RC & R/L Time Constant

Discussion in 'Homework Help' started by Zan, Nov 26, 2005.

  1. Zan

    Thread Starter New Member

    Nov 26, 2005
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    I came across this problem and cannot solve it. I have been on it for hours with no luck, yet I know it's simple.

    A 100-V source is in series with a 1-M Ω R and a 4-μ F C. a) How much time is requried for Vc to be 63V?
    T= RxC 1-M Ω R X 4- μ F C = 4s. This part I get.

    b)How much is Vc after 4s? The anwser is supposed to be 86V. But how do I get it? What formula do I use? I have the course book but it doesn't explain how it got 86.

    I would appreciate if somone would explain- in detail- how to solve these problems. and/or summarize how you solve for voltage in R/C and R/L.

    Thanks
     
  2. hgmjr

    Moderator

    Jan 28, 2005
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    Review the material contained in the tutorials located on this website in the chapter on
    RC and RL circuits. See if that helps you give you a better understanding of the subject.

    hgmjr
     
  3. richbrune

    Senior Member

    Oct 28, 2005
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    I'm having trouble too-my browser or something won't let me get to chapter 16.
     
  4. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    I've just reviewed all of the link. Everything worked just fine.
     
  5. Zan

    Thread Starter New Member

    Nov 26, 2005
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    I reviewed that material before I made this post and didn't get it. I reviewed it again and it all came through. I misunderstood how to plot one of the formulas.
     
  6. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    Zan,

    Was this a question from a book? Because for it to be 86V, it requires more than 1 time constant. One time constant is 63% of the voltage, and each successive time constants is 63% of the remaining voltage ad infinitum, although we say after five time constants it's considered 100%.
     
  7. m-rice

    New Member

    Nov 25, 2005
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    To get an answer of 86V the question should have been "What is the voltage after 8 seconds (two time constants)?"

    Check it out: V = 100(1 - exp(-8/4)) gives 86.47 volts

    Martin
     
  8. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    The question could have said ... what would the voltage be after an additional 4 seconds?

    I wanted to know if he reported the question as written in the book or did he paraphrase and lost something in the translation.
     
  9. Zan

    Thread Starter New Member

    Nov 26, 2005
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    This is how the problem is in the book, word for word:

    A 100V source is applied in series with a 1-MΩ R and a 4-μF C that has already been charged to 63V. a) How much is Vc after 4 sec?

    First I did: T=RC which equals 4 seconds.
    Then I used the Universal Time Constant Formula that is presented on this board because the formula given in the book is different. And when I plug in the numbers to the book formula I get something else.

    I first did the (e^(t/t) = (e^(4/4) = 2.718281828 then: 1- 1/2.718281828 = .6321205588 X 37V = 23V.

    I got the 37 from: 100V - 63v)

    Vc= 37V + 23V = 86V. Of course if you want to be exact it's like 86. 43V. The book isn't that exact on this problem.


    Is there an easier(shorter) way to do these problems?
     
  10. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    OK ...

    You know the Capacitor was already charged to 63 volts ... and you know that in 1 time constant the capacitor charges to 63% of the source voltage.

    You figured out the time constant was 4 seconds and the question wanted to know what would the voltage be 4 seconds after the 63% charged capacitor. Well, 4 seconds was an additional time constant. The fastest way, since you were working with 100 Volts would be to convert the percentage to voltage from a handy monograph or if you memorized the percentages for the first five time constants.

    As you do more problems, you will instinctively figure these out while saving the formula for the future problems.
     
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