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LR Circuit inductance

  1. Nov 13, 2015 #1
    1. The problem statement, all variables and given/known data
    An inductor is connected to the terminals of a battery that has an emf of 12.0Vand negligible internal resistance. The current is 4.86mA at 0.940ms after the connection is completed. After a long time the current is 6.45mA.

    2. Relevant equations
    V = IR
    V(t) = V0 (exp(-t/τ))

    3. The attempt at a solution
    At infinite time, an inductor appears like a wire. Therefore, 12V = 6.45mA * R
    R = 1860 Ohms

    V(t) = V0 exp(-t/(L/R))
    (4.86mA * 1860 Ohm) = 12V exp(-t/(L/R))
    .753 = exp(-t/(L/R))
    natural log both sides...
    .28 = t / (L/R)
    L/R = 0.0033
    L = 6.18 H

    However, MasteringPhysics does not accept this solution.
     
  2. jcsd
  3. Nov 13, 2015 #2
    Derp. should've been using 1- exp(-t/(L/R))
     
  4. Nov 13, 2015 #3

    ehild

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    Homework Helper
    Gold Member

    What is V(t)? The voltage across what?

    You completely ignored the battery. It is present. You can consider the coil as an ideal inductor connected in series with its resistance and connected to the battery. The voltage across the inductor is not the same as the voltage across the resistor.
     
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