laplace forum, help, tutor says it wrong, but i dontknow where

Discussion in 'Homework Help' started by ninjaman, May 11, 2015.

  1. ninjaman

    Thread Starter Member

    May 18, 2013
    306
    1
    hello

    i have attached my laplace assignment and my tutor has told me its wrong. im not sure where though
    upload_2015-5-11_10-18-33.png
    solve for the current
    upload_2015-5-11_10-19-11.png
    upload_2015-5-11_10-19-25.png
    upload_2015-5-11_10-19-48.png
    upload_2015-5-11_10-20-4.png

    is any of this correct?
    thanks
    simon
     
  2. MrAl

    Well-Known Member

    Jun 17, 2014
    2,438
    492
    Hi,

    You may want to check your work a little more. The coefficient looks wrong because 6/2000=0.003 and that is the max current (cap shorted out), and the exponential factor just 'modulates' that current over time.

    You might be better off just using Laplace substitutions right from the start, as long as you are using Laplace anyway.
    For example from your very first equation we derive:
    (i/C)*(1/s)+R*i=6/s

    and solve for 'i'.

    Unless of course you are told to do it the longgggg way :)
     
    Last edited: May 11, 2015
  3. ninjaman

    Thread Starter Member

    May 18, 2013
    306
    1
    hello MrAI
    i think the longgggg way is what is wanted, thanks for your help!!!
     
  4. ninjaman

    Thread Starter Member

    May 18, 2013
    306
    1
    sorry, im not too good at this. i have an example and i followed the example through and i dont know what i did wrong.
    how do i rearrange the coefficient. i followed the example almost exactly on the web site. my tutor took the question from the site and changed the values. i followed the order that the site used to solve the problem. though i cant get the right answer.
    this is the web site, its the first example http://www.intmath.com/laplace-transformation/10-applications.php
    any help would be great!

    thanks

    simon
     
  5. MrAl

    Well-Known Member

    Jun 17, 2014
    2,438
    492
    Hello again,

    The example on the web site has an initial voltage across the capacitor. Did you notice that? That changes the coefficient because the capacitor voltage acts with the source voltage at t=0.
    So the coefficient (for cap voltage polarity in the example) changes from 5/R to (5-1)/R=4/R because the cap voltage 'subtracts' from the source voltage.

    The web site first writes out the equation, then transforms it, then does a little algebra to get the correct Laplace form, then finally transforms the result to the time domain.

    What you can do is follow each step, number the steps from 1 to N, then state which step you did not get the same result as the web site did. We can then take a look at that.

    Dont forget though that an initial capacitor voltage changes things a little, as shown on the web site. You have to solve for the second term 10^-6/s from 1/C*Integral(i)dt.

    In your original example, if there is no initial cap voltage then the coefficient is 6/2000=0.003, but if the cap voltage is 1 volt with the same polarity as shown across the cap, then the coefficient will be (6-1)/2000 which equals 0.0025 not 0.003, so you see the difference here. Check to see if there might be an initial cap voltage in your tutor's example.

    I almost forgot to mention that the initial voltage transform comes from an application of one of the Laplace operations (you can look these up on the web) for which the that part would be expressed as:
    (1/s)*Integral[-infinity to 0] f(t) dt

    so we end up with the addition of:
    (1/s)*Vc(0)*C

    which accounts for the initial cap voltage.
     
    Last edited: May 11, 2015
  6. ninjaman

    Thread Starter Member

    May 18, 2013
    306
    1
    hello

    this is the web site questions
    upload_2015-5-11_18-1-56.png
    this is the assignemtn question
    upload_2015-5-11_18-2-21.png
    the question looks the same just the values have changed, the voltage across the capacitor is the same(i think)
     
  7. WBahn

    Moderator

    Mar 31, 2012
    17,757
    4,800
    You can't ignore the initial conditions. You did the same thing in another thread recently.

    Does it make sense that you would see the exact same voltage versus time curve if the capacitor were initially discharged versus if it were initially charged to 2000 V or to -2000 V? If you would expect those to be different, then isn't it reasonable to expect it having an initial charge of 1 V to have an effect that can't be ignored?
     
  8. MrAl

    Well-Known Member

    Jun 17, 2014
    2,438
    492

    Hello again,

    Well now that you have shown the ENTIRE problem question, we find that you had the right answer all along. I wish you had done that from the start :)

    It appears that you actually DID follow the example pretty well, even with the initial voltage. Also, the final answer was pretty small on my monitor so it looked like 2.5e-5 not the correct 2.5e-3, but magnifying it i see it really is 2.5e-3 which after all is correct.

    As i was saying before, if you take 6 volts and subtract 1 volt we get 5 volts, and 5 divided by the resistance of 2k gives us 2.5e-3, so that coefficient is correct. The exponential exponent is -t/RC which is of course -50*t, so you have that right too. So the correct answer is <drum roll please>
    0.0025*e^(-50*t)

    What i am guessing now is that the tutor did not want you to use an initial voltage across the capacitor, or they just got it wrong altogether. You'll have to discuss this with them. Perhaps you can get the result from them and we can compare, or else ask them for another SOLVED problem of the exact same nature supplied with the correct result and we can check that just to make sure THEY are doing it right.
     
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