inductors.effect of supply current on induced voltage

Discussion in 'Homework Help' started by milly molly mandy, Mar 29, 2009.

  1. milly molly mandy

    Thread Starter Member

    Sep 4, 2008
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    Over the past 2 or 3 weeks I’ve been getting all tangled up in inductors,blimey , it’s like being forced to watch two different tennis matches simultaneously! Very confusing for a feeble scientific brain like mine. However, the mists are clearing and I think I’m beginning to get the hang of it at last.
    As I am speaking now, I AM LOOKING AT A PICTURE OF ONE CYCLE OF A SINE WAVE, WHICH IS STUCK ON MY WALL, AND I AM IMAGINING THAT THIS SINE WAVE REPRESENTS THE SUPPLY CURRENT THAT IS GOING THROUGH AN INDUCTOR.
    I understand that it is a changing magnetic field that causes a voltage to be induced in the inductor, a voltage which opposes the supply voltage. I understand now that this induced voltage is at its strongest when the current wave is on its steepest parts, because that is when the current/magnetic field is changing most rapidly.

    WHAT I AM NOT TOO SURE ABOUT IS EXACTLY WHAT IS HAPPENING TO THE MAGNETIC FIELD AROUND THE INDUCTOR (& THUS ALSO OF COURSE THE INDUCED BACK-VOLTAGE) DURING EACH STAGE OF THE CORRESPONDING CURRENT SINE WAVE.

    That is, on the steepest part of the leading edge (I think that’s the right expression)of the current sine wave, I suppose the INDUCED VOLTAGE will be rising.
    But how about when the wave is falling from its 1st +ve peak, going down towards 0? What is happening to the magnetic field (and thus also, to the induced back-voltage) at this stage? I mean, does the induced magnetic field change its polarity or anything, when the current is falling?
    And how about the part of the current sine wave where it cuts through the horizontal axis at 0, and changes direction? What happens to the magnetic field (and thus also to the induced back-voltage) at this point, when the current has changed direction, but is rising?
    In short, I suppose I am a little confused because the relationship between SUPPLY CURRENT AND the magnetic field/induced voltage is tricky because it’s so complex:

    SUMMARY OF WHAT I DON'T UNDERSTAND
    AN INCREASED CHANGE IN ONE THING (current level/magnetic field) CAUSES A SIMPLE INCREASE IN ANOTHER (nduced voltage)
    BUT WHAT DOES A SIMPLE INCREASE/DECREASE IN ONE THING (current level/magnetic field)CAUSE TO HAPPEN TO ANOTHER?(induced voltage)
    AND WHAT DOES A CHANGE IN DIRECTION IN ONE THING (current/magnetic field) CAUSE TO HAPPEN TO ANOTHER? (induced voltage)


    I’m sorry this is so long. If anyone could deal with any of my points at all, or perhaps point me in the right direction, I’d be very grateful. I may not completely understand any answers, it takes time with me, but hopefully I will progress a little more in understanding.
     
    Last edited: Mar 29, 2009
  2. steveb

    Senior Member

    Jul 3, 2008
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    To find out exactly what is happening to the magnetic field requires solving Maxwell's field equations for the particular geometry of inductor. You end up with a magnetic field distributed in space, and the magnetic field lines which cut through the loop of your circuit, when added up (integrated) over the entire loop area, represent the total flux. Now this is probably going into too much detail at this stage, but it's good to understand the basic background. The point is that calculating the exact magnetic field is not that easy. But conceptually it's not that hard to imagine that there is a magnetic field directly generated by the current, and that some of that field cuts through the circuit, and that you can add all magnetic field lines to get flux  \Phi, and Faraday's Law says the negative rate of change of flux is the electro motive force (voltage) that results. Application of Faraday's law leads directly to the well known inductor equation

     EMF=-V=-{{d\Phi}\over{dt}}= -{{dLI}\over{dt}} Faraday's Law

     V=L{{dI}\over{dt}} inductor equation (if inductance L is constant)

    So now if we simplify, both the magnetic field and the total flux are proportional to the current that is flowing. So, yes, the magnetic field and flux will change polarity when the current direction changes. When the sine wave of the current reaches its maximum (or minimum) value, the rate of change of field (flux) is zero and no voltage is generated. When the current is at the maximum slope (rate of change), the voltage is maximum. This is why the current and voltage are 90 degrees out of phase with each other in an inductor (with sine waves).

    I expect my explanation is not the best, but hopefully some other people can expand on this and make it more clear.
     
    Last edited: Mar 29, 2009
  3. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    Hook an inductor in series with a 100\Omega resistor. Put a squarewave across the inductor/resistor pair. View voltage across inductor with an oscilloscope. Do the same with a sinewave input.

    If you do not have a scope, many schools/university science labs would allow a 'demonstration' of that type if you ask.

    Looking at a static sinewave won't tell you much unless you've seen how they react. There are equations that will tell you exactly what will happen, but once you've seen it, many other things make sense as well.

    If you have no access to a scope, resistor, or inductor, download a free spice program and simulate the circuit with a virtual scope.
     
  4. milly molly mandy

    Thread Starter Member

    Sep 4, 2008
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    Thanks for your replies SteveB and ThatOneGuy, which I've printed off and am going to read on a bus journey this morning
     
  5. milly molly mandy

    Thread Starter Member

    Sep 4, 2008
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    hello Steve B and ThatOneMan,

    Thanks very much for your replies.

    Re your helpful reply Steve B, I find a combination of stretching myself a bit further blended with small, easy steps, works well for me, (I enjoy background reading), which is pretty much what you did in your explanation. eg, I am pleased to have learned about the symbol for magnetic flux[​IMG].(where's it gone? I tried to paste it)

    I found it helpful to go through your step by step explanation SteveB (last 8 lines)while looking at the sine wave on my wall. Hopefully, this complex information will have thus embedded itself a little more firmly in my brain.

    I would be grateful if you could answer the following further questions:

    i) With regard to the “back-voltage” that the changing magnetic field causes – does this back-voltage exist, ALL THROUGH THE COURSE OF ONE CYCLE OF THE SINE WAVE OF THE SUPPLY CURRENT? Or is the back-voltage only there during the NEGATIVE PART OF THE CURRENT SINE WAVE CYCLE? If so, (and I know I might be wrong about this!), then, during the positive part of the current sine wave cycle, is the induced voltage of the same polarity as the supply voltage?
    (That thing you said about "Faraday's Law says the negative rate of change of flux is the electro motive force (voltage) that results." made me think about this actually.)

    ii) Re the formula you provided: EMF = -V = ..etc (sorry, I can't copy the rest of it) which relates to Faraday's Law
    What do the little ‘d’ and the little ‘t’ represent? (I guess ‘t’ might be ‘time’)
    Actually, when you mentioned "Faraday's Law" I remembered reading about it in my Electronics for Beginners textbook, so I had another look, and found that it was mentioned with regard to that experiment where you move a magnet in and out of a coil to induce a voltage in the coil. This beginners book of course didn't mention your formula, it just said "The greater the rate at which the magnetic field changes, the greater the induced voltage"

    iii) In the said formula, is it ‘minus’, as in eg, “-V”, because it’s something to do with it being a formula for the back-voltage?
    Re your reply, That One Guy
    Luckily for me, I have loads of resistors, breadboard etc, because fortunately, there’s a big electronics store in Birmingham city centre where you can buy these things individually. I bought myself a lovely new regulated power supply from there a few weeks ago, in fact, which is great, much better than batteries. I’ve never possessed an inductor though, I must get one. Does it matter how many Henrys it is, for your suggested experiment?
    I like the sound of this “free spice program”. Can I download it, and if so, how?
    Thanks.
     
    Last edited: Apr 1, 2009
  6. vvkannan

    Active Member

    Aug 9, 2008
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    As you see from the formula we have a voltage across the inductor whenever the current changes(di/dt) and it doesnt matter wether it is positive or negative half cycle

    This might help you
    http://www.tonmeister.ca/main/textbook/node115.html
    Have you been introduced to differention concept?
    'dt' according to differential calculus means small incremental change in time .
    di/dt simply gives you the change of current with respect to time or simply the rate of change of current.
     
  7. steveb

    Senior Member

    Jul 3, 2008
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    I plan to read carefully through your questions and will get back with some answers. As far as your question about the notation, this dx/dt notation indicates the derivative of x. In calculus they make a big deal about the "derivative". It looks confusing at first but is really very simple. The derivative is basically the ratio of a small change in a variable (in this case x) to a corresponding small change in another variable ( in this case t, which does stand for time in this case). The small change is considered to be infinitely small, or really they take the limit (of the ratio) as the change goes to zero.

    In the context of your question, the rate of change of flux is the derivative of flux with respect of time  {{d\Phi}\over{dt}}. This can be made more clear with a simple example.

     \Phi_2 is the flux at time  t_2 which is a short time after time  t_1 when the flux is  \Phi_1.

    The rate of change of flux is approximately  {{d\Phi}\over{dt}}={{\Phi_2-\Phi_1}\over{t_2-t_1}}.

    The meaning of the little "d", which is really just a notational symbol, can be explained as follows. In mathematics they often use the symbol delta \Delta to indicate a change. For example,

     \Delta t=t_2-t_1 and  \Delta \Phi=\Phi_2-\Phi_1

    So the symbolic notation with 'd' resembles this "Delta" notation

    {{d\Phi}\over{dt}}\approx {{\Delta\Phi}\over{\Delta t}}={{\Phi_2-\Phi_1}\over{t_2-t_1}}.

    But, the part I've left out is that the true derivative needs to take the limit as  \Delta t goes to zero. This is why they use the notation with "d" and not with  \Delta.

    Here is a web-link to a better explanation.

    http://www.sosmath.com/calculus/diff/der00/der00.html
     
    Last edited: Apr 1, 2009
  8. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    The inductor does generate a "kick-back" through a single phase of a sine wave.

    Inductors have a magnetic field around them when current is going through them. The larger the value of inductor, the larger the field. The larger this field is, the more enhanced the effect of the inductor "not wanting" to change current. When the voltage passes the peak of the sinewave, current will start reducing. When that happens, the magnetic field will collapse, inducing current into the inductor, trying to keep the amount of current flowing the same.

    After the zero crossing point, the current reverses direction, and the magnetic field grows with increasing voltage. The field generated impedes current from increasing, again, at the negative peak of the sinewave, the magnetic field is at its maximum value, and when the voltage starts to increase toward zero, the field again collapses, which induces current.

    The oversimplified process above is called Reactance, and is measured in Henries (usually milli Henries or microHenries). Inductors work with current similar to the way capacitors work with Voltage. Since capacitors are cheaper from MUCH easier to build from tiny to large values, and measure and test the results of, capacitors are more common in filter circuits where either an inductor or capacitor of equal impedance at a given frequency would work.

    Capacitors "store" voltage, Inductors "Store" current. Due to the relationship of voltage and current, in a Filter Circuit, an inductor in series does roughly the same function as a capacitor in parallel.

    --ETA: When I mention "Kick-Back" above, It is NOT related to the "Inductive Kickback" which occurs when suddenly removing power from an inductor. Doing that creates high voltages, as the inductor "wants" to continue conducting, and as the magnetic force around it collapses, the voltage increases rapidly, as there isn't a "place for the current to go".
     
  9. KL7AJ

    AAC Fanatic!

    Nov 4, 2008
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    Hello, 3M:

    Always remember ELI the ICE man. In an inductor, (L), the Current (I) leads the voltage (E) by 90 degrees. That will help you keep the math straight.

    Now...for the physical interpretation. Imagine a heavy weight suspended by a string. Attached to that weight is a stretchy spring. If you yank on the spring horizontally, it will take a while for the weight to actually move. If you imagine the yank with your hand as being equivalent to voltage (force or potential) and the actual movement of the weight (current flow) you can get a pretty good picture of what's happening with the inductor. The MOVEMENT of the current lags the force applied to the current (EMF)

    Hope this helps a bit.

    Eric
     
  10. steveb

    Senior Member

    Jul 3, 2008
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    I've been contemplating how to answer these two questions (i and iii) for a while. I believe that I answered question ii as best I could. However, these two other questions had me worried that I would create confusion if I don't answer them clearly. I can't claim that I have clear answers now, but waiting any longer wont fix that.

    The sign of the voltage shown as -V was just to be consistant with the way that people typically define voltage on an inductor. So it's basically a convention that is established. Faraday's law says that the EMF is of opposite sign to this voltage (V as established). The logic here is that an external changing magnetic field would act like an input energy source, like a generator. However, the self inductance is different in that it can not generate energy. The current in the inductor generates the changing magnetic field that couples into it's own circuit. However, if this generated voltage as a positive EMF, it would violate conservation of energy. So the EMF is negative. We could choose to say the V=EMF, but this would result in a new coil equation V=-L{{dI}\over{dt}} which is different than the established standard.

    This answers question iii. You should now be thoroughly confused. Welcome to the world of electromagnetics!

    For question i, if we are talking about sine waves, the current and voltage will not always be the same sign, nor will they always be of opposite signs. Both the voltage and current are sine waves but they are shifted by one quarter cycle. So when one hits a peak, the other hits the midvalue, and vice versa.

    To visualize this, you could make two sinewaves for your wall, and displace them by one quarter cycle. You would see that half the time the current and voltage have the same sign, and half the time, they have different sign.

    I will now stick my head in the guillotine, for proper punishment of the worst intellectual crime. "Needlessly confusing a good student". :rolleyes:
     
  11. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    steveb, nice work with \TeX! I do enjoy seeing non ASCII mangled formulas!

    Apologies for off topic.
     
  12. wr8y

    Active Member

    Sep 16, 2008
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    OOOPS!!! ELI means that currrent LAGS voltage in an inductor (E is electromotive force in volts and it comes before I (current) in an inductive (C) circuit.) Likewise for ICE....

    (I know you meant to say that, but your fingers disobeyed your mind).
     
  13. wr8y

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    Sep 16, 2008
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  14. milly molly mandy

    Thread Starter Member

    Sep 4, 2008
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    Thank you very much everyone, for your time and your answers. I am about to sit back and read carefully through the whole thing.

    This reply of mine seems to have ended up tagged on to the wrong thread! Hopefully, the moderators will spot it and relocate it.
     
  15. milly molly mandy

    Thread Starter Member

    Sep 4, 2008
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    hello everyone,

    I'm very sorry about the delay in acknowleding your helpful answers, but my life has been hectically busy over the past few days (including a French exam coming up soon)

    I did send an acknowledgement a few days ago, but I don't know what's happened to it.

    I've just read through all your great and fascinating answers again on my laptop, and in a moment, I'm going to print them off. I am very pleased to have become a little more knowledgeable about inductors, and their background.

    Steve B, I have found the information you have kindly provided in your replies very useful, and have learned a lot from it. You don't need to worry about confusing me, as now that I am 53 I seemed to have acquired a tendency to hang doggedly on to anything (worthwhile), until I have got it sorted. I can now see the complex background of this topic must make simplying it quite a challenge

    I'm looking forward to reading through all your answers again in a minute, when I've printed them. Hopefully, on the 2nd reading, even more of their content will be its way into my brain, and organise itself there.

    I suspect that I may have some further questions on this topic, I hope that's OK



    I
     
  16. milly molly mandy

    Thread Starter Member

    Sep 4, 2008
    25
    0
    hello everyone,

    I'm very sorry about the delay in acknowleding your helpful answers, but my life has been hectically busy over the past few days (including a French exam coming up soon)

    I did send an acknowledgement a few days ago, but I don't know what's happened to it.

    I've just read through all your great and fascinating answers again on my laptop, and in a moment, I'm going to print them off. I am very pleased to have become a little more knowledgeable about inductors, and their background.

    Steve B, I have found the information you have kindly provided in your replies very useful, and have learned a lot from it. You don't need to worry about confusing me, as now that I am 53 I seemed to have acquired a tendency to hang doggedly on to anything (worthwhile), until I have got it sorted. I can now see the complex background of this topic must make simplying it quite a challenge

    I'm looking forward to reading through all your answers again in a minute, when I've printed them. Hopefully, on the 2nd reading, even more of their content will be its way into my brain, and organise itself there.

    I suspect that I may have some further questions on this topic, I hope that's OK



    I
     
  17. b.shahvir

    Active Member

    Jan 6, 2009
    444
    0
    Hi, :)

    To visualize the concept of inductance is pretty difficult. I have been dabbling with it for quite sometime now! In order to explain you the working of an ‘ideal’ inductor, I will have to make certain assumptions;
    1) the coil of inductor should possess zero resistance
    2) The inductor should have a magnetic material core (iron, ferrite, etc…. but not air-core). In an air-core inductor, the voltage and current waveforms will be practically in-phase with each other thus defeating the purpose.
    3) The magnetic core should be ‘Lossless’
    4) The input voltage to the ideal inductor is purely sinusoidal.

    The voltage and current waveforms do not immediately phase shift by 90 deg as depicted by textbook diagrams. Consider positive half cycle of ‘voltage versus current’ waveform for an ideal inductor. At ‘t (time)’ = 0, as the voltage wave just rises above zero, the current wave rises in-phase with it. However as the voltage wave further progresses at ‘t (time)’ > 0 towards it’s peak value, the rate of change of magnetic flux becomes more pronounced ad the current wave now starts going out of phase w.r.t the voltage wave. Eventually as the cycle progresses, the current wave now completely goes 90 deg out of phase w.r.t the voltage wave (assuming an ideal inductor).

    If you concentrate on the negative half cycle of the voltage vs. current waveform, you will find that the phase shifting is due to the ‘discharge’ of electrical energy by the inductor back to the power source as the magnetic field collapses (or rises, whatever may be the case), which precisely occurs when the input voltage wave has already reversed itself. This charge/discharge action of an inductor occurs due to the finite time taken my the magnetic field to rise or collapse……as force fields cannot change it’s states abruptly at ‘t = 0’ as per the laws of Physics or Mother Nature (if it would, the consequences would be disastrous). This in turn causes a delay in the flow of current thru the inductor and results in the phase shift between the voltage and current waveform. This delaying property of an inductance is also sometimes known as ‘Magnetic Inertia’.

    I suggest you have a look at the link below;

    http://forum.allaboutcircuits.com/showthread.php?t=21616

    Although the query deals with magnetic core design pertaining to transformer, I have attempted to intuitively explain the concept of inductance and the term ‘Magnetic Inertia’. I hope my explanation helps you visualize inductors better! ;)

    Kind Regards,
    Shahvir


     
  18. David Lewis

    Member

    Jun 29, 2009
    12
    0
    Thatoneguy wrote: "The oversimplified process above is called Reactance, and is measured in Henries (usually milli Henries or microHenries)."

    I think reactance is measured ohms and inductance is measured in henries.
     
  19. David Lewis

    Member

    Jun 29, 2009
    12
    0
    milly molly mandy wrote: "But how about when the wave is falling from its 1st +Ve peak, going down towards 0? What is happening to the magnetic field (and thus also, to the induced back-voltage) at this stage?"

    The strength of the magnetic field is decreasing.

    milly molly mandy wrote: "I mean, does the induced magnetic field change its polarity or anything, when the current is falling?"

    Not necessarily. No quantity changes polarity until it crosses an axis on the graph.
     
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