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RL Circuit

  1. Dec 13, 2003 #1
    I want to check my answers for this problem: Part c) below doesn't agree with the book. And even though part a) agrees, I'm not sure about my method.

    I've attached a figure for reference.

    For the attached figure, the switch is closed fo t<0, and steady-state conditions are established. The switch is thrown open at t = 0.
    a) Find the initial voltage across L just after t = 0. Which end of the coil is at higher potential: A or B?

    c) How long after t = 0 does the current in the 6kOhm resistor have the value 2.00 mA.

    For part a), I used Kirchhoff's loop rule to get the currents at steady state.

    6kOhmI1 = 18V
    I1 = 3mA

    2kOhmI2 - 0.4H(di2/dt) = 18V ==> di2/dt = 0 at steady state.
    I2 = 9ma

    Itotal = 12mA

    Once the switch is thrown open I have an RL series circuit.
    In a series circuit I should be the same across both resistors so:

    I1 goes to zero, so I total in the circuit is 9mA.

    6kOhmI + 2kOhmI - 0.4H(di/dt) = 0

    0.4(di/dt) = 8kOhmI = 8kOhm(9mA) = 72Volts; B has the higher potential.

    If my reasoning above is correct then for part c)

    I = Iinitial(e^(-t/T)) where T = L/R = .4/8000 = 50 microseconds

    2mA = 9mA(e^(-t/50us)

    ln(1/6) = -t/50us
    t = 50ln(1/6) = 75.2 microseconds. The answer is the book is 75.2us

    Thanks in advance for the help.

    Attached Files:

    Last edited: Dec 16, 2003
  2. jcsd
  3. Dec 14, 2003 #2
    Hmmmmmm? 24 hrs and no response. No ideas or suggestions on why the answer for the second part is different than what I came up with?
  4. Dec 15, 2003 #3
    Wrong picture. Can't tell what you're talking about. :smile:
  5. Dec 16, 2003 #4
    Sorry about that.

    I made some errors in my posting. The diagram should show the emf as 18V. There was also a multiplication error that I corrected.

    What had me confused is that the current in the loop on the left goes to zero after the switch is opened. I was thinking (or wasn't thinking) that it did not immediately go to zero.

    Anyway I see now that it does, so the only current in the right side loop is 9mA and everything works out.


    I'll take more care in the future when posting my questions.
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