Ok I have utterly confused myself and I need some help. Why is it that we see amps go down as voltage goes up, such as motors, ballasts?
I am looking over a math book from my college days, and it says that voltage and amperage are directly proportional.

I thought the whole reason for increasing voltage to an end product was to reduce amps saving on wire size, etc?

Am I missing something stupid? I feel like I have writers block, you know when you can almost remember something? it is driving me nuts!!

P=IE

The way it was explained to me long years ago, Amps and volts are directly proportioal and opposite like being on either end of a see-saw when one goes up the other goes down. hope this clears things up

Thanks guys. I didnt read far enough into the book. Once I got to the Watts section it all came back. Thanks again

You should think of the current and voltage as being inversely proportional in cases where the power is the constant. That would include changing winding configurations in motors or transformers and switching power supplies. Think of a 24 volt control transformer that can have the primary windings in parallel for 120 VAC or in series for 240 VAC. Whether it might draw 1 amp in the first case or .5 amp in the second case, each winding still has 120 volts across it and .5 amp flowing through it. But yet the second case will likely be more efficient as a system because of less I squared R loss in getting to the point of use.

In the case of motor starting currents you're waiting for the motor to spin up and for the counter EMF to reduce winding currents. Here you shouldn't think of the voltage at the motor where you are observing it but at some point upstream before all the I^2 R loss that is the cause of the lower measurements.

Now if you have a purely resistive load, the current will vary directly with the voltage. Just don't think of nichrome heating elements as fitting into that classification since the resistance increases with temperature.
Happy New Year!
Joe

Just remember power factor. Motors run most efficiently at the voltage for which they were designed, not at +/- 5%.

And...with nominal shaft torque, I think.
The nominal rating is the optimal using of el. energy.

I=E/R
For a fixed resistance, as voltage increases, current also increases. However, the reason that we increase voltage in order to decrease current is that with multi-tap ballasts or motors, the higher voltage windings have 4 times the resistance. This decreases the current by half.

In my opinion we have to consider that motor try to maintain the nominal revolution.
And the all machine load the motor the constant mechanical power. If we lower the voltage motor try still maintain the mechanical power and the current has to rise and vice versa.
It is connected with E inducted in the motor in opposite to the U supply voltage.
It is most observed when the motor starts, current is the biggest. I=(U-E)/R, R is summary windings and line supply.
Sorry for my language, I am Polish.

Oh come now, guys- you all know better than to look at a meter and believe P=IE and E=IR, that's as much a half-true myth as the Bohr model of the atom or newtonian physics. Not to mention the OP apparently fully understood the answer after the first response- that's just not right! This is the theory forum, needs more differential equations!

dP=dIdE
dE=dIdR

Or at least use the AC version instead of DC for rule-of-thumb calculations:
P=I^2R
P=E^2/R

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

I'm new to this forum. Lot's of interesting and good stuff here. A moot point, but, as I remember, and I AM a dinosaur, current is not the speed at which electrons travel, but a measure of the number of electrons that are going by in one second, where one amp is about 6.3 X 10**18 electrons per second.

Cheers,

Andy

Andy,
Welcome along mate!
There are no dinosaurs here, man.
Yeah, you are right, I sort of mucked that explanation up a tad.