Basic electronics student befuddled by volt/Coulomb relationship

[blingo]

I'm trying to get my head around the definition of a volt as it relates to a Coulomb. To my understanding a Coulomb of charge is defined as 6.25 x 10 to the 18th excess electrons or protons. A volt is defined as the potential difference present when 1 joule of force moves 1 Coulomb through a resistance of 1 ohm in 1 second. Does the charge of 1 Coulomb inherently produce 1 joule of attractive or repulsive force? Does 1 Coulomb of charge equal 1 volt? If someone understands this it would be a tremendous help.

 

[BeeBop]

I'm trying to get my head around the definition of a volt as it relates to a Coulomb. To my understanding a Coulomb of charge is defined as 6.25 x 10 to the 18th excess electrons or protons. A volt is defined as the potential difference present when 1 joule of force moves 1 Coulomb through a resistance of 1 ohm in 1 second. Does the charge of 1 Coulomb inherently produce 1 joule of attractive or repulsive force? Does 1 Coulomb of charge equal 1 volt? If someone understands this it would be a tremendous help.

Think of a Coulomb as a quantity, and Voltage (Electromotive Force) as a pressure, or measure of pressure.

One coulomb of charge Needs a pressure of one volt, to move it through a one ohm resistance in a time of one second.

 

[Torben]

BeeBop took the words out of my mouth. The analogy isn't quite perfect but it's good enough to go with for now. Coulombs are more like gallons and voltage is more like PSI.

Say you have a hose. That's like a resistor. The pressure (voltage) of the water combined with the diameter of the hose (resistance) determines how much water (current) you can push through the hose in one second. If you make the hose diameter bigger (lower the resistance) you can push more water with the same pressure, or else push the same amount of water with less pressure.

If you need to use the smaller hose, you can raise the pressure if you need to push more water through.

This is why we wind up with Ohm's Law and its various rearrangements: voltage = current times resistance; current = voltage divided by resistance, etc. You can't change one value without the others changing. If you keep the resistance the same but raise the voltage, the current rises. If you keep the voltage the same but raise the resistance, the current drops. If you keep the current the same but lower the voltage, the resistance must drop.

Jeez. I guess BeeBop didn't steal all my words or I'd have shut up by now. I hope I haven't made this more confusing. Again, the analogy isn't perfect but it works for me (but I'm a hobbyist, not an EE).


Cheers,

Torben

 

[BeeBop]

Torbin, I didn't take anything...

Very well put!

 

[Lagroht]

I am a bit puzzled about this subject myself, what do you guys think of this line of reasoning?

electrons and protons attract each other, by convention 1 electron has a charge of -1 and a proton has a charge of +1.
(wikipedia: elementary charge)

This attraction is a force, which can be denoted in Newton (force required to give 1 kg an acceleration of 1 m/s^2).

1 joule = 1 N * M, in other words the amount of energy to accelerate an object with a force of 1 Newton, over a distance
of 1 meter. (of course if the actual object you apply the force is heavier than 1 kg you will not get the 1 m/s^2
acceleration, but if the object is in fact 1 kg, you will)

and if you keep on supplying energy to fuel your force application you can express this in watts (1 w = 1 joule /s)

now note that the charge of 1 coulomb equals the charge of 6.25 * 10^18 electrons and that 1 volt equals 1 joule / 1 coulomb.

since 1 electron has a given charge, so do 6.25 * 10^18 electrons, but what can be different is the amount of force applied to them;
this is determined by the existence of an equal or unequal amount of protons on the other size of the conduit, this is what determines
the amount of potential energy of a given charge point. (in other words the number of joules per coulomb).

"One Coulomb (C) is the amount of charge such that a force of 9.0×10^9 N occurs between two point like objects with charges of 1 C separated by a distance of 1 m."
(from wikipedia)

So if you look at 1 joule / 1 coulomb = 1 volt, what are we talking about if we talk about 14 volts potential difference between points A and B?

I guess we talk about 12 joules of potential energy spread out over 6.25 * 10^18 electrons caused by a surplus of electrons at 1 of the points
relative to the number of protons at the other point. In really the point probably will not contain exactly 6.25 * 10^18 electrons, but are we only talking
about a fraction of joules per coulomb which yields the same voltage.

Also note that this indicates that given a fixed number of elektrons in a given point the voltage associated with that point is strictly relative to any other point.

In other words: the number of joules per coulumb of a given point is different given the charge of whichever other point you compare that point with.

I guess the wikipedia definition is talking about the situation where 1 point has 6.25 * 10^18 electrons and the other has 6.25 * 10^18 protons. Apparently this
generates 9.0×10^9 N of force. Quite massive, so I guess in normal electronics we are talking about much lower potential differences.

 

[MikeMl]

I'm trying to get my head around the definition of a volt as it relates to a Coulomb. ...

All of the water analogies aside, a Coulomb is only indirectly related to a Volt; it is directly related to the Amp as in: A Coulomb is the amount of charge that passes a point in a circuit carrying one Amp each second.