Ohm's Law

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Jraef said:
Actually from a basic level, you can't "increase" current. By that I mean that current is a function of the load. A load PULLS current, you don't PUSH current into something.

However in the resistive circuit examples above, if the resistance stays the same, then increasing the voltage will cause the load to pull more current.
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I dissagree the voltage pushes the current and the load restricts the flow in proportion to ohms law.
 
memyselfandI said:
I meant no harm folks, I just asked a basic Ohm's Law question. It's good to think about things like this...it keeps us sharp.
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I truly believe in accuracy through volume and lead poisoning isnt allways through paint chips. M16A2 manual.
 
barbeer said:
Not that I know of......I was always taught that voltage goes up, current down.
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(I wrote this a couple of months ago, in response to a similar question.)

This is a common error, because of the way we explain Ohm's and Watt's laws. For a given amount of power, it's true that voltage and current vary inversely. That takes making power a constant, and voltage a variable.

However, that's just math. You must remember what are and are not considered the variables in the real world. To keep power constant as voltage changes, you must change the impedance of the load as the voltage changes.

The equipment impedance and the system voltage are usually considered constants. But, when discussing a variable voltage and a constant impedance, current will change proportionately with voltage, not inversely.
 
memyselfandI said:
Can you increase voltage and current simultaneously?
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I would say no to the "simultaneously" part of this question. Current can increase as a result of a voltage increase, however they are not simultaneous. I would consider this a trick question. The chicken or the egg?

This kinda gets into the discussion about voltage pushing the current or the load pulling the current. One causes the other, they do not happen simultaneously.
 
Nuh-uh!

Nuh-uh!

ELA said:
I would say no to the "simultaneously" part of this question. Current can increase as a result of a voltage increase, however they are not simultaneous. I would consider this a trick question. The chicken or the egg?

This kinda gets into the discussion about voltage pushing the current or the load pulling the current. One causes the other, they do not happen simultaneously.
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In a resistive circuit, current is proportional to the voltage across the resistance at all times--simultaneous. That is,

i(t) = v(t)/R

In a reactive circuit, current leads or lags the voltage--not simultaneous.

Now, there may be miniscule delays in a resistive circuit, say in the femtosecond range, but we can't measure them, so for all practical purposes these delays are zero.
 
ELA said:
I would say no to the "simultaneously" part of this question. Current can increase as a result of a voltage increase, however they are not simultaneous. I would consider this a trick question. The chicken or the egg?

This kinda gets into the discussion about voltage pushing the current or the load pulling the current. One causes the other, they do not happen simultaneously.
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Voltage does indeed push current, and in resistive circuits, the resultant current change is considered instantaneous.

Unless, of course, you consider the speed of light to be a relatively slow event.
 
I find the water in pipeline analogy very useful...

Water pressure = Voltage level

Water flow rate = Current

Frictional loss and pipe size = Resistance...., (mind you lesser diameter pipe offers more resistance)

Too less the pipe size ,...the user at the ned does not get enough water ( low voltage for sparkies)

Now you can do the rest....

Cheers.
 
LarryFine said:
Unless, of course, you consider the speed of light to be a relatively slow event.
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That must account for rattus's femtosecond. Two turtles crossing the road were killed in a head-on collision. The cause of the accident couldn't be determined because the only witness, a snail, provided the following statement: "I'm so confused, it all happened so fast!"

chaterpilar said:
I find the water in pipeline analogy very useful...
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chaterpilar: Location: Saudi Arabia
What do you know about water out there in Saudi Arabia?:grin:
 
rattus said:
...In a reactive circuit, current leads or lags the voltage--not simultaneous. ...
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So, one might say that with a capacitive reactance load, the current would have to be a "relative" of the voltage so as the arrive the previous day.:roll:

carl
 
rattus said:
Now, there may be miniscule delays in a resistive circuit, say in the femtosecond range, but we can't measure them, so for all practical purposes these delays are zero.
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I agree that a femtosecond is a short time and that delays can be near zero.
LarryFine said:
Voltage does indeed push current, and in resistive circuits, the resultant current change is considered instantaneous.

Unless, of course, you consider the speed of light to be a relatively slow event.
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I also agree that the speed of light is fast.



Not to dwell on this but my point is that if you believe, as you stated, that voltage pushes current then you cannot have it both ways.

If one is dependent upon the action of the other then they cannot be simultaneous.

The time may be so short that we cannot measure it "with todays instrumentation" but that does not make them simultaneous.

In order for them to be simultaneous they would have to be independent of each other.

I do not think either of us has some deep understanding that the other is missing. We just view things differently.
 
All Ohm's Law states is a relationship between (at the basic level) volts, amps and Ohms.

*Any* of them can go up or down in any manner, simultaneous or not, and the equation still holds. The manner is dictated by the system or the device and is only explained by the Law.

In most real applications however, as the current increases the voltage will decrease. The reason the current increases is due to a reduction in resistance and (so long as the source and keep up) an increase in current flow will be, for the most part, proportional to the increase in Watts produced.

Although it is rare, it is possible to increase voltage and current simultaneously. It probably is rare because without a corresponding change in resistance, such a simultaneous increase would cause damage to the system.

We like it when the volts go up and the amps go down. Conductors are sized in amps and the smaller the conductor, the less money needed to buy them.
 
Guys -
I suggest it is time to get a little closer to the physics:-?

In a few applictions, and even then, only over a very limited voltage range does the current go down when the voltage go up. Jack up the voltage on a motor very far and the current definitely goes up - way up.

Voltage goes up, current goes up - is not rare. Most devices work that way. For example, all of the resistors in K8's receiver/transmitter/linear, all electric heaters, diodes, SCRs, vacuum tubes - okay not tunnel diodes:cool: (again only over a narrow range), DC motors.

If you look at Vrms/Irms, even reactive loads act the same way.

V = IR is true. V = IZ is true, but they are now vectors.

carl
 
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mivey said:
Two turtles crossing the road were killed in a head-on collision. The cause of the accident couldn't be determined because the only witness, a snail, provided the following statement: "I'm so confused, it all happened so fast!"
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Q: What did the snail say after he crawled onto the turtle's back?

A: Wheeeee!!!
 
080619-2204 EST

Normal electrical circuit analysis will assume steady-state conditions, and a linear invartient load unless otherwise stated. This means as the source voltage increases so will the source current increase. The same is true if the source is a current source instead of a voltage source.

Motors with a mechanical load are not generally a linear load, but resistors are. Switching power supplies are not generally a linear load. Incandescent lamps are not a linear load, but do exhibit an increase in current with an increase in applied voltage. Somewhat tend to be a constant cuurent load. A negative temperature coefficient thermister is not a linear load, its resistance decreases with increasing current.

A number of persons that responded to this question nead a basic course in linear circuit theory. Also the assumptions to a question need to be defined.

.
 
Much Ado!

Much Ado!

crossman said:
I'm not sure that is a good analogy at all. To me, it is more correct to say that voltage "pushes" current than to say that a load "pulls" current. The voltage is active, whereas a resistor is passive. I guess the whole thing is rather arbitrary though.
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Does it really matter whether the electrons are pulled or pushed? Maybe both?
 
Yes - resistance is futile ...

Yes - resistance is futile ...

I'm really tired, so I didn?t read anything since page 2, but this thread intrigues, I?m hear to learn.

What about getting onto impedance rather than resistance to apply Ohm's law to AC circuits? And stick to 3 phase? These lighting panel references are dreary. So is DC (the motors have out priced themselves anyway).

Then bring torque into the equation; example - applying voltage to the stator in a squirrelly inductance motor at any given frequency to match the torque it demands.

Too much ? maybe amps go up (saturation ? where else are you going to put the energy)
Too little ? maybe amps go up (starvation - for want of a better term)

Why don?t we invent a term just for fun? (Is there one Jreaf?)

Maximum power flow? Nah ?. Doesn?t work, already assigned

Optimal power flow? ?Hmmmm ? help me out, the correct voltage at any instant matching the torque requirements at any given instant in a load that transduces that voltage to optimal stator flux thereby minimizing current (in amperes which although interpreted as flow is truthfully a measure of heat not unlike a calorie or joule perhaps and ? and ?. ZZZZZZZ
 
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