Volts

100,000 lines of force, cut once each second by one conductor, will generate a potential of one volt.
Reply to
Nog
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Originally from comparison to a standard chemical cell. Now, in the metric system as a relationship between the fundamental units of charge and energy (a given charge acquires a given energy when it falls through a volt).

Their internal resistance varies. As current passes through the battery, its internal resistance uses up some of the battery voltage by ohm's law. 9 volt alkaline batteries have an internal resistance on the order of ohms, so can dump only an ampere or so before their internal resistance uses up all of the 9 volts. A big lead acid battery has internal resistance on the order of milliohms, so it takes hundreds of amperes to drop all of its voltage.

The chemistry produces volts, the area of the plates determines the internal resistance.

Reply to
John Popelish

Volt: A unit of electromotive force. It is the amount of force required to drive a steady current of one ampere through a resistance of one ohm. Ref:

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All baterries have a voltage rating and a energy rating (typically measured in Amp-Hours or milliAmp-Hours). The energy rating will determine whether a

6V battery power a torch/flashlight or a motorcyle starter.

Current and power are needed to determine if a particular voltage is dangerous. For example, compare the jolt from a 3KVA transformer with 230V on the primary side (source) and 2300V secondary (load) to a charged up person at 23,000V (walking on carpet in winter) touching a door knob. You'll feel a shock if you touch the door knob (23,000V), but will survive the experience. If however you grab onto the 2300V terminals of the transformer, you would be medium-well in a matter of seconds.

Dwayne D. Chrusch

Reply to
Dwayne

How is a volt defined. If a 9 volt battery can start a car,creating a very large current how does it differ from a 9 volt battery? Is it due to the amount of ionised material available? Is it the amount of P.D between both terminals that creates the amount of voltage(pressure) available? If an object has a large amount of free electrons or extra ones does that mean the voltage potential would be higher? As for static electricity,is this a state of an object being ionised?

Thanks

Reply to
Music Man

Strictly speaking, the Ampere is the primary *defined* SI unit of electricity. All other units follow from there.

Although 1 Amp = 1 Coulomb / sec, that's a statement of equivalency to charge carriers. The older standard definition is (although I believe it's getting updated)

1 Amp is that current, when flowing in two rectilinear conductors of negligible cross section, 1 metre apart in vacuuo, will produce a force (due to magnetic fields, my comment) of 2 x 10^-7 Newtons between them.
Reply to
PeteS

That's one thing I never understood -- why the SI comittees never chose the Coulomb as the "fundemental" unit, but instead chose the Ampere. The Amp is a derived unit equivalent to the rate of charge, while charge itself is the 'more' fundemental unit.

They need to define a new unit of charge that is a mole of electrons (or charge), also -- but since all other common useage units (Farad, Henry, Ohm, etc.) are already in use, this new unit would only be academic.

JMO

Reply to
tlbs

It's one Joule per Coulomb. Imagine placing two plates out in the vacuum of space and providing a potential of one volt between them. Now take a cup with one Coulomb of electrons and gently empty it out very close to the negative plate. The electrons will accelerate away from the negative plate towards the positive one. When they are just about to hit the positive plate, their total potential energy (relative to the fixed plate) will be one Joule. Works the same if you used protons (much more massive) released at the positive plate, which will accelerate more slowly towards the opposite plate. Distance between the plates isn't important.

Given this, we have:

Volt = Joule/Coulomb Amp = Coulomb/Second Ohm = Joule-second/Coulomb^2 (a Joule-second is angular momentum) Henry= Ohm-second = Joule/Amp^2 = Joule-second^2/Coulomb^2 Farad= Coulomb/Volt = Coulomb^2/Joule

A nice result is that:

R*C = (Joule-second/Coulomb^2) * (Coulomb^2/Joule) = seconds

L*C = (Joule-second^2/Coulomb^2) * (Coulomb^2/Joule) = seconds^2

Which is why you look at the timing constants of RC as proportional to just that and the timing constant of an LC as proportional to SQRT(LC).

It's nice when the units work out.

Jon

Reply to
Jonathan Kirwan

I know. I've read the CRC handbook where it describes this. However, we humans do like to think in countable things and Coulombs are countable.

In any case, I wasn't trying to stay in SI units as should have been obvious from the various ways I wrote what I did. I was merely trying to get the point across that one doesn't have to accept that the units in electronics are defined only in terms of the other units, in some kind of circular definition. I didn't like the idea of defining volts in terms of amps and ohms, where ohms is defined in terms of volts and amps. Neither ohms nor volts gets any clearer, that way.

The electric units really do have physical meaning.

I remember reading something like that. Good to see it written. Thanks. So we'll call a volt, Joule/Amp-second, and when trying to determine volts as Joules/Coulomb, put our cup at one end of those wires and let the electrons drain into it for one second.

I wonder if the more "natural" idea of counting electrons as the primary SI unit (or protons) as being a Coulomb fell away to Amps because electrostatic methods of establishing the value depend too much on hard to control shapes and intervening materials.

In any case, thanks.

Jon

Reply to
Jonathan Kirwan

1 Volt is the results of 1 amp of current in a 1 ohm Resistor.

i also have explain this in different terms when educating youths

Voltage is like water moving at a set rate of speed. current is like the amount of water moving.

water coming out of a small pipe at high pressure is like having high voltage but low current! you can stop the water with out to much problem because the cubic feet per second isn't that great ! thus the weight of the water is not that heavy. now change the pipe to a large size, you now have a large volume of water! even at a slow moving rate (low voltage), it can knock you over due to the cubic feet of water hitting you! thus the weight factor. so this all comes down to this. given enough current (volume of the water) it can cause harm even at slow moving speeds (voltage).

the speed (Volts) and cubic feet of water(current) is the Electromotive force.

i have used this to explain to young one's about electricity and many to this day have learned well on this theory the difference between VOLTS and AMPS!

hope that helps you also.

Reply to
Jamie

^^^^^^^^^^^^^

** ** ******** *** *** ** ********** ***** **** ** ** ** ***** ** ** ** ** ** ***** ** ** ** ** ** *** ** ** ** ** ** * ** **** ** ** ** *** ********** *** ** ** ******** ***

NO!!! NO!!! NO!!! NO!!! NO!!! NO!!! NO NO NO NO NO!!!!!!!!!!!!

Voltage is _PRESSURE_ - it has nothing to do with "speed" (other than that more pressure can force more flow through a given resistance. See "Ohm's Law".)

Now, if you have a PSI gauge at the faucet, its pressure reading might decrease when you open the valve - that's because of the PRESSURE DROP CAUSED BY THE INTERNAL RESISTANCE OF THE WATER SUPPLY.

Another thing - a water pressure gauge measures PSIG (pounds per square inch gauge) or PSIA (pounds per square inch absolute) and its reference is intrinsic, or built-in - it's either atmospheric pressure, 15 PSIA, or zero (vacuum). With voltage, the reference is usually Earth ground or circuit ground or circuit common - a voltmeter measures the voltage between two points (AKA potential, or electromotive force (EMF)). Period. A water pressure measures the difference in pressure between the water in the pipe, and the air, which doesn't seem obvious at first, because we're embedded in it. The "other lead" of a water meter is the outside of the diaphragm, which is exposed to air. This is where the water pipe model of electricity kind of breaks down.

Yes, Now we're getting a little closer.

^^^^^^^^^^^^^

^^^^^^^^^^^^

Yes. Voltage is pressure. Rate of flow is current, like the current in a river.

NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO!!!!!!!!!!

Speed is just another aspect of flow rate. VOLTAGE IS PRESSURE!!!

Well, you've been wrong.

Well, STOP!

Don't worry, now that you've been corrected, it might.

Thanks, Rich

Reply to
Rich Grise

At the time, it was much easier to measure, and create repeatable standards for.

Probably still is.

It used to be defined electrolytically in terms of mass of silver deposited under defined conditions in a given time.

--
"Electricity is of two kinds, positive and negative. The difference
is, I presume, that one comes a little more expensive, but is more
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Reply to
Fred Abse

Electromotive FORCE, actually.

Force

----- = pressure area

Electricity has no analogous equivalent for pressure.

Reply to
JeffM

Rich, i bet you call you DICK in for short.

gets to the point very fast.

Reply to
Jamie

-------------------- Except that they don't exist.

-Steve

--
-Steve Walz  rstevew@armory.com   ftp://ftp.armory.com/pub/user/rstevew
Electronics Site!! 1000\'s of Files and Dirs!!  With Schematics Galore!!
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Reply to
R. Steve Walz

Voltage is a 'potential function'. It is a scalar function of position. Force, on the other hand, is a vector function of position.

If you think of a wire, and a voltage difference along the wire, you can see that the electric field is really being directed along the wire. So, if you take a charge, and move it along the wire from one point to another point, you'll do work on that charge because of the force imparted on the charge by the field. The amount of work you do will be the energy required. However, this amount of energy depends on the amount of charge you move. More charge = more force. Thus, by dividing out the amount of charge you used, you get a value which is independent of the amount of charge. This is the voltage.

It's formally defined as the negative of the path integral of the electric field between any two points. Since the electric field is simply the vector equal to force on a charge divided by the charge, it's easy to see that voltage is really just the difference in potential energy, divided by the charge used to measure the energy. It's units are joules/coulomb, which makes sense.

As far as analogies go, you think of the electric field as being like the acceleration of gravity. Gravity is a vector field, pointing downwards (the gravitational field). The resultant force on an object is the field at that point, times the mass of that object. The 'voltage' between two points, in this domain, would be the path integral of the field (ie, acceleration of gravity) along the path between those two points. In the gravitational realm, voltage is thus simply the difference in height times a constant (assuming a constant gravitational field). Given the right units, it would just be difference in height. The potential energy difference between these two heights is the mass of the object times this potential difference. Letting an object fall through this height will give you a corresponding amount of kinetic energy.

Thus, voltage is neither pressure nor force. It's a way to predict the amount of work you can get from an electric field.

The water analogy is that the electric field is like gravity, charge is like mass, the rate of change of mass along some path is current, and voltage is the difference in height between different pools. Voltage differences don't make the mass want to move; it's paths along the gravity vector that makes water move. Resistance is flow restriction of the path. Power obtainable is head (which is height) * flow rate. etc, etc.

--
Regards,
  Bob Monsen
Reply to
Bob Monsen

If you had equated pressure to voltage, your analogy would be quite a lot better. Not great, not true, but commonly used. You can have a voltage with no flow. This can be high or low. If you have no water flow- there is no "speed". Your analogy breaks down right away. Note that you are also involving a speed when you talk about cu ft/sec which comes down to mass*velocity at a given measurement point. You can also have high or low water current at a given high or low voltage. Your analogy breaks down immediately ,and, if the student has to learn more, he or she will have to unlearn a lot. The old, commonly used analogy is better than what you are presenting. It has its faults but all analogies do.

While Bob Monsen is right it is not absolutely necessary to start with field concepts, and basic circuit theory doesn't actually require these concepts so the pressurevoltage , current flow analogy goes a lot further conceptually than what you are using.

--
Don Kelly
dhky@peeshaw.ca
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Reply to
Don Kelly

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