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Homework Help: Electric circuit - time constant to charge

  1. Feb 21, 2010 #1
    1. The problem statement, all variables and given/known data
    Four identical capacitors are connected with a resistor in two different ways. When they are connected as in part a of the drawing, the time constant to charge up this circuit is 0.91 s. What is the time constant when they are connected with the same resistor as in part b?
    http://edugen.wiley.com/edugen/courses/crs2216/art/qb/qu/c20/ch20p_99.gif

    2. Relevant equations

    Q=Q0*(1-e-t/RC)
    Q0=CV
    t=RC

    3. The attempt at a solution
    a) Q0=Q/(1-e-ta/RC) = 4C/V(1-e-ta/RC)
    b) Q0=Q/(1-e-tb/RC) = 5C/2V(1-e-tb/RC)

    4C/V(1-e-ta/RC) = 5C/2V(1-e-tb/RC)

    4/(1-e-ta/(4ta)) = 5/2(1-e-tb/(5/2tb))

    e-2/5 = 3/8 + 5/8 * e-1/4

    I'm stuck there.. Not sure if what I did is right.. If it is, I can't find a way to use the natural logarithm
     
    Last edited: Feb 21, 2010
  2. jcsd
  3. Feb 21, 2010 #2

    ideasrule

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    You don't need the equation giving the charge at a certain time because the question doesn't ask you to find the charge at a certain time. Just write out an expression for the time constant for (a), then do the same for (b). Solve the two resulting equations.
     
  4. Feb 21, 2010 #3
    I tried using the equation t=RC because Ra = Rb; Ca = 4C and Cb= (5/2)C
    R= ta/4C = 5tb/2C
    tb = (5/8)ta
    Why doesn't this give me the right answer?

    I can't find another equation to solve for tb.

    4C/V(1-e-ta/RC) = 5C/2V(1-e-tb/RC)

    4/(1-e-ta/RCa) = 5/2(1-e-tb/RCb)

    e-2tb/5RC = 3/8 + 5/8 * e-ta/4RC

    I'm stuck there again..
     
  5. Feb 21, 2010 #4

    ehild

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    Recalculate Ca.

    Again: you do not need the time dependence of charge. By the way, you use wrong formulas. The time constant is T=RC. Ta= RCa, Tb=RCb. Ta is given: 0.91 s. Tb/Ta=Cb/Ca.

    ehild
     
  6. Feb 21, 2010 #5
    When the capacitors are connected in parallel the total C is C1+C2+...+Cn which in case a) gives 4*C..
    And i did use tb/ta = cb/ca
    tb= taCb / Ca = ta(5/2)C / 4C = 5ta / 8

    I used the equation with the charge because if there's a way to take the natural logarithm the charge would cancel.
     
  7. Feb 21, 2010 #6

    ehild

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    The four capacitor are not connected in parallel in case "a".

    Q(t)=Q0*(1-e-(t/RC) ) = Q0*(1-e-(t/T) ) is the formula for the time dependence of the charge on the capacitor. t is the time after connecting the capacitor to a voltage source. T is he time constant .

    ehild
     
  8. Feb 21, 2010 #7
    ah.. got it.. thanks.. that was easy...
     
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