I may be asking something trivial, but unfortunately, I could not find an answer so far.

Suppose an AC circuit. The voltage of the capacitor is given by

$$v_C(t)=Q(t)/C = 1/C [ \int_0^t i_C(τ)dτ + v_C(0)]$$

and the current is

$$i_C(t) = C \frac{dV_C(t)}{dt}$$

Their Laplace transforms are: $$V_C(s) = \frac{I_C(s)}{Cs}+\frac{V_C(0)}{s}$$ and $$I_C(s)=sCV_C(s)-CV_C(0)$$ According to Ohm's Law: $$Z_C(s) = V_C(s)/I_C(s)$$ By substituting, $$Z_C(s)=\frac{1}{Cs-\frac{V_C(0)}{V_C(s)}}$$

It is known that $$Z_C(s)=\frac{1}{Cs}$$ which obviously occurs from the above equation if \$\small V_C(0)=0\$, but that is not always the case. Could someone explain this issue? What happens if my capacitor's initial condition is not zero?

  • 1
    \$\begingroup\$ Are the square brackets in the first equation correct ? the LHS is volts. the RHS has a term \$ v_C (0) * 1/C\$ which doesn't have the units of volts. \$\endgroup\$
    – AJN
    Commented Jun 17, 2020 at 13:28
  • \$\begingroup\$ When we were taught solving circuits using Laplace txform, we first transformed the capacitor (or inductor) into a capacitor with zero initial voltage and a voltage source connected in series (inductor with current source in parallel). You have effectively found the impedance of a compound device which is a combination of a capacitor (with zero initial voltage) in series with a voltage source (representing the initial charge). See this link for example (Page 25 and 26). ius.edu.ba/sites/default/files/u772/ee202laplacetransform.pdf \$\endgroup\$
    – AJN
    Commented Jun 17, 2020 at 13:42
  • \$\begingroup\$ In addition to AJN response and in a more general case, think in the procedure used to determine the Thevenin impedance seen from a two terminal complex circuit possibly containing voltage and current sources. These generators are replaced by short circuit or open circuit, respectively. \$\endgroup\$ Commented Jun 17, 2020 at 14:12
  • \$\begingroup\$ Frequency response assumes zero initial conditions. \$\endgroup\$
    – Chu
    Commented Jun 17, 2020 at 15:41

1 Answer 1


View 1

When solving circuits using Laplace transform, one method commonly taught is to replace a capacitor with an initial voltage with a capacitor with zero initial voltage and a special voltage in series with it. Slide 25 and 26 give an example. The equivalent circuit in \$s\$ domain has a capacitor \$C\$ with impedance \$1/(sC)\$ and a voltage source \$v(0)/s\$ in series. This equivalent circuit has the same characteristics as the original capacitor.

equivalent circuit for a capacitor with initial voltage

(Image taken from above link.)

View 2

For AC circuits, the AC impedance is given as \$dv/di\$ (ratio of change in voltage to change in current). A similar argument is used to arrive at the formula \$1/sC\$ in the Wikipedia article for impedance

As such, I do not see any problem in expression you have derived; It's just that the initial condition is handled separately and impedance of capacitor is taken as \$1/sC\$. This turns out to be easier to work with in practice.


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