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Finding Vc

  1. Sep 23, 2009 #1
    http://img41.imageshack.us/img41/416/32650113.png [Broken][/URL]

    Relevant equations ?

    I know area under the curve is 660
    But I dont know what the question is asking
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 23, 2009 #2
    find Vc at t = 10 microseconds?
  4. Sep 24, 2009 #3
    Yeah, is it just 44*.10?
  5. Sep 24, 2009 #4
    No, consider what you are doing with the integral - what are the units of the area under the curve? Knowing that, recall the relation between capacitance and charge.
  6. Sep 24, 2009 #5
    Yeah Q=VC
    As for the units under the area of curve, I dont know mA/microsecs?
  7. Sep 24, 2009 #6
    The area is under the curve represents mA * uS

    but an amp A = C / s, so the area is the charge being deposited in the capacitor, so you got Q.
  8. Oct 1, 2009 #7
    Look at the Hint carefully. It says that you should integrate I(t) over a period of time to find the voltage. If you want to integrate you should have a function, i.e. current as a function of time.
    So...obtain the relation between the current and the time. Then integrate it over the desired period to obtain the voltage.
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