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Capacitor transient charging equation of an RC series circuit

  1. Aug 29, 2012 #1
    1. The problem statement, all variables and given/known data
    I would like to derive the equation Vc=Vss+(Vi-Vss)*e^(-t/RC)
    Vss is the steady state voltage
    Vi is the initial capacitor voltage
    Vc is the capacitor voltage
    3. The attempt at a solution

    I tried to find solution using laplace transform. E=iR+1/c∫idt.
    Taking laplace, E/s=I(s)R+1/(cs)*I(s)-q(0+)/(cs)
    Put q(0+)=cVc(0+)=cVi
    Then taking inverse laplace
    i(t)=(E-vi)/R *e^(-t/RC)
    Vc=1/c∫idt
    =(E-vi)(1-e^(-t/RC))
    What is wrong? why am i not getting the correct answer? Please help...
     
  2. jcsd
  3. Aug 29, 2012 #2

    berkeman

    User Avatar

    Staff: Mentor

    Welcome to the PF.

    It looks like your two starting equations are fundamentally different. Could you show the circuit along with labels for the variables (what is Vss versus Vc for example?).

    Vc=Vss+(Vi-Vss)*e^(-t/RC)

    E=iR+1/c∫idt
     
  4. Aug 30, 2012 #3
    |-------E volt dc----------|
    (+)___^^^^______||_____|(-)
    ..................(+)|---Vc---|(-)
    Input is a dc voltage(polarities shown).
    Vc=capacitor voltage
    Before applying input, capacitor voltage Vc=Vi
    After applying input and reaching steady state, Vc=Vss
    But here in this case, steady state capacitor voltage Vss=E
     
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