Hi I'm doing an online course, and some of it covers basic electrical theory. Now, I'm not an educator, or an engineer, but please, can somebody tell me if this text seems to be written by a 12 year old: "Resistance is the real part of impedance, where as reactance is the imaginary part." Have a look at the clip to get the full paragraph.

l Sounds right to me. The word real in the context of impedance and AC circuits is not the same as "the opposite of fake." Likewise the word imaginary in AC theory is not the same as saying, "make believe". Theory question about "j" imaginary unit (AC circuit analysis)

okay, this is new to me! Thanks for the enlightenment. (I owe someone an apology, if only an imaginary one.)

I have a very feeble grasp on these things. Can you give a dummies definition of real/imaginary in AC theory? oopps, just noticed your link! you rock!

On a related note, I excitedly sat down last night to watch a documentary on Nikola Tesla. It ended up being lame. -Mike

Is that the one that is more about the troubles he had through his life, and less about some of his accomplishments? That guy doesn't get anywhere near the credit he deserves. When he invented or developed something, he did the heavy lifting. Some other famous, or infamous "inventors" of the day, had people they paid to do that for them.

Impedance is a complex number consisting of a "real" part which is straight up resistance (sometimes called DC resistance) and an "imaginary" part which is reactance. Reactance is only applicable to AC current and is the "resistance" of the capacitance and inductance in the circuit to AC signals How much reactance a capacitor or inductor presents is frequency dependent. Don't get caught up in the "imaginary" terminology, reactance is quite real, "imaginary" numbers are just the mathematical representation of the effects of reactance in a circuit (the amount the current in the circuit is out of phase with the voltage). The i (or j if you're an engineer) represents the square root of -1 and is just a way of expressing complex numbers. You can use the pythagorean theorem to convert a complex number to a polar representation (magnitude and phase angle). In terms of cable impedance you don't really need to worry about the complex representation as you can easily figure out the impedance of a cap or inductor at a particular frequency. So, as a quick example let's ignore inductance and look at a pickup/cable combination. We can also assume that the impedance of whatever the cable is plugged into on the other end is high enough to not affect the pickup/cable combination much. In this situation the impedance of the circuit is the series DC resistance of the pickup plus the capacitive reactance caused by the separation of the conductors in the cable. If the DC resistance of the winding is 10k Ohms and the capacitance is, say 2nF we can figure out the impedance at a particular frequency via Zcap= 1/(2*pi*f*C). The 2*pi*f term is often shown as omega (like a small w, can't figure out how to insert that). So at 20Hz, the cap is 1/(6.28*20*2exp-9) = 3.98M Ohms; at 2kHz it's 39.8k and at 20kHz it's 3.98k. When you look at how the signal is transmitted through the cable, the series resistance of the pickup winding and the shunt to ground of the cable capacitance creates a voltage divider that varies with frequency. The more the impedance of the cap drops (with increasing frequency) the more of the signal is shunted to ground by the cap (or dropped by the series resistance, can look at it both ways). So if we have a 1V signal, at 20Hz it is basically unaffected by the capacitance. By the voltage divider rule Vout= Vin*(R2/R1+R2) where R2 is the cap's impedance we see that Vout at 20Hz is 0.997V. At 2k it is 0.8V and at 20k it is 0.28V. So you can see how that capacitance acts as a low pass filter. It progressively shunts off more and more signal as the frequency goes up. You can also just use the resistance and capacitance to come up with the corner frequency of the filter (where it is -3dB down from Vin (approx .707V in this example). f=1/(2*pi*R*C) so the -3dB point of that filter is at 7960Hz. I think I may have overcomplicated this. Does it make any sense?

Another difference between the real and imaginary impedance components is that resistance dissipates power, reactance does not. This leads to an additional term for power: When a load is resistive, the power is given by voltage x current, and is termed real power. When a load is reactive, the product of voltage x current is termed apparent power. DC resistance often refers to "resistance as measured by an ohmmeter". With the very small diameter of magnet wire used to wind pickups, the AC resistance is virtually the same as the DC resistance all the way up to 20 KHz. In the case of heavier gauge wire, the AC resistance can be significantly higher than the DC resistance. Either way, AC resistance is still resistance, ie, the real part of impedance. -

The capacitive/resistive voltage divider described here also explains how that tone control on your bass works. I always find it helpful if I can relate the theory back to something practical.