What's a Joule?

[Chris Williams]

I am teaching myself from Thomas L. Floyd's "Principles of Electric
Circuits". So far (chapter 4) it seems that he has done a good job
describing all of the units and how they should be visualized except
for Joules (energy.) This he pretty much just says is "the amount of
available energy" but without making it any clearer.

It would appear that fundamentally everything boils down to 1.) the
quantity of electrons (Q), 2.) time (t), and 3.) energy (W.)

Q = Total quantity of electrons
t = time spent doing something with the electrons
W = "the amount of available energy"

So current is easy:
I = (number of electrons) / (time)
Rate at which electrons were moved (through a wire)

But voltage is problematic:
V = (W?) / (number of electrons)
Rate of "the amount of available energy" to the number of electrons
....which isn't terribly clear

Reading over the text a couple of times, and I am thinking that
voltage is the electrical version of "pressure"...?
So in water/piping terms, pressure would be a function of a) the
amount of water and b) the amount of "push" it had to make it want to
go into a pipe. Push would then be a product of gravity, mechanical
pumping, etc. So assuming that this is true then W == "Push."

Pressure = "Push"(kg) / Water(cm^2) for instance

So this allows me to understand a Volt

Voltage = "Push"(joules) / Electrons(coulomb)

But what is the electric version of "push"?
My current guess is that this is the total "negativity" or
"positivity" of a terminal on a power source. Which--in the case of a
battery say--would be the ratio of positively charged atoms to neutral
atoms in a positive electrode, or the number of free electrons to the
total number of neutral atoms in the negative electrode...?

Is this reading correct?

Thank you,
Chris Williams

 

[Jonathan Kirwan]

On 27 Oct 2004 02:07:56 -0700, thesagerat@yahoo.co.jp (Chris Williams) wrote:

snip
It would appear that fundamentally everything boils down to 1.) the
quantity of electrons (Q), 2.) time (t), and 3.) energy (W.)

Q = Total quantity of electrons
t = time spent doing something with the electrons
W = "the amount of available energy"

So current is easy:
I = (number of electrons) / (time)
Rate at which electrons were moved (through a wire)

or, the way I prefer to think of it,

I = dQ/dt

Instantaneous change in Coulombs divided by the instantaneous change in time.
(When it is expressed using finite values, then these must by definition amount
to average quantities and are never really precise for any instant in time.)

But voltage is problematic:
V = (W?) / (number of electrons)
Rate of "the amount of available energy" to the number of electrons
...which isn't terribly clear
snip

Joules per Coulomb, yes. But... Think of it this way....

Set up two plates in a vacuum, somewhere in an otherwise empty universe.
Provide 1 Volt of potential difference across these plates (we'll figure out
what that is in a moment) and set a Coulomb's worth of electrons over near the
surface of the one that is more negative, so that they are accelerated towards
the other plate. Eventually, they will travel across the intervening space and
will strike the other plate. When they do, the net energy of the impact will be
1 Joule. Release two Coulombs to start and the net energy of the impact of that
many will be two Joules. And so on. And it doesn't matter how far apart the
plates are moved. If they are close together, the electrons will be accelerated
more and take less time to arrive, but their final velocity will be the same
when they hit and the resulting energy of the impact will be the same. Further
away, they will take much longer and be accelerated more slowly, but they have
much further distance over which to gain their velocity and they will arrive
again with the same impact energy. Hence, Volts is expressed as Joules/Coulomb
and doesn't need to include the distance between the plates.

Don't forget that:

Ohm = Joule-second/Coulumb^2 = Watt/Amp^2 = Henry/second

Watt = Joule/second

Henry = Joule-second^2/Coulomb^2 = Joule/Amp^2 = Ohm-second

Farad = Coulomb^2/Joule = Coulomb/Volt

If you then look at an "RC time constant," this is:

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

Nice, eh? Time, as you might expect. Also, take a look at the SQRT(L*C), which
is:

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

Once again, time! Dimensional analysis to the rescue.

Jon

 

[John Larkin]

On 27 Oct 2004 02:07:56 -0700, thesagerat@yahoo.co.jp (Chris Williams)
wrote:

But what is the electric version of "push"?
My current guess is that this is the total "negativity" or
"positivity" of a terminal on a power source. Which--in the case of a
battery say--would be the ratio of positively charged atoms to neutral
atoms in a positive electrode, or the number of free electrons to the
total number of neutral atoms in the negative electrode...?

That's a good model: the voltage on a conductive object represents the
density of electrons, where zero volts means the number of electrons
just equals the number of protons.

So voltage is charge density on a conductor. Measured in volts.

Current is the rate of flow of charge from one conductor to another,
in amps. If there's a conductive path between the charged things, the
voltage difference tends to push charges along that path; more voltage
difference, more push.

Power is the current flow multiplied by the voltage difference between
the conductors involved; watts.

Energy is the time integral of power; watt-seconds, aka joules.

Two nearby conductors at different voltages constitute a charged
capacitor, which stores some joules of energy. If current is allowed
to flow between them long enough for the voltage difference to vanish,
the stored joules will have been dissipated as power over the time it
took to discharge the system. If a resistor was used as the conductive
path, those joules would have heated up the resistor, which is where
the stored energy went.


John

 

[William J. Beaty]

thesagerat@yahoo.co.jp (Chris Williams) wrote in message news:<da271e02.0410270107.6175782c@posting.google.com>...

Pressure = "Push"(kg) / Water(cm^2) for instance

So this allows me to understand a Volt

Voltage = "Push"(joules) / Electrons(coulomb)

Nope. Electromagnetic energy (Joules) are nothing like a "push."

Voltage is not pressure, but it's connected. Voltage is more like
the depth of water. Potential difference between conductors is
like the height difference between two lakes: connect them and
there will be a flow. Voltage is also like the height of a hill
up which we roll a boulder, with the charge being the boulder's mass.
In that case the potential energy of the boulder is "Joules."
The higher the boulder (i.e. the higher the voltage,) the more
Joules are stored as potential energy... and the greater the
mass, the greater the energy. Joules = height * mass,
and Joules = Volts * Coulombs.


Another note: whenerver circuits are involved, the energy is
electromagnetic energy, and it's made of e-fields or b-fields
(or both.) In other words, the Joules have a particular location.
The Joules stored in a capacitor are stored in the insulating gap.
The Joules transmitted along a wire are in the space surrounding
the wire.

 

[Don Kelly]

"Jonathan Kirwan" <jkirwan@easystreet.com> wrote in message
news:6upun09l2r38i0al3bmveo8k8mg6v3hj61@4ax.com...

On 27 Oct 2004 02:07:56 -0700, thesagerat@yahoo.co.jp (Chris Williams)
wrote:

snip
It would appear that fundamentally everything boils down to 1.) the
quantity of electrons (Q), 2.) time (t), and 3.) energy (W.)

Q = Total quantity of electrons
t = time spent doing something with the electrons
W = "the amount of available energy"

So current is easy:
I = (number of electrons) / (time)
Rate at which electrons were moved (through a wire)

or, the way I prefer to think of it,

I = dQ/dt

Instantaneous change in Coulombs divided by the instantaneous change in
time.
(When it is expressed using finite values, then these must by definition
amount
to average quantities and are never really precise for any instant in
time.)

But voltage is problematic:
V = (W?) / (number of electrons)
Rate of "the amount of available energy" to the number of electrons
...which isn't terribly clear
snip

Joules per Coulomb, yes. But... Think of it this way....

Set up two plates in a vacuum, somewhere in an otherwise empty universe.
Provide 1 Volt of potential difference across these plates (we'll figure
out
what that is in a moment) and set a Coulomb's worth of electrons over near
the
surface of the one that is more negative, so that they are accelerated
towards
the other plate. Eventually, they will travel across the intervening
space and
will strike the other plate. When they do, the net energy of the impact
will be
1 Joule. Release two Coulombs to start and the net energy of the impact
of that
many will be two Joules. And so on. And it doesn't matter how far apart
the
plates are moved. If they are close together, the electrons will be
accelerated
more and take less time to arrive, but their final velocity will be the
same
when they hit and the resulting energy of the impact will be the same.
Further
away, they will take much longer and be accelerated more slowly, but they
have
much further distance over which to gain their velocity and they will
arrive
again with the same impact energy. Hence, Volts is expressed as
Joules/Coulomb
and doesn't need to include the distance between the plates.

Don't forget that:

Ohm = Joule-second/Coulumb^2 = Watt/Amp^2 = Henry/second

Watt = Joule/second

Henry = Joule-second^2/Coulomb^2 = Joule/Amp^2 = Ohm-second

Farad = Coulomb^2/Joule = Coulomb/Volt

If you then look at an "RC time constant," this is:

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

Nice, eh? Time, as you might expect. Also, take a look at the SQRT(L*C),
which
is:

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

Once again, time! Dimensional analysis to the rescue.

Jon
------------

I note that now the ampere is now defined as the basic element in the SI
system - the coulomb is then an ampere-second. I got jumped on once for
using the coulomb.
Check out:
http://tcaep.co.uk/science/siunits/index.htm

--
Don Kelly
dhky@peeshaw.ca
remove the urine to answer

 

[Chris Williams]

Ok I believe I understand correctly now--I was a bit off.

Floyd's description of "potential energy" was throwing me off as, how
do you evaluate the total potential energy of a mass? If you have a
boulder just sitting there on the ground, how much energy does it
"potentially" have? Well that kind of depends on what you do with the
boulder afterwards (like rolling it up a hill.)
I was mistakenly thinking of the boulder (or water) in it's starting
state, instead of it's final moment of impact.

So:

Voltage = Current * Resistance
Voltage is the product of the speed and the resistance. So in water
terms, you will have greater pressure by using a smaller pipe
(increasing resistance) just as you will need a greater amount of
pressure to increase the flowthrough if the size of the pipe stays the
same.

Voltage = Energy / Quantity
Energy = Voltage * Quantity
The greater the amount of force on a certain body, the greater the
amount of energy we will have _at the time of use_. Similarly, a
larger mass being delivered at the same amount of force will have more
energy _at the time of use_.

Does this seem correct?

-Chris

 

[Don Kelly]

I like the stone, furlong, fortnight, standard stroke of a standard glass
rod on a standard cat. system or the
SFF Damncatclawedme system of units
(actually I am old enough to prefer using the Coulomb to the ampere in
dimensional analysis)
Take care,
--
Don Kelly
dhky@peeshaw.ca
remove the urine to answer
"Jonathan Kirwan" <jkirwan@easystreet.com> wrote in message
news:4vv0o0php9jqolgnq4fugerr3l79qen4tu@4ax.com...

On Thu, 28 Oct 2004 03:31:23 GMT, "Don Kelly" <dhky@peeshaw.ca> wrote:

I note that now the ampere is now defined as the basic element in the SI
system - the coulomb is then an ampere-second. I got jumped on once for
using the coulomb.

Yes, I'm aware of it. I still prefer to think in terms of countable
things,
though. Penchant of mine. Still, there are a variety of commensurate
systems
to the SI system. One I like is a relativistic mapping, which makes
length and
time the same, describing everything in a 'pure' number of light-seconds.

Jon

 

[Don Kelly]

"Chris Williams" <thesagerat@yahoo.co.jp> wrote in message
news:da271e02.0410272154.180dc061@posting.google.com...

Ok I believe I understand correctly now--I was a bit off.

Floyd's description of "potential energy" was throwing me off as, how
do you evaluate the total potential energy of a mass? If you have a
boulder just sitting there on the ground, how much energy does it
"potentially" have? Well that kind of depends on what you do with the
boulder afterwards (like rolling it up a hill.)
I was mistakenly thinking of the boulder (or water) in it's starting
state, instead of it's final moment of impact.

So:

Voltage = Current * Resistance
Voltage is the product of the speed and the resistance. So in water
terms, you will have greater pressure by using a smaller pipe
(increasing resistance) just as you will need a greater amount of
pressure to increase the flowthrough if the size of the pipe stays the
same.
-------

Basically OK
----------

Voltage = Energy / Quantity
Energy = Voltage * Quantity
----

OK but "quantity" needs further refinement.
-

The greater the amount of force on a certain body, the greater the
amount of energy we will have _at the time of use_. Similarly, a
larger mass being delivered at the same amount of force will have more
energy _at the time of use_.

Does this seem correct?

-Chris
-----

You are getting closer but you are mixing mechanical and electrical terms-as
well as mixing up energy and power concepts. You can have a force on a body
but if it doesn't move- no energy is involved. You can have a voltage source
but if it is open circuit (no current) then there is no energy involved. If
you have a current due to this voltage, then you will have power =V*I but
power is the time rate of change of energy so the energy will depend on both
power and time. Mechanically a force times a velocity is power but velocity
is distance/unit time so energy includes a time factor.

Voltage or potential difference between a and b is defined in terms of the
energy per unit charge to move the charge from a to b in an electrical
field. This is energy or work and is expressed in Joules (1 watt-second=1
Joule). It can exist whether or not there is a current. Certainly, in a
given field, it will take 2 times as much energy to move 2 Coulombs as to
move 1.

The analogy would be the energy needed per unit mass to move a mass from
point a to point b in a gravitational field. This also is expressed in
Joules. More energy is needed to lift 2kg than to lift 1 kg.

In either case, it doesn't matter how long you take or what path you take-
the energy depends on the end points. In the case of a mass being lifted
from a to b, the mass could be lifted up directly against gravity or slid up
a long ramp- the energy change depends on the initial and final heights in
the gravitational field. Ditto for the movement of a charge in an electrical
field. Small force over a long time or higher force for a shorter time =same
energy

Current at a point is the rate of change of charge passing that point.
Coulombs/second
A voltage or potential is needed to produce a current (except in true
superconductors which aren't available at Radio Shack)

An analogy is the total flow rate in a river at some measuring point- say
gallons/minute. This flow is produced by a pressure difference or head (eg.
potential difference).

--
Don Kelly
dhky@peeshaw.ca
remove the urine to answer