ECN Forum Forums Technical Discussion Electrical Theory and Applications something i thought i knew

dP=dIdE (??????????), to simply.

1. P=EI
1. P+dP = (E+dE)(I+dI) = EI + EdI+IdE+dEdI so:
dP=EdI+IdE+dEdI am I wrong?

I'll freely admit, I'm a 3rd year student and I'm just starting to understand all of this theory. I know that there is inductive and capacitive reactance in a circuit that is shown by power factors. I have also never seen any of these differential equations or the AC equations shown by Steve. My question is this: Do the DC equations still correctly show the basic relationship between Power, Voltage, Current and Resistance, or does AC change all of this as Steve seems to be suggesting? If I'm missing something big, I would welcome someone pointing me in the right direction.

You are right. We only try very, very simply to describe, to understand these problems shortly. In real equations of course You have to take complex, resistance and reactance as circuits parameters and then count current i and angle(u,i), and then power=P+Q. But, hmm,..., how to that all describe clearly and shortly?
May be for a start it is easier to describe DC motors?

No Trekkie, you are not having a Blonde moment.
Why is it that PoCo's the world over use HV and EHV transmission and distribution systems?.
Simple, higher voltage equals lower current and that in turn means lower VxI losses, it also means that smaller CSA conductors can be used for a given line length (measured in miles, not feet).
One thing that is often forgotten at a basic level, is the fact that voltage (Or potential difference) is the pressure that drives the electrons through the conductor, the current is merely the speed at which them electrons travel.
This may sound crazy, but more push, less speed!.

One other thing, the use of the letter E as voltage, makes equations hard to read from this end, I suppose it's what you're used to.
I guess it comes back to EMF

Watt's Law and Ohm's Law hold true, but only for instantanous measurements we never make in the real world- whenever you start dealing with harmonics and power correction, the easy rules of thumb (EG, taking RMS voltage and multiplying by peak current) are not accurate. In practice, you won't actually need to integrate or use any calculus because we're working with constant frequencies like 60Hz (and 60Hz harmonics) with GREATLY simplifies things! (And yes, that wiseass diffeq I put down was wrong. Shame on me- should all be simply dt. If you have to work in the wonderful world of RF, acoustics, vibration, etc, all the frequencies are different and it can get excrutiatingly painful very quickly!)

For instance, lets say you had a resistive load of 1 Ohm and a capacitive load of 0.5j Ohms (j=i= the square roof of negative 1) in series. This would give you an equivilent complex impedance of (1+.5j) which would be 1.118<26.5° in phasor notation.

Now, the apprentice might put his handy-dandy fluke on this and say "aha, I've got 107A, 120V x107A = 12.8kW!" but he'd be wrong. In this case, you'd have 12.8kVA, which is simply RMS volts x peak amps taken from your meter, but to actually determine power, phase angle has to be considered.
E = 120<0° Volts
R = 1.118<26.5° Ohms
I=E/R= 120<0 / 1.118<26.5 = 107<-26.5° Amps

-26.5° is a PF of .88 leading.

E = 120<0° Volts
I = 1.118<-26.5° Amps
P=ExR= 120<0° x 1.118<-26.5° = 12.8 x cos(-26.5°) = 12.8 x .88 = 11.2kW.

Oops, stupid edit timeout. That last line should be I, not R:

P=ExI= 120<0° x 107 <-26.5° = 12.8k x cos(-26.5°) = 12.8k x .88 = 11.2kW.

Shouldn't that be RMS volts x RMS amps?*

*Assuming the meter is a true RMS reading meter.

Larry C

Yes, should be RMS amps, not peak.

Edit: see, this is why I like being able to go back and edit posts. grrr, stupid edit time-limit!

Thank you for taking the time to explain this to me! I just wanted to say how much I appreciate the guys that make up this forum. I've only been an electrician for 3 years now, and I've learned a lot about being one since I stumbled onto this forum.

If you stop learning, you are no longer an electrician