Physical meaning of Impedance

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!).

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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.

Keep in mind that the capacitor is made of 2 conductive plates that are separated by an insulator. When the two plates are at the same potential, the density of charges is the same on the two side, thus no charges want to move. When there is a difference of potential between the two plates, charges are arriving on 1 side, or leaving the other. But no charges are moving through the insulator. It is like the charges on the highest plate are frightening the charges on the other one! But they never cross the insulator. And the thinner the insulator, the higher the capacitor effect. Now, what about the opposition current is facing in a capacitor. It´s quite simple. Imagine for a given voltage, there would be a given number of seats for charges for each plates (addition or suppression). As soon as the doors open, charges are randomly going (or leaving) to a seat. When there is choices in seats, charges flow is high. But when there are less free seats, charges are flowing slower. That´s how the opposition current faces is changing over time for a given voltage. The complex notation is the solution to write such changes in time. Finally, I suggest you try to imagine the physical meaning of capacitor impedance with a step in voltage instead of a sine. I hope my explanation helps you. If not, don´t give up.