Confusion about voltage in a circuit

[x86]

So basically, I am inferring the following based off of what I've learned.

When we hook a battery up to a circuit, this creates a potential difference. Say we use a 9V battery. Here is a picture:

From what I know, the horizontal is a equipotential surface, and the vertical has different voltages associated with different points. For example, the top point is 9V, middle 4.5V, bottom 0V.

I have some questions:

(1) Now, let's assume we placed a capacitor on the circuit in the position of the 4.5V dot. Then why do we say the voltage across the capacitor is 9V? Shouldn't the voltage across the capacitor be very tiny, depending on the capacitors size? I.e., the voltage at one plate minus the voltage at the other plate? This certainly wouldn't be 9V.

(2) When we have electrical components on the circuit- why is it that we can say they consume voltage and cause a voltage drop? Isn't this impossible? Isn't it defined that the top is 9V, the bottom is 0V. Therefore, shouldn't it be impossible for this voltage to drop, unless we change the batteries voltage?

(3) If the red line is an equipotential surface, exactly what encourages the current to flow? Is it that the electrons will gather up along this equipotential surface, and eventually end up repelling each other? Thus causing the current to flow?

 

[davenn]

From what I know, the horizontal is a equipotential surface, and the vertical has different voltages associated with different points. For example, the top point is 9V, middle 4.5V, bottom 0V.

That's incorrect ... Why would you think that ?

You have not indicated the presence of a resistor, so assuming you are indicating a normal piece of copper wire, there will be a short circuit across the battery and the voltage will be zero or something close to it

(1) Now, let's assume we placed a capacitor on the circuit in the position of the 4.5V dot. Then why do we say the voltage across the capacitor is 9V? Shouldn't the voltage across the capacitor be very tiny, depending on the capacitors size? I.e., the voltage at one plate minus the voltage at the other plate? This certainly wouldn't be 9V.

incorrect ... see my comments above

(2) When we have electrical components on the circuit- why is it that we can say they consume voltage and cause a voltage drop? Isn't this impossible? Isn't it defined that the top is 9V, the bottom is 0V. Therefore, shouldn't it be impossible for this voltage to drop, unless we change the batteries voltage?

No, we don't say, "they don't consume voltage".

again, refer to my above comments

(3) If the red line is an equipotential surface, exactly what encourages the current to flow? Is it that the electrons will gather up along this equipotential surface, and eventually end up repelling each other? Thus causing the current to flow?

no ... again refer to my first commentsDave

 

[jim hardy]

Edit: oops - i see Dave already answered while i was typing

sorry, guys.. X86:
Have you really thought about the definitions of potential and voltage? You seem to be mixing the concepts.

An electric potential (also called the electric field potential or the electrostatic potential) is the amount of electric potential energy that a unitary point electric charge would have if located at any point of space, and is equal to the work done by an electric field in carrying a unit positive charge from infinity to that point.

Nobody is going out to infinity to make that measurement for us.
So we always use the difference of potential between two local points.
That's the definition of voltage, potential difference between two points..

So whenever you state a voltage you must make it clear between what two points.

(1) Now, let's assume we placed a capacitor on the circuit in the position of the 4.5V dot. Then why do we say the voltage across the capacitor is 9V?

I wouldn't say that at all.
A capacitor has two ends not one. Where do they go? You only mentioned one place. It takes two places to have a potential difference.

I think your questions will answer themselves when you get your basics straightened out.

old jim

 

[davenn]

The ONLY way you would get voltage reading like you have indicated would be if you had 2 resistors of identical value in series between the positive and negative terminals of the battery
like this ...

so ...
--- if you put the negative lead of your voltmeter at point C and the positive lead on point A, you would measure 10V
--- if you put the negative lead of your voltmeter at point C and the positive lead on point B, you would measure 5V
--- if you put the negative lead of your voltmeter at point B and the positive lead on point A, you would measure 5V

5V is dropped across each resistor
This is called a voltage divider. if the resistances are identical, the centre voltage at point B will always be 1/2 the full supply voltage
if the resistor values are different, then the voltage at point B will be some value relating to the voltage drops across each resistorDave

 

[davenn]

Now there is one proviso (caveat) to this
I won't mention it yet and confuse you, I want to make sure you have got the basics that Jim and I have stated above
once those are sorted I will comment on the other "option"

Dave

 

[x86]

I think after reading the above posts, I now understand where my confusion was coming from. I know that voltage is potential difference (we must measure the potential in regards to two points)

But I'm a little confused about one concept regarding the below circuit diagram:
a) What is the potential difference betwen A and B? Isn't it zero, because it's an equipotential surface?
b) What is the potential difference between B and C?
c) What is the potential difference between B and D?

I'm thinking b and c have the same answer, but this is kind of confusing to me, because they have different lengths.

This also kind of relates to my capacitor confusion. If we put a capacitor at C then the potential difference between its plates will be 9V. But why? This part confuses me, because it feels like B--->C should have a different potential difference than B--->D (related to the wire length)

 

[davenn]

hi x86

Am not sure why you are still confused ? did you study that circuit I posted with the 2 resistors ?

did you not understand the text I posted at the top of my first reply ?...

You have not indicated the presence of a resistor, so assuming you are indicating a normal piece of copper wire, there will be a short circuit across the battery and the voltage will be zero or something close to it regardless of which points you measure between

in your latest diagram, there is no difference between any of those points ( when dealing with a length of copper wire a few metres or so long )IM thinking your confusion stems from the fact that part of the wire is level and part if vertical ... is that correct ?

if so you are making incorrect assumptions

Dave

 

[x86]

hi x86

Am not sure why you are still confused ? did you study that circuit I posted with the 2 resistors ?

did you not understand the text I posted at the top of my first reply ?...
in your latest diagram, there is no difference between any of those points ( when dealing with a length of copper wire )

Dave

Yes. I did read your above post. I am just confused about the voltage between different points in the above diagram (copper wire only).

I still don't understand what you mean by "there will be a short circuit across the battery and the voltage will be zero or something close to it."

My confusion is stemming from being unsure what the potential difference in a battery hooked up to a copper wire is, at different points.

For instance, let's say that A and B have different potentials (somewhere in space). Then their potential difference is B-A

But how does this relate to the circuit? Now the points A,B are no longer in space, but on the circuit. I know the battery provided potential difference. But I'm confused about the potential difference on different parts of the circuit.

 

[davenn]

Yes. I did read your above post. I am just confused about the voltage between different points in the above diagram (copper wire only).

I still don't understand what you mean by "there will be a short circuit across the battery and the voltage will be zero or something close to it."

OK

consider a short to say a few metres length of copper wire ... it's going to have close to zero Ohms resistance ( copper is a GOOD conductor)

if you put that bit of wire across your battery, its going to short circuit the battery, the wire is going to get VERY hot and the battery is going to go flat quickly
as there will be a large current flowing Even putting your voltmeter directly across the battery terminals you will read zero ( of something very close to) volts

My confusion is stemming from being unsure what the potential difference in a battery hooked up to a copper wire is, at different points.
For instance, let's say that A and B have different potentials (somewhere in space). Then their potential difference is B-A
But how does this relate to the circuit? Now the points A,B are no longer in space, but on the circuit. I know the battery provided potential difference. But I'm confused about the potential difference on different parts of the circuit.

BECAUSE that length of wire is very low resistance, ALL the battery voltage will be dropped across that bit of wire

Now here's part of that proviso I hinted at earlier ...

That bit of copper wire ISNT a perfect conductor it has a VERY SMALL resistance over its length. therefore if you have a good voltmeter that can read small voltages, you will likely read a difference in voltage between 2 points, but it will only be a few micro to milliVolts depending on the length of the wire and how apart your measurement points are

But for practical purposes, for a length of wire under a few metres, you don't need to take that into consideration and it can be ignored, as other components in the will produce much larger voltage drops

 

[davenn]

have a look at this pic I drew ...

The voltage measured between points A and B in any of those 3 circuits will ALL be 10V

 

[jim hardy]

I know that voltage is potential difference (we must measure the potential in regards to two points)

But I'm a little confused about one concept regarding the below circuit diagram:
a) What is the potential difference betwen A and B? Isn't it zero, because it's an equipotential surface?
b) What is the potential difference between B and C?
c) What is the potential difference between B and D?

A question well stated is half answered.

You are sending us mixed messages.
Your horizontal wire AB you say is an equipotential surface.
Then you say your vertical wire is not an equipotential surface.
What is the difference between those two wires? Electric potential is unaffected by gravity.

If there is no potential difference between A and B there'll be no current flow from A to B.
If AB BC and CD are all made from the same kind of wire, and are as near the same length as they appear in your drawing,
then why do they (edit have) behave so differently when connected in series?

In circuit analysis we usually assume our conductors to have zero resistance hence zero voltage drop.
But we remain aware that's an exaggeration, in actuality there is small voltage required to push current along a wire.
I've measured the millivolts along printed circuit tracks to see where current was going.
In power distribution we aim for less than 3% voltage drop in our feeder.

Now - a 9 volt transistor radio (006P) can make only about an amp or two
so in your circuit above
i'd expect an amp or two to flow
the battery to get very hot
and potential difference of only millivolts per inch along your wire.

If you apply enough current to drop volts per inch along the wire it will quickly glow red hot and melt. That's how a soldering gun works.

Dave has given great direction.
You need only to slow down and realize that you already know these answers.
We live in such a rushed and instant-response world that it is difficult to make ourselves think in the small incremental steps requisite for figuring things out.
Much of learning is discovering what we already know.

If we put a capacitor at C then the potential difference between its plates will be 9V. But why?

Once again you have failed to define to what two points your capacitor's two plates are connected.
Which plate is "put at C"?
Where's the other plate connected ?
How can you conceive of a potential difference between only one point?
It is unclear what you are thinking.

"Slow down, you move too fast" simon&garfunkel

 

[x86]

hi x86

Am not sure why you are still confused ? did you study that circuit I posted with the 2 resistors ?

did you not understand the text I posted at the top of my first reply ?...
in your latest diagram, there is no difference between any of those points ( when dealing with a length of copper wire a few metres or so long )IM thinking your confusion stems from the fact that part of the wire is level and part if vertical ... is that correct ?

if so you are making incorrect assumptions

Dave

Thank you guys for being so patient with me, and I am sorry for sending mixed signals. After pondering for a while, I've realized that I've been unnecessarily overcomplicating the problem. Let's assume the circuit is a superconductor and the wire has a resistance of 0 V.

You are correct, I am confused about the fact that some of the wire is level and some of the wire is vertical. I was told that in circuits, the level wire will always be an equipotential surface, and that the vertical wire will always have a voltage equal to that of the battery. I do not understand why this is the case. The only thing that I know is that it will take 9J of work to move a charge of 1 C from the (+) terminal to the (-) terminal. Everything I said in my first post were things that I think, not necessarily things that are true.

 

[davenn]

its OK

As Jim said ... and one of his fav saying on the forums ... "A question well stated is half answered"

another words, if you sit and think about how to ask a good question ( well constructed), chances are you already have a good part of your answer

If you don't mind me asking, what is your age and what school level are you at ?
It can be difficult on a forum like this sometimes, because we can't see the person or know their background education
its sometimes difficult to know how easy/complex to make the answers

cheers
Dave

 

[davenn]

I was told that in circuits, the level wire will always be an equipotential surface, and that the vertical wire will always have a voltage equal to that of the battery.

That is totally incorrect and no wonder you were getting confused with our responses.

think about the wiring in just about any electronics ... inside a TV, a computer, a car, a plane

the wiring curls all over the place as it makes its way from power sources to control panels/displays to the sensors etc being controlled.
The potentials are not affected by that at all. What affects the voltage drop on a wire is its length.
As the wire gets longer, its resistance increases and this produces a voltage drop

D

 

[x86]

its OK

As Jim said ... and one of his fav saying on the forums ... "A question well stated is half answered"

another words, if you sit and think about how to ask a good question ( well constructed), chances are you already have a good part of your answer

If you don't mind me asking, what is your age and what school level are you at ?
It can be difficult on a forum like this sometimes, because we can't see the person or know their background education
its sometimes difficult to know how easy/complex to make the answers

cheers
Dave

Yes, I apologize if my questions are not well stated. I really need to work on my communication skills. I am a first year computer engineering student taking an introduction to electricity course. Basically, I learned about the basics such as coulombs law, electric fields, gausses law, amperes law, capacitors, electric flux, magnetic flux, electric potential,capacitance, dialectrics, current, resistance, current density, magnetic fields, induced magnetic fields, faradays law, lenz law. And a few other basic stuff. It is probably a very basic physics course, using calculus. Now we are on circuit analysis.

 

[davenn]

cool ... lots of good learning ahead of you

never be afraid of asking for help, all of us on here do our best to help, but what we really like doing is teaching people to help themselves
ie. guide you so you can work out the answers for yourself ( but sometimes just giving a straight answer to clear confusion works well)

 

[x86]

cool ... lots of good learning ahead of you

never be afraid of asking for help, all of us on here do our best to help, but what we really link doing is teaching people to help themselves
ie. guide you so you can work out the answers for yourself ( but sometimes just giving a straight answer to clear confusion works well)

Ah okay. Thank both of you for your help, and sorry about the confusion. It is actually pretty simple when I think about it. It takes 9 J to pump a 1 C charge from the (+) terminal to the (-) terminal in the above diagram, and if the wire has resistance, then we lose some of this energy to heat, and of course how much we lose depends on the length of the wire and the resistance of the wire.

 

[davenn]

It takes 9 J to pump a 1 C charge from the (+) terminal to the (-) terminal in the above diagram

from the (-) terminal to the (+) terminal

electrons are the charge carriers and they move from negative to positive

Have you been told about the difference between conventional current flow + to - and electron flow - to + and why there is a difference ?

Dave

 

[x86]

from the (-) terminal to the (+) terminal

electrons are the charge carriers and they move from negative to positive

Have you been told about the difference between conventional current flow + to - and electron flow - to + and why there is a difference ?

Dave

Yes. Current is how a positive charge would flow in the circuit. It moves from high potential to low potential, whereas electrons move from low to high potential. Of course, the only thing flowing is electrons, so current seems like something more imaginary. In my example above, I was referring to an imaginary +1C charge.

 

[jim hardy]

I was told that in circuits, the level wire will always be an equipotential surface, and that the vertical wire will always have a voltage equal to that of the battery. I do not understand why this is the case.

That is good drafting procedure to draw them that way and nothing more.
In vacuum tube days we drew + at top of page and signal flow left to right, top to bottom just like we read..

Today's computer drawings are nightmares- signals go every which way.

really need to work on my communication skills.

Lots of people do.
I quote this guy below often, so do not be offended if i seem repetitious.
Forty five years ago i bought a book at a library yard sale.
The first essay in it was Lavoisier's "Introduction to his treatise on Chemistry".
I started it and found, though the language seems archaic (translated from 1700's French)
it is a great primer on analytical thinking and clear communication.
Over the years i came to appreciate just how rare those skills are. But they can be learned, in fact my company gave courses in "analytical troubleshooting".

Pay special attention to his quote near the end starting with "Instead of applying observation to things we wish to know..."
you'll see it a lot.
Here it is:
https://web.lemoyne.edu/giunta/lavpref.html

Here's the last line:

"But, after all, the sciences have made progress, because philosophers have applied themselves with more attention to observe, and have communicated to their language that precision and accuracy which they have employed in their observations: In correcting their language they reason better."

and that's why engineers need to take some English credit hours. (English speaking engineers, that is)

good luck, happy to see you working your way out of confusion.
thanks for the kind words.old jim

 

[CRT]

do really voltmetres measure joule required to move a coulomb of charge?

 

[sophiecentaur]

So basically, I am inferring the following based off of what I've learned.

When we hook a battery up to a circuit, this creates a potential difference. Say we use a 9V battery. Here is a picture:

The problem with the OP (and diagram, in particular, is that it assumes ideal components (without extra information) and so it is an impossibility. Bad start for a technical discussion. A real battery will have internal resistance and so will a real wire so the 9,4.5,0 figures make assumptions about the nominal battery collage (emf) and the resistance of the components (including wire).
Equipotential Surfaces are very hard to define in circuits with ideal conductors because the potential is the same all over any particular wire and undefined in the space around. Using the term just makes a discussion more difficult. You will notice that circuit theory books don't do it that way -= for a good reason.

 

[davenn]

do really voltmeters measure joule required to move a coulomb of charge?

no, a voltmeter measures the potential difference between 2 points in a circuit

Equipotential Surfaces are very hard to define in circuits with ideal conductors because the potential is the same all over any particular wire and undefined in the space around. Using the term just makes a discussion more difficult. You will notice that circuit theory books don't do it that way -= for a good reason.

You also missed the point that got clarified earlier that x86 thought that the horizontal section was equipotential and the vertical section of the circuit had the voltage drop corresponding to the figures on that circuit that you quoted
This is of course incorrect and got sorted out Dave

 

[CRT]

It was clear to me from the beginning what *86 has asked for but the way you explained it with the three circuits was incredable. It Polished my understanding.Sir
a big thanks or answering my question too.But i wanted to know a little bit more.
On what basis are the batteries labeled as 6v or 9v or 12v.I mean to say do manufacturer measure it with voltmeter after making the battery or they use different technique.or more simply i want to ask sir i have 2 plates one with 10 c of positive charge and 2nd having neutral,can i say i have a cell. if yes how can i know its voltage.

 

[sophiecentaur]

no, a voltmeter measures the potential difference between 2 points in a circuit

You also missed the point that got clarified earlier that x86 thought that the horizontal section was equipotential and the vertical section of the circuit had the voltage drop corresponding to the figures on that circuit that you quoted
This is of course incorrect and got sorted out Dave

But yes - by implication and the definition of the Volt. The same argument could be applied to all physical quantities and their measurement.

I take your point - to some extent. But I can't find any statement to the effect that the concept of equipotential surfaces has no place in (normal) circuit theory. It was sort of implied but, in the interests of newcomers to the thread and the subject, I think it cannot be stressed too much. The thread meandered in a different direction when it perhaps should have been terminated and re-started or had a massive signpost somewhere in it. I know diagrams can be a pain to draw but a suitable set could have helped a lot here, to show the basic problem with the original approach. (I do make efforts in that direction on occasions so I am suggesting 'do as I do" and not 'just as I say' ) The same problem arises when people try to 'explain' electricity in terms of the forces and fields between electrons. Answers to these sorts of questions do not tend to go back far enough to square one and 'sorted out' can easily be in the mind of the answerer and not necessarily the questioner or bystander.

 

[sophiecentaur]

It was clear to me from the beginning what *86 has asked for but the way you explained it with the three circuits was incredable. It Polished my understanding.Sir
a big thanks or answering my question too.But i wanted to know a little bit more.
On what basis are the batteries labeled as 6v or 9v or 12v.I mean to say do manufacturer measure it with voltmeter after making the battery or they use different technique.or more simply i want to ask sir i have 2 plates one with 10 c of positive charge and 2nd having neutral,can i say i have a cell. if yes how can i know its voltage.

The unloaded Voltage output (emf) from a cell will depend upon the technology (Chemistry) - for instance, the two different metals used in a simple cell will define the basic emf. Just how much current you can draw from a cell whilst maintaining this voltage and its charge capacity will depend upon the details of the design - such as the total area of the electrodes and the sort of electrolyte used.
Cells from all manufacturers have the same nominal output volts for a given 'type'. All Alkaline cells are the same and so are the old Leclanche types and Lithium primary cells. You don't need to measure it - except to check that there is continuity in there. It could be necessary to give batteries a check to make sure there is no internal leakage - but these days, the production standards are probably too good for that to be needed.

 

[CRT]

A lot of thanks for explaining sir.

 

[davenn]

I take your point - to some extent. But I can't find any statement to the effect that the concept of equipotential surfaces has no place in (normal) circuit theory.

you still misunderstand
Its a circuit drawing, he thought the horizontal parts of the drawing were equipotential and the vertical parts defined a voltage drop
This as I said is totally incorrect ... its just the way circuits need to be drawn
Its JUST a drawing, a bunch of lines, connecting a collection of components Dave

 

[sophiecentaur]

you still misunderstand
Its a circuit drawing, he thought the horizontal parts of the drawing were equipotential and the vertical parts defined a voltage drop
This as I said is totally incorrect ... its just the way circuits need to be drawn
Its JUST a drawing, a bunch of lines, connecting a collection of components Dave

I don't misunderstand. I just make the point that people could read this thread and not realize what you and I know - because the concept of equipotential lines is totally misplaced here.
This (anonymised) post :
"do really voltmetres measure joule required to move a coulomb of charge?" shows the possible level of confusion that the thread can be generating.
But you and I agree, obviously, on the facts.

 

[CRT]

i am extremely sorry to both of you sir.it was me i think who led to mesh.i will not post here my comments again.sorry dave sir and Sophiecentaur sir.

 

[x86]

I don't misunderstand. I just make the point that people could read this thread and not realize what you and I know - because the concept of equipotential lines is totally misplaced here.
This (anonymised) post :
"do really voltmetres measure joule required to move a coulomb of charge?" shows the possible level of confusion that the thread can be generating.
But you and I agree, obviously, on the facts.

But isn't that the definition of voltage? For instance, if the potential difference between two points is 9V (A-B), doesn't that mean it takes 9J of energy to move a +1C charge from point A to point B by the electric field?

 

[davenn]

i am extremely sorry to both of you sir.it was me i think who led to mesh.i will not post here my comments again.sorry dave sir and Sophiecentaur sir.

You didn't do anything wrong

 

[sophiecentaur]

i am extremely sorry to both of you sir.it was me i think who led to mesh.i will not post here my comments again.sorry dave sir and Sophiecentaur sir.

Hey - don't apologise. One of the main reasons for the existence of PF (apart from letting some of us show off) is to straighten out misconceptions and help people see things in more productive ways. For heaven's sake don't stop posting (as long as you don't mind people disagreeing with you occasionally. ) It does everyone good to go over old ground and get things straight in their own minds.

 

[sophiecentaur]

But isn't that the definition of voltage?

Yes, that is a definition of the Voltage but that is not the same as actually measuring a voltage using those two quantities. In basic terms you could say that a voltmeter does determine the energy per coulomb but it really only does this by an indirect means - for instance by measuring the Current through High resistance. I think am making a genuine distinction here.

 

[jim hardy]

@CRT

Just another old timer here saying you needn't apologize at all

do really voltmetres measure joule required to move a coulomb of charge?

That question really interested me - i was trying for days to figure out an answer.

As Mr Sophie said, the plain vanilla everyday meter measures how much current an unknown voltage pushes through a known resistance.
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/galvan.html

Microamps flowing through the red coil create torque.
The needle deflects until the torque from current just balances the torque from the spring.
The scale is marked in volts rather than micro-amps.
You'd switch in various series resistor values to make a multi-range voltmeter.

There does exist a device called "Electrostatic Voltmeter" that measures potential more directly
http://en.wikipedia.org/wiki/Electrostatic_voltmeter

Electrostatic voltmeter utilizes the attraction force between two charged surfaces to create a deflection of a pointer directly calibrated in volts...
The pivoted sector NN is attracted to the fixed sector QQ. The moving sector indicating the voltage by the pointer P and is counterbalanced by the small weight w. In newer instruments the weight is replaced by a spring, thus allowing the meter to be used both in horizontal and vertical positions.

Hmmmm at least it uses Newtons...

An electroscope can measure charge...
http://physics.kenyon.edu/EarlyApparatus/Static_Electricity/Electroscope/Electroscope.html

The two electroscopes below are equipped with the bottom plate of Volta's condenser. The upper plate is of the same dimensions as the lower one, and has an insulating glass handle attached, and its lower surface is covered with varnish for insulation.
The condenser allows small quantities of charge to be detected. It is charged by touching the object under test to the lower plate, while grounding the top plate. If the lower plate has been positively charged, negative charge will be transferred to the upper plate. The ground connection is then broken, followed by disconnecting the test object. The capacity, C, is then decreased by lifting up the upper place. Since the potential across the capacitor, V, is related to the constant charge Q by Q = CV, the potential increased and the electroscope leaves diverged.

REF: Thomas B. Greenslade and Richard H. Howe, "A Modern Use of Volta's Electroscope", The Physics Teacher, 19, 614-615 (1981)

Now it strikes me that the act of separating the plates involves overcoming the attraction between them
so in principle one exerts some minute amount of Newton-meters in that experiment.

My meager brain isn't deep enough to figure out how one would make a joules-per-coulomb meter from an electroscope
but it looks like an interesting thought experiment.

So, Mr CRT, I thank you for your thought provoking question. It's given me something to think about while the TV is on..

old jim

 

[jim hardy]

@CRT Will you pardon an old guy a bit of nostalgia ?

This digression has a purpose, so please bear with me.

This is why kids should be shown how to work on old cars. It teaches them a lot.

I knew i'd encountered a simple, earthy "Joules per second " meter someplace in my distant past.
Just now came to me.

My 1953 Ford Stationwagon (wish i still had it)

had gages operated by "Heated Wire" technology of the 1930's. Rugged and reliable, if less sophisticated than today's computer dashboards.
I can't find a Ford shop manual explanation but this old Chrysler one will suffice to make the point:
http://www.allpar.com/history/mopar/electrical2.html
Basically a wire is heated by a current proportional to what it is you wish to measure.
The wire expands with temperature and that mechanical movement is transmitted to the indicating needle.

Now think about this --- the temperature of that wire is in proportion to the rate at which heat is put into it.
That'd be watts, or joules per second.

Deflection then is in proportion to Joules per second.

GIBRALTAR: the original Chrysler gauge design

Virtually all Mopar RWD passcars used thermal-type instrumentation (shown at right). Simply put, this means that the pointer of each gauge (excluding the ammeter, of course) is mechanically linked to a bimetallic strip. The strip is wrapped with resistance (heating) wire, just like what’s in old toasters, etc. As the current passed through this wire is increased, the wire gets hotter, transferring this heat to the bimetal strip, which bends more and more as the temperature is increased, deflecting the pointer. As you probably guessed, this gauge design is inherently very well damped and very slow to respond - which is probably a good thing. Nobody wants to see the gas gauge, for instance, swing wildly as the fuel in then tank sloshes around. And the designers clearly were just as happy that Vern can’t see the temp gauge fluctuate as the thermostat opens and closes, or the oil pressure drops down quickly to 20 PSI in traffic on a hot day.

One end of the Nichrome (resistance) wire is connected to the sending unit for that particular gauge: a thermistor (temperature-variable-resistance solid-state device, see above) in the case of the water-temperature gauge; a simple variable resistor linked to a float in the example of the gas gauge (fig. 2 in the photo section), and a variable resistor linked to a diaphragm in the case of the oil gauge (“A” in fig. 4, photo section.) (Cars without an oil gauge have a warning light, which is activated by means of a simple switch - “B” in fig 4).

The other end of all these gauges need a reference voltage supply. Something rock-steady and unwavering. If this voltage varies, so will the gauge readings. Take a look a the crude regulator in fig. 6, at right. This relic of the 1930s is all that’s there for your gauges to work off of. It’s junk!

That gets me one step closer to your direct indicating "Joules per coulomb" voltmeter.

Old brain hasn't clicked onto it just yet..
BUT
This beats watching TV commercials !

Those old gages were well suited to their task
and provoked thought - as one drove down the road, all gages would shift slightly when you hit a shady part of road and the dashboard cooled a degree or two.

As i said - an old car is good for a kid. But he needs the shop manual and some tools.

thanks mentors for indulging me...

old jim

 

[CRT]

hats off jim hardy sir your enormous experience can do anything impossible...

 

[CRT]

(joule per second)/ampere gives joule per Coulomb can the readings from that old car metre be callibrated to give j/c

 

[Orien Rigney]

Capacitors placed across a D.C. battery will allow no fluctuation other than a steady rise to the maximum pressure (potential) a battery has stored in it. But! While a battery does not pass D.C. current, it will pass a cyclical maximum of A.C. potential.

 

[CRT]

yes you are correct.its only due to frequency of the supply.a capacitor after full charged behaves as open circuit due to infinite capacitive reactance when frequency is 0 say in dc

 

[jim hardy]

(joule per second)/ampere gives joule per Coulomb can the readings from that old car metre be callibrated to give j/c

i don't think so. I can't make it work in my head, at least.

But keep up the 'thought experiments'. And try some out with real parts.

 

[zoki85]

With voltmeters/ampermeters it is possible to measure various things, including the value of gravitational acceleration (g) near Earth surface.

 

[Orien Rigney]

I'm familiar with gravitational potential energy but have never tried using a voltmete to measure this energy. Please explain how it is done.

 

[zoki85]

I'm familiar with gravitational potential energy but have never tried using a voltmete to measure this energy. Please explain how it is done.

Some methods how g can be measured:

1. Climb the building 50 meters high. Drop the instrument and measure with a stop-watch how long it takes to hit the ground level. Then calculate g≈2h/t²
2. Hang the instrument to the ceiling by rope of length l. Measure time of a period of small oscillations of the pendulum. Then calculate g≈4π²l/T²
3. Read the manual of the instrument if the mass is listed there. Hang the instrument by dynamometer hooked to the ceiling and measure force. Then g=G/m

I would prefer methods 2 and 3 over 1 which is destructive and more expensive methods. And there are certainly other, more rafined methods of measurement

 

[jim hardy]

Fun with physics !

That's the theoretical approaches.

A practical fellow would go to the nearest university, find the geophysics department , and inquire of the lab supervisor:
"I'll give you this fine voltmeter if you'll tell me what is the precise value of local gravity."

Where i worked local gravity was ~0.17% less than standard
and it made that U of Miami geophysicist's day when i asked
because he'd just recently measured it to be around 978._something gils at Dodge Island seaport using a pendulum ..
Disclaimer - no bribery was necessary. You know what a treat it is when somebody shares your interest...

@CRT everyday things are tremendously interesting. Indulge that natural curiosity.Sigh again I've digressed
But "...all work and no play..."
we learn from play, too.

old jim

 

[Orien Rigney]

I believe dropping a one kilo weight from a height of one meter onto a well calibrated strain gage or scale would also suffice?

 

[zoki85]

Read this anecdote: http://www.spiritsound.com/bohr.html
He,he

 

[donpacino]

Ah okay. Thank both of you for your help, and sorry about the confusion. It is actually pretty simple when I think about it. It takes 9 J to pump a 1 C charge from the (+) terminal to the (-) terminal in the above diagram, and if the wire has resistance, then we lose some of this energy to heat, and of course how much we lose depends on the length of the wire and the resistance of the wire.

It appears that others have answered your questions. I just want to offer some prospective

You're studying computer engineering right?
While looking at energy and charge are important in some aspects, for the most part you will likely not work with them in your career as a computer engineer. You will most likely only work with voltage, current, and power.
When working with those, for the most part, you do not need to concern yourself with factors such as energy and charge. Most of the time they won't be important when looking at circuit analysis and design from a higher level.