Laplace transform in circuits

Discussion in 'Homework Help' started by thisonedude, Sep 12, 2014.

  1. thisonedude

    Thread Starter Member

    Apr 20, 2014
    52
    0
    So i'm working on this problem. I don't undersntad how they got 20 and 80 volts. i know the voltage across both capacitors is equal to 100V since it's been a long time. But how did they get that .2V1=.8V2? and how did they solve for each voltage? the Answer is at the bottom. I just need help with a. 1.JPG 2.JPG
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    784
    The charge on the two capacitors must be the same...

    Q1=C1V1=Q2=C2V2

    and so on....
     
  3. dalam

    Member

    Aug 9, 2014
    58
    6
    For any capacitor the basic equation is Q=C*V.
    So V=Q/C, thus, V is inversely proportional to C.
    [​IMG]
    Hence the division.
     
  4. MrAl

    Distinguished Member

    Jun 17, 2014
    3,758
    791
    Hi,

    Another approach is as follows...

    You know the total Z presented to the input source in position 'a' is:
    Z=R1+1/s/C1+1/s/C2

    and you know the voltage is 100 and so you know the current is:
    I=100/Z

    and so you know the voltage across C2 is:
    vC2=I*(1/s/C2)

    and the voltage across C1 is:
    vC1=I*(1/s/C1)

    Now if we calculate out vC2 we get:
    vC2=I*(1/s/C2)=100/Z*(1/s/C2)=12500/(s+625)

    Now using the Laplace final value theorem:
    "The limit of f(t) as t approaches infinity is equal to s*F(s) as s approaches zero (all poles of s*F(s) in LHP)"
    we get:
    lim s*F(s) (as s--.0)=lim 12500/(s+625) (as s-->0)=12500/625=20

    and doing the same for vC1 we get:
    lim 50000/(s+625) (as s-->0) =80

    so we get 20 and 80 volts for vC2 and vC1 respectively.

    We can also do it symbolically and then we get the formula:
    vC2=Vs*C1/(C1+C2)
    and
    vC1=Vs*C2/(C1+C2)

    (Vs is the source voltage which is 100 for this exercise)

    From inspection we can see that the final voltage denominator is the sum of the two capacitors, and the numerator is the opposite capacitor. Another thing you will notice is that the resistance is not present because it disappears in the limit.
     
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