capacitor discharging

Discussion in 'Homework Help' started by Vanush, Apr 19, 2008.

  1. Vanush

    Thread Starter Active Member

    Apr 19, 2008
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    [​IMG]

    hi all. nice site you have here

    i was wondering if any of you can help me with this problem.

    In the circuit shown above, the swtich has been in the upper position for a long time and moves to the lower psoition at t = 0. Find i(t).

    basically from my understanding, in the upper position, the capacitor gets charged by the voltage source to (20/3) V (i got this from the thevenin equivalent cct). So at t=0, it goes into the lower position, then what happens? Will any current flow? I dont think so... cuz theres no difference between the above and below circuits. lol. thats my theory - but i cant find any theory regarding its correctness - so forget that. the capacitor will discharge, but through where? Is it given by the textbook formula I = [Vc*e^(-t/RC)]/R? If so, what's R -> will the current flow to the mesh on the left with the voltage source?

    textbooks always deal with the case where capacitor discharging circuit just consists of the capacitor and a resistor in series, but this isnt
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
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    During the time between switch contacts the capacitor will discharge through the 20K resistor in parallel with it.
     
  3. Vanush

    Thread Starter Active Member

    Apr 19, 2008
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    Do you mean, BETWEEN the time the switch is in the upper and lower position, as in, in mid-air? or do you mean t>= 0
     
  4. hgmjr

    Moderator

    Jan 28, 2005
    9,030
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    vanush,

    You have noticed that the two power sources are reversed from one another, right?

    hgmjr
     
  5. Vanush

    Thread Starter Active Member

    Apr 19, 2008
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    im not sure what you mean?
     
  6. hgmjr

    Moderator

    Jan 28, 2005
    9,030
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    What I mean is the 10V dc voltage source associated with the switch in the upper position is oriented in a direction to charge the cap with a negative voltage with respect to the reference point which I take to be the bottom end of the capacitor and the 10V dc voltage source associated with the switch in the bottom position is oriented in a direction to charge the capacitor with a positive voltage with respect to the same reference point.

    hgmjr
     
  7. rwmoekoe

    Active Member

    Mar 1, 2007
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    at t=0, the voltage is -6.66v.
    when the switch touches the lower pole, the voltage source become +6.66v, with a 6.66 k ohm resistor in series. the voltage at c is -6.66, so the total voltage difference is 13.33v, hence i(t)=13.33/6.66k.
    what happens is the capacitor descharging it's -6.66v, all the way through 0v until it reaches 6.66v.
     
  8. Vanush

    Thread Starter Active Member

    Apr 19, 2008
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    but is i(t) the same as the current derived from the thevenin equivalent cct as u have just posted
     
  9. rwmoekoe

    Active Member

    Mar 1, 2007
    172
    0
    yes it is. just simplify it.
     
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