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something i thought i knew #173035 12/31/07 07:25 PM
Joined: Jun 2004
Posts: 220
T
trekkie76 Offline OP
Member
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!!

Tools for Electricians:
Re: something i thought i knew [Re: trekkie76] #173043 01/01/08 06:00 AM
Joined: May 2005
Posts: 247
T
techie Offline
Member
P=IE

Re: something i thought i knew [Re: techie] #173046 01/01/08 07:46 AM
Joined: Jul 2007
Posts: 265
W
wire_twister Offline
Member
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


Jimmy

Life is tough, Life is tougher when you are stupid
Re: something i thought i knew [Re: wire_twister] #173049 01/01/08 09:37 AM
Joined: Jun 2004
Posts: 220
T
trekkie76 Offline OP
Member
Thanks guys. I didnt read far enough into the book. Once I got to the Watts section it all came back. Thanks again

Re: something i thought i knew [Re: trekkie76] #173052 01/01/08 11:00 AM
Joined: Nov 2005
Posts: 823
J
JoeTestingEngr Offline
Member
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

Last edited by JoeTestingEngr; 01/01/08 11:01 AM.
Re: something i thought i knew [Re: JoeTestingEngr] #173084 01/02/08 09:33 AM
Joined: May 2005
Posts: 41
L
Last Leg Offline
Member
Just remember power factor. Motors run most efficiently at the voltage for which they were designed, not at +/- 5%.

Re: something i thought i knew [Re: Last Leg] #173098 01/02/08 05:00 PM
Joined: Oct 2007
Posts: 8
G
gabrielpacyna Offline
New Member
And...with nominal shaft torque, I think.
The nominal rating is the optimal using of el. energy.

Re: something i thought i knew [Re: gabrielpacyna] #173102 01/02/08 06:38 PM
Joined: Sep 2006
Posts: 8
B
bb Offline
Junior Member
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.

Re: something i thought i knew [Re: bb] #173104 01/02/08 07:18 PM
Joined: Oct 2007
Posts: 8
G
gabrielpacyna Offline
New Member
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.

Re: something i thought i knew [Re: gabrielpacyna] #173126 01/03/08 07:27 AM
Joined: Mar 2005
Posts: 1,213
S
SteveFehr Offline
Member
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

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