# Capacitor transient charging equation of an RC series circuit

• Engineering

## 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))

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berkeman
Mentor

## 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))
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

|-------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