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Current and inductor

  1. Apr 20, 2012 #1
    1. The problem statement, all variables and given/known data

    Your circuit has a resistance of 175Ω and an ideal battery with an emf of 6.3V. What value of inductor should you add to the circuit to ensure the current increase in the first 58μs is less than 4.9mA?

    2. Relevant equations

    I think

    E=L(di/dt)


    3. The attempt at a solution

    E=L(di/dt)

    6.3=L(4.9*10^-3/58*10^-6)

    L = 0.07457H

    But the answer is 0.069H

    Thanks.
     
  2. jcsd
  3. Apr 20, 2012 #2
    I believe you want to use the equation for current growth through a inductor. Also di/dt doesnt equal (4.9*10^-3/58*10^-6)
     
    Last edited: Apr 20, 2012
  4. Apr 21, 2012 #3
    WOW you're right! Thanks!!

    Then the formula would be I=Io(1-e^(-Rt/L))

    Io=6.3/175=0.036A

    Solving for L, L=0.069H.

    But why is my method wrong? and why is di/dt not (4.9*10^-3/58*10^-6)? di is 4.9*10^-3 and dt is 58*10^-6 right?

    Thank you!
     
  5. Apr 21, 2012 #4

    NascentOxygen

    User Avatar

    Staff: Mentor

    When current is flowing, KVL says that the 6.3V is shared (unevenly) between the resistance and the inductance. Your "approach" assumes circuit resistance = 0 and all 6.3V is applied across the inductor.
     
  6. Apr 21, 2012 #5
    Ah, I see, thank you NascentOxygen!!
     
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