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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?
  22. Mar 12, 2010 #21


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    Because of the differential equation that I'm showing you, the current sinusoidal waveform lags the voltage sinusoid by 90 degrees. That's what happens when you differentiate the current sinusoidal waveform. Are you familiar with differentiation? Does the equation that I wrote make sense to you?

    In vk6kro's simulation, the voltage and current are starting at 0 at t=0. That's the initial condition for the simulation. Since the voltage swings one way first, that provides an offset in the current sinusoid. Notice how the current sinusoid waveform is slowly coming down near the right side of the plot? It will eventually settle down to where it has zero offset, just like the driving voltage sinusoid.

    It may be helpful for you to think about how the current and voltage act in a parallel LC "tank" resonant circuit. In that circuit, the energy stored in an oscillating waveform goes back and forth between the inductor (its current), and the capacitor (its voltage). At the time when the inductor current is max, the capacitor voltage happens to be zero. When the capacitor voltage is max, the current in the inductor (and in the cap ovbiously) is zero.
  23. Mar 12, 2010 #22

    The Electrician

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    Yes, it does.

    If there is any resistance in the circuit at all, and there always is in the real world, then the differential equation for the circuit becomes:

    [tex]V(t) = L* \frac{di(t)}{dt}+R*i(t)[/tex]

    The solution to this equation has two parts: a transient part and a steady-state part.

    The transient part is a decaying exponential of current, and its magnitude depends on when you suddenly connect a sine wave to the inductor (which also has some resistance, remember). I've attached an image with three plots showing what goes on.

    The first plot shows a sine wave of voltage suddenly applied to an inductor plus resistance. It shows two cases; the blue and red traces are the applied sine wave of voltage. The blue shows the sine wave being applied just as it goes through zero volts with a positive slope. The red trace shows the applied sine wave being applied at the positive peak of the sine wave.

    I have inverted (multiplied by -1) the current traces to make them easier to distinguish from the voltage traces.

    The green trace shows the current due to the application of the blue voltage. The magenta trace shows the current due to the application of the red voltage.

    The second plot is the same as the first plot, but showing a longer time.

    Notice that the green trace doesn't have equal swings about the zero axis at first. This is because it's not possible for the current in an inductor to change instantaneously, so when the blue voltage is suddenly applied, the current in the inductor must be zero to start with.

    A transient current, shown in black, is present in the inductor to make this true. If you've ever studied differential equations, you will recognize this as the initial condition, which results in the transient. This transient current gradually dies out, as can be seen clearly in the second plot. After it does die out, the inductor has equal swings above and below the zero current axis.

    The third plot shows what would happen if there were no resistance in the circuit. The transient current would never die out, and the green current wave would never become re-centered on the zero current axis. This can never happen in the real world, of course. It's only possible mathematically.

    The transient current is zero if you apply the sine wave of voltage at the right time, as in the red trace (and it corresponding current, the magenta trace).

    So, when you first apply the sine wave of voltage, the current may not be such that V(t) = L*di/dt at first, depending on just when you apply the voltage. The transient current (if there is one) must die out first.

    This is all just plain old circuit theory.

    Search on bing.com with the search terms: "transient analysis" inductor "natural response"
    and you'll find lots of discussion of transient analysis.

    Attached Files:

  24. Mar 12, 2010 #23
    I understand that vk6kro's post was based on a circuit with an inductor,plus some resistance,after comparing it to The Electrician's plots.

    I didn't understand why the nature of the current in the circuit varies with the instant you apply voltage,though...

    Also,before I go on to a circuit with both inductor and reistance,I'd like to know more about an ideal purely inductive circuit,that has reached its steady state....like the red and magenta voltage-current plot pair.

    As I said earlier,In the differential equation describing the ideal inductive circuit i.e

    [tex]V(t) = L \frac{dI(t)}{dt}[/tex]

    the applied voltage is equal to the change of the current w.r.t time......by Ohm's law,the current should be due to the applied emf..not the change of current.

    Also,as I asked earlier...what is the current 'I' in this equation?

    Is it the net current due to applied emf and inductor back emf?
  25. Mar 12, 2010 #24

    The Electrician

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    Because the current through an inductor cannot change instantaneously. The rate of change of current in an inductor is proportional to the applied voltage. In order for the current to change instantaneously, you would need to apply infinitely great voltage for a short time--an impulse of voltage, in other words. That's not what you're doing when you suddenly apply a voltage of finite value.

    If the applied sine wave of voltage is anything other than its peak at time tzero, then there will be a transient component of voltage, because in the steady state, at any time other than the peak of the applied voltage, the current in the inductor will be non-zero. Since the current in the inductor was zero just before we applied the voltage, it can't change instantaneously to a non-zero value just after we apply the sine of voltage.

    How does it follow that if:

    [tex]V(t) = L \frac{dI(t)}{dt}[/tex]


    "the current should be due to the applied emf..". Do you mean that you think that inductor current should be proportional to the applied voltage? Algebraically, something like this:

    [tex]i(t) = K*v(t)[/tex]

    where K is a constant of proportionality? If that's what you think, you are mistaken.

    The three common circuit components, resistors, inductors and capacitors have three different terminal characteristics:


    [tex]v(t) = R*i(t)[/tex]


    [tex]v(t) = L \frac{d\;i(t)}{dt}[/tex]


    [tex]v(t) = \frac{1}{C}\int i(t)\;dt[/tex]

    If you want current in terms of voltage, then the relationships become:


    [tex]i(t) = \frac{v(t)}{R}[/tex]


    [tex]i(t) = \frac{1}{L}\int v(t)\;dt[/tex]


    [tex]i(t) = C \frac{d\;v(t)}{dt}[/tex]

    The current in the inductor is the integral of the applied voltage, in the steady state.

    If a sine wave of voltage is suddenly applied to the inductor, then there may be a transient component of current which will eventually die out, but this transient component will be just what is needed to keep the inductor current from changing instantaneously, while allowing L*di/dt to be equal to the applied voltage at time t=0.

    In the steady state, when the sine wave of applied voltage is going through zero in the positive direction, the inductor current will be a negative peak. But that can't happen when we first apply the sine wave of voltage at that point in the cycle, because the current in the inductor was zero just before we applied the voltage (tzero minus) and it must also be zero just after we apply the sine of voltage (tzero plus). The way this can happen is to have a transient current as shown in my plots. We must have both i(t) be zero (at time tzero),
    and L*d i(t)/dt equal the suddenly applied voltage. The transient current bridges the gap between the initial conditions and the steady state.

    But that transient isn't needed if the sine wave of voltage is applied at its peak, because then the current in the steady would normally be passing through zero at this time. No transient is needed to make both i(t) be zero and L*d i(t)/dt be equal to the suddenly applied voltage.
    Last edited: Mar 12, 2010
  26. Mar 13, 2010 #25

    When we first did inductors....plain and simple,we said that there is a back emf at the instant that the current through the inductor changes....here the current means the applied voltage's current.....like,for example in a dc circuit,when we switch the power on....at the first few instants,when the current is trying to reach a certain constant value,.....thus,in the equation
    [tex]Vapplied= L \frac{d\;i(t)}{dt}[/tex]

    for an inductor in a dc circuit(in the situation I referred to about),the i is the current that the applied voltage establishes...and that changes value....hence changing the value of magnetic flux linked to the inductor...and the back emf in the inductor is a result of this...is that okay?

    This is exactly what I've been getting at.....when the sine wave of applied voltage is going through zero,the inductor current is at the peak.....so the current in the plot,at steady state,when Vappl. is zero is due to the inductor,and its back emf.

    So it must be similar throughout the rest of the cycle also...i.e the current on the plot must be due to the inductor.
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