DiverDan,

I hope that it doesn't look like we are forming a bandwagon to call you 'wrong'; your notes are overall quite informative and useful. I think that several of us are picking up on points that we disagree with to a certain extent; I just wanted to note the overall positive nature of your writing...and then dive into one of those little points *grin*

I am going to call that an obfuscatory equation, math which hides the truth behind algebraic manipulations. It might be the right equation for a computer program to use, but it doesn't make things clear to humans.

Also, I take issue with calling that equation 'exact', even if the IEEE does so. At best it is a much better approximation to reality. But it presumes sinusoidal applied voltage and linearity of the various interacting materials, and can only be close, not exact. But that is me picking very small nits [Linked Image]

The most basic voltage drop equation is simply Ohm's law, E=I*R where E is the voltage dropped across the resistance, and I is the current flowing. You simply substitute in the resistance of the wire in question, and this gives you the voltage drop. The resistance of the wire is determined by the resistance per unit length of the wire and total length of the run, and is the same resistance parameter used in the IEEE equation. This basic equation is accurate enough for most voltage drop calculations that an electrician will encounter, but it misses on significant points.

In particular, in AC circuits, the voltage drop will include both the DC resistance of the wire and the AC inductance of the wire, and the current flow is not necessarily in phase with the voltage.

But with a simple little trick, we can rescue Ohm's law. We simply replace all of the 'real' quantities (single numbers representing values) with 'complex' quantities (numbers of the form x + jy where j is the square root of -1) The combination of resistance and inductive reactance (R and X) becomes the 'complex impedance' R + jX. The combination of in phase current (D) and out of phase current (Q) becomes the complex current D + jQ. Q is positive for _leading_ reactive current, eg in capacitive circuits.

Now we simply apply Ohm's law using these 'complex' quantities. To deal with 'complex' multiplication, remember that j is the square root of -1, so whenever you get j^2 just replace it with -1

E = I * R becomes E = (R + jX) * (D + jQ) = RD - XQ + j(RQ + DX)

Getting more complex, but we see that the core is still E = I * R, ordinary Ohm's law, but when we expand out the calculation we see the reactive component of impedance and the power factor of the current flow. By writing the equation in complex form, we get the intuitive: Voltage drop is caused by the current flowing through the impedance.

If I have gotten my signs correct, then the above equation should be reducible to the IEEE equation (meaning that they should be mathematically the same). All those trig functions (cos(theta), sin(theta), arccos(power factor) )?? they come out of the fact that rather than reporting complex current in the x + jy form, current is described in terms of total current and power factor. The trig functions are essentially converting amps and power factor into the D + jQ representation of current flow. I have not checked in detail, but the IEEE equation might collapse if the load has a _leading_ power factor; eg current flowing into a capacitor. The arccos(PF) equation would treat a leading PF the same as a trailing PF, but the real circuit would respond differently.

Sorry to run-on. Here is a nice tutorial that I found on using complex math to describe impedance: http://www.ibiblio.org/obp/electricCircuits/AC/AC_11.html
The entire textbook is worth a look.

-Jon

P.S. On the voltage drop heating the wire issue: since both voltage drop and wire heat are caused by the same resistance, consideration of one is closely related to consideration of the other. Most of the time I consider them separate issues, since voltage drop depends upon the _total_ resistance of the wire, but doesn't care about resistance per unit length, whereas wire heating depends upon resistance per unit length, but doesn't care about the distance or _total_ resistance.

A 1 foot length of 12ga wire has a resistance of 1.589 milli-ohms. Run 500A through this wire and you have a voltage drop of 0.8V, generally considered acceptable in a 120V circuit. Of course, not only would the insulation degrade on this wire, but the copper itself would quickly fail [Linked Image]