Impedances

If you connect a capacitor to a DC voltage source, a large current flows at first, but this quickly comes to a standstill. If, on the other hand, an AC voltage source is used, periodic charging and discharging occur, which results in a continuous current flow. The applied voltage corresponds to the voltage across the capacitor at all times: $$U(t) = U_C(t) = \frac{Q(t)}{C}$$ When applying a sinusoidal AC voltage $$U(t) = U_0\sin\omega t$$ one obtains the following differential equation $$U_0\sin\omega t = \frac{Q(t)}{C}$$ In order to calculate the current from the charge, this equation must be derived from $t$: $$U_0\omega\cos\omega t = \frac{I(t)}{C}$$ The cosine function, with which the course of the current can be described, equals a phase shift of $90^\circ$ between voltage and current: $$\boxed{\varphi = +\frac{\pi }{2}}$$ In a capacitor, the current is said to lead the voltage. You can now assign an AC resistance $X_C$ to the capacitor. Analogous to Ohm's law, the peak voltage is divided by the peak current: $$X_C = \frac{U_0}{I_0} = \frac{U_0}{U_0\omega C}$$ After shortening $U_0$ you get the AC resistance of a capacitor $$\boxed{X_C = \frac{1}{\omega C}}$$ The peak current can then be calculated using Ohm's law as follows: $$I_0 = \frac{U_0}{X_C}$$ The same relationship applies accordingly to the effective values: $$I_\mathrm{eff} = \frac{U_\mathrm{eff}}{X_C}$$ Due to the phase shift of $90^\circ$ between current and voltage, the power given off in the form of heat is exactly 0 at all times. This resistance is therefore also referred to as reactance. A similar consideration can be made for a coil. The coil is completely transparent to DC voltage. The flow of current would only be impeded by the ohmic resistance of the wire and would otherwise be infinitely large. In the case of an AC voltage, however, the beginning current flow induces an induction voltage due to Lenz's law, which initially inhibits the current flow. If you apply a sinusoidal AC voltage to the coil, as with a capacitor, then the voltage across the coil at any point in time is: $$U(t) = U_L(t) = -L\dot{I}(t)$$ In order to get the course of the current over time, $t$ must be integrated according to the time. This then delivers $$I(t) = \frac{U_0}{\omega L}\cos\omega t$$ The opposite sign and the cosine function result in a phase shift of $-90^\circ$ between current and voltage: $$\boxed{\varphi = -\frac{\pi}{2}}$$ Hence, the current lags behind the voltage. Analogously to the capacitor, the reactance is then: $$X_L = \frac{U_0}{I_0} = \frac{U_0}{\frac{U_0}{\omega L}}$$ Forming and shortening then yield for the resistance: $$\boxed{X_L = \omega L}$$