In general, the voltage across and current through a capacitor or inductor do not have the same form:
$$i_C(t) = C \dfrac{dv_C}{dt} $$
$$v_L(t) = L \dfrac{di_L}{dt} $$
Thus, in general, the ratio of the voltage across to the current through is not a constant.
However, recalling that:
$$\dfrac{d e^{st}}{dt} = s e^{st} $$
where s is a complex constant $s = \sigma + j \omega$, we find that, for these excitations only:
$$i_C(t) = (sC) \cdot v_C(t)$$
$$v_L(t) = (sL) \cdot i_L(t)$$
In words, for complex exponential excitation, the voltage across and current through are proportional.
Now, there are no true complex exponential excitations but since:
$$e^{j \omega t} = \cos(\omega t) + j \sin(\omega t) $$
we can pretend that a circuit with sinusoidal excitation has complex exponential excitation, do the math, and take the real part of the solution at the end and it works.
This is called phasor analysis. The relationship between a sinusoidal voltage its phasor representation is:
$$ v_A(t) = V_m \cos (\omega t + \phi) \rightarrow \mathbf{V_a} = V_m e^{j \phi}$$
This is because:
$$v_A(t) = \Re \{V_me^{j(\omega t + \phi)}\} = \Re\{V_m e^{j \phi}e^{j \omega t}\} = \Re\{\mathbf{V_a} e^{j \omega t}\} $$
Since all of the voltages and currents in a circuit will have the same time dependent part, in phasor analysis, we just "keep track" of the complex constant part which contains the amplitude and phase information.
Thus, the ratio of the phasor voltage and current, a complex constant, is called the impedance:
$$\dfrac{\mathbf{V_c}}{\mathbf{I_c}} = \dfrac{1}{j\omega C} = Z_C$$
$$\dfrac{\mathbf{V_l}}{\mathbf{I_l}} = j \omega L = Z_L$$
$$\dfrac{\mathbf{V_r}}{\mathbf{I_r}} = R = Z_R$$
(Carefully note that though the impedance is the ratio of two phasors, the impedance is not itself a phasor, i.e., it is not associated with a time domain sinusoid).
Now, we can use the standard techniques to solve DC circuits for AC circuits where, by AC circuit, we mean: linear circuits with sinusoidal excitation (all sources must have the same frequency!) and in AC steady state (the sinusoidal amplitudes are constant with time!).
So my question is why does the magnitude of the ratio of the complex
voltage to the complex current now suddenly carries a physical meaning
(if my understanding is correct, it is the resistance which can be
measured in ohms, just like in the resistor).
Remember, the complex sources are a convenient fiction; if there were actually physical complex sources to excite the circuit, the phasor representation would by physical.
The physical sources are sinusoidal, not complex but, remarkably, we can mathematically replace the sinusoidal sources with complex sources, solve the circuit in the phasor domain using impedances, and then find the actual, physical sinusoidal solution as the real part of the complex time dependent solution.
Here's an example of the physical content of impedance:
Let the time domain inductor current be:
$i_L(t) = I_m \cos (\omega t + \phi)$
Find the time domain inductor voltage using phasors and impedance. The phasor inductor current is:
$\mathbf{I_l} = I_m e^{j\phi}$
and the impedance of the inductor is:
$Z_L = j \omega L = e^{j\frac{\pi}{2}}\omega L$
Thus, the phasor inductor voltage is:
$\mathbf{V_l} = \mathbf{I_l} Z_L = I_m e^{j\phi}e^{j\frac{\pi}{2}}\omega L = \omega L I_m e^{j(\phi + \frac{\pi}{2})}$
Converting to the time domain:
$v_L(t) = \omega L I_m \cos (\omega t + \phi + \frac{\pi}{2})$
Note that the magnitude of the impedance shows up in the amplitude of the sinusoid and the phase angle of the impedance shows up in the phase of the sinusoid.
I'd like to ask what is the meaning of the value obtained from X(jω)
with certain frequency ω
Consider for a moment, the synthesis equation where we 'construct' $x(t)$ out of a weighted 'sum' (integral) of the orthonormal basis functions of time: $e^{j\omega t}$
$$x(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty}\mathrm{d}\omega\:X(\omega)\:e^{j\omega t}$$
Here, it is clear that $X(\omega)$ is the amount or weight of $e^{j\omega t}$ that goes into 'constructing' $x(t)$ from these basis functions. We haven't specified what $X(\omega)$ is but we assert that, for a large class of $x(t)$, there is an appropriate $X(\omega)$ such that the above holds.
As you may have already concluded, there is a way to find $X(\omega)$ given an $x(t)$ which is
$$X(\omega) = \int_{-\infty}^{\infty}\mathrm{d}t\:x(t)\:e^{-j\omega t}$$
which (left as a fun exercise for the reader) can be verified by substituting the later expression into the former expression.
Since, in the first expression, we're integrating with respect to $\omega$ which has units of $\mathrm{s}^{-1}$, it must be that $X(\omega)$ has the units of $x(t)$ multiplied by $\mathrm{s}$.
Why there is a factor of $1/T$ difference between the units of Fourier
series and Fourier transform ?
If we were to make a discrete approximation of the first expression, it would be something like
$$\tilde{x}(t) = \frac{1}{2\pi}\sum_{n = -\infty}^\infty \frac{2\pi}{T} \:X(n\frac{2\pi}{T})e^{jn\frac{2\pi}{T}t} = \sum_{n = -\infty}^\infty \:\frac{X(n\frac{2\pi}{T})}{T}e^{jn\frac{2\pi}{T}t} = \sum_{n = -\infty}^\infty \:a_n e^{jn\frac{2\pi}{T}t}$$
which is periodic with period $T$ (thus the tilde over). Using the usual method of integrating the product of both sides with $e^{-jm\frac{2\pi}{T}t}$ over a period $T$, we arrive at
$$\int_0^T\mathrm{d}t\:\tilde{x}(t)\:e^{-jm\frac{2\pi}{T}t} = Ta_m$$
since the integral on the right hand side vanishes for $n \ne m$. And so,
$$ a_n= \frac{X(n\frac{2\pi}{T})}{T} = \frac{1}{T}\int_0^T\mathrm{d}t\:\tilde{x}(t)\:e^{-jn\frac{2\pi}{T}t}$$
So, the $\frac{1}{T}$ comes from the fact that the $a_n$ are the 'sampled' Fourier transform divided by the period $T$ as shown above.
Best Answer
For concreteness, assume a linear circuit element or network with driving point impedance $Z_a$:
$V_a(j \omega) = Z_a(j \omega) \cdot I_a(j \omega)$
Let:
$I_a(j \omega) = 1 \Leftrightarrow i_a(t) = \delta(t)$.
Now, it is easy to see that the $z_a(t)$ is the time domain voltage response (divided by 1A) due to a current impulse.
This is analogous to the impulse response of a system, $h(t)$, in linear time invariant (LTI) systems theory.