you insist on believing.

And don't send me a resume.

John

16. ### Robert BaerGuest

I hate to tell you, but ohm's law is not trashed.
Please consider the resistance of the coil and the current thru the
coil, along with the (remote?) load.

17. ### Phil AllisonGuest

"John Larkin"

** You did notice the date on the paper the OP quoted - didn't you ?

........ Phil

18. ### Winfield HillGuest

You won't be able to maintain a steady-state DC voltage drop
across a conductor with zero DC resistance, because that would
imply infinite current. However, the transient or AC-voltage
case the story is entirely different. Consider, for example,
my 10kV 300kHz ferrite-core transformer, which has 22 turns on
its secondary. The single-turn primary has 450 volts across
its ends. Since it's a thick band of 1"-wide copper, with for
all practical purposes, zero DC resistance, by your obviously-
incorrect understanding this is not possible. BTW, the current
in my transformer's single thick strip of copper is modest and
due almost entirely to activities outside the copper strip.

19. ### Fred BloggsGuest

That lecture note is poorly written, confused, and useless. A voltage
drop, or more accurately potential drop, is understood to be a measure
of energy expenditure per unit of charge injected through the component.
There is no such energy expenditure in a pure inductance, it is all
storage, and therefore there is no potential drop across it. Faraday's
law introduces the concept of electromotive force due to induction, or
e.m.f., and quantifies its interaction with the electric field in closed
loops. It is a statement about the net potential drops around the loop
external to the inductor and not the inductor itself. You cannot have a
di/dt in the inductor without an external circuit loop, the loop will
always be present, but because it can take many forms, it is simpler to
consider that the algebraic sum of the external loop drops as appearing
across the inductor in the sense that this is where one would place the
measurement probes. Get real.

20. ### Phil AllisonGuest

"Fred Bloggs"

** You did notice the date on the paper the OP quoted - didn't you ?

Wake up.

.......... Phil