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RC Circuits and Capacitance

  1. Apr 20, 2013 #1
    1. The problem statement, all variables and given/known data
    "A capacitor is being charged from a battery and through a resistor of 20kΩ. It is observed that the voltage on the capacitor rises from 10% of its final value to 90%, in 15 seconds. Calculate the capacitor's capacitance.

    t = 15s
    R = 20kΩ
    Vi = 10% of max
    Vf = 90% of max

    2. Relevant equations
    Vc(t) = Vs(1-e^(-t/τ))
    τ = RC

    3. The attempt at a solution
    Vc(t) = 90% = .9
    Vs = 100% = 1.0
    .9 = 1.0(1-e^(-t/τ))
    .9 = 1-e^(-t/τ)
    e^(-t/τ) = .1

    From there it has been a while since I took calculus, but I continued with

    ln(e^-t/τ) = ln(.1)
    -t/τ = ln(.1)
    -15/20000C = ln(.1)
    -.00075 = ln(.1)C
    C = .000326 F

    I got an answer, but I feel like Im missing something to do with the voltage on the capacitor at t = 0 being 10%. I have no way to check if this answer is correct, so I guess I'm really just asking for someone to check my work (and if it's wrong, give me a push in the right direction)
     
    Last edited: Apr 20, 2013
  2. jcsd
  3. Apr 20, 2013 #2

    gneill

    User Avatar

    Staff: Mentor

    Hi Kashuno, Welcome to Physics Forums.

    Yes, something's missing. You're given the time between two points on the charging curve, but you're not given an absolute time for either point. In other words, you can't fix either the start or the end of the interval in question without doing a bit of fancy footwork.

    Why don't you write expressions for both of the given amplitudes, letting the first occur at time t and the second at time t + 15s. How many variables and equations will that leave you with?
     
  4. Apr 20, 2013 #3
    Wow! I can't believe I didn't see that! I was having a hard time finding an equation that incorporated percentages that I didn't even think of multiple equations. :rolleyes:

    Set it up and get two equations with two unknowns, solve and substitute, done. Thank you so much!
     
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