# Circuits of AC - impedance

I have a question regarding impedances. While it is intuitive that $R$-element has impedance $R$, it's not so easy for me to figure out impedances of $L$ and $C$.
Here is my, probably incorrect, reasoning. For element $C$ following differential equation is satisfied: $i(t) = C \frac{\mbox{d} u}{\mbox{d} t}$. Our input is $u$, and utput is $i$. By transforming this equation we have
$$\frac{U(s)}{I(s)} = G(s) = \frac{1}{sC}$$
Since we cosider only sinusoidal inputs we know, that output will also be sinusoidal of the same frequency, but different magnitude and phase, so it's fine to use Fourier instead of Laplace, then let $s:=j \omega$, then
$$G(s)=\frac{1}{j \omega C} = -j \frac{1}{\omega C}$$, which is indeed equal to impedance of element $C$.

Now - is that a coincidence, or the actual reasoning behind impedances and whole method of solving circuits based on them? If so, why take $u$ as an input, and $i$ as an output? Is it to make analogy to well known $R=\frac{U}{I}$?

If I tried to use not necessarily sinusoidal voltage and current, would I use the more general form: $\frac{1}{sC}$? Would it work?

• The R element has resistance, L and C have impedance Feb 1 '17 at 17:23
• One could say that impedance is a complex number which consints of resistance - $$Re(Z)$$ and reactance $$Im(Z)$$.
– gabe
Feb 1 '17 at 17:37
• Impedance is defined as V/I. So the result is no coincidence. It appears that you are studying this in a formal setting (university). As you continue on, I think everything will make more sense. Feb 1 '17 at 21:17
• Yes, the Laplace form (s domain) can be used for transient analysis as well as steady state analysis. At least that is what I remember from my university days. Feb 1 '17 at 21:20
• Impedance of a pure inductance or capacitance has no resistance, its only when you combine inductance or capacitance or both with resistance that you get a complex number with real (resistance component) and an 'imaginary' (+/- j) component. Feb 2 '17 at 12:57