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Finding Current after 50s Capacitors

  1. Jun 29, 2011 #1
    1. The problem statement, all variables and given/known data
    I have managed to calculate the first part of the question but the second part b I keep getting wrong. I was wondering whether you might give me some guidance. I am trying to find the current after 50s in part b below.

    2. Relevant equations
    time constant = RC

    3. The attempt at a solution

    a) Time constant = RC = (50 x 10-6 F) x (100, 000Ω)
    Time constant = (5 x 10-5 F) x (100, 000Ω)
    Time constant = 5s Correct

    b) T1/2 = 5 x In 2 = 3.4657...
    50 seconds corresponds to 50/3.4657 = 14.41 half-times.
    Δ changing current 14.41t1/2 = 90µA/2^14.41 = 4.13 x 10^-3 µA = 4.13 x 10^-6A
    Current after 50 seconds is 4.13 x 10^-6A

    This is wrong. Help. Thanks.
  2. jcsd
  3. Jun 29, 2011 #2
    Your setup looks okay to me, but the arithmetic in your last step is wrong. Try the last calculation again.

    A couple of miscellaneous pointers:

    1. A handy number to remember is that 10 half-lives corresponds to a decrease of about 1000 (actually 1024). So after more than 14 half-lives, we expect a decrease by a factor of about 10,000. Thus we can see that your answer is much too large; it's about 1/20 of the initial current, not 1/10,000! Quick estimates like this are very useful for finding calculation mistakes.

    2. To cut down on calculation, you can use the number e as an exponential base instead of 2. The time-constant RC is actually the "1/e-life" (as opposed to the 1/2-life). So once you calculate RC, you don't need to translate into half-lives, you can just get the answer immediately. Numerically, e = 2.718... but your calculator may have a button labeled "e^"

    In this case, RC = 5, so we have,

    I (50s) = (90)e^(-50/5) = (90)e^(-10)

    And you're done.
  4. Jun 29, 2011 #3


    User Avatar

    Staff: Mentor

    50/3.4657 is 14.427
  5. Jun 29, 2011 #4
    Thanks a lot.

    I cannot believe I missed that.
  6. Jun 30, 2011 #5
    Also, a µA is 10^(-6) A, not 10^(-3) A
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