Exam Help

Discussion in 'Homework Help' started by hitmen, Dec 5, 2008.

  1. hitmen

    Thread Starter Active Member

    Sep 21, 2008
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    Well, my exams are over and here I am going to post the questions I have difficulty with.

    In the first question, I am able to derive that Ix(0+) = (4-Vc(0+))/6 and that Ix(t) = (1+2e-1000t) / 3 for t>0. However, I am not sure how to calculate the energy absorbed by capacitor over 0<t<∞ and the amount of charge deposited into the capacitor.

    Can someone explain how the first 2 equations relate to the 3th part because I only know how to solve them blindly.

    Do kindly ignore my markings on the exam paper.
     
    Last edited: Dec 5, 2008
  2. hgmjr

    Moderator

    Jan 28, 2005
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    The image is a bit fuzzy. Can you post a clearer picture?

    hgmjr
     
  3. studiot

    AAC Fanatic!

    Nov 9, 2007
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    Both answers require calculus can you do this?
     
  4. hitmen

    Thread Starter Active Member

    Sep 21, 2008
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    Ummm... I am only interested in the Energy absorbed and the charge deposited in the capacitor. I cant do it in the exam. I can do calculus but I am not sure of how to even START doing the problem. how about giving me the first step or a rough concept?
     
  5. hitmen

    Thread Starter Active Member

    Sep 21, 2008
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    Do you mind giving the first step?
    Thanks.
     
  6. studiot

    AAC Fanatic!

    Nov 9, 2007
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    Having some trouble posting here.

    OK now we have done the diode problem, we can go on with this one. Sorry I couldn't post anything earlier.

    I would guess from this that you have been given the formulae energy = 1/2CV^2 and charge q = CV for a capacitor.

    So noting that initially the capacitor appears as a short circuit and finally as an open circuit calculate the voltages using normal theorems (egKVL) for these two conditions and apply the formula to the voltage change across the capacitor.

    I note by inspection that the capacitor is at -2 volts initially.

    The charge is easier as it is just the voltage rise times the capacitance.

    An alternative method is to use you currents to calculate the current through the capacitor using the differential equation

    i = C dVc/dt where Vc is the capacitor voltage, seen as a node voltage in KVL

    Then you integrate this to find the same solution

    Oh and well done for getting as far as you have. Most courses skate over this stuff quickly because there are better more comprehensive mathods you will come on to.
     
  7. Bigcountry

    Active Member

    Jul 4, 2008
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    In reference of the clearer picture. Since it looks like you took a picture of it You could put it in paint and trace over the numbers and schematic. Just idea though.
     
  8. hitmen

    Thread Starter Active Member

    Sep 21, 2008
    159
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    I know the formula i = C dVc/dt but how I calculate Δt when it is from 0 to infinity?
     
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