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Current through an inductor connected directly to a battery

  1. Nov 3, 2013 #1
    1. The problem statement, all variables and given/known data

    A long, fine wire is wound into a coil with inductance 5 mH. The coil is connected across the terminals of a battery, and the current is measured a few seconds after the connection is made. The wire is unwound and wound again into a different coil with L = 10 mH. This second coil is connected across the same battery, and the current is measured in the same way. Compared with the current in the first coil, is the current in the second coil:

    a) twice as large
    b) one-fourth as large
    c) unchanged
    d) half as large
    e) four times as large

    2. Relevant equations

    (1) εL = -L(dI/dt)
    (2) ε - IR - L(dI/dt) = 0

    3. The attempt at a solution

    So, coming to a symbolic solution here is no problem. The answers include no exponential term, so the wire is obviously assumed to have zero resistance. Therefore, |ε| = εL, and I = ε/L * t. Therefore, the current in the second case is half as large as the current in the first case.

    What bothers me here is this: R = 0, so by equation (2) the back-emf is equal to the battery's emf at all times. This would imply zero current at all times, wouldn't it? But if current is zero at all times, then the equation would false because dI/dt would be zero.

    Could someone clear this up for me? How would an ideal inductor interact with an ideal source of emf with no resistors in between?
     
  2. jcsd
  3. Nov 3, 2013 #2

    gneill

    User Avatar

    Staff: Mentor

    The current starts off at zero if the battery potential is suddenly applied to the coil. But in order to maintain the constant back-emf the coil follows the rule: E = L dI/dt (for a suitable choice of direction of current). Now, what does that tell you about dI/dt if E is to remain constant for all time? What's the eventual fate of the current?

    Now consider the possibility that the wire is not a superconductor but has some very small but finite resistance. How would your answer change?
     
  4. Nov 3, 2013 #3
    By the equation E = L dI/dt, current would approach infinity for a superconductor. If wire had resistance, the current would asymptotically approach E/R, with E being the emf. These results are what I expect, but I am having trouble understanding how current begins to flow in the first place if E = L dI/dt at all times. It seems, to me, like this circuit:

    circuit.png

    I know qualitatively that this is a different matter, but it illustrates the point that is bothering me. The two equal and opposite batteries should completely cancel one another, resulting in zero current. The differential equation (equation 2) implies that the inductor and battery act as equal and opposite sources of EMF, and yet current is increasing (and must be for the back-emf to set up in the first place).

    Is it perhaps that the back-emf is only extremely close to the potential difference across the battery, but not exactly the same?
     
  5. Nov 3, 2013 #4

    gneill

    User Avatar

    Staff: Mentor

    The only way dI/dt can be constant is if the current is changing. The coil can retain its back-emf only if it draws current, and that current must steadily increase with time.

    A lot of feedback systems exhibit this same property, where it appears that the stasis enforced by the feedback should actually prevent the stasis from coming into being in the first place. The thing is, at the macroscopic level it appears that the potentials are precisely equal and that no charge should flow, but at the 'differential' level, there is a vanishingly small but sufficient difference to get the process started. Blame it on the finite speed of light :smile:

    Once a single charge carrier crosses the line, a change in current has happened and the process has begun.
     
  6. Nov 3, 2013 #5
    That's what I needed to hear, thank you!
     
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