How much voltage is dropped over a capacitor with resistor over time

Discussion in 'Homework Help' started by bookstoreboy, Feb 26, 2016.

  1. bookstoreboy

    Thread Starter New Member

    Feb 15, 2016
    8
    1
    This should be easy but I've spent too much time on it.
    "How much voltage is dropped over a 150milliFarad cap after 3.33 seconds if it's charged by a 50v source through a 22K resistor?"

    It's actually answered over here already.
    https://answers.yahoo.com/question/index?qid=20130625103720AALu4qM

    The textbook says Vc=Vie^(-t/RC)
    I don't get why the Yahoo answer has a 1-e in it. What is the 1 doing there? If I copy Yahoo's work, it comes out ok, but if I go with the textbook I get 18v or something wrong.

    Thanks for any help on this.
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    This is Homework Help, not Homework Done For You. YOU need to show YOUR best attempt to answer YOUR homework problem. Going out and finding someone else's attempt does not qualify as YOUR best attempt.

    What level of mathematics are you expected to use to solve the problem? How much calculus have you had?
     
  3. WBahn

    Moderator

    Mar 31, 2012
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    Have you asked yourself if the textbook answer makes any sense fundamentally?

    What would it yield for the initial voltage across the capacitor?

    What would it yield for the voltage across the capacitor after a very long time?
     
  4. WBahn

    Moderator

    Mar 31, 2012
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    How did you get 18 V? That result makes no sense at all.

    The time constant of your system is nearly an hour, so in 3.33 seconds it will have charged hardly any at all.

    Anything I did to try to figure out what you did wrong would be guessing unless and until you show your work.
     
  5. WBahn

    Moderator

    Mar 31, 2012
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    Is that what the textbook says is the answer for THIS problem? Or is it just one of many formulas in the textbook?
     
  6. bookstoreboy

    Thread Starter New Member

    Feb 15, 2016
    8
    1
    Thanks for the walk, but why bring up calculus? Just multiply and divide. Don't worry, I figured it out, I'll try to write out the formulas:

    When charging an 'empty' cap, use [​IMG]
    Where [​IMG] is the voltage you're charging to and T is how many seconds you gave it.

    To discharge, do [​IMG]
    Where Vi is volts you started with T is how long you let it discharge.

    That extra '1' I was hung up on flips the curve.
     
  7. WBahn

    Moderator

    Mar 31, 2012
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    I brought up calculus because both of these equations do not just magically appear in your text -- they are the result of applying Kirchhoff's Laws and Ohm's Law to the circuit and then using calculus to solve the resulting differential equations. If you have a calculus background, then you are in a position to derive them and understand them at a much deeper level. If you don't have that background, then you can only accept them at face value and memorize them -- but you can still sanity check whether you are using the right one or not.

    What would your equation look like if the system starts at some non-zero initial voltage, say Vi, at time t=0 and then changes exponentially to some other non-zero final voltage, say Vf ?

    And I still haven't seen how you got that 18 V, or what your final answer is. For your circuit, the voltage at 3.33 s is well under 100 mV.
     
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