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Homework Help: Capcitance again FTL

  1. Oct 31, 2007 #1
    capcitance again FTL :(

    1. The problem statement, all variables and given/known data

    All capacitors of the open circuit in Figure 3 are discharged at time t = 0, when the switch
    S is turned on and capacitors starts to be charged. What is the time constant [tex]\tau[/tex] of the charging process? What is the charge on capacitor C2 at an arbitrary time t > 0? If the resistances R1 and R2 are increased by a factor of 2, while capacitances C1 and C2 are decreased by a factor of 2, would the charging process occur faster or slower?

    2. Relevant equations
    Resistors in sereis = R1 + R2 +....
    capacitors in series = C1 + C2 + ....
    capacitors in parallel = 1/C1 + 1/C2 + ....
    Voltage across a capacitor = V(o)e^-(delta t/[tex]\tau[/tex])
    capacitor discharging = I(o)e^-(t/[tex]\tau[/tex]) = Q(o)e^-(t/[tex]\tau[/tex])

    3. The attempt at a solution
    ok, bak with another one of these variable infested things, lol. First, i find R(eq) by adding the resistors in series and find Ceq by first combining the capacitors in series and then adding them in parallel. Now i have R(eq) and C(eq). Multiplying these together nets me [tex]\tau[/tex], correct?

    if that's right, i then take [tex]\tau[/tex] and place it in the equation V(o)e^-(delta t/[tex]\tau[/tex]) where my voltage is half of C(eq). This is because both pieces of parallel capacitors get equal voltage.

    As for the final part of the question, i would assume that the charge occurs faster due to the lower capacitance, but im not sure by how much. since the resistors add in series, they simple become 2R(eq) while C(eq) goes up much more, or is this wrong to assume? any direction would be awesome.

    ps. you guys are THE WIN...end of story
  2. jcsd
  3. Oct 31, 2007 #2


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    It would be great if you could post your Figure 3.
  4. Oct 31, 2007 #3
    doh! sry, i meant to put it as an attachment, but i guess my comp skills are down there with my physics ability, lol. here it is.

    Attached Files:

  5. Oct 31, 2007 #4


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    Charge constant is given by RC, where R is the total resistance in the circuit and C is the total capacitance in the circuit.
    For the second part, at any instant voltage across the parallel combination of capacitors is the same. Using charge distribution formula you can find the charge on C2.
    For third part, again find the total resistance and capacitor in the circuit and hence find the time constant.
  6. Oct 31, 2007 #5
    You switched the series/parallel capacitance equations, careful. I don't know if it was a typo or not, but I thought I would point that out.
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