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Simple RL Circuit

  1. Jan 28, 2010 #1
    1. The problem statement, all variables and given/known data

    The switch has been closed for a long time.

    a) Current flowing through inductor. Voltage across it. IS any current flowing through the resistor.

    b) The switch opens at t = 0s. Find an equation for the inductor voltage as a function of time for t > 0. How long does it take for the inductor to reach 10% of its maximum value?

    2. Relevant equations

    Kirchhoff's Voltage Law (KVL)

    3. The attempt at a solution

    a) 1A is flowing throw through the inductor as the circuit is stable and the inductor is replaced with a short circuit this causes the resistor to have no current through it and the inductor to have no voltage across it.

    b) This is where I am not sure about:

    t > 0


    I used the KVL equation around the circuit to get:

    [tex]-(100)10^{-3}\frac{di_{L}}{dt} + 1i_{L} = 0[/tex]

    I am not sure if this is the right way to do this.
  2. jcsd
  3. Jan 30, 2010 #2


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    Homework Helper

    -L *di/dl is the emf arising if the current change. So your equation is

    -0.1\frac{di_{L}}{dt} = 1i_{L}

    The starting condition is [tex]
    I_{L}(0)=1 A

    The equation is easy to solve.

  4. Jan 31, 2010 #3
    Ok I attempted this and here is what I got:


    [tex]iR + v = 0[/tex]

    [tex]iR + L\frac{di}{dt} = 0[/tex]

    [tex]\frac{di}{dt} = -\frac{iR}{L}[/tex]

    [tex]-\frac{di}{i} = \frac{R}{L}dt[/tex]

    [tex]-lni + C = \frac{R}{L}t[/tex]

    [tex]I_{L}(0)=1 A[/tex]

    [tex]-ln1 + C = \frac{R}{L}0[/tex]

    [tex]C = 0[/tex]

    [tex]-lni + 0 = \frac{R}{L}t[/tex]

    [tex]0 = \frac{R}{L}t + lni[/tex]

    [tex]1 = e^{\frac{R}{L}t} + i[/tex]

    [tex]i = 1 - e^{\frac{R}{L}t}[/tex]

    [tex]v = L\frac{di}{dt}[/tex]

    [tex]v = L\frac{d(1 - e^{\frac{R}{L}t})}{dt}[/tex]

    [tex]v = -Re^{\frac{R}{L}t}[/tex]
  5. Jan 31, 2010 #4


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    Homework Helper

    Is is correct up to here, but wrong from here.

    You have a "* " instead of "+"

    [tex]i = e^{-\frac{R}{L}t}[/tex]

    [tex]v = \frac{di}{dt}[/tex]

    [tex]v = L\frac{d(e^{-\frac{R}{L}t})}{dt}[/tex]

    [tex]v = -Re^{-\frac{R}{L}t}[/tex]

  6. Jan 31, 2010 #5
    Ok I got it thanks for the help.
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