# Derivation of capacitive reactance

Let $Q(t)$ be the charge stored in a capacitor $C$ in $t$ time for alternating current. Now, $$\displaystyle Q(t)=Q(0)+\int_{0}^{t}I(t)dt\\=Q(0)+\int_{0}^{t}I_oe^{j\omega t}dt\\=Q(0)+\frac{I_oe^{j\omega t}-I_o}{j\omega}\\=Q(0)+\frac{I(t)-I_o}{j\omega}$$

It yields, $$\displaystyle C=\frac{Q(t)}{V(t)}= \frac{Q(0)}{V(t)}+\frac{I(t)}{j\omega V(t)}-\frac{I_o}{j\omega V(t)}.$$

But, reactance of capacitor, $$\displaystyle X_c=\frac{V(t)}{I(t)}=\frac{1}{j\omega C}.$$ Constant term $\left(\frac{I_o}{j\omega}\right)$ of the integration seems to be unexpected here. What's wrong in this approach?

• {Cheat}: If you just calculate the indefinite integral, there will not be a constant offset. – honeste_vivere Oct 6 '15 at 15:12
• Probably because you need to be careful with the definitions and taking the Real Part when you want physical quantities. The actual sinusoidal current is not I=I_0 e^{iwt}, it is the real part of that. – hft Oct 6 '15 at 20:50

When you find the response of a linear circuit for a frequency $\omega$ you are taking the differential equations that describe it and looking for the homogeneous solution. The true response of the circuit is this homogeneous solution plus a particular solution which depends on the initial conditions. That is the missing term you found.

The particular solution is disregarded because it (in many cases) decays exponentially and you're interested in the steady-state. If you are interested in the transitory behavior, look into the Laplace Transform.

If this capacitor is a normal one, I wouldn't expect the current to have a constant amplitude. The amplitude could be changing over time, so your integral should treat $I_0$ as a variable as well? Some other properties should be used to derive the current instead.