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_knight_
10-03-2006, 09:47 PM
I was imposed with a question, I hope someone can enlighten me on. It goes as follows...

If you decrease resistance in a circuit will the amperage increase or decrease? I answered decrease, and to my surprise I was wrong.

Now the reasoning for my answer is: If you have a loose connection or a charred wire your going to have increased resistance on this circuit, along the same circuit with this issue, you will also have increased amperage due to the increased resistance on the line.....this even goes (to my understanding) along the same line as with windings and so forth, if I'm checking a compressor and I have a high resistance reading due to a bad connection or contaminated windings I'm going to have an increased amp draw across this circuit. Like I said this is to my understanding...please fill me in otherwise.

Now....The reason I was wrong is because according to Ohm's law if you have a higher rated resistor in a circuit according to the formula (example 12volts/10ohms that equals 1.2amps change that resistor to 20ohms and you have .6amps) than your amperage rating goes down.

Now please, for the love of God what am I missing...is there a difference between a resistor and resistance in itself, please help.

cg2
10-03-2006, 10:41 PM
Simplest way ive found to keep it straight: The least possible resistance is dead short. Whats the amperage going to be?

_knight_
10-03-2006, 10:49 PM
Originally posted by cg2
Simplest way ive found to keep it straight: The least possible resistance is dead short. Whats the amperage going to be?

Right so if you keep lowering your resistance until you have none at all your going to have a dead short, if you keep adding resistors to a circuit it shouldn't keep lowering your amp rating....right? I mean this is part of ohms law, so I'm trying to figure out where they're wrong and where I'm right or vice versa.

-80guru
10-03-2006, 10:54 PM
http://www.csgnetwork.com/ohmslaw.html. The link did'nt work dont go to it.

bdclark
10-03-2006, 11:27 PM
what you are missing is Ohm's Law itself.

Simply stated: "The current in a circuit is directly proportional to the applied voltage and inversely proportional to the circuit resistance.

bdclark
10-03-2006, 11:28 PM
what you are missing is Ohm's Law itself.

Simply stated: "The current in a circuit is directly proportional to the applied voltage and inversely proportional to the circuit resistance."

In series cicuits if you keep adding resistors, then yes your amperage will continue to drop for the entire circuit.

In parallel circuits the circuits total resistance is always lower than the value of any of the resistors in any leg off of that circuit. A parallel circuit has more paths for the cuurent to flow, thus you have less resistance to the cicuits total current flow. Make sense?

[Edited by bdclark on 10-03-2006 at 11:37 PM]

_knight_
10-03-2006, 11:42 PM
having the parallel resistors makes sense, but having them in series does not.

But again my question is....why is my amperage through the roof when I have a high resistance across a bad connection, I thought that the resistance caused the high amperage, or is it the lack of emf flowing through thats causing that? thanks for all of the replies

[Edited by ididntdoit on 10-03-2006 at 11:49 PM]

bdclark
10-03-2006, 11:48 PM
well, in series you only have one path that the current can flow. If you stack resistor upon resistor you increase the resistance within that single path circuit because the resistance can only be applied to that one path of current, thus lowering the total amount of current within that path. any clearer or are my words still muddy?

_knight_
10-03-2006, 11:50 PM
Originally posted by bdclark
well, in series you only have one path that the current can flow. If you stack resistor upon resistor you increase the resistance within that single path circuit because the resistance can only be applied to that one path of current, thus lowering the total amount of current within that path. any clearer or are my words still muddy?

loud and clear, thank you

bdclark
10-03-2006, 11:53 PM

[Edited by bdclark on 10-03-2006 at 11:59 PM]

bdclark
10-04-2006, 12:11 AM
yes, a lack of emf/ voltage causes electrons not to flow.
no, high resistance does not cause high amperage.
yes there is a difference between resistor and resistance.

I just thoroughly read. Are you reading the resistance with the disconnect off? I hope you are. Not to insult your intelligence, but you should be taking amp reading with the unit on, and ohm readings with the unit off. And the way you are stating this makes it sound as if you are taking ohm readings with the disconnect closed.

[Edited by bdclark on 10-04-2006 at 12:19 AM]

keepitsimplestupid
10-04-2006, 12:28 AM
Let me try:

R = rl/A;

r = rho = a material property called resistivity
l = length
A = cross-sectional area
R = Resistance

As the cross-sectional area of the conductor gets smaller the resistance increases. As the length gets longer the resistance increases. r for copper is lower than r for Aluminum.

resistance is a property of a "resistor" and other things that you can measure. It has units of ohms.

Now lets look at the current draw stuff. If you had a 3 phase motor and measured the current in each leg they would be equal usually. If one leg had a corroded terminal, that leg would read a lower current and the others higher becaus e it isn't using it's fair share. This assumes that the motor is turning.

Motors have a property called impeadance (Z) because the voltage is AC and the motor has inductance. Z = Sqrt ( R^2 + (Xl - Xc)^2)), so that when there is no capacitance and no inductance Z = R. So you are comparing apples to oranges when you comapre a DC "resistance" of a motor to an AC impedance of a motor. In other words Vac/Iac of a running motor will not equal the DC Resistance of the winding.

Ohms law, where V=IR has to be applied carefully in AC circuits because your assuming V and I are both sinusoidal and in phase. That equation is really V = I*cos(theta) * R for sinusoids. If V is an arbitrary waveform and I is an arbitrary waveform and the circuit is purely resistive then V and I must be True RMS values. Most meters are average respnding and TRUE RMS reading which is not the same as a TRUE RMS meter.

If you used what you were taught, you would get the right answer. Elementary courses use simplifications. Knowing too much in an elementary course is a bad thing.

Your problem using V=IR and solve for I, you get I=V/R, then you look at it and say as the denominator gets smaller, I gets bigger. done.

Are you sorry you asked? Did I confuse you more?

bdclark
10-04-2006, 12:44 AM
Correct me if I am wrong but I thought that inductive reactance canceled out capacitive reactance due to one leading 90* and one lagging 90*...or is that what you are saying and I just didnt understand what i just read?

jc47
10-04-2006, 12:49 AM
But again my question is....why is my amperage through the roof when I have a high resistance across a bad connection, I thought that the resistance caused the high amperage, or is it the lack of emf flowing through thats causing that? thanks for all of the replies

'Ohm's law' only works for circuits (or parts of circuits) that contain ONLY linear resistors. If you add anything else to the circuit besides linear resistors, then the answer to the question "If you decrease resistance in a circuit will the amperage increase or decrease?" could go either way.

When you say "why is my amperage through the roof when I have a high resistance across a bad connection," I am guessing that you are talking about a circuit with a three-phase motor in it. This happens because the resistance in the circuit causes the voltage across the motor to drop; in most situations, three-phase AC motors will draw more current as the supply voltage drops. As more current flows through the motor the voltage drops across the bad connection increases proportionally and the problem gets worse.

keepitsimplestupid
10-04-2006, 12:50 AM
You have it backwards. Ever been in a building which has a power factor correction system, it uses banks of capacitors to cancel out the motor loads. (X subscript l - X subscript c)

bdclark
10-04-2006, 01:00 AM
guess we learn something everyday

_knight_
10-04-2006, 07:10 AM
Originally posted by bdclark
yes, a lack of emf/ voltage causes electrons not to flow.
no, high resistance does not cause high amperage.
yes there is a difference between resistor and resistance.

I just thoroughly read. Are you reading the resistance with the disconnect off? I hope you are. Not to insult your intelligence, but you should be taking amp reading with the unit on, and ohm readings with the unit off. And the way you are stating this makes it sound as if you are taking ohm readings with the disconnect closed.

[Edited by bdclark on 10-04-2006 at 12:19 AM]

lol...I know I know a dumb question groups me with the dumb people, but no, I dont have the power on when checking resistance, and yes I do have the power on when checking amperage, and yes i am making sure that the motor is running before I check amperes...thanks though

_knight_
10-04-2006, 07:13 AM
Originally posted by jc47

But again my question is....why is my amperage through the roof when I have a high resistance across a bad connection, I thought that the resistance caused the high amperage, or is it the lack of emf flowing through thats causing that? thanks for all of the replies

'Ohm's law' only works for circuits (or parts of circuits) that contain ONLY linear resistors. If you add anything else to the circuit besides linear resistors, then the answer to the question "If you decrease resistance in a circuit will the amperage increase or decrease?" could go either way.

When you say "why is my amperage through the roof when I have a high resistance across a bad connection," I am guessing that you are talking about a circuit with a three-phase motor in it. This happens because the resistance in the circuit causes the voltage across the motor to drop; in most situations, three-phase AC motors will draw more current as the supply voltage drops. As more current flows through the motor the voltage drops across the bad connection increases proportionally and the problem gets worse.

Thank you, that is absolutely where I'm getting confused at, that makes perfect sense now, thanks

RichardL
10-04-2006, 04:08 PM
This is one "Cool" question.....
But I must digress from the Ohm's law application because it is superceeded he by "Other" factors..
The question as posed, involves a given motor that is running, and I will assume it is also under a load..
This is "Not" a strip-heat bank which would of course involve a unity power factor..
It is an "Inductive-Motor", running at a given load, and it "Will" require a "Given amount of "Power" to achieve this....
Assume that you install a resistor of infinite value (Hi-Ohms)...
The motor will continue to run albeit pulling considerably greater amperage. The motor output would be reduced to 66% of rated Horsepower and if load exceeds that, overheating and burnout is immenent...
"However"....Resistance less than infinite would still "Decrease" the amperage in the "One" leg involved, but the "Power" to the motor would remain the same which would involve an increase in the 2 legs not having the resistor installed...
Remember the definition of how an induction motor operates by the very man that invented them...(Nick-Da'man)
"An induction motor is nothing more and nothing less than a transformer with a rotating secondary"
Sooooo...
The answer to the question as posed is that the current "Will" increase in the legs without the added resistor, and will decrease in the leg with the added resistor....
And yes...3ph. motors run quite well when one leg is disconnected after motor is up to speed and the motor is loaded to <66% of its rated HP.