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LR circuit energy dissipation

  1. Dec 17, 2006 #1
    1. The problem statement, all variables and given/known data
    A LR circuit has a 10V battery, 5.50H inductor, and a 6.7 Ohm resistor. The battery is closed at t=0. (a) What is the time constant of the circuit? (b) How much energy is delivered by the battery during the first 2 seconds? (c) How much of this energy is stored in the magnetic field of the inductor? (d) How much of this energy is dissipated in the resistor?

    2. Relevant equations


    3. The attempt at a solution

    a) This is simple, time constant = R/L
    b)my book never defines an equation for the total work done by the battery excpet dW/dt=IE. When I use this equation with the current value at t=2 (and dt=2), and solve for dW with the initial battery voltage, I get an answer of about 27 Joules. However, I don't think this is right since the next two parts don't agree with this answer.
    c)The energy in an inductor is stored in the magnetic field created by the inductor. So UL=1/2LI^2. When using the given values and the current solved for at t=2 i Get around 5.10 Joules.
    d) I would assume that if the two previous values are correct, then the difference between the two would be the answer. However, looking for a way to confim this, I did an intergral of I^2R over time. So I had (E/R[1-e^(-tR/L)])^2Rdt and integrated from t=0 to t=2. When I solved for this value I got a little less then 14 Joules.

    14+5 != 27. So what's wrong here?
    Last edited: Dec 17, 2006
  2. jcsd
  3. Dec 17, 2006 #2
    It is part your part b. Since energy is equal to

    [tex] \int_{0}^{t}i(t)*v(t)dt=power [/tex]

    You know the voltage supplied by the source and you should have your function of current by using the step-by-step method for transient circuits. Now just solve!

    In your relevent equations you have power as I^2*R which is true but only in steady state conditions not for transient.
    Last edited: Dec 17, 2006
  4. Dec 18, 2006 #3
    I don't quite follow what you did. So if I initgrate the fuction of current times the function of voltage with respect to time i'll get power. Then i divide that vaule by that change in time to get the total work done by the battery? then it is this value subtracted from the value of part c to get part d?

    Is there a way to write an equation for the power dissipated by the resistor as a function of time?
  5. Dec 18, 2006 #4


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    I think Valhalla intended to say something a bit different

    The equation should have read

    [tex] \int_{0}^{t}i(t)*v(t)dt=energy [/tex]

    The product [itex] i(t)*v(t) [/itex] is the instantaneous power, and its integral over time is the energy delivered. If your voltage source is ideal the voltage is constant at E.

    The energy disipated in the resistor is the time intergral of the instantaneous power dissipation. The voltage is not constant for the resistor, but it is proportional to the current.
  6. Dec 18, 2006 #5
    what do you mean constant at E?

    so v(t)=E(1-e^(-Rt/L))
    and i(t)=(E/R)(1-e^(-Rt/L))
    therefore, i(t)*v(t)=(E^2/R)(1-e^(-Rt/L))^2
    integrate with respect to time and that is the total energy delivered into the circuit, correct?

    EDIT: That integral is what I original intergrated and I got 13.57 Joules; however, I was solving for the dissipated energy of the resistor and used I^2R=P=IE.
    Last edited: Dec 18, 2006
  7. Dec 18, 2006 #6


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    I mean the voltage difference across the battery is E = 10V. That is what is given in the problem. E is not a function of time except for the step at t = 0 when the battery switch is closed. The current is a function of time, and it is common to all elements of the circuit. The total energy deliverd to the circuit is the time integral of the power produced by the battery, which is P(t) = E*i(t). The voltage across the resistor is a function of time and it is proportional to the current through the resistor: v(t) = i(t)*R. The energy dissipated in the resistor is the integral of i(t)*v(t). That has to be less than the total energy delivered by the battery. The energy that is not dissipated in the resistor is stored by the inductor.
  8. Dec 18, 2006 #7
    and that energy in the inductor is 1/2LI^2 correct
  9. Dec 18, 2006 #8


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    That's what everybody says, where I is the current that is flowing at the end of the time interval. I'll leave it to you to prove it by calculating the difference between the energy delivered and the energy dissipated.

  10. Dec 18, 2006 #9
    awsome, i did the intergral for the total work done and it fits with my other answers. Thanks you so much guys, you all where a huge help.
  11. Dec 18, 2006 #10
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