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RL Circuit

  1. Mar 7, 2014 #1
    1. The problem statement, all variables and given/known data

    For the circuit shown below, the current through the inductor is defined as follows:

    [tex]
    i(t)=\begin{cases}
    0 & \text{if } -\infty <t<0 \\
    1-e^{-2t}& \text{if } 0\leq t < \infty
    \end{cases}
    [/tex]

    L = 1H, R2 = 2 ohms

    rlcircuit.png

    For t=> 0 find:

    a) VL(t)
    b) Vr(t)
    c) V1(t)
    d) Power absorbed by the inductor
    2. Relevant equations

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



    3. The attempt at a solution

    As t goes to infinity, current goes to 1A and the inductor becomes a wire. So V1 = 2V

    But i don't think this is what it is asking for.
    From the relevant equation section, solving for V would leave only 2

    W = 0.5Li^2

    W = 0.5(1)(1-e^{-2t})^2

    Vr = (1-exp(-2t))/2

    Vl = 0V
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 8, 2014 #2

    Simon Bridge

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    Looks like it wants t→0 ... for very small times, the exponential term is approximately linear.
     
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