# 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. Oct 6, 2015 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, 2015 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.
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.