What's the physical meaning of the imaginary component of impedance?

Question

As you know, impedance is defined as a complex number.

Ideal capacitors: $$ \frac {1} {j \omega C} \hspace{0.5 pc} \mathrm{or} \hspace{0.5 pc} \frac {1} {sC} $$

Ideal inductors: $$ j \omega L \hspace{0.5 pc} \mathrm{or} \hspace{0.5 pc} sL $$

I know that the reason why they 'invented' the concept of impedance is because it makes it easy to work with circuits in the frequency domain (or complex frequency domain).

However, since in real-life circuits both voltages and currents are real numbers, I'm wondering if there is any actual physical meaning behind the imaginary component of impedance.

Answer 1

The physical 'meaning' of the imaginary part of the impedance is that it represents the energy storage part of the circuit element.

To see this, let the sinusoidal current $i = I\cos(\omega t)$ be the current through a series RL circuit.

The voltage across the combination is

$$v = Ri + L\frac{di}{dt} = RI\cos(\omega t) - \omega LI\sin(\omega t)$$

The instantaneous power is the product of the voltage and current

$$p(t) = v \cdot i = RI^2\cos^2(\omega t) - \omega LI^2\sin(\omega t)\cos(\omega t) $$

Using the well known trigonometric formulas, the power is

$$p(t) = \frac{RI^2}{2}[1 + \cos(2\omega t)] - \frac{\omega LI^2}{2}\sin(2\omega t) $$

Note that the first term on the RHS is never less than zero - power is always delivered to the resistor.

However, the power for the second term has zero average value and alternates symmetrically positive and negative - the inductor stores energy half the time and releases the energy the other half.

But note that $\omega L$ is the imaginary part of the impedance of the series RL circuit:

$$Z = R + j\omega L$$

Indeed, via the complex power S, we see that the imaginary part of the impedance is related the reactive power Q

$$S = P + jQ = \tilde I^2Z = \frac{I^2}{2}Z = \frac{RI^2}{2} + j\frac{\omega L I^2}{2} $$

Thus, as promised, the imaginary part of the impedance is the energy storage part while the real part of the impedance is the dissipative part.

Answer 2

There is a physical meaning behind the imaginary component of the impedance. You can re-cast the complex impedance $Z = R + jX$ (using engineering's notation $j$ for the imaginary unit) in polar form to get $Z = |Z|\exp(j\phi)$. $|Z|$ is the magnitude of the impedance, and scales the amplitude of the current to get the amptlitude of the voltage. $\phi = \arctan(X/R)$ is the phase shift by which the current lags the voltage.

The current and the voltage are themselves expressed as complex quantities. The voltage and the current at any given point are real numbers, but in an A/C circuit, they will both oscillate in magnitude. The amplitude I referred to in the previous paragraph is the amplitude of that oscillation. Those two oscillations are typically not in phase with each other: the current won't reach its maximum value at the same time as the voltage. You can usually take the zero-crossing of your voltage as the reference point in time, and describe the phase shift of the current as being relative to that.

Answer 3

Imaginary components in physics often mean phase shifts. In this case the impedance is like a resistance, but it kicks in when the current is changing by messing with its phase.

Answer 4

In this case, the magnitude is telling you how to scale your input signal, and the argument is telling you how to phase shift it.

Complex numbers usually represent 'amplification' and 'twist'.

So, say, 1 means 'leave it the same', 2 means 'double it', 0.5 means 'halve it', i means 'one quarter turn', -1 means 'one half turn', -3i means 'triple it and give it a three-quarter turn'. (1+i)/sqrt(2) means 'one eighth of a turn', etc.

Incidentally, this is why i*i = -1 in the first place. Two quarter turns successively are a half turn.

And the famous formula e^i*pi=-1 is really saying 'grow at right angles to yourself for as long as it takes to make a half turn, and you'll have turned round'!

Answer 5

The imaginary impedance as mentioned above, is the energy storage part. When a circuit element has a purely imaginary impedance, like, an inductor or a capacitor, in a harmonic AC circuit, the current through these elements is out of phase of the voltage across them by 90 degrees.

Now, power dissipated by a circuit element is simply $V\cdot I$ (the dot product of the two phasors). Since $V$ is perpendicular to $I$, power dissipated is $0$. What that means is that imaginary impedence does not dissipate energy outside the circuit.