Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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.

share|improve this question
3  
Converting to polar notation instead of real/imaginary is easier to interpret. The angle in polar notation tells you the amount of phase picked up at a given frequency. The phase picked up in a circuit is very important in control theory. –  Chris Mueller Apr 23 at 22:48
1  
However, since in real-life circuits both voltages and currents are real numbers... I really don't want to sound like Wikipedia, but I think that this is a great place to say "citation needed". –  AndrejaKo Apr 24 at 6:40

4 Answers 4

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.

share|improve this answer

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.

share|improve this answer

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

share|improve this answer

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'!

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.