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A doubt about inductors and capacitors

  1. Jul 5, 2012 #1
    [/itex]Hi I am new here :smile:

    I know (correct me if wrong please :smile:) that inductors store current in its magnetic field, and capacitors store voltage in its electric field

    But, why when disconnecting a inductor charged from a DC circuit, a spark is produced at the disconnected point and it loses its energy while the capacitor not and the electricity is stored like a battery? I understand that [itex]I = \frac{V}{R}[/itex] and the spark in inductors are produced because Voltage rises to compensate the resistance when opening the circuit but, why does a capacitor voltage stay the same despite no current and infinite resistance and a inductor discharges immediately at the same circumstances?
    Many thanks.
     
  2. jcsd
  3. Jul 5, 2012 #2

    Simon Bridge

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    That would be a very confused way of putting it yes.

    Capacitors store energy in their electric fields.
    Inductors store energy in their magnetic fields.

    Inductors cannot become "charged" in a DC circuit ... in a steady-state DC circuit they look like a short circuit and act as an electromagnet. If you disconnect one, the collapsing magnetic field induces further current in the coil.

    Where inductors are just a wire (short circuit) in DC, capacitors are a gap (open circuit). The power supply just moves charge from one side of the gap to the other. Disconnect the supply and there is no path for the charges to get back ... so the charges get stored.

    In the fluid-analogy for electric circuits, a capacitor is a water-tank. I don't think there is an equivalent to an inductor.
     
  4. Jul 5, 2012 #3

    AlephZero

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    If a flow or water corresponds to the current, a sort of analogy would be a propellor in the water pipe that all the water has to flow through, turning a heavy wheel. When you apply some water pressure, the flow/current can only increase slowly because it takes time to get the wheel spinning. If the flow is steady, the wheel doesn't affect it (ignoring friction, which corresponds to the resistance of the wire in a real inductor).

    If you try to stop the water flow suddenly, you get a huge pressure (voltage) because the kinetic energy in the wheel is still trying to pump the water through the propellor, but the water has nowhere to go - except possiblly to burst the pipe (send a spark across the electrical switch).

    OK, that's not so good as some of Maxwell's mechanical analogies for EM fields built from infinte arrays of gyroscopes coupled together with linkages - but it's a start.

    It could be fun making a real device like this plus a water tank to simulate an LC circuit - a project for a science fair, maybe? Demonstrate resonance by rasiing and lowering the tank a bit, and sloshing water everywhere if you do it at the right frequency....
     
    Last edited: Jul 5, 2012
  5. Jul 5, 2012 #4

    Simon Bridge

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    That's not bad: water-wheels are usually used to represent light-bulbs. But you are right - when the flow stops the wheel wants to keep going.
     
  6. Jul 5, 2012 #5
    A collapsing magnetic field induces an emf. There can only be an induced current if there is a complete circuit.
     
  7. Jul 5, 2012 #6

    Simon Bridge

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    Well OK - though you can have a current without a complete circuit, the cited example did involve a spark so presumably the emf was strong enough to push charges across the gap.
     
  8. Jul 5, 2012 #7

    mfb

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    The analogy of a disconnected inductor (static case: no current flow) is a short-circuit for a capacitor (static case: no voltage).
    And a disconnected capacitor (infinite resistance) corresponds to a inductor with a short-circuit (no resistance).
     
  9. Jul 5, 2012 #8
    Faradays laws state how to calculate induced emf not induced current.
     
  10. Jul 5, 2012 #9

    rbj

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    not the best fluid analogy for a capacitor. a better analogy is a cylinder with a spring-loaded piston inside.

    sure there is. a water turbine with the turbine shaft connected to a flywheel.
     
  11. Jul 5, 2012 #10
    Yes dualogs and analogs - I think you mean dualog. Analogs rather refers to system elements of the same type.

    Wikipedia must have a good long discussion for any reader wishing to expand their knowledge.
     
  12. Jul 5, 2012 #11
    We usually think of a DC circuit being connected to a voltage source. A capacitor will charge up only to the value of the voltage of the battery and then no more current will flow. For an inductor connected to a battery the rate of change in current is what gives the 'resistance' to the battery, so if you have a strong enough battery and your inductor is constructed well enough, then very high currents can be achieved.

    But, if you consider a current source, then since the current is not changing then no voltage will be observed across the inductor. On the other hand, the current source is continiously pumping current into a capacitor which can then charge up to very high potential. You can then witness sparks across a gap when the voltage becomes high enough. I am not absolutely sure but storm clouds would acquire the high potentials this way and you would call the discharge lightning and thunder.
     
  13. Jul 5, 2012 #12
    I don't wish to be rude, but that is only partially correct. Faraday's Law, herein "FL", relates magnetic flux to voltage. Specifically, FL relates net flux to net voltage. When a time-varying magnetic flux is present in the vicinity of a conductor, a Lorentz force acts on the conductor's free electrons, resulting in charge motion, which is current. If the path is open, current does indeed flow, but it is called a displacement current, which can be small, or at very high frequencies, quite large.

    If the loop is open, charges accumulate at each end, and there is an emf across the gap between the 2 ends of the conductor. As the mag flux varies with time, these charges move and displacement current continues. Even with an open path there is current as well as voltage.

    If the loop is closed with a resistance R, obviously we have conduction current as well as displacement current. Assuming it is low enough frequency where conduction is much greater than displacement. The current in the loop has its own magnetic flux encircling the conductor. Per law of Lenz, LL herein, its polarity is opposite to that of the external field. So the actual flux enclosed by the loop is the sum total of the external plus the internal flux associated with the current in the wire.

    We have mutual & self induction going on simultaneously. FL states that the voltage one time around the loop equals the negative of (LL) the time derivative of magnetic flux. But we must remember that the voltage, or emf if you prefer, is a net quantity, as is the flux.

    A good example is to let the loop be superconducting. The voltage once around the loop becomes zero. This is due to the current in the loop having a magnetic flux around the wire that cancels the incoming external flux. Hence we have zero emf.

    If the loop resistance R, is not zero, but very small, we have about the same current, and a very small voltage. Again, per LL, cancellation takes place. But if the loop R is made larger than a critical value so that the loop current is too weak to produce a magnetic flux that cancels the external flux, then the flux "phi" in FL, is essentially the external flux. Increasing R results in a loop emf that hardly changes. When R is large, the internal flux generated by the loop current is very small compared to the external incoming flux, and the net flux is very nearly that of the external flux. Under these conditions, we can state that the induced voltage, i.e. work per unit charge once around the loop, is the negative of the time derivative of the external flux.

    But for low values of R, we cannot make that assumption. I don't wish to nitpick, but that is a point that should be made. I will elaborate if needed. Best regards.

    Claude
     
  14. Jul 5, 2012 #13
    I have a limited range of about 12 A level text books and none of the explanations therein relate to anything posted here. I know that an emf involves charges separated and that means there must have been a 'current' for however long.
    I will confine my interest to the text book explanation of faradays law in terms of induced emf and rate of change of flux linkage.
    I am vaguely interested in how you would calculate the currents you allude to (or the voltages) but it looks like it will be beyond me.
    The original post is from someone new here and I hope that he is able to pick his way through everything posted and come out wiser.
     
    Last edited: Jul 5, 2012
  15. Jul 5, 2012 #14
    https://www.physicsforums.com/showthread.php?p=2163546#post2163546

    The above link is to a computation sheet I posted a few years ago. A loop which has a resistance R, inductance L, frequency omega, & external flux phie, can be computed for current, voltage, & internal flux phii. I will elaborate if needed.

    Claude
     
  16. Jul 5, 2012 #15
    I read your computation sheet which is interesting. Is it published in any recognised text books?
    I cannot find a recognised text book that states anything other than
    Induced emf = rate of change of flux linkage. none state any reference to induced current as a version of faradays law. There must be a good reason for that !
    It is misleading to confuse the 'displacement' current, ie the movement of electrons giving rise to the induced emf with the induced current in a complete circuit resulting from said induced emf.
    anyone wanting to get to grips with faradays law should stick to the text book explanations.
     
  17. Jul 5, 2012 #16
    It depends on the text you are reading, as most refer to devices powered from a CVS (constant voltage source). Transformers, motors, & generators are often explained using FL. A xfmr is a good starting point. A CVS energizes the primary & flux is established in the core. The flux relation with input emf is given by FL. An important condition is that the primary flux couples the secondary winding almost at 100%, & vice-versa.

    The secondary voltage is also given by FL, & from FL we get that Vs/Ns = Vp/Np. The secondary & primary voltages scale according to the turns ratio. If a load is connected to the secondary, current will flow. A magnetic flux will be present which tends to cancel the core flux due to the primary emf source. The secondary voltage will tend to drop. As you draw more secondary current, the secondary voltage goes down. But before that can happen, something else takes place.

    When the secondary is open, the core flux induces a counter-emf, cemf herein. This cemf opposes the source emf. The exciting current in the primary is the difference between the CVS value & the cemf, divided by the primary impedance. This exciting current consists of loss current plus magnetizing current in quadrature. The magnetizing current times the primary turns is the mmf.

    When secondary current is drawn due to loading, the magnetic flux of the secondary current induces a counter-counter-emf, or ccemf herein. The ccemf opposes the cemf which opposes the source emf. The secondary current times secondary turns is the counter-mmf or cmmf. The primary current is now as follows. The ccemf adds with the source emf, and the cemf subtracts. Hence the primary current goes up so that its amp-turns balance the loaded secondary amp-turns.

    But the primary amp-turns produces counter-counter-mmf, ccmmf, & counter-counter-counter-emf, cccemf. The result is that the secondary voltage drops only a small amount, owing to resistance & leakage reactance in the windings. Hence the secondary voltage is very close to what FL predicts w/o the need to bother with internal/external flux components cancelling.

    The cancellation of fluxes is mitigated by the primary current increasing so that constant secondary voltage is maintained. That is why you can simply compute the secondary voltage per FL even though loading currents are present.

    But now, let us change conditions. A magnetic flux is radiated from a radio station antenna. A loop located miles away encloses this flux & induction takes place. The coupling from the loop back to the broadcast tower antenna is nowhere near 100%, typically a fraction of 1%. So now if a loop encloses a flux, & the loop is open, or has a very high R value closed, the inducred voltage can be computed per FL, w/o the need to consider internal flux & cancellation.

    But if the R is low enough, the loop current magnetic flux cancels the external flux & the voltage drops. Because of loose coupling to the radio station source antenna, the current in the antenna DOES NOT INCREASE as a result. The counter-counter-emf in the loop due to load current does not get mitigated by increased current back at the source.

    That is my point. As a check, take my equations, & use the condition that R is much greater than omega*L. The quantity omega*L is the inductive reactance XL. Let R >> XL. Notice that the voltage induced is independent of load & exactly as stated per FL. I'll leave that as an exercise for anyone interested. Also, I, the loop current, is simply V/R.

    But should R fall in value, or XL increase in value due to increased frequency, then we must use the full expression, the more complicated computation. Again, I will elaborate if needed.

    Claude
     
  18. Jul 6, 2012 #17
    I don't wish to be rude but an analysis of transformer action and radio wave propagation does not clear up any misunderstandings about Faraday's law.
    My reading of text books is that a changing magnetic flux linkage produces an emf in a conductor. There is no 'induced current' unless there is a complete circuit (I accept that a 'spark' does constitute a complete circuit of sorts).
    There are countless published questions at A level (that is what I teach) along the lines of : A plane coil of 10 turns and area 0.1m^2 is placed with its plane at right angles to a magnetic flux of 2T. The flux decreases uniformely to zero in 0.8s, calculate
    a) the emf induced across the ends of the coil
    b) the current induced in this coil.
    I wonder how you would respond to a question of this sort and what use you would make of the laws of electromagnetic induction.
    I bet a pound to a penny you would get the same answer as me and there would be no mention of 'displacement' current
     
  19. Jul 6, 2012 #18
    I just gave you 2 lengthy explanations. Is there something I stated that needs correction or clarification? As far as displacement current goes, I only brought that up to demonstrate that even under open conditions, it exists. At the frequencies we deal with such as 50/60/400 Hz power, the displacement current is much smaller than conduction current, so that it need not be considered when sizing conductors. If the load current (conduction), is 4.0 amps, with a displacement current of 4.0 milliamps, they are 3 orders of magnitude, apart, 6 orders when you consider that power varies as I2.

    Sizing the wire for 4 amps is just fine. Now as to how I would compute the induced quantities per your example, I would do this. Compute the inductance based on turns, area, & height of coil. Then I would compute the rate of change, i.e. radian frequency, which you gave in terms of the rate which the flux decreased to zero. This is "d(phi/dt)". You did not state an R value, nor did you state if the loop is open or shorted.

    I would find out what R is, open, short, or in between. I would then enter the values into the equation sheet I posted earlier. Would I get the same answer as you did? As I said, if XL is smaller than R, then yes, you and I would get the same answer. But if R is smaller than XL, the induced emf will be smaller than what you say it is.

    I don't understand why anybody would dispute what I'm saying. It's too easy to see. Take the coil in your example above. Short it using a total resistance of 0.010 ohm. Measure the current with an ammeter, clamp on type. Now short the loop with 0.020 ohm, 0.10 ohm, 0.50 ohm, etc. You will see that the current hardly changes. But the voltage once around the loop is changing, increasing as R increases.

    If the R value is increases to a value where R >> XL, things change. When the induced current's magnetic flux is too small to cancel the external flux, then the emf is simply -N*d(phi/dt). FL is always in effect, but we must remember that FL applies for mutual as well as self induction. Increasing R results in the same emf, but current decreases as R increases. Have you examined my equations, letting R >> XL? My result is identical to what you're saying. So why do you take issue?

    When R is small, the net emf once around the loop is obtained by superposing internal & external flux change wrt time. But per Lenz' law they are opposite in polarity, hence they cancel each other out. Faraday's Law holds, but there are 2 terms, not just 1. If you were correct that the induced emf is just -N*d(phi)/dt, then as R decreases, I would continually increase. Do you believe in conservation of energy law, CEL?

    The power in the loop is P = V2/R. As R decreases, P increases. But the magnetic flux has a finite energy/power. Integrating the field over its volume results in a value of power. The loop power can never exceed the field power. So as the R value drops, the V value must eventually drop as well, else power increases without limit. Make sense?

    I will elaborate, but please examine what I've said carefully. Please take my equations & make the substitution I stated. It will agree with what you've been saying for those specific conditions.

    Regarding your question, here is my answer. First condition is that R >> XL, so that phi = A*B (flux linkage = area * flux density) = 0.1 m^2 * 2 T = 0.20 weber.

    d(phi)/dt = 0.20 weber/0.8 seconds = 0.25 volts/turn.

    V = -N*d(phi)/dt = -10 turns * 0.25 volts/turn = -2.50 volts per right hand rule, or just 2.50 volts per left hand rule. BR.

    Claude
     
    Last edited: Jul 6, 2012
  20. Jul 6, 2012 #19
    Well done !!!! your answer for the induced emf would be accepted by an exam board (at A level)
    These days students are encouraged to think for themselves and if they are not given the inductance of a coil they assume it is zero (after all it is only 10 turns.... a reasonable assumption). 15 years ago they would have been told to take the inductance to be zero.
    You are not given the resistance of the wire.... presumably it is not important (and it is not when it comes to calculating induced emf....that is what is taught !!!!)..... presumably the coil is 'open'..... you would/should have been told that is the question.
    One criticism of your answer.....flux linkage is Nθ (should be phi....cant get it) = NBA not BA.
    It makes no difference to the answer to say that is 0.25V/turn then to multiply by the number of turns but it is a point of principle and (I think) definition. Flux linkage still has a meaning when the flux is not changing and there is therefore no V.
    Flux linkage = NBA.
    The second part of the question....induced current ....would usually show as a ....'comment on your answers' in an exam these days and the students would be expected to realise (because no resistance is specified !!!!) that resistance is unimportant and that the coil is therefore 'open' and that there is no induced current in an open circuit (sparks aside) as given in faradays laws.
    If the resistance of the coil was given as ... say 10ohms then students woulds be expected to calculate the 'induced' current to be 2.5/10 = 0.25A for 0.8s
    I hope this clears up some of the finer points of the application of Faraday's law for you.
    All the explanations I ,and my students, need are in countless A level text books.
    I don't understand why anybody would dispute what is written in school text books.
    I just gave you more than 2 lengthy explanations.
    I will elaborate, but please examine what I've said carefully.
     
  21. Jul 6, 2012 #20
    But without knowing the inductance L, how can you say that I is simply 0.25A. If L*omega = XL is larger than R, then V is NOT 2.50V. That is my point. I assumed the loop was open since you stated "the ends of the wire". That sounds open to me. Of course if the loop is open, then the resistance of the wire is not relevant since this resistance is in series with an open, which is very large. For an open I can assume R >> XL. So in this narrow case, we need not consider internal flux. Since the displacement current is very small, its flux cannot cancel the external, so that FL can be applied in 1-term form.

    But you just stated that if R= 10 ohm, that V would remain 2.5V, so that I = 0.25A. My caveat is as follows. Only, only, only if R >> XL. Should that NOT be the case, V will be less than 2.50V, & I will be less than 0.25A. Of course the V/I ratio must always be 10 ohms per Ohm's law. If XL was 35 ohms, & R = 10 ohms, then to obtain V you must use the equation I gave on that sheet I posted. Please see what you get. Thanks.

    By the way, what is "A" level teaching? Engr college, physics college, high school, tech school, trade school, vocational, etc.?

    Claude
     
    Last edited: Jul 6, 2012
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