It goes into the field! At least thats what field theorists say. They say
the field is what has the energy... but of course they only say this because
the that is how they interpret the field equations ;/ (has to do with the
fact that potential energy depends only on the relative positions)
But to answer your question, when lift up a weight in a gravitational field
you are supplying work, i.e. energy, to the weight giving it potential
energy... you did that by first giving it kinetic energy to move it. So you
have actually increased it's potential energy... hence it's not "where did
the potential energy go" but "where did it come from" ;)
It might bring some comfort to you to know that the equation to find
the amperes that a conductor can safely carry without overheating
comes from Mechanical Engineering.
It is the Fourier heat transfer equation. Mechanical engineers know a
whole lot more about this than electrical engineers.
The equation is (TC - TA) = I**2 R (RCA)
Solving I = SRT((TC-TA)/(R*RCA))
I in amperes, TC is maximum conductor insulation temperature in
degrees C, TA is ambient temperature in degrees C, R is dc resistance
in ohms of conductor, RCA is thermal Resistance in thermal ohm feet.
I square R is the heat generated in the conductor when I amperes flows
through the conductor with resistance R in ohms. The amperes flow
because of a potential difference in voltage that exist between
Variations of this equation were used by Rosch in 1938 and by Msgs
Neher and McGrath in 1957 to develop ampacity tables found in the
National Electrical Code.
This does not tell you what voltage is but it does put some rather
elite electrical engineers that like to poke fun at mechanical
engineers in their place.
Voltage is work (or energy) per unit charge required to move a unit charge
from a to b in an electrical field. An electrical field is produced by the
presence of other charges.
Think of mechanical potential energy per unit mass required to move a
unit mass from point a to b in a gravitational field. A gravitational field
is produced by other masses.
The two are analogous. In both cases it doesn't matter what path you take
from a to b.
Don Kelly email@example.com
remove the X to answer
I kinda forgot to answer your questions:
Since you said you understand current, you do realize that I = Q/t? and
that if Q goes up and t goes down in proportion then I doesn't change? The
same thing is going on with voltage.
Analogy: If you have 10 N weight lifted up 3 feet then it has the same
potential energy as a 5 N weight lifted up 6 feet? (ok, not exactly since
the gravitational field is a bit weaker but close enough)
Remember that voltage isn't a fundamental quantity but is a function of more
V = J/C is one expression of V but it's also V = W/A as it is A*Ohm. (we get
the last two from ohms law)
I think you really need to think about it more. Take 2 C of charge and place
them at some distance apart, say it has 10 joules of energy, if you now move
them apart you increase the energy to maybe 20 joules? Or equivalently, but
with less information, we have went from 5V to 20V.
It is exactly analogous with many other physical quantities that depend on
more than one thing. If I have 80 C moving past a point in 2 second then
that is 40A but so is 40 C moving past a point in 1 second. I could also
have one electron moving past that point in 1/40C of a second and it would
also be equivalent to 40A.
Also if I reduce the time I move the 80C then from 2 seconds to 1 second(if
I slow them down) then I cut the current. In fact if I "freeze" everything,
even though I have 80C of charge sitting there, I have no current at all!
"The difference in voltage measured when moving from point A to point B is
equal to the work which would have to be done, per unit charge, against the
electric field to move the charge from A to B. "
Note that it is a "per-unit" quantity that depends only on the distance in
the electric field. It doesn't depend on the amount of charge which is why
it is being divided out.
Think of it as energy per unit charge. Current gives us the other half of
the equation. If we know the energy per unit charge, or the voltage
difference, and we know the charge per second(along some path), or the
current, then if we multiply them we have
Energy/charge * charge/time = energy/time = power = work/time
i.e., V*I = P
If we had some idea of the time involved we could get the energy too.
If you want a microscopic concept then it is the energy contained per
electron in the electric field... (do you understand the electric field?)
The macroscopic concept is one of "force" or some "ability to do work"...
note that it is a mixed up concept because it's not fundamentally correct
but it is understanding by "consequence". (kinda like understanding anything
we have to associate it with things we know)
Just like the electric field is force between charges divided by
charge, electric potential (ie, voltage) is potential energy divided
You aren't having trouble with voltage, though. You are having issues
with energy. What is it? Think about that for a bit before
My answer is that it is the potential to move something. When you hold
a hunk of matter above the ground, it has the potential to start
accelerating when you let go of it. Thus, it has potential energy,
given to it by the attraction of gravity. How much? It depends on how
high you hold it.
Same thing for voltage. It is the potential to move charged particles
around (ie, create a current).
Just like the attraction of gravity gives a hunk of matter different
'potential energy' at different altitudes above the ground. So, to get
the hunk of matter up there, somebody had to give it some energy.
To get 10 volts out of 2 coulombs, you need to put in 20 joules of
energy to separate the charges.
Voltage, as used in electronics, is a relative measure. You pick some
place in the circuit, and say "that is ground", meaning that is where
you measure the rest of the voltages from. Then, take a particle, like
an electron, and integrate the force it takes you to move it to some
other place in the circuit with respect to distance. Thankfully, it
doesn't matter how you go, any path will do. Now, divide by the charge
of the particle. That number will equal the voltage at the destination
point, relative to the ground of the circuit.
In physics, there are actually two potential fields, the electrostatic
potential, and the vector potential. Those fields, the first a scalar
field, and the second a vector field, influence how a charge will
move. A charge will move along the gradient of the electrostatic
field, and a moving charge will turn to align its motion with the
That's making it too bloody complicated.
Absolutely basic that:
Voltage (Or EMF, electro motive force, or potential or whatever you
want to call it) is the pressure that can push an electric current
through a circuit.
The source of the voltage can be various devices, such as a battery, a
generator or a storage device such as capacitor.
'Voltages' can be DC (Direct current) or AC (Alternating current).
Take a pencil and draw a square to represent 'the source'.
Then draw a circuit from and external to the source comprising wires
(which have virtually no resistance in most practical applications)
and a load (which could be say a single heating resistor of R ohms).
Electric current (amps) will flow in the above circuit. The higher the
voltage the greater the current that will traverse the circuit.
The formula; Ohm's Law is Voltage/Circuit Resistance = Current flow.
A practical example being 230 volts, a 20 ohm resistor, and a
resulting current flow of 230/20 = 11.5 amps
If you want to get into the amount of power (watts, or watts per hour)
how many coulombs of energy are being transferred you can make further
But the above is basic.
PS. Working in telecommunications for some 40 years we once had a boss
who was an 'Industrial Engineer'. We (experienced subordinates) always
gave him a hard time saying "Well who can expect an Industrial
Engineer to understand electricity with more than one frequency!"
So congratulations to the OP on wanting to understand electricity.
There's a fascinating bit of philosophy hidden in this inquiry.
Voltage and charge are two completely different kinds of
physical quantities, and the distinction between these
is repeated in many different ways over all disciplines (which
is why so many analogies are offered when the question comes up).
Voltage is an example of an intensive quantity. Charge is an
example of an extensive quantity. If you consider a system (like,
let's take a battery/bulb flashlight), the voltage of that battery is
an intensive quantity, and the charge that the battery can deliver is
an extensive quantity. Double the flashlight, and there are two
batteries and two bulbs,twice the charge, but the voltage is the same.
Double the dimensions of the flashlight, the bigger battery has
eight times the charge, but the voltage is STILL the same.
Extensive quantities include mass, charge, cost of a bag of potatoes.
Intensive quantities include density, voltage, cost per pound of
Voltage, in particular, is the ratio of two extensive quantities,
electrical energy and stored electric charge, in the sense of taking
a derivative of energy with respect to charge. Just like the
cost per pound of potatoes, it's intensive.
The implications of this include another check you can perform on
equations: you can't add or equate intensive and extensive quantities,
just like quantities with different units.
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