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Resistance, Capacitance,

  1. Nov 23, 2005 #1
    this seems to be a simple question
    for the diagram at http://img.photobucket.com/albums/v11/biggm/ohm.jpg

    a)the circuit has been connected a long time...what is the voltage across the capacitor?

    this one i calculated and got the right answer of 6V.

    b)if the battery is disconnected how long does it take to discharge the capacitor to 1/10 of its initial voltage.

    i have the equation t=-RC ln (q/Q)

    where R= resistance, C=capacitance, q= final charge and Q=initial charge, i have calculated the initial and final charges using q=CV and final voltage of 3/5V. Capacitance is given...what value do i use for resistance?
    i can see given the answer that it should be 18/5...which is the product of the initial and final voltage although this could be coincidental....any help here??
     
  2. jcsd
  3. Nov 23, 2005 #2

    Astronuc

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    Staff: Mentor

    With the voltage source removed and the capacity supplying the voltage, which capacitors are in series?
     
  4. Nov 23, 2005 #3
    the 1ohm and 4 ohm are in series, and the 8ohm and 2ohm are in series.
    do i just add these as in series and then....if i add them together as two then in parallel it does not work out...
     
  5. Nov 23, 2005 #4

    Astronuc

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    Staff: Mentor

    The 1 and 4 ohm resistors are in series and the 1 and 8 ohm resistors are with respect to the 10V source.

    However when 10V source is removed and the capacitor provides the potential difference, resistors 4 and 2 ohm are then in series and resistors 1 and 8 ohm are in series.
     
  6. Nov 23, 2005 #5
    ah thanks for that tip...it works out now but i fail to see how the 4 and 2 combination and the 1 and 8 ohm combination are in series when the potential is removed.
     
    Last edited: Nov 23, 2005
  7. Nov 24, 2005 #6

    mezarashi

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    Homework Helper

    Some quick photoshopping reveals

    [​IMG]
     
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