colin mac

I've long been confused by Lenz's law and I'm looking here trying to make sense of what is going on. http://www.allaboutcircuits.com/vol_1/chpt_15/2.html

It explains that if you have an inductor and pot in series and the current is constant, there is no voltage dropped across the coil. That's fine. It then says
that if the resistance of the pot is reduced so to increase the current, a voltage
is induced in the coil. My first question on that is, does the voltage across the inductor plus the voltage across the resistor add up to more than the battery voltage then?
It then says that the polarity of the induced voltage is such that it tries to keep
the current constant. Does that mean if you started out with say 12V battery and 1K pot resistance and then reduced the resistance the current stays at 12mA? If so, would that mean the voltage across the resistor decreases? Could anyone explain please?

3iMaJ

If you can handle the math, Lenz's law is a consequence of conservation of energy and is treated rigorously through Maxwell's equations.

Circuit theory is a simplification of electromagnetic theory and I've attached a link that gives your question a much more rigorous treatement so long as you can handle the mathematics.

http://en.wikipedia.org/wiki/Electromagnetic_induction

But the quick answer is no, your example does not violate the law of conservation of energy. That is a fundamental inviolatable law of physics.

ericgibbs

hi colin,
The way that article describes the Faraday effect is confusing.

Look at the attached clip. NOTE: the 'minus' sign before the equation.!

Credits to this link.

http://hyperphysics.phy-astr.gsu.edu...ic/farlaw.html

Attached Images

screenhunter3.jpg (51.0 KB, 20 views)

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colin mac

I think I understand. Does Lenz's law say that the magnetic flux produced by the current through the coil changes (gets bigger and smaller) so that when added to the magnetic flux from the moving magnet, the overall flux stays constant?

ericgibbs

hi,
Think of it in terms of 'magnetic fields'.

When a current is flowing in a coil a magnetic field is created around the coil.
Its like a electromagnet.

When the current stops flowing the 'magnetic field' just dosn't disappear, it collapses, the collapsing field generates a current in the coil.

The current flowing in the coil, as the coil has inductance, generates a voltage across the coil which
acts in the same direction as the original voltage which produced the current in the first place.

Consider that I have a coil with a very large inductance, say its wound on a iron former.

In series with coil is a ammeter, I connect a 12V battery across the meter and coil.
The current that starts to flow in the coil creates a magnetic field, this increasing magnetic field generates a current in the coil
which opposes the current from the battery.

The current flowing from the battery thru the coil will increase exponentially until the coil/core saturates and the magnetic field cannot increase so the 'back emf/voltage' is no longer created and the current is limited only by the resistance of the coil winding.

If I now disconnect the battery, the magnetic field will collapse onto the coil and induce a current into the coil,
the polarity is in the same 'sense' at the battery.

The faster I disconnect the battery the higher will be the 'back emf/voltage'

Emf = -L [inductance] * [rate of change of current with respect to time]

This is the reason its necessary to fit a clamp diode across relay coils being switched by transistor, this back emf can be VERY high.
REM: the car ignition coil.!

Do you follow.?

__________________
Eric
"Good enough is Perfect"

PIC tutorials:
Nigel's: www.winpicprog.co.uk/
Gramo's: www.digital-diy.net/
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colin mac

Ok thank you eric.

crashsite

(My emphasis)

Is that true? If it's a pure inductance does the current increase in a linerar manner?

3iMaJ

No, its a first order differential relationship.

V = L*di/dt

crashsite

I've always thought about inductors and capacitors as, "1's". The voltage across a 1F capacitor, charging at a rate of 1A, increases by 1V per second. The current through a 1H inductor, with 1V applied across it, will increase at a rate of 1A per second.

Unless I'm missing the whole point, that seems pretty linear.

ericgibbs

hi,
Perhaps this link will help explain.

Googled for Inductor Current

Inductor transient response : RC AND L/R TIME CONSTANTS

__________________
Eric
"Good enough is Perfect"

PIC tutorials:
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crashsite

Yes, that explains what happens when you add a resistance in series (which, in a practical sense there always is since the very wires of the circuit and inductor have resistance). I was referring to a pure inductance or capacitance.

The formula for RC and LR time constants include the natural logarithm, e. My math skills are really bad so I'm forced to mentally work these things down to almost mechanical models where it seems to satisfy most people in this business to accept and use the math formulas.

When I took electronics in college (not for an engineering degree), they just gave the formula and the value of about 62% of the applied voltage or current for 1 time constant (and 62% of the remainder for each succeeding time increment).

I like the way the guy on the link described the time constant and it's the same process I had to work through when I figured this out. But, I don't think the time constant applies for pure reactance.

ericgibbs

[QUOTE
But, I don't think the time constant applies for pure reactance.[/quote]

Hi crashsite,
The problem would be with an ideal reactance, ie: no added resistor It wouldn't be easy to meaure the inductor voltage..

Did you college in the UK.?

__________________
Eric
"Good enough is Perfect"

PIC tutorials:
Nigel's: www.winpicprog.co.uk/
Gramo's: www.digital-diy.net/
Bill's: www.blueroomelectronics.com/

3iMaJ

The definition of a linear relationship is 2*f(x) = f(2*x). For your example where everything is 1 this relationship holds, however in a more complex relationship such as:

i(t) = cos(2*t)
v(t) = L*di(t)/dt
v(t) = -2*L*sin(2*t)

definitely does not hold.


No where in there did I mention resistance.

In an LC circuit, in some cases we would like that circuit to be resistance free as we would have a perfect oscillator.

I'm sorry I used so much math, but I don't know another way of explaining it.

The short answer is, mathematics don't care whether or not there is a resistance present, that simply changes the solution of the differential equation created by the inductor and capacitor.

Edit: In your case when the current is i(t) = t, the relationship is absolutely linear. However, that does not mean that the V-I relationship is linear in all cases as I pointed out above.

crashsite

[quote=ericgibbs;312954The problem would be with an ideal reactance, ie: no added resistor It wouldn't be easy to meaure the inductor voltage..

Did you college in the UK.?[/QUOTE]

The voltages across inductors are measured all the time but, usually with an oscilloscope since they are changing at the time of measurement.

No, I didn't college in the UK. I have been to the UK a few times since I had a job that took me to various satellite tracking stations (including the one in Oakhanger...not the NATO stuff just down the road). I would usually stay in Alton. England is nice...especially if you get there during your two weeks of summer. My college "career" was at a 2 year junior college in Central Oregon.

crashsite

Wow, I'm not even sure where to start and stop quoting here.

It's certainly not my intent to make light of the complexity of electrical reactance. My comments were specific to pure reactance in a DC environment. If, instead for 1H the inductance is 1uH, the time/voltage/current realationship holds but with one factor being changed by a factor of one million. ie: 1uH, 1V, 1A and 1uSec.

In an AC or dynamic DC environment things get a lot more complicated fast and phase angles and delta values and all sorts of "wierd" math comes into play and generally bamboozles me. But, at the same time I know that the basic priniples still apply and the circuit actions can be visualized in that context rather than on interpretation of mathematical fomulae.

The one that still freaks me out, even though I know it works because I've done it and know the explanation for it, is getting high voltages across the L and C of a series resonant circuit. Even though it doesn't really, because for the phase angles, at first glance it seems to defy Kirchoff's laws since the measured voltages across the L and C are much greater than the applied voltage at the resonant frequency.