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Energy dissipated in resistor

  1. Mar 3, 2014 #1
    1. The problem statement, all variables and given/known data

    In the circuit shown in the figure, both switches operate together; that is, they either open or close at the same time. The switches are closed a long time before opening at t=0. (Figure 1)

    Figure_P07.25.jpg

    How many microjoules of energy have been dissipated in the 12 kΩ resistor 22ms after the switches open? How long does it take to dissipate 24% of the initially stored energy?

    2. Relevant equations

    P=i(t)v(t)
    E=0.5*C*v^2
    i = C * dv/dt
    KVL

    3. The attempt at a solution

    First I found the energy in the capacitor:

    E = 0.5*C*v^2 = 0.5 * (10/3)*(10^-6)*(120)^2 = 0.024 J

    I then wrote a KVL equation around the center loop (only closed loop remaining with switches open) using passive sign convention to get:

    -V_C + V_R = 0, V_R = V_C
    Using i=C*dv/dt, V_C = (1/C) * ∫i dt
    V_R = C*dv_C/dt
    (1/C)*∫i dt = iR
    (1/C)*i = R*di/dt
    i = K * e^(t/(RC))

    I'm not exactly sure where to go from here, obviously I need to solve for K but I'm not sure what I can use for the initial condition. After solving for that, I assume you can plug that value into p=i^2*r, and w=[itex]\int[/itex][itex]\stackrel{t2}{t1}[/itex]p dt.
     
  2. jcsd
  3. Mar 3, 2014 #2

    BvU

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    Initital voltage is determined by the circuit with the switches closed.
     
  4. Mar 3, 2014 #3
    I know that much, but I'm not exactly sure where that fits into my equation.
     
  5. Mar 3, 2014 #4

    CWatters

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    The voltage on the capacitor at t=0 is not 120V. There is a potential divider.
     
  6. Mar 3, 2014 #5

    CWatters

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    When the switches open the circuit simplifies. I would write an equation for the capacitor voltage vs time. That will allow you to work out the voltage and hence the energy left in the capacitor at any given time. Where did the rest go?
     
  7. Mar 3, 2014 #6
    I think I understand what you're saying. So far I've found the equivalent resistance of the 3 resistors using source transformation (with the switches closed obviously), then left the circuit in the form:

    hmd59SJ.png

    To find the voltage across the capacitor and resistor using V=IR. I got V=102.5 Volts. Now I believe the energy at time t=0 should 0.5*C*v^2 = 0.0175J. Assuming this is correct, I wrote what I think should be capacitor voltage vs time as:

    V_C = V_0 * e^(-t/(RC)), meaning V_C at 0.022 seconds should be 59.14 V. Then the energy at that time would be 0.5*C*(59.14)^2 = 0.00583J, so the change in energy is then -0.0117J, which should be equal to the energy dissipated in the resistor. Does what I did sound follow what you suggested?
     
  8. Mar 3, 2014 #7

    BvU

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    Looks a bit strange to me. 12 kΩ parallel with 68 kΩ can't give 1530 Ω.

    Could you show what you did there ? Is there a reason to change to a current instead of a voltage ?
     
  9. Mar 3, 2014 #8

    SammyS

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    I get V = 102 Volts, exactly. So you're pretty close.
     
  10. Mar 3, 2014 #9
    I changed to a current so I could find the equivalent resistance across all three resistors, not just the 12kΩ and 68kΩ. The resulting circuit was much simpler than just combining the resistors in parallel.

    Good to know. I submit the answer I posted earlier which turned out to be correct. Thanks all.
     
  11. Mar 3, 2014 #10

    BvU

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    Amazing, but I believe it. Then the meaning of the 1530 Ω is like all three in parallel ?

    Wouldn't it be wiser (safer, more conservative, perhaps) to lead towards 120 V * (12 kΩ // 68 kΩ) / (1.8 +12 kΩ // 68 kΩ) ?
     
  12. Mar 3, 2014 #11
    Not sure I follow what you're saying, I just did what I was comfortable with and knew would work.
     
  13. Mar 3, 2014 #12

    BvU

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    Great! I'll read up on source transformation, looks nifty (but I never used it thus far)
     
  14. Mar 4, 2014 #13

    CWatters

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    Yes but I got very slightly different numbers..

    Using potential divider I got 102V exactly...

    V_0 = 120 * (12//68) / ((12//68)+1.8)
    = 120 * 10.2/(10.2+1.8)
    = 102

    So the starting energy is 17.34mJ

    Then using..

    V_C = V_0 * e^(-t/(RC))

    with

    t = 0.022
    R= 12K
    C= 10/3 uF

    V_C = 102 * e^-0.55
    = 58.85V

    Which gives a final energy of 5.78mJ so the change is 17.34-5.78 = 11.56mJ. Very close to your answer.

    I haven't bothered to work out the last part of the question but it's straightforward.
     
  15. Mar 4, 2014 #14

    CWatters

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    PS. I don't see the need to do any transforms to solve this one. Mainly because it says the switches are closed for a long time before t=0 so the capacitor can be assumed to be at the potential divider voltage. I suppose had they said the switches were closed for time t=??mS it would be different.
     
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