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Power in RL circuits

  1. Nov 13, 2009 #1
    1. The problem statement, all variables and given/known data

    At t = 0, a battery is connected to a series arrangement of a resistor and an inductor. If the inductive time constant is 42.1 ms, at what time (in ms) is the rate at which energy is dissipated in the resistor equal to the rate at which energy is stored in the inductor's magnetic field?
    2. Relevant equations

    power dissapated by resistor: P=(I^2)R

    Power stored by inductor: P=IL(di/dt)

    Current in an RL cuircuit: I=V/R(1-exp(-t/T))

    where T is the time constant tau = (L/R)

    3. The attempt at a solution

    I just set them equal, then simplified.

    (I^2)R = IL(di/dt)

    I=(L/R)(di/dt)

    replace L/R with Tau and Current with above equation

    V/R(1-exp(-t/T)) = T (di/dt)

    take derivative of current

    V/R(1-exp(-t/T)) = T (V/R)(1/T)*exp(-t/T)

    cancelations

    1-exp(-t/T)=exp(-t/T)

    add exp(-t/T) to both sides and take natural logs

    Ln(1/2) = -t/T

    so t = -T*Ln(1/2)

    I plug this in to the online homework and it keeps telling me I'm wrong, but I can't see
    what I'm doing incorrectly, any ideas?

    (sorry I don't know LaTex yet :P)
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Nov 13, 2009
  2. jcsd
  3. Nov 13, 2009 #2

    rl.bhat

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    Homework Helper

    The instantaneous rate at which the source delivers energy to the circuit is P = V*I. This is equal to
    V*I = I^2*R + LI*dI/dt
    Put I^2*R = LI*dI/dt . So
    V*I = 2*LI*dI/dt
    dI/dt = V/L*e^-t/T
    Put this in the above equation and solve for t.
     
  4. Nov 13, 2009 #3
    Hi, I am a little confused. Where did you get the equation:

    dI/dt = V/L*e^-t/T? My book says it is (V/R) instead of (V/L)?

    Also I plugged your equation for di/dt into the equation above it.

    And it reduced to:

    1/2= exp(-t/T)

    or

    -T*Ln(1/2) = t, which is what I also got.
     
  5. Nov 14, 2009 #4

    rl.bhat

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    Homework Helper

    Instantaneous current is given as
    i = V/R*[1 - e^-(R/L)t]
    di/dt = V/R*(R/L)*e^-(R/L)t
    = V/L*e^-(R/L)t
     
  6. Nov 16, 2009 #5
    ah sorry I've been so late, heh and it seems I can't even take a derivative anymore.

    thanks for the help.
     
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