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Capacitor transient charging equation of an RC series circuit

  • Engineering
  • Thread starter priya.k
  • Start date
  • #1
2
0

Homework Statement


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

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...
 

Answers and Replies

  • #2
berkeman
Mentor
56,448
6,365

Homework Statement


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

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...
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
 
  • #3
2
0
|-------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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