Why do capacitors and inductors have imaginary impedance?

[Hybird]

Consider an AC source with series elements (resistor,inductor, capacitor). In order to determine information from the circuit it is useful to consider impedance. So here's the question, total impedance is a complex number, and that is because the impedance of the cap and inductor are imaginary numbers. I'm tryin to find out why they are imaginary.

Any help?

 

[OlderDan]

Consider an AC source with series elements (resistor,inductor, capacitor). In order to determine information from the circuit it is useful to consider impedance. So here's the question, total impedance is a complex number, and that is because the impedance of the cap and inductor are imaginary numbers. I'm tryin to find out why they are imaginary.

Any help?

In a complex number representation, sinusoidal functions are written as complex exponentials. The imaginary numbers for the impedence of capacitors and inductors indicates that the voltage and current in these devices is 90 degrees out of phase. Voltage leads the current in an inductor and lags the current in a capicitor. The catchy little phrase "ELI the ICE man" can help you remember which is which.

 

[Corneo]

Well let's just talk about the impedance of an induction. The impedance is defined as followed

Z_L \equiv \frac {\mathbf{V}_L}{\mathbf{I}_L}

where V_L and I_L are phasors. Consider v_s(t) = A \cos (\omega t), then \mathbf{V}_L = A \angle 0. The current through an inductor is given by

i_L(t) = \frac {1}{L} \int_{t_0}^t v_L ~dt + I_0

If you work that integral out will you get i_L(t) = \frac {A}{\omega L} \sin(\omega t) = \frac {A}{\omega L} \cos(\omega t - \frac {\pi}{2}). So I_L = \frac {A}{\omega L} \angle -\frac {\pi}{2}

Then


Z_L \equiv \frac {\mathbf{V}_L}{\mathbf{I}_L} = \frac {A \angle 0}{\frac {A}{\omega L} \angle - \frac {\pi}{2}} = \omega L \angle \frac {\pi}{2} = j \omega L

The complex number j appears in the impedance of an inductor when converting from polar form to rectangular form. Hope that answered your question.