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Homework Help: Question about RL circuits

  1. Feb 27, 2005 #1
    The resistance of a superconductor. In an experiment carried out by S. C. Collins between 1955 and 1958, a current was maintained in a superconducting lead ring for 2.50 yr with no observed loss. If the inductance of the ring was 4.44*10^(-8) H, and the sensitivity of the experiment was 1 part in 10^9, what was the maximum resistance of the ring? (Suggestion: Treat this as a decaying current in an RL circuit, and recall that e^(-x) = 1-x for small x.)

    I know that I = ((emf)/R)*(1-e^(-Rt/L)). How do I get the maximum resistance from this? Also, what does the sensitivity of the experiment mean?
    Any help would be great! thx in advance!
  2. jcsd
  3. Feb 28, 2005 #2
    use [tex] I(t) = I_0 e^{-Rt/L} [/tex] instead, as the problem suggested, for small x, e^(-x) = 1-x
    [tex] I(t) = I_0 (1-Rt/L))[/tex]
    [tex] I_0 - I(t) = Rt/L [/tex]
    the sensitivity is 1 part in 10^9 implies
    [tex] \frac{I_0-I(t)}{I_0} < 10^{-9} [/tex]
    the rest is simple algebra
  4. Feb 28, 2005 #3
    [tex] R = \frac{L} {t}(1-\frac{I} {I_0}) [/tex]
    and [tex] \frac{I} {I_0} < 10^{-9} [/tex].
    Do I substitute [tex] \frac{I} {I_0} [/tex] with 1^(-9)?
  5. Feb 28, 2005 #4
    you substitute the whole [tex] \frac{I_0-I} {I_0} [/tex] with 10^-9
  6. Feb 28, 2005 #5
    wait a minute, where did [tex] R = \frac{L} {t}(1-\frac{I} {I_0}) [/tex] came from????
    The unit is not even correct.... how did you get this?
  7. Feb 28, 2005 #6
    [tex] I(t) = I_0 e^{-Rt/L} [/tex]
    [tex] I(t) = I_0 (1-Rt/L)[/tex]
    [tex] \frac{I(t)}{I_0} = 1-Rt/L[/tex]
    [tex] \frac{Rt} {L} = 1-\frac{I} {I_0}[/tex]
    [tex] R = \frac{L} {t}(1-\frac{I} {I_0}) [/tex]

    am I wrong?
    and then
    [tex] \frac{I_0-I} {I_0} [/tex]
    simplifies to [tex] -\frac{I} {I_0} [/tex] so [tex] -\frac{I} {I_0}<10^{-9} [/tex]
    Last edited: Feb 28, 2005
  8. Feb 28, 2005 #7
    oh yeah, you are right, my mistake
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