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Inductors Charging

  1. Mar 1, 2004 #1
    Most people understand (gut understanding) the charging of capacitors more than that of inductors. They just apply the duality to inductors and move on (without seeking the same understanding of inductors).

    Teachers of my past never could explain exactly why the charging of a capacitor slowed down with time, but I figured it out by thinking it over. It's because the more electrons that move from one plate to the other, the harder it is for the ones next in line to do so. Why harder? Because there are now more -ve charges repelling them at the destination plate, and more +ve charges attracting them at their originating plate.

    Thus, the charging of capacitors slows down with time (universal time constant curve).

    But what about inductors? They're current graph (charging of flux) is exactly the same shape as a capacitor's voltage graph (charging by electrons). WHY?

    Without using differential equations and any sort of method that just points at the duality with caps, can someone explain why the initial FLUX into a inductor has a much easier time building up, than the later FLUX (FLUX is proportional to current through).
  2. jcsd
  3. Mar 3, 2004 #2
    I think of it more simply as the voltage difference between the source and the capacitor is decreasing so the current decreases so the number of charges per second flowing onto the cap decreases.

    Electrical eng's don't think of flux at all :) But now that I'm a physicist I'll have to think about it. From zero current to some current is a big change, so there is a large voltage across the inductor. As the current builds up, going from some current to slightly more is a smaller change so the voltage decreases allowing more current to flow. Eventually, the current is limited by the resistance in the circuit with zero volts across the inductor. Is that an answer?
  4. Mar 6, 2004 #3

    It's true that the voltage across the inductor sinks as more current is supplied through it (because that current will also drop across the R in the circuit.. and to obey KVL, inductor voltage sinks!)

    But why does the current's rate of change slow down? (without calculus or anything)

    And why is the charge rate of an inductor increased when the R in the circuit is increased? Intuitively, it's not clicking.

    Capacitors, for some reason, make much more intuitive sense to all of us (my teachers, authors, etc.).

    Why aren't there any good explanations for inductors?
  5. Jun 9, 2006 #4
    regarding inductor I have one more question, that why in purely inductive sircuits, current lags by voltage by exactly 90 degree? again mathemetical explaination is not enough. what exactly happens when, ac supply voltage is given to purely inductive circuit?
  6. Jun 9, 2006 #5


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    You folks need to all be more explicit about what you are driving the test inductor with. You can't talk about the current through and voltage across the inductor as it responds to some nebulous drive source.

    The current through an "ideal" inductor (zero resistance) will not level off if driven by a voltage source with a zero output impedance. It *will* level off if driven by a real voltage source with a set output resistance. The leveling off is due to the stabilization of the voltage across the inductor at zero Volts, and all the supply voltage dropping across the output resistance. The final DC current through the inductor is V/R.

    Capacitance is defined by the voltage that you get in some geometry when you charge it up with some amount of charge [tex]Q = C V [/tex] The relationship that you refer to ("no calculus") for current and voltage in a capacitor is [tex]I = C \frac{dV}{dt}[/tex] The current is proportional to the change in voltage with respect to time, and that proportionality coefficient is the capacitance.

    Inductance is defined by the flux linkage that you get in some geometry when you pass a current through it [tex]\wedge = L I [/tex] The relationship that you refer to ("no calculus") for current and voltage in an inductor is [tex]V = L \frac{dI}{dt}[/tex] The voltage across an inductor is proportional to the change in current with respect to time, and that proportionality coefficient is the inductance.

    So to answer pankti's question (with a quiz question), if you put a voltage across an inductor where [tex]V = sin(\omega t)[/tex], then what is the equation for the current, and what is the phase shift between the sin and cos functions?

    And to the OP's question, if the calculus is not intuitive enough for you, think of it this way:

    -- You put a voltage across an inductor
    -- That voltage causes a current to start to flow
    -- The change in current generates a back EMF voltage to be generated
    -- The higher the inductance, the higher the back-EMF that is generated
    -- The back-EMF voltage opposes the externally impressed voltage somewhat, which slows down the possible rate of changes in the current.
    -- The change in current is proportional to the voltage across the inductor, and when there is a series resistance in the overall circuit, the current increases (more and more slowly) to the point where all the voltage drop of the source power supply is across that series resistor (and any real coil resistance of course).
    Last edited: Jun 9, 2006
  7. Mar 7, 2010 #6
    Sorry to pull this thread out after such a long time....but I've been thinking and thinking about what the op said about the capacitor...
    It seems to me the 'levelling off' of the current curve i.e its sinusoidally varying nature is a direct consequence of how the voltage applied (which is sinusoidally varying) demands the current to change in order to match up the potential accross the capacitor in order to equal the applied voltage.
    If I'm right,there is no place for the description that the op gives. Please tell me if I'm wrong.
  8. Mar 7, 2010 #7
    Perhaps this post supports my point??

  9. Mar 7, 2010 #8


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    Yes, that is the better way to think of it. The voltage source by itself is happy to supply all the "force" you need to herd those pesky electrons. But when the delta-V gets lower and lower, you get less current, so less electron movement...
  10. Mar 7, 2010 #9
    That's a relief!
  11. Mar 8, 2010 #10
    Hi everyone..
    I think I'll make use of this thread to get myself clear about inductors in ac....
    I've been thinking and I've come up with an explanation...ot's just that I'm not sure about it.....could someone tell me if my explanation is okay?

    I'll put it in the next post.
  12. Mar 8, 2010 #11
    Referreing to my diagram,the horizontal axis is time t and the verical axis is a meaure of the voltage applied and current I. (PLease take some time to go through my post)

    Now, at t=o,the applied voltage is 0 but at that instant,it is changing to a higher value dv.

    At the point dv,a potential has been established in the direction of dv and so instantly, a current is produced.

    However,due to the inductor in the circuit,senses this rise in current from 0 to a certain value corresponding to dv,as instantly produces a back emf equal to dv,and so there is no net current in the circuit.

    at V applied= dv+dv' (dv' is the increment on dv after a time dt..and its not equal to dv due to sinusoidal nature of curve),again the applied potential has increased,which is supposed to increase the current due to Vappl(Applied voltage)...howevever,again,due to the inductor,there is no net current in the circuit.

    At t= 2T-dt,the voltage is instantaneously decreasing at the greatest rate,so induced emf is largest but value of Vappl is smallest(zero)..so the maximum induced emf has the dominant effect,and the current in the circuit is maximum.

    This continues till T. At T-dt,there was again a certain value of Vappl,but it failed to produce any current due the equal back emf by the inductor.

    At T,there is no change in cuurent(The V curve is flat),so there is no tendency of back emf in the inductor.

    However,at T+dt, there is a lower value of Vappl= Vmax-dv,so there is a reduction in current in the circuit due to the applied voltage,so there is a reduction in flux linked to the coil due to the applied voltage...hence,the inductor now tries to push current in the direction of the applied voltage,to prevent this reduction in fluxed lined to it (Lenz's law).

    Now,there is a small current in the direction of the applied voltage due to both the inductor's induced emf and the Vappl.

    At T+2dt,the value of Vappl is further decreased to Vmax-dv-dv'. Thus,there is a larger decrease in Vappl from T + dt to T+2dt than there was from T to T+dt(due to sinusoidal nature of curve,for equal intervals dt,there are uneven changes in Vappl).

    Due to this larger drop in Vappl,there is a larger induced current in coil at T+2dt than at T+dt.

    At successive increments of dt,the drop in dv is greater (due to sinusoidal nature of curve) and so induced emf is greater at each successive instant.

    However,there is decrease in Vappl at each point,but due to the increasing induced emf,the net current in direction of applied voltage increases.

    Attached Files:

  13. Mar 9, 2010 #12
    Okay,if noone wants to read my huge post,then could someone atleast clarify this one point..then I could try to check my understanding on my own...

    Suppose we have an ac supply,and only an inductor is in the circuit....initially,when the applied voltage is switched on,there should be no current (due to cancellation of applied emf by back emf).....this 'no current' condition continues till the appl. voltage becomes maximum(peak of voltage sinusoid).

    Now,when the applied voltage starts decreaseing(from peak value of voltage applied),a certain value of cuurent is seen in the circuit....is this current solely due to the induced emf due of the inductor?.....that is,does the appl. voltage not have any role in producing this current?

    (Note: in the graph of current and applied voltage for ac circuit with only inductor,the current reaches its maximum value when the appl. voltage is zero! So I guess whatever current there is,must be due to the induced emf of the inductor).
  14. Mar 9, 2010 #13


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    Yeah, sorry that the post was a bit too long for my limited attention span :blushing:

    But what you are saying here in the quote seems wrong. The equation relating the current and applied voltage is always true for an inductor:

    [tex]v(t) = L \frac{di(t)}{dt}[/tex]

    When you say

    that doesn't fit with the equation, does it?
  15. Mar 11, 2010 #14
    I was referring to the situation in which the applied voltage is just switched on.

    I found this in http://www.tpub.com/neets/book2/4.htm

    Again,since I'm trying to explain the equation you gave to myself (might be quite useless,all the same it's absorbing!),I found that it the fact that the applied voltage is zero at the point the current is maximum must mean that the current in the circuit must be due to induced emf due to the coil.

    However,at this point(V appl. is zero),the rate of change of applied voltage is maximum on the sinusoidal graph,so the current due to it should be changing at the max. rate,thus induced emf due to coil is max.(please confirm)
    Last edited: Mar 11, 2010
  16. Mar 12, 2010 #15
    The only clue left to find the 'truth behind inductors in an ac circuit' seems to be as to whether the current flow at the instant of zero applied potential is only due to the inductor (which continues for the rest of the tme too).....could someone just say yes or no?
  17. Mar 12, 2010 #16


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    I tried a Spice simulation of the current in an inductor.

    http://dl.dropbox.com/u/4222062/current%20in%20inductor.PNG [Broken]

    The current scale is at the right of the graph.

    It seems the current in an inductor is zero when power is applied (assuming it is applied at the zero crossing point of the sinewave) and it continues for several cycles to be offset from the zero current line in the direction of the first input cycle.
    Eventually, the current swings equally on either side of the zero current line.

    This was done with a 1 KHz sinewave input voltage and a 1 mH inductor. I also put a very small resistor (0.1 ohm) in series with the inductor.

    However, I am cautious (suspicious) of simulators and reproduce this result to see if others get similar results. I never liked the idea of a current flowing before there was a voltage, though, so this may be more elegant.
    Last edited by a moderator: May 4, 2017
  18. Mar 12, 2010 #17
    Thanks for the post vk6kro ...but I'm still at a loss....as I said,I was trying to explain the graph for myself...and the only conclusion I could reach (especialy to address the fact that at all points for which voltage is zero,the current is max).....is that the current is due only to the induced emf of the inductor.

    It would be really nice if you explained on the lines I'm already on,as I've got used to it.
  19. Mar 12, 2010 #18
    By the way,does the fact that the current in the first few cycles does not swing equally in both directions have any special significance in the idea of what actually happens in the circiut?
  20. Mar 12, 2010 #19


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    You are being confused by incomplete graphs (the ones you posted, not vk6kro's). They are not showing the full inductor current waveform. It does not magically start at the point where the voltage is max at 90 degrees. If they drew that graph completely, it would have been swinging up from a negative value and crossing zero when the voltage is max. It's unfortunately that a poorly drawn I-V diagram has caused you so much confusion. The equation that I wrote holds true for all time. Just do the differentiation to get the I(t) waveform.

    [tex]V(t) = L \frac{dI(t)}{dt}[/tex]
  21. Mar 12, 2010 #20
    Even in vk6kro's graph,it shows the current to be max when applied voltage is zero....also I don't uderstand what the 'I' in the equation is....its equating the applied voltage to the change in current (not the absolute value of the current)....so is that current the current due to the net effect of the applied voltage and the inductor's emf?
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