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Capacitance - finding the voltage across a capacitor given a time varying current

  1. Jul 11, 2012 #1
    1. The problem statement, all variables and given/known data

    IMG.jpg

    referring to the attatchment. the current consists of two semi circles. the question asks me to find the voltage across a 47-uF capacitor when t = 2ms.

    2. Relevant equations

    v(t) = 1/C ∫ i(t)dt +v(t0) {I realise their has to be limits for the integrand,I just can't type them}

    A = [1/2.π.r^2]

    3. The attempt at a solution
    now after a while of thinking i realised i could find the area under the curve of this circle with the area formula multiplied by the capacitnce which gives me the right answer.

    However, in my first attempt i assigned an equation to the first semi circle. this was

    (2√(1 - (t - 1)^2))

    then i integrated this by using trigonometric substitution.

    I came up with this

    arcsin(t - 1) + (t - 1)*(√(1 - (t - 1)^2))

    i then evaluated at the upper limit t giving me the same formula.

    then evaluated at t0 = 0 giving (-π/2). i then subtracted this away from the top formula.

    so my entire formula is arcsin(t - 1) + (t - 1)*(√(1 - (t - 1)^2)) + (π/2).

    substituting t = 2ms i'm left with (π/2) + (π/2) = π.

    i multiply by 10^-6 because i have to account for the axes being both in ms and mA.

    then i divide by 47-μF giving me double the correct answer. roughly 66.48mV. however the answer is half that. where did is screw up with the integral evaluation???
     
  2. jcsd
  3. Jul 11, 2012 #2

    tiny-tim

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    Hi Basher! :smile:

    (try using the X2 button just above the Reply box :wink:)
    nooo … √(4 - (t - 1)2)) :wink:
     
  4. Jul 11, 2012 #3
    thanks. Did you have a crack at the problem?
     
  5. Jul 12, 2012 #4

    tiny-tim

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    or a bash? :wink:

    no … i was perfectly happy with …
    :smile:
     
  6. Jul 12, 2012 #5
    hahaha. fair call. thanks
     
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