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| Volts, amps, watts |
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| Cold Fusion |
 Posted: Mon May 12, 2008 11:11 pm Post subject: Volts, amps, watts |
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Forum Ph.D.
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Why cant any website give a real definition of these? They always avoid the main question of, what exactly are these things. So, what are they? I want to know this in terms of the electrons and what they are doing.
Also, in a parallel circuit, the more resistors you add the lower the total resistance gets; why is this exactly?
More questions to come....
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| bit4bit |
| Posted: Tue May 13, 2008 4:17 am Post subject: |
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They are simply units/dimensions same as metres, seconds or kilograms.
Volts are a unit of potential difference (voltage) and are a measure of the amount of energy per unit charge. p.d.=E/q
Amps are a measure of current flow, which is the rate of change of charge in a circuit, having the instantaneous value of i=dq/dt. or for a constant (time invariant) signal i=q/t for any time interval.
Watts are a measure of electrical power. They can be given by P=vi, since power is the rate of change of energy, and:
v*i=(E/q)*(q/t)=Eq/qt=E/t
For the parallel circuit question, resistors in parallel share a common potential difference, since each one is connected atone end to the positive rail, and the other end to the negative rail (or ground). Crrent flows in each branch, and the total current flowing through the network is the sum of currents in each branch. By Ohm;s law, i=v/r, so if:
itotal=i1+i2+i3 + ...
and the voltage is the same across each branch, then:
V/rtotal=V/r1 +V/r2 +V/r3 + ...
The voltages cancel, and you are left with
1/rtotal=1/r1 +1/r2 +1/r3 + ...
So the resistance decreasing is a result of this equation.For a qualitative idea, the more branches you have, the more paths current has to flow.
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| Harold14370 |
| Posted: Tue May 13, 2008 8:27 am Post subject: |
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This is probably best explained by analogy to water flowing in a pipe. In this analogy, the electrical charge, which is measured in coulombs, is like the volume of water. One coulomb is equal to 6.24E18 electrons. An ampere is 1 coulomb per second. If you have a wire that is carrying an ampere of current, there are 6.24E18 electrons flowing past any given point in the wire every second.
If you pump water up a hill to a different elevation, it gains energy. This energy is equal to the pressure difference from the lower to higher elevation, multiplied by the volume. Energy=Pressure * Volume. If you pump the water in a continuous stream, the power required is the energy per unit time, or Power = Pressure * Volume/time = Pressure * flow rate.
In our electrical analogy, voltage is like pressure, and flow rate is like current. Power, as measured in watts, is equal to Voltage *current, since current is a measurement of the flow rate of electrons. So one watt is equal to one volt multiplied by one ampere.
The reason electrical resistance goes down when you have parallel paths is the same reason the resistance in a piping system would go down if you add parallel flow paths. |
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| William McCormick |
| Posted: Tue May 13, 2008 6:45 pm Post subject: Re: Volts, amps, watts |
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| Cold Fusion wrote: |
Why cant any website give a real definition of these? They always avoid the main question of, what exactly are these things. So, what are they? I want to know this in terms of the electrons and what they are doing.
Also, in a parallel circuit, the more resistors you add the lower the total resistance gets; why is this exactly?
More questions to come.... |
Volts are the pressure of electrons, trying to get to the opposite polarity pole. Amps are the intensity/induction of electrons in a flow caused by volts. That is why volts times amps equals watts.
It is the same as air pressure. If you know you have a certain air pressure, and you have a 3/8 line, you know how many cfm (cubic feet a minute). CFM would be the watts in the analogy.
Amps create more volume of electrons through intensity and because of ambient radiation, induction. The field around the wire gets larger as amps go up. That is how an amp meter works, by reading the induction field created by the higher voltage potential, created by the higher amperage.
Watts are the volts times amps, through what ever medium or device you are powering or flowing through, at the given ohms. Ohms limit volts and amps.
Ohms can change, upon the addition of higher or more voltage to a circuit.
There are different types of heating elements. Some heating elements are constant ohm heating elements, others are variable ohm heating elements.
The constant ohm heating elements tend to blow the fast acting fuses that should be used on heating circuits, quickly and effectively. The reason is that if you get a voltage increase. That can easily occur during the loss of a building neutral. The fast acting fuses will blow, due to an increase in amperage, due to an increase in voltage. This will protect the heaters the equipment and the circuits.
Induction motors will often only read a fraction of an ohm resistance when you place your ohm meter across the wires feeding the induction motor.
Sometimes it is because of a start winding engaged by the centrifugal start winding mechanism that disengages the start winding after a certain rpm is reached and the run winding can take over. Upon starting the start winding is engaged electrically. So you are reading the ohms of two circuits in parallel. Once started or with the start winding disengaged you only read the run windings.
Sometimes just the run windings that are the only windings in three phase motors, will show almost no ohms as well.
However once the motor is started, the momentary almost infinite draw of the magnet coil, is reduced by increased run ohms. Once the rotor lock situation is remedied by the motor spinning, and becoming in sync with the hertz that set the rpm. The ohms reduce. (Edited Ohms increase).
Some heaters actually lower (edited raise) in ohms as they heat up. They are usually marked with the start amperage that they will draw. However once they are heated you can check the amperage and it is often a fraction of the start amperage.
I wrote this program some years ago. You are welcome to use it and share it. It is really cool because you can enter in any two values and see how they effect the other two values, that are automatically given.
I saw one similar but it was not setup the way I wanted it. I sometimes create twenty occurrences of the program while I am working out something.
http://www.Rockwelder.com/Electricity/setupwir.exe
Sincerely,
William McCormick
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Warning: Any information contained in this post could be part of a conspiracy to make you stupid. So only use it if you understand it. Use at your own risk.
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Last edited by William McCormick on Tue May 13, 2008 8:21 pm; edited 2 times in total |
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| Harold14370 |
| Posted: Tue May 13, 2008 7:27 pm Post subject: |
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| You would do well to ignore everything Mr. McCormick has written. About the only thing he got right was volts times amps equals watts. CFM is not like watts, it is like amps. Amps are not induction. Ohms do not limit volts. When a motor gets up to speed the ohms do not reduce; the current does. You do not get a voltage increase when you turn on a heater. |
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| William McCormick |
| Posted: Tue May 13, 2008 8:06 pm Post subject: |
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| Harold14370 wrote: |
| You would do well to ignore everything Mr. McCormick has written. About the only thing he got right was volts times amps equals watts. CFM is not like watts, it is like amps. Amps are not induction. Ohms do not limit volts. When a motor gets up to speed the ohms do not reduce; the current does. You do not get a voltage increase when you turn on a heater. |
Harold if you increase the pressure in a 3/8" air line, that will increase the CFM. Even though the ohms, or the restricting 3/8" line has not changed. In this scenario the amperage would change. Because the 3/8" line did not change.
If you increase the volts to a circuit with set ohms. The amperage will go up as well as the voltage, and that of course will increase the wattage.
Or in the analogy the CFM. Wattage is total throughput. CFM is total throughput.
This is because of intensity/induction in the wire. The wire literally expands. There is a field around the wire that grows. An induction field. It inducts electrons into the wire from ambient radiation. And intensifies the flow. With a wider and wider field of induction.
What in particular do you not agree with?
Sincerely,
William McCormick
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| Edmund |
| Posted: Wed May 14, 2008 2:03 am Post subject: |
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| Harold14370 wrote: |
| You would do well to ignore everything Mr. McCormick has written. |
I agree with you
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| William McCormick |
| Posted: Wed May 14, 2008 3:04 pm Post subject: |
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| Edmund wrote: |
| Harold14370 wrote: |
| You would do well to ignore everything Mr. McCormick has written. |
I agree with you |
I am going to venture that you will not dispute a single thing I wrote and corrected last night. I just switched up two terms. A cut and paste error. You can see that the rest of the post explains what I meant.
I used to manufacture three phase heating equipment. So I have some experience in it.
That is exactly how it is. Therefore for whatever reason you dispute it, you will be fighting reality.
I only have to fight unreality.
Sincerely,
William McCormick
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| Edmund |
| Posted: Wed May 14, 2008 3:26 pm Post subject: |
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Why would i bother enquiring into parts of what you wrote.
I have asked questions in other topics about thigsn you said but you fail to give answers, I just agree that people should ignore everything you say.
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| Cold Fusion |
| Posted: Wed May 14, 2008 3:54 pm Post subject: |
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Its funny that I understand quantum theory yet do not understand electric circuits so well.
I am curious about the usage of watts because the audio industry mainly uses it to describe the volume output of its equipment. Wouldn't it be more useful to use amps instead? (As in, this is a 6 amp speaker)
Thanks for the EQ's and definitions.
My physics teacher used the water/ pipe example, but I see problems with it. Doubling the number of signal paths does not reduce the resistance considerably. Having two resistors instead of one would only appear to lighten the load per resistor, and divide the work in two, which would still allow you to come out with the same resistance. There must be something else to it.
I found a website that seems to make sense of this. Can you verify its accuracy?
http://www.eskimo.com/~billb/miscon/voltage.html
If you increase the voltage, are you also increasing the speed of the electrons? (I heard this once)
Why is it that by having a very high voltage in power lines, you are creating less resistance than one with a lower voltage?
Why does AC cause less resistance than DC?
_________________
Forget all that you know, achievements can only be accomplished by starting from nothing and selectively applying facts that are purely objective and absolutely necessary.
"Two things are infinite: the universe and human stupidity; and I'm not sure about the universe."
"Great spirits have always found violent opposition from mediocrities. The latter cannot understand it when a man does not thoughtlessly submit to hereditary prejudices but honestly and courageously uses his intelligence"
-Einstein
http://boinc.berkeley.edu/download.php
Use your computing strength for science! |
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| Harold14370 |
| Posted: Wed May 14, 2008 4:42 pm Post subject: |
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| William McCormick wrote: |
| Harold if you increase the pressure in a 3/8" air line, that will increase the CFM. Even though the ohms, or the restricting 3/8" line has not changed. In this scenario the amperage would change. Because the 3/8" line did not change. |
Yes, but you said cfm would be the watts in the analogy. It's not. Watts is a unit of power. CFM is a measurement of flow rate. To get energy you would have to multiply the CFM by the pressure.
| Quote: |
| This is because of intensity/induction in the wire. The wire literally expands. |
No, not unless it heat up, and that would only be thermal expansion.
| Quote: |
| There is a field around the wire that grows. An induction field. It inducts electrons into the wire from ambient radiation. |
Where did you get this bizarre idea? The wire is already full of electrons. The ambient radiation does not have electrons in it. How many times do you have to be told that? |
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| William McCormick |
| Posted: Wed May 14, 2008 4:42 pm Post subject: |
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| Harold14370 wrote: |
| You would do well to ignore everything Mr. McCormick has written. About the only thing he got right was volts times amps equals watts. CFM is not like watts, it is like amps. Amps are not induction. Ohms do not limit volts. When a motor gets up to speed the ohms do not reduce; the current does. You do not get a voltage increase when you turn on a heater. |
Harold when you talk about watts. You are talking about a load. I would agree with you that in an open circuit a resistor will not lower volts.
However under a load, the volts are lowered and this causes the amps to lower. As well. Volts control amps.
Let me give you a real example. I have a large elemental resistor. It has 38.4 ohms, resistance across its leads. It is feed from a 480 volt power supply. So each of its leads are receiving 480 volts. It will draw 12.5 amps and output 6000 watts of heat.
Now I add in another identical resistor in series. It would be the same as adjusting a rheostat to control a heating load. Except it is simpler and more clear to just add in another resistor.
When I add in this second identical resistor into the circuit. The total load will be 3000 watts, at 480 volts, at 6.25 amps, and the ohms created will be 76.8 ohms.
But that means now that each resistor/heater is receiving 240 volts, across its leads, each heater has a resistance of 38.4 ohms each. Each heater is pulling 6.25 amps each, and outputting 1500 watts each.
So by adding a resistor or increasing the resistance you lower the voltage.
If one of the those resistors was a constant ohm heater, you can see how it would cut the volts in half to the heater.
Sincerely,
William McCormick
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Warning: Any information contained in this post could be part of a conspiracy to make you stupid. So only use it if you understand it. Use at your own risk.
http://www.Rockwelder.com |
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| William McCormick |
| Posted: Wed May 14, 2008 4:43 pm Post subject: |
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| Edmund wrote: |
Why would i bother enquiring into parts of what you wrote.
I have asked questions in other topics about thigsn you said but you fail to give answers, I just agree that people should ignore everything you say. |
Where is this mythical question?
Sincerely,
William McCormick
_________________
Warning: Any information contained in this post could be part of a conspiracy to make you stupid. So only use it if you understand it. Use at your own risk.
http://www.Rockwelder.com |
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| Edmund |
| Posted: Wed May 14, 2008 4:55 pm Post subject: |
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| William McCormick |
| Posted: Wed May 14, 2008 5:02 pm Post subject: |
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| Harold14370 wrote: |
| William McCormick wrote: |
| Harold if you increase the pressure in a 3/8" air line, that will increase the CFM. Even though the ohms, or the restricting 3/8" line has not changed. In this scenario the amperage would change. Because the 3/8" line did not change. |
Yes, but you said cfm would be the watts in the analogy. It's not. Watts is a unit of power. CFM is a measurement of flow rate. To get energy you would have to multiply the CFM by the pressure.
| Quote: |
| This is because of intensity/induction in the wire. The wire literally expands. |
No, not unless it heat up, and that would only be thermal expansion.
| Quote: |
| There is a field around the wire that grows. An induction field. It inducts electrons into the wire from ambient radiation. |
Where did you get this bizarre idea? The wire is already full of electrons. The ambient radiation does not have electrons in it. How many times do you have to be told that? |
Cubic feet a minute of air is measured at sea level atmospheric pressure.
So when you increase pressure in a 3/8" air line. You get more atoms of air per linear section/unit lenth of air line.
That means that with the higher pressure, it is going to move air faster. That is accounted for in the pressure part of the formula.
But you also get an increase in intensity due to compaction of molecules of air. So that each linear inch of air in the 3/8" air line, that passes through the outlet, will carry a higher intensity of air atoms with it. At whatever the velocity caused by pressure.
That intensity or compaction caused by pressure/volts in electricity, is the amperage. Amperage is the density of electrons in the wire.
Watts tells the whole story of volts, pressure/velocity, and amps intensity/induction.
I would not make this stuff up.
If there was no intensity there would be no need for amps in the formula.
Sincerely,
William McCormick
_________________
Warning: Any information contained in this post could be part of a conspiracy to make you stupid. So only use it if you understand it. Use at your own risk.
http://www.Rockwelder.com |
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