This how I think of voltage / resistance / current

[vulturebetrayer]

Hello everyone.

I wrote the following up last night.  It's how I would explain the relationship between voltage, resistance and current.  Note, I am using Intensity of Current when discussing this because I feel it's important to hammer home the relationship between the variables in Ohms equations. 

Understanding the relationship of variables has been of the major issues I have run into when attempting to learn electronics equations.

I am concerned that maybe I'm missing something.  Maybe, I'm totally wrong and need a huge correction. 

I'd like to share this on my site with some examples and decent anecdotes that give real world problems new electronics folks may run into when attempting their first projects.  One example I can think of is using a resistor to limit current so as not to blow an LED;  maybe changing voltages using a buck converter or voltage regulator.

So all this started out as notes and then I added some structure to explain this to myself.
This is how I learn, by writing everything out and then going over it meticulously and finally showing someone who knows their stuff and having them explain to me where I'm horribly wrong.

So please take a look at this info and tell me what you would like changed / added / corrected:


V is Voltage measured in Volts
I is Intensity of current measured in Amps
R is Resistance measured in Ohms

•   Ohms Law:

•   Volts = Intensity times Resistance
•   V=IR

•   Intensity = Volts divided by Resistance
•   I = V/R

•   Resistance = Volts divided by Intensity
•   R = V/I

Practical Nomenclature of Ohms Law:
•   Volts Amps Ohms
•   Volts = Amps times Ohms
•   Ohms = Volts divided by Amps
•   Amps = Volts divided by Resistance

Think of voltage like like the 'speed' of electricity.  It's the difference in electrical potential.  This basically is the path the electrons you are pushing are going to follow.  Think of it as a moving pathway the electrons hitch a ride on, the more Volts the faster the electrons move.

Think of Amps like the intensity of the electricity or the amount of electrons being pushed at a time.  Think of the electrons on the moving pathway (Voltage) piling on more and more, you can move more electrons at a time if there are more amps.

Think of Ohms as amperage absorption.  The circuit uses a certain amount of amps to be functional, yet sometimes it is important to limit the amount of ohms.

So you must supply the proper voltage and amperage.  Your circuit may contain components that are not self-limiting in amperage; meaning it may want to pull more amps than it can handle, (which is a wonderful way to release magic smoke).  This is why you will need to calculate resistance and add a resistor if necessary to the positive side of the component if necessary.


After this I'd like to add some circuit examples, and real life equations.

I haven't considered explaining rectification because I don't fully understand the equations involved in the calculations.  I have built a rectifier using diodes, but don't understand how I'd choose different components other than using a generic schematic (still trying to wrap my brain around 'flutter' too with caps), so any resources on rectifiers and capacitance and flutter, and I'd be very thankful!

Thanks in advance!

[IanB]

So please take a look at this info and tell me what you would like changed / added / corrected:


V is Voltage measured in Volts
I is Intensity of current measured in Amps
R is Resistance measured in Ohms

•   Ohms Law:

•   Voltage = Intensity times Resistance
•   V=IR

•   Intensity = Voltage divided by Resistance
•   I = V/R

•   Resistance = Voltage divided by Intensity
•   R = V/I

Voltage is the quantity, volts is the unit of measure. If you are writing Intensity and Resistance, then you should write Voltage for consistency.

Practical Nomenclature of Ohms Law:
•   Volts Amps Ohms
•   Volts = Amps times Ohms
•   Ohms = Volts divided by Amps
•   Amps = Volts divided by ResistanceOhms

Think of voltage like like the 'speed' of electricity.  It's the difference in electrical potential.  This basically is the path the electrons you are pushing are going to follow.  Think of it as a moving pathway the electrons hitch a ride on, the more Volts the faster the electrons move.

Not quite. Voltage is like the 'pressure' of the electricity. It is how hard it pushes. If there are more volts the electricity will try harder to break through an obstacle and more current will flow.

Think of Amps like the intensity of the electricity or the amount of electrons being pushed at a time.  Think of the electrons on the moving pathway (Voltage) piling on more and more, you can move more electrons at a time if there are more amps.

Think of Ohms as amperage absorption.  The circuit uses a certain amount of amps to be functional, yet sometimes it is important to limit the amount of ohms.

No, current is never absorbed. Current is always conserved and never disappears. This is a very important law of electricity.

So you must supply the proper voltage and amperage.  Your circuit may contain components that are not self-limiting in amperage; meaning it may want to pull more amps than it can handle, (which is a wonderful way to release magic smoke).  This is why you will need to calculate resistance and add a resistor if necessary to the positive side of the component if necessary.

[edy]

I would alter this conceptualization quite a bit. For example, voltage is not the "speed" of the electricity nor the path it wants to push or follow. Ohms is not amperage absorption, and the idea of what Amps are is a bit hard to follow. While I appreciate your explanation, and it seems you did a lot of thinking about it, I want to clarify it based on my experience and what may be easier to teach people.

VOLTAGE:

Voltage is the difference in potential energy between two points. Just like gravitational field creates different potential at different heights above the ground. What happens is when there is a difference in energy levels, a FORCE is created that tries to equalize the difference (make it zero). For people jumping off buildings, the potential difference in gravitational energy turns into MOTION and you fall, picking up speed until you hit the ground. For electrons, the difference in potential energy causes EMF or electromotive force which moves electrons.

Now, if you are using the water analogy, you can also think of voltage as say the height of a waterfall. You can have a very tall waterfall, or a short waterfall. The taller waterfall is like higher amount of voltage. Obviously the water analogy only goes so far... it is an analogy. But essentially when you have a battery, or a capacitor, or static electricity building up between 2 objects.... There is a separation of CHARGES which creates a voltage or potential energy difference between 2 points in space. At some point, whether you have a conductor between these points or not, if the voltage difference will get big enough it will want to jump across the path.

In the case of a battery, if you connect the 2 ends with a wire (conductive path) the electrons will flow quite easily from one side to the other. Now, if you have static electricity you will need very large voltages to build up before a spark is created across the gap... because air is a poor conductor.

So in summary, voltage is when you have a CHARGE DIFFERENCE across something. That's all. How you choose to connect things across the charge difference (what materials, objects, etc.) will result in properties that you can measure called AMPERAGE (the number of electrons passing by per second) and RESISTANCE (an intrinsic property of the material to allow electrons to pass through it).


RESISTANCE:

Materials have the ability to allow electrons to flow through easily (conductor) or more difficult (insulator). Metals typically are better conductors because of the presence of "free" electrons, so they are not as tightly bound to the atoms and can move around through the metal, exchanging places with the ones beside them, so that the metal overall remains the same but yet there is the ability for electrons being supplied on one side of the metal able to make their way to the other side. That's what happens in a wire.

When you have an insulator, the electrons cannot pass through easily. They are more tightly bound to their atoms. So in order to make the electrons flow, you will need a HIGHER VOLTAGE. The bigger the potential difference between the 2 sides of the insulator (or the thinner the insulator) at some point electrons will flow. But not many.... that's when AMPERAGE comes in, because Voltage, Amps and Resistance are all related by Ohm's law.

AMPERAGE:

So now imagine you have a VOLTAGE being applied across a material with a certain intrinsic property called RESISTANCE. The voltage creates a potential energy difference across the material, so electrons wish to move across the material. There is a force generated, wishing to move the electrons across. That force causes movement of the electrons but has nothing to do with the speed of the electrons, because if you have a large resistance, those electrons may still move with much difficulty through the material, if any move at all.

The amperage is the net amount of electrons moving per second across the wire, or the net amount of charge in Coulombs. Given the same amount of resistance in a wire, if you increase voltage across it by 10x, you will get 10x increase in the net number of electrons travelling across the wire as well. So in general, the more voltage you apply to something, the more amps will flow across it.

Using the water analogy, if you think of a water pipe..... If the resistance is the size of the pipe (diameter), then voltage is the pressure being applied to the water to move, and the amperage is how much water is flowing through the pipe per second. So imagine a very tiny pipe (high resistance), you apply pressure to one end. Call that pressure P (this is like the voltage). Now you measure the number of gallons flowing through the pipe past a certain point.... that's like your amperage.

Now imagine you increase the diameter of your pipe, make it 2x wider. You need less pressure to push the water through... less voltage is needed to get the same amount of flow! So if your resistance drops to 1/2, keeping same voltage (pressure), you increase amps 2x (more current). Now imagine you have the same pipe but all you do is increase pressure by 2x, then your flow rate or amperage will also increase by 2x.

Anyways, I hope this makes more sense.... Electricity is not always as intuitive, and we can't use macroscopic analogies to always understand microscopic atomic/quantum phenomenon. But this is a good start.

 

[WaveyDipole]

Think of voltage like like the 'speed' of electricity.  It's the difference in electrical potential.  This basically is the path the electrons you are pushing are going to follow.  Think of it as a moving pathway the electrons hitch a ride on, the more Volts the faster the electrons move.

Voltage is 'the difference in electrical potential' but it would be incorrect to relate it to the 'speed' of electricity. The speed of electricity is related to the dielectric constant of the medium that it travels in. In a vacuum, electricity travels at the speed of light, which is about 300,000 kilometers per second. This is not related to the voltage, so it is incorrect to say that the more Volts the faster the electrons move. Voltage is a measure of potential difference, or electrical tension or pressure. Think of it like water pressure. The greater the pressure, the greater the POTENTIAL there is to release a greater or more powerful flow of water.

Think of Amps like the intensity of the electricity or the amount of electrons being pushed at a time.  Think of the electrons on the moving pathway (Voltage) piling on more and more, you can move more electrons at a time if there are more amps.

The flow of electricity is referred to as 'current' and amps are a measure of the intensity of that current flow. The greater the flow, the greater its measurement in amps. This is related to potential (the Volt is a measure of potential) because the greater the potential, the greater current will flow. For example, the higher the water pressure, the greater the volume of water that will flow in the same time frame when the tap is opened.

Think of Ohms as amperage absorption.  The circuit uses a certain amount of amps to be functional, yet sometimes it is important to limit the amount of ohms.

Ohms is a measure of resistance. The greater the resistance the more your current flow is impeded. Its a bit like the tap - the tighter it is closed, the higher the resistance to the water flow and therefore less water flows.  In electrical circuits, resistance restricts current flow and will often cause some energy to be given off as heat. While a circuit is working it consumes a certain amount of POWER which is measured in Watts and is a function of amps times volts (W = VA). I guess you could characterize this rather loosely as 'amperage absorption' as while current is flowing the circuit is carrying out work and consuming power. Current itself is not absorbed, just like water flowing through the pipe - it isn't absorbed anywhere as it has nowhere to go, except when it leaves the tap at which point it is expended.

So you must supply the proper voltage and amperage.  Your circuit may contain components that are not self-limiting in amperage; meaning it may want to pull more amps than it can handle, (which is a wonderful way to release magic smoke).  This is why you will need to calculate resistance and add a resistor if necessary to the positive side of the component if necessary.
[/tt][/b]

While it is true that in order to energize a circuit you need to supply the 'voltage' or electrical potential (e.g. a battery or AC power), 'amperage' is a measure of current flow, which is what happens when you activate the circuit by throwing the 'On' switch. You don't exactly supply it. Its a bit like opening a tap. The water is under pressure and is released and starts to flow. You can regulate the flow depending on how wide you open the tap. In a similar way, electricity starts to flow around your circuit. So with your example with the resistor and LED, you calculate the resistor such that sufficient current flows to light the LED but not burn it out. Its not so much that you supply the amps, but that you design the circuit to regulate the current flow, which in the case of a LED will be measured in milliamps.

No component is 'self-limiting in amperage'. It will have certain design characteristics which mean it will have minimum requirements in order to work satisfactorily and specified maximum limits that must not be exceeded otherwise the component will suffer damage - and possibly release 'magic smoke' in the process. So yes, one calculates the circuit design in order to avoid that.

PS. A couple of replies came in while I was typing this so apologies for any duplication.

[Ratch]

Hello everyone.

I wrote the following up last night.  It's how I would explain the relationship between voltage, resistance and current.  Note, I am using Intensity of Current when discussing this because I feel it's important to hammer home the relationship between the variables in Ohms equations. 

Understanding the relationship of variables has been of the major issues I have run into when attempting to learn electronics equations.

I am concerned that maybe I'm missing something.  Maybe, I'm totally wrong and need a huge correction. 

I'd like to share this on my site with some examples and decent anecdotes that give real world problems new electronics folks may run into when attempting their first projects.  One example I can think of is using a resistor to limit current so as not to blow an LED;  maybe changing voltages using a buck converter or voltage regulator.

So all this started out as notes and then I added some structure to explain this to myself.
This is how I learn, by writing everything out and then going over it meticulously and finally showing someone who knows their stuff and having them explain to me where I'm horribly wrong.

So please take a look at this info and tell me what you would like changed / added / corrected:

Yes, you need lots of correction.

V is Voltage measured in Volts
I is Intensity of current measured in Amps
R is Resistance measured in Ohms

You don't define quantities by reciting their measurement units.  Volts is not defined by voltage, amps is not defined by amperage, and resistance is not defined by ohms (notice that when used as a unit, "ohm" is not capitalized). 

•   Ohms Law:

•   Volts = Intensity times Resistance
•   V=IR

•   Intensity = Volts divided by Resistance
•   I = V/R

•   Resistance = Volts divided by Intensity
•   R = V/I

Practical Nomenclature of Ohms Law:
•   Volts Amps Ohms
•   Volts = Amps times Ohms
•   Ohms = Volts divided by Amps
•   Amps = Volts divided by Resistance

The good and correct physics books will tell you that Ohm's law is not V=IR and all its variations. Ohm's law refers to a material's electrical linearity property.  If a material, like most metals, conducts current in direct proportion to the voltage applied across it, it is considered "ohmic" and follows Ohm's law.  Materials, such as gases or components such as junction diodes, have very nonlinear voltage vs current curves, and do not follow Ohm's law.  V=IR and its variations is very correct, because it is the definition of resistance.  Again, it is not Ohm's law.  This misnomer has been propagated in many textbooks and other technical literature.  V=IR should be referred to correctly as the definition of resistance, and Ohm's law as a property of a material.

Think of voltage like like the 'speed' of electricity.

What is the speed of "electricity"?  Anyway, voltage is not a speed measurement.

It's the difference in electrical potential.

The potential of what?

This basically is the path the electrons you are pushing are going to follow.  Think of it as a moving pathway the electrons hitch a ride on, the more Volts the faster the electrons move.

Charge carriers (electrons in metals) follow a conduction path, not the voltage.  Units of "volts" are not capitalized.

Think of Amps like the intensity of the electricity or the amount of electrons being pushed at a time.  Think of the electrons on the moving pathway (Voltage) piling on more and more, you can move more electrons at a time if there are more amps.

Amps, like volts is not capitalized.  "Intensity" in physics means a quantity of something per unit.  A better word for the amount of amps is "value".  Amps is the charge flow per unit time.  "Electricity" of a vague term that can mean almost anything pertaining to electrical science.

Think of Ohms as amperage absorption.  The circuit uses a certain amount of amps to be functional, yet sometimes it is important to limit the amount of ohms.

Ouch!  That explanation hurts.

Resistance does not "absorb" amperage.  Resistance is an impediment to charge carrier flow.  Voltage is the energy density of electrical energy per unit charge.  That is why voltage has MKS units of joules/coulomb.  Look at it this way.  It takes energy to bring two electrons together, because they want to repel each other.  It takes more energy to bring many electrons together.  It takes still more energy to bring all the electrons closer together.  If you have a higher energy per unit of electrons (voltage) at one end of a wire, the electrons are going to move to the other end of the wire where the energy per unit of electrons (voltage) is lower.  This is because they naturally want to spread away from each other, so they go from high energy density per unit to the lower voltage.  During their trip through the wire they encounter resistance which causes them to lose energy as heat.  Therefore, they arrive at the lower voltage end of the wire with less energy density per unit charge (voltage).

So you must supply the proper voltage and amperage.  Your circuit may contain components that are not self-limiting in amperage; meaning it may want to pull more amps than it can handle, (which is a wonderful way to release magic smoke).  This is why you will need to calculate resistance and add a resistor if necessary to the positive side of the component if necessary.
[/tt][/b]

After this I'd like to add some circuit examples, and real life equations.

I haven't considered explaining rectification because I don't fully understand the equations involved in the calculations.  I have built a rectifier using diodes, but don't understand how I'd choose different components other than using a generic schematic (still trying to wrap my brain around 'flutter' too with caps), so any resources on rectifiers and capacitance and flutter, and I'd be very thankful!

Thanks in advance!

Ratch

[Stephan_T]

Hi,
just as brief thoughts about the basic electric terms:

CURRENT is what flows. The flow (transfer) of electric charge.

RESISTANCE is the property of an electric conductor, which hinders the current flow.

VOLTAGE, I find a bit more difficult to explain.

I would like to encourage you to look at other languages. The French term for voltage is "tension électrique".
In German it is called "elektrische Spannung". Looking at other english translations for this German word you find "electric tension", but also concepts like:
strain (on rope), pressure, eagerness, etc.
see http://dict.leo.org/?search=Spannung

So VOLTAGE is more of an electric push/pull kind of idea. I think of it as the electric tension that may cause a current to flow, as long as there is some kind of conductor available.

My most important thought about Ohms law is to remember that it only applies to conductors. You may have heard of "isolation resistance", but that is following Ohms law.

I remember Ohms law always in the form

R = V / I
where R is constant, while V and I may be turned up or down.

The basic idea is to observe the relation (ratio) between voltage and current at a given (passive) conductive component.
[while I was writing this, Ratch has given the better explanation of Ohms law]
The number of Ohms is the basically the number of volts that it takes to push one ampere through the resistor. Especially when looking at small resistance values (for example contact resistance of plugs and switches), I am using a 1A constant current source and measure the voltage drop across the resistor/resistance. The mV values directly translate to mOhms.

Included here is another important concept: The voltage drop. When current flows through a resistor/resistance, we speak of the voltage drop. In fact, I think of every voltage measurement actually as a measurement of the current through the (high) internal resistance of the meter.

[onlooker]

With all the more precise definitions and explanations, I expect the OP would get more confused. Based on the OP's writing, I think the better intuitive description or correction is to use the  analogy of water pipe network.
Nothing precise here, but can relates to other daily experience better.

Voltage     = pressure difference between any two points in the pipe network.
Current     = water flow (per second water volume flowing through a cross-section of a pipe)
Resistance = pipe resistance (e.g. thin or long pipe tends to restrict/slow down water flow more)
Battery     = water pump (to create pressure difference in the pipe network)
Capacitor   = Water tank with flexible membrane separate the inlet/outlet of the tank into 2 halves.
Inductor    =  water inertia (that create water hammers) 
Diode        = check valve/one-way valve.
...

[amyk]

Hi,
just as brief thoughts about the basic electric terms:

CURRENT is what flows. The flow (transfer) of electric charge.

RESISTANCE is the property of an electric conductor, which hinders the current flow.

VOLTAGE, I find a bit more difficult to explain.

I would like to encourage you to look at other languages. The French term for voltage is "tension électrique".
In German it is called "elektrische Spannung". Looking at other english translations for this German word you find "electric tension", but also concepts like:
strain (on rope), pressure, eagerness, etc.
see http://dict.leo.org/?search=Spannung

So VOLTAGE is more of an electric push/pull kind of idea. I think of it as the electric tension that may cause a current to flow, as long as there is some kind of conductor available.

It's interesting that you bring up the other languages, because in Chinese, the words for voltage is literally "electron pressure", current is "electron flow", and resistance is something like "electron obstruct".

In English, "current" should be interpreted not as in "new", but as in the movement of water in a river. The water analogy is probably the most intuitive and closest approximation to understanding; and it's been around for over 100 years, so you would be unlikely to find a better analogy.

[Ratch]

Hi,
just as brief thoughts about the basic electric terms:

CURRENT is what flows.

No, the charge carriers are what flow.  Current is the amount of charge carriers per unit time.

The flow (transfer) of electric charge.

Only means movement unless a rate is specified.

RESISTANCE is the property of an electric conductor, which hinders the current flow.

No, resistivity is a property of a material (electrical conductor).  Resistance is determined by the resistivity and the shape of the material.

VOLTAGE, I find a bit more difficult to explain.

I already did in my previous post.

I would like to encourage you to look at other languages. The French term for voltage is "tension électrique".
In German it is called "elektrische Spannung". Looking at other english translations for this German word you find "electric tension", but also concepts like:
strain (on rope), pressure, eagerness, etc.
see http://dict.leo.org/?search=Spannung

So VOLTAGE is more of an electric push/pull kind of idea. I think of it as the electric tension that may cause a current to flow, as long as there is some kind of conductor available.

If they don't describe voltage as the energy density per unit charge, then they are misdefining voltage.  The MKS units of voltage are joules/coulomb.  Current does not flow, by the way.  Charge flow rate is current.  Current flow means charge flow rate flow, which is redundant and ridiculous.  Current should be described as being present or existing.

My most important thought about Ohms law is to remember that it only applies to conductors. You may have heard of "isolation resistance", but that is following Ohms law.

Yes, you have to have conduction before you can determine whether the conduction is linear or not.

I remember Ohms law always in the form

R = V / I
where R is constant, while V and I may be turned up or down.

That is a false definition for Ohm's law.  That is the definition of resistance.  Ohm's law pertains to electrical linearity.

The basic idea is to observe the relation (ratio) between voltage and current at a given (passive) conductive component.
[while I was writing this, Ratch has given the better explanation of Ohms law]
The number of Ohms is the basically the number of volts that it takes to push one ampere through the resistor. Especially when looking at small resistance values (for example contact resistance of plugs and switches), I am using a 1A constant current source and measure the voltage drop across the resistor/resistance. The mV values directly translate to mOhms.

I specifically said that V=IR was not Ohm's law, but instead the definition of resistance.

Included here is another important concept: The voltage drop. When current flows through a resistor/resistance, we speak of the voltage drop. In fact, I think of every voltage measurement actually as a measurement of the current through the (high) internal resistance of the meter.

Which is an application of the definition of resistance, not Ohm's law.

Ratch
Hopelessly Pedantic

[Ratch]

With all the more precise definitions and explanations, I expect the OP would get more confused. Based on the OP's writing, I think the better intuitive description or correction is to use the  analogy of water pipe network.
Nothing precise here, but can relates to other daily experience better.

Does the analog water pipe circuit leak?

Voltage     = pressure difference between any two points in the pipe network.
Current     = water flow (per second water volume flowing through a cross-section of a pipe)
Resistance = pipe resistance (e.g. thin or long pipe tends to restrict/slow down water flow more)
Battery     = water pump (to create pressure difference in the pipe network)
Capacitor   = Water tank with flexible membrane separate the inlet/outlet of the tank into 2 halves.
Inductor    =  water inertia (that create water hammers) 
Diode        = check valve/one-way valve.
...

Will you get wet?

Ratch
Hopelessly Pedantic

[BobsURuncle]

Quote from: vulturebetrayer on Yesterday at 06:19:12 PM

Hello everyone.

I wrote the following up last night.  It's how I would explain the relationship between voltage, resistance and current.  Note, I am using Intensity of Current when discussing this because I feel it's important to hammer home the relationship between the variables in Ohms equations. 

>Understanding the relationship of variables has been of the major issues I have run into when attempting to learn electronics equations.

I am concerned that maybe I'm missing something.  Maybe, I'm totally wrong and need a huge correction. 

I'd like to share this on my site with some examples and decent anecdotes that give real world problems new electronics folks may run into when attempting their first projects.  One example I can think of is using a resistor to limit current so as not to blow an LED;  maybe changing voltages using a buck converter or voltage regulator.

So all this started out as notes and then I added some structure to explain this to myself.
This is how I learn, by writing everything out and then going over it meticulously and finally showing someone who knows their stuff and having them explain to me where I'm horribly wrong.

So please take a look at this info and tell me what you would like changed / added / corrected:


V is Voltage measured in Volts
I is Intensity of current measured in Amps
R is Resistance measured in Ohms

•   Ohms Law:

•   Volts = Intensity times Resistance
•   V=IR

•   Intensity = Volts divided by Resistance
•   I = V/R

•   Resistance = Volts divided by Intensity
•   R = V/I

Practical Nomenclature of Ohms Law:
•   Volts Amps Ohms
•   Volts = Amps times Ohms
•   Ohms = Volts divided by Amps
•   Amps = Volts divided by Resistance

Think of voltage like like the 'speed' of electricity.  It's the difference in electrical potential.  This basically is the path the electrons you are pushing are going to follow.  Think of it as a moving pathway the electrons hitch a ride on, the more Volts the faster the electrons move.

Think of Amps like the intensity of the electricity or the amount of electrons being pushed at a time.  Think of the electrons on the moving pathway (Voltage) piling on more and more, you can move more electrons at a time if there are more amps.

Think of Ohms as amperage absorption.  The circuit uses a certain amount of amps to be functional, yet sometimes it is important to limit the amount of ohms.

So you must supply the proper voltage and amperage.  Your circuit may contain components that are not self-limiting in amperage; meaning it may want to pull more amps than it can handle, (which is a wonderful way to release magic smoke).  This is why you will need to calculate resistance and add a resistor if necessary to the positive side of the component if necessary.


After this I'd like to add some circuit examples, and real life equations.

I haven't considered explaining rectification because I don't fully understand the equations involved in the calculations.  I have built a rectifier using diodes, but don't understand how I'd choose different components other than using a generic schematic (still trying to wrap my brain around 'flutter' too with caps), so any resources on rectifiers and capacitance and flutter, and I'd be very thankful!

Thanks in advance!

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

You have a number of conceptual errors here, as do some of the other contributors to this thread.  Watch this video, it will set you down the correct path.

[IanB]

The good and correct physics books will tell you that Ohm's law is not V=IR and all its variations. Ohm's law refers to a material's electrical linearity property.  If a material, like most metals, conducts current in direct proportion to the voltage applied across it, it is considered "ohmic" and follows Ohm's law.  Materials, such as gases or components such as junction diodes, have very nonlinear voltage vs current curves, and do not follow Ohm's law.  V=IR and its variations is very correct, because it is the definition of resistance.  Again, it is not Ohm's law.  This misnomer has been propagated in many textbooks and other technical literature.  V=IR should be referred to correctly as the definition of resistance, and Ohm's law as a property of a material.

If you want to say a material conducts current in direct proportion to the voltage across it, then

   I = (1/R) · V

is a precise mathematical statement of the same thing, where (1/R) is the constant of proportionality. The only difference is that the law of proportionality is in the first instance expressed in words and in the second in mathematical notation. It is the exact same statement either way.

If they don't describe voltage as the energy density per unit charge, then they are misdefining voltage.  The MKS units of voltage are joules/coulomb.

In physics, pressure has dimensions of energy per unit volume, but this is not a commonly useful way to think about pressure. Hence 1 Pa is usually defined as 1 N/m² rather than 1 J/m³.

Similarly, it is much more intuitive to think about voltage as an electrical pressure than as an energy density, especially for people who are trying to grasp fundamentals. There is a place to get into deep philosophical debates, but teaching is not it.

[Ratch]

-----------------------------------------------------

You have a number of conceptual errors here, as do some of the other contributors to this thread.  Watch this video, it will set you down the correct path.

[url=https://www.youtube.com/watch?v=yRLuZg5dm-E]https://www.youtube.com/watch?v=yRLuZg5dm-E

It was humorous seeing a disembodied head and hand doing the talking and gesturing.  I think analogs are OK for illustrating and explaining a particular point, but are confusing when coupled to another physics discipline.   Do hydraulic engineers learn their craft by studying electrical technology?  If not, then why should electrical students study hydraulics?  The professor seemed to imply that positive charges were present in a electrical circuit.  That, of course is not true.  Only negative charges are present in wires. Why do the analog folks equate pressure (force per unit area) with voltage.  I would think they would choose force for voltage.   After all, the electric field does exert a force on the conducting electrons.  The prof equated physical height with voltage, which I thought was not necessary.  Voltage is really an energy density, and is not a difficult concept to understand.

It would be instructive if you pointed out any errors and misconceptions you find in this thread.

Ratch
Hopelessly Pedantic

[Ratch]

The good and correct physics books will tell you that Ohm's law is not V=IR and all its variations. Ohm's law refers to a material's electrical linearity property.  If a material, like most metals, conducts current in direct proportion to the voltage applied across it, it is considered "ohmic" and follows Ohm's law.  Materials, such as gases or components such as junction diodes, have very nonlinear voltage vs current curves, and do not follow Ohm's law.  V=IR and its variations is very correct, because it is the definition of resistance.  Again, it is not Ohm's law.  This misnomer has been propagated in many textbooks and other technical literature.  V=IR should be referred to correctly as the definition of resistance, and Ohm's law as a property of a material.

If you want to say a material conducts current in direct proportion to the voltage across it, then

   I = (1/R) · V

is a precise mathematical statement of the same thing, where (1/R) is the constant of proportionality. The only difference is that the law of proportionality is in the first instance expressed in words and in the second in mathematical notation. It is the exact same statement either way.

I agree with the above statement, but how does that apply to the definition of Ohm's law?

In physics, pressure has dimensions of energy per unit volume, but this is not a commonly useful way to think about pressure. Hence 1 Pa is usually defined as 1 N/m² rather than 1 J/m³.

Nope, those are different units.  I challenge you to show me a link or other proof describing pressure defined that way.

Similarly, it is much more intuitive to think about voltage as an electrical pressure than as an energy density, especially for people who are trying to grasp fundamentals. There is a place to get into deep philosophical debates, but teaching is not it.

Intuitive?  I cannot conceive what "electrical pressure" is.  How would anyone put the definition of basic physical units into a philosophical concept?

Ratch
Hopelessly Pedantic

[IanB]

Ratch
Hopelessly Pedantic

You are messing up your quoting, making it hard to follow your posts.

Initially I thought from your signature that you were being deliberately obtuse. But it seems you genuinely lack understanding.

Rather than point out your mistakes right away, I will leave you some time to review what you have posted and see if you can work them out for yourself.

For now, I'll just state that voltage is indeed the direct electrical analog of hydraulic pressure.

[Ratch]

You are messing up your quoting, making it hard to follow your posts.

I try to eliminate some repetitive material that has been posted before, and break up the posts to answer individual points.  Sorry if some material is missing.

Initially I thought from your signature that you were being deliberately obtuse. But it seems you genuinely lack understanding.

Say on.  I am always ready to defend my positions.

Rather than point out your mistakes right away, I will leave you some time to review what you have posted and see if you can work them out for yourself.

I already have before I posted.  I just looked again and cannot see anything amiss.  Please list the items you don't agree with so I can elucidate my statements.

For now, I'll just state that voltage is indeed the direct electrical analog of hydraulic pressure.

So what if it is?  Like I said before, why study hydraulics instead of electronics?

Ratch
Hopelessly Pedantic

[IanB]

I try to eliminate some repetitive material that has been posted before, and break up the posts to answer individual points.  Sorry if some material is missing.

Please look back at your posts and see what you are doing. You are placing your words inside a quote block as if you are quoting me. This is not cool. Please fix it.

[Ratch]

IanB,

Well, I gave it my best shot.  It looks a little different, but I don't think the clarity of the narrative is much different.

Ratch
Hopelessly Pedantic

[IanB]

Well, I gave it my best shot.  It looks a little different, but I don't think the clarity of the narrative is much different.

Thanks, that makes it clearer who is being quoted 

[IanB]

If you want to say a material conducts current in direct proportion to the voltage across it, then

   I = (1/R) · V

is a precise mathematical statement of the same thing, where (1/R) is the constant of proportionality. The only difference is that the law of proportionality is in the first instance expressed in words and in the second in mathematical notation. It is the exact same statement either way.

I agree with the above statement, but how does that apply to the definition of Ohm's law?

If two statements are equivalent, and one statement is a statement of Ohm's law, then the other statement is also a statement of Ohm's law.

In physics, pressure has dimensions of energy per unit volume, but this is not a commonly useful way to think about pressure. Hence 1 Pa is usually defined as 1 N/m² rather than 1 J/m³.

Nope, those are different units.  I challenge you to show me a link or other proof describing pressure defined that way.

Well, you can work it out: 1 J/m³ = 1 Nm/m³ = 1 N/m². The units are identical. The identity of pressure with energy per unit volume is very important in physics and engineering.

Similarly, it is much more intuitive to think about voltage as an electrical pressure than as an energy density, especially for people who are trying to grasp fundamentals. There is a place to get into deep philosophical debates, but teaching is not it.

Intuitive?  I cannot conceive what "electrical pressure" is.  How would anyone put the definition of basic physical units into a philosophical concept?

When you said voltage was best understood as energy per unit charge I thought you were trying sow confusion, hence my misplaced comment about philosophical debates. Sorry about that.

But "electrical pressure" is nothing but voltage. You will find voltage described as electrical pressure in physics texts.

It works like this. Pressure is energy per unit volume (J/m³) and voltage is energy per unit charge (J/C). Both pressure and voltage are forms of energy density. Both pressure and voltage are also forms of motive force or potential (hence voltage differences are often referred to as EMF--electromotive force, or as PD--potential difference).

Continuing the similarity, if I wish to move a unit volume of fluid across a pressure gradient I have to do work on the fluid proportional to the difference in pressure (and in SI units the work performed is measured in joules). If I wish to move a unit of charge across a voltage gradient I have to do work on the charge proportional to the difference in voltage (and in SI units the work performed is also measured in joules).

Such similarities between potentials and flows in different domains are very useful in physics as an aid to understanding concepts. They are something students should learn to appreciate early in their studies.

[helius]

That was very good; basic, but excellent pedagogy. The professor has obviously had many experiences trying to explain electrical concepts to resistant (lol) students and thinking of better ways to do it. I think it's very easy to forget how we acquired our own basic knowledge and make the assumption that what seems obvious to us should be obvious to others, but finding clear, correct ways to explain our knowledge is not trivial.

The idea that voltage IS height, not a metaphor but literally, is fine if that is how you choose to think of it. You do need to be careful, however, because while a bare statement like that is not an analogy, the mind quickly jumps to analogies when using it. Later in the video the professor says "and if this colomb rolls down from 3V to 0V..." which sneaks across the analogy of rolling. Care is needed because there is no analog of inertia in the electrical world.

The professor seemed to imply that positive charges were present in a electrical circuit.  That, of course is not true.  Only negative charges are present in wires.

It's fine to be pedantic, but don't give that as an excuse for being dead wrong. If only negative charges were present there would be a huge net negative charge. The charge on a 1 square foot double-sided PCB would be -100 kilocoulombs, making it explode instantly at near the speed of light 
(With a relative permittivity of 4.7 across a 1.57mm thick FR-4 substrate, the force pulling on it would be .3 zettanewtons.)

[LvW]

Watch this video, it will set you down the correct path.

Thanks for the great video.
To me, the most important part of the video is mentioning the role of the electric field within the conducting material (wire or resistor or semiconductor).
This makes clear that the existence of a current (movement of charges) within each conducting material requires as a precondition an electric field - that means: An electric voltage!

Hence, from the physical point of view it is not correct (as we very often can hear or read) to say "a current is injected into..." or even "we pump a current into...".
It is always the voltage which allows a current - not vice versa. (And it is not a "chicken/egg problem" as I have heard very often).
I think, these facts are important for a good understanding of the relations between voltage and current.
As another example: In electronics, a so-called "current source" is always a voltage source equipped with a very large internal dynamic source resistance. 

Of course, for analyzing/designing circuits it is common practice to assume that a current can create a voltage V=IR across a resistor (simple example: resistive voltage divider).
But we always should keep in mind that such a mathematical relation does not imply any statement regarding cause and effect.       
(Similar considerations apply, of course, for other simple practical relations as Ic=B*Ib.) 

[Ratch]

If you want to say a material conducts current in direct proportion to the voltage across it, then

   I = (1/R) · V

is a precise mathematical statement of the same thing, where (1/R) is the constant of proportionality. The only difference is that the law of proportionality is in the first instance expressed in words and in the second in mathematical notation. It is the exact same statement either way.

I agree with the above statement, but how does that apply to the definition of Ohm's law?

If two statements are equivalent, and one statement is a statement of Ohm's law, then the other statement is also a statement of Ohm's law.

V=IR is not equivalent to Ohm's law.  The definition of resistance allows "R" in the equation to vary widely depending of the voltage and current.  Ohm's law insists that a material's voltage and current must be in direct proportion to each other.  For instance, the current vs voltage of a junction diode is a crooked curve.  It does not follow Ohm's law.  Yet it follows V=IR at all points on the curve.  Therefore, V=IR is not a definition of Ohm's law.  It is instead a definition of resistance.

In physics, pressure has dimensions of energy per unit volume, but this is not a commonly useful way to think about pressure. Hence 1 Pa is usually defined as 1 N/m² rather than 1 J/m³.

Nope, those are different units.  I challenge you to show me a link or other proof describing pressure defined that way.

Well, you can work it out: 1 J/m³ = 1 Nm/m³ = 1 N/m². The units are identical. The identity of pressure with energy per unit volume is very important in physics and engineering.

That is a perfect example of sophistry.  Just because two physical quantities have the same units does not mean they are equivalent.  For instance, torque has units of newton-meters, which is the same as units of translational energy. No one would say that torque is energy just because it has the same physical units, would they?  Identical physical quantities must have the same units, but two quantities having the same units does not necessarily mean they are equivalent.

But "electrical pressure" is nothing but voltage. You will find voltage described as electrical pressure in physics texts.

It works like this. Pressure is energy per unit volume (J/m³) and voltage is energy per unit charge (J/C). Both pressure and voltage are forms of energy density. Both pressure and voltage are also forms of motive force or potential (hence voltage differences are often referred to as EMF--electromotive force, or as PD--potential difference).

There is a lot of misinformation in textbooks and tech literature.  As I said before, just because two quantities have the same units does not mean they are the same.  Trying to equate characteristic electrical quantities to characteristic mass quantities adds another layer of confusion to the issue.

Continuing the similarity, if I wish to move a unit volume of fluid across a pressure gradient I have to do work on the fluid proportional to the difference in pressure (and in SI units the work performed is measured in joules). If I wish to move a unit of charge across a voltage gradient I have to do work on the charge proportional to the difference in voltage (and in SI units the work performed is also measured in joules).

I agree that energy is exchanged when moving a charge across different electrical energy densities.  I don't agree that moving a liquid between two different pressures causes any energy exchange.   Liquids are incompressible, so no energy is involved.  Now, if you specified a gas, then I would agree with your senario.

Such similarities between potentials and flows in different domains are very useful in physics as an aid to understanding concepts. They are something students should learn to appreciate early in their studies.

I think it is just a source of confusion.

Ratch

[helius]

Just because two physical quantities have the same units does not mean they are equivalent.  For instance, torque has units of newton-meters, which is the same as units of translational energy. No one would say that torque is energy just because it has the same physical units, would they?  Identical physical quantities must have the same units, but two quantities having the same units does not necessarily mean they are equivalent.

To be exact, torque has units of Newton × meters. It is a derived unit that is only meaningful when applied to vectors, and you cannot derive this unit from N·m because exchanging a dot for a cross product is not an allowed algebraic manipulation.
The derivation of N/m² from N·m/m³ is an allowed manipulation because a dot product is a scalar quantity.

[amyk]

Discussion about the basic principles always seem to attract a bunch of "my way is right, everyone else is wrong" arguments. I think there's a bit of bikeshedding going on. The other questions which have had similar effects here are "Which way does the current flow?" and "How does a transistor work?"

[Ratch]

Just because two physical quantities have the same units does not mean they are equivalent.  For instance, torque has units of newton-meters, which is the same as units of translational energy. No one would say that torque is energy just because it has the same physical units, would they?  Identical physical quantities must have the same units, but two quantities having the same units does not necessarily mean they are equivalent.

To be exact, torque has units of Newton × meters. It is a derived unit that is only meaningful when applied to vectors, and you cannot derive this unit from N·m because exchanging a dot for a cross product is not an allowed algebraic manipulation.
The derivation of N/m² from N·m/m³ is an allowed manipulation because a dot product is a scalar quantity.

Aren't you doing the same thing as above?  You are saying that energy/volume, which are both scalar quantities with no direction, is equivalent to pressure, which is a vector quantity, and thus has a direction.

Ratch
Hopelessly Pedantic

[Ratch]

Discussion about the basic principles always seem to attract a bunch of "my way is right, everyone else is wrong" arguments. I think there's a bit of bikeshedding going on. The other questions which have had similar effects here are "Which way does the current flow?" and "How does a transistor work?"

Sure, why not?  We can't all be right, can we?  Only by presenting informative facts for evaluation can we sort out which one of us has the correct perspective.

Current flow is an oxymoron.  I will be happy to explain that if your are interested.

To which type of transistor are you referring?  FETs, BJTs?  Each type works in different ways.

Ratch
Hopelessly Pedantic

[vulturebetrayer]

Wow guys.  Thanks for all the discussion on this topic!
I have learned a lot about the holes in my thinking and am going to go back to the drawing board before publishing anything.
The blow up on this thread is precisely why I posted here.
I don't want to publish anything that sets someone on the wrong track and from what I've seen here, I certainly would have.
Please keep the discussion going, it's very interesting.
 

[IanB]

Ratch, you have so many gaps in your knowledge and misunderstandings about basic concepts that you really should not be trying to teach others.

It is one thing to lack knowledge and be trying to learn. It is quite another to be ignorant and to be insisting you are correct as you are doing here. Claiming that you know better than physics texts and technical literature is the height of foolishness.

If you continue in the same way there are many knowledgeable people here who will rapidly lose patience with you.

[Ratch]

Ratch, you have so many gaps in your knowledge and misunderstandings about basic concepts that you really should not be trying to teach others.

With respect to knowledge of electronics and physics, I did not fall off the turnip truck yesterday.

It is one thing to lack knowledge and be trying to learn. It is quite another to be ignorant and to be insisting you are correct as you are doing here. Claiming that you know better than physics texts and technical literature is the height of foolishness.

I asked you before to show and prove that the knowledge I disseminated is incorrect.  Thus far, you have not done so.

If you continue in the same way there are many knowledgeable people here who will rapidly lose patience with you.

I would hope they would instead set me straight.

Ratch
Hopelessly Pedantic

[Ratch]

Wow guys.  Thanks for all the discussion on this topic!
I have learned a lot about the holes in my thinking and am going to go back to the drawing board before publishing anything.
The blow up on this thread is precisely why I posted here.
I don't want to publish anything that sets someone on the wrong track and from what I've seen here, I certainly would have.
Please keep the discussion going, it's very interesting.
 

It is good to see that you have a positive mental attitude.  Most folks would be intimidated and drop out from the discussion.  I applaud you.

Ratch
Hopelessly Pedantic

[Ratch]

Quote from Batch:
It was humorous seeing a disembodied head and hand doing the talking and gesturing.  I think analogs are OK for illustrating and explaining a particular point, but are confusing when coupled to another physics discipline.   Do hydraulic engineers learn their craft by studying electrical technology?  If not, then why should electrical students study hydraulics? 

The analogy is meant to foster understanding by drawing analogies with systems that are more commonly familiar to the student, i.e less abstract than the ideas of electric fields etc. If hydraulics confuse you then ignore it.     Analogs are also useful tools for solving real world engineering problems. Sometimes mechanical problems, for example, can be more easily solved by setting up an analogous electrical circuit and solving the circuit equations or merely observing the circuits behaviour.  In fact this was often done physically with analog computers back in the mid 20th century before digital computers were prevalent.

To each his own method.  But, it is one thing to use certain methods to solve problems, and another thing to teach and understand the science involved.

The professor seemed to imply that positive charges were present in a electrical circuit.  That, of course is not true.  Only negative charges are present in wires.

Of course there are positive charges in electrical circuits or there would be a humongous net negative charge.  The positive charge carriers are called protons.

Tilt!  There are no positive charge carriers in metals.  The negative charge carriers (electrons) in metals are loosely held to the metal atoms, are mobile, and contribute to the conductivity of the metal.  The positive ions of the metal are stationary, and do not transport any charge.


Why do the analog folks equate pressure (force per unit area) with voltage.  I would think they would choose force for voltage.   After all, the electric field does exert a force on the conducting electrons.

Voltage is not a force.  A force has direction, voltage does not have a direction.  Voltage is measure of potential energy difference.  Electric fields are the source of force on a charge.

Correct, but I was suggesting the analogy folks use force for voltage instead of pressure.  I never said that voltage was force.


The prof equated physical height with voltage, which I thought was not necessary.  Voltage is really an energy density, and is not a difficult concept to understand.

Voltage is not an energy density.  It is a measure of potential energy per unit of charge (Joules/Coulomb), or the amount of energy it takes to move 1C of charge against a 1 N/C E-field.  It is a scalar quantity, i.e. it has no direction.  This can be understood by remembering that Work (or energy) = force * distance, so for a coulomb of charge in an E-field the change in potential energy of the charge = E-field * distance moved, or in SI units:   V = (Newtons/Coulomb)*meters = N*m/C = J/C = Volts.

I said that voltage was the energy density per unit charge.  See reply #4 of this thread.  I agree with, and have known the facts in the above paragraph since the year one.

In the water pipe analogy:  (I use ? to represent the nabla symbol, i.e. the upside down delta)

Yes, the symbol for the gradient in vector calculus.
 

Pressure Gradient (?P) is in units of Newtons/meters^3;  it is analogous to E-field  which is in Newtons per unit of charge. These are a vector quantities.

Use whatever analogy you want.  Agreed, the E-field is a vector quantity.  Its field strength is measured in units of newtons/charge or volts/meter.  That is old news to me.
 

Pressure (P) is a measure of potential energy per unit volume of a fluid (joules/meter^3), or the amount of energy it takes to move 1 cubic meter of fluid against a pressure gradient of 1N/m^3. This is a scalar quantity just like the analogous Volts. Again, Energy = Force * distance, so for a cubic meter of fluid we have P = ?P * distance, or in SI units  (N/m^3)*m = N/m^2 = N*m/m^3 = J/m^3. (in the second term you see the more familiar definition of pressure as force per unit area: a Pascal in SI units).

If the "fluid" were a gas, I would agree with you.  But a liquid is incompressible, so so no work is done.  It is like applying a large force against a immovable object.  No work is done in that case, either.  The energy transfer to compress/expand a gas is well known in stoichiometic chemistry as PV.  No change in volume, no energy exchanged.
 

Summary of the analogy:
 
 E-field (N/C)  => Pressure Gradient (N/m^3)
 Volts  (J/C)    => Pressure                (J/m^3)

Say what you want about those analogies, I never had any use for them.

Ratch
Hopelessly Pedantic

[IanB]

Care is needed because there is no analog of inertia in the electrical world.

I'm curious why you do not think inductance counts?

Consider a short hydraulic pipe compared to a long hydraulic pipe. Increasing the flow rate in the long pipe requires work to be done to overcome the inertia of the fluid. Once the flow is established it resists change due to the stored kinetic energy (and this gives rise to water hammer effects with large pressure spikes if you try to interrupt the flow). The inertial effects are proportional to the length of the pipe.

Compare this with a long, straight wire. Increasing the current in the long wire requires work to be done to establish the magnetic field around the wire. Once the magnetic field is established it resists change due to the stored magnetic energy (and this gives rise to a back EMF with large voltage spikes if you try to interrupt the current). The inductance is proportional to the length of the wire.

The similarity here seems very direct. The main difference is that you need a very long straight wire to have a readily observable effect (unless you coil the wire around a magnetic core--but you can coil up a pipe too).

[BobsURuncle]

I said that voltage was the energy density per unit charge.  See reply #4 of this thread.  I agree with, and have known the facts in the above paragraph since the year one

Your definition is still wrong. Energy per unit charge is not the same as energy density or energy density per unit of charge.  I have two 12V batteries, which by definition each have 12V of potential energy across their terminals , but my 12V lithium ion battery has a much higher energy density than my 12V lead acid battery.

If the "fluid" were a gas, I would agree with you.  But a liquid is incompressible, so so no work is done.  It is like applying a large force against a immovable object.  No work is done in that case, either.  The energy transfer to compress/expand a gas is well known in stoichiometic chemistry as PV.  No change in volume, no energy exchanged.

A fluid can be a gas or a liquid.  You are confused between potential and kinetic energy: static pressure which has potential energy and a pressure differential which clearly can do work moving a compress-able or in-compress-able fluid through a pipe.

[LvW]

Claiming that you know better than physics texts and technical literature is the height of foolishness.

IanB - surely, I do not want to jump into your discussioin with Ratch. However, I cannot resist to comment your above quoted sentence.
Because within the last 30 years I have seen so many errors and false explanations in the techical literature, I only can warn anybody to blindly rely on statements and claims to be found in printed form or in the internet. 

[IanB]

Claiming that you know better than physics texts and technical literature is the height of foolishness.

IanB - surely, I do not want to jump into your discussioin with Ratch. However, I cannot resist to comment your above quoted sentence.
Because within the last 30 years I have seen so many errors and false explanations in the techical literature, I only can warn anybody to blindly rely on statements and claims to be found in printed form or in the internet.

I understand that mistakes can be made and that corrections are sometimes published, but broad and general errors across a body of literature are surely rather rare when describing fundamental concepts, especially in university textbooks?

I do agree with you however that blind and uncritical acceptance of anything written is something to be avoided.

[Ratch]

I said that voltage was the energy density per unit charge.  See reply #4 of this thread.  I agree with, and have known the facts in the above paragraph since the year one

Your definition is still wrong. Energy per unit charge is not the same as energy density or energy density per unit of charge.  I have two 12V batteries, which by definition each have 12V of potential energy across their terminals , but my 12V lithium ion battery has a much higher energy density than my 12V lead acid battery.

Lets get this cleared up.  Voltage is electrical energy per charge, which is an energy density.  Your example does not prove me wrong.  A Li-ion might have a larger energy capacity, but all that means it can sustain an voltage at a particular current for a longer period of time.  One does not define energy storage of a battery by its output voltage.  Voltage is still the energy density of the charge.

If the "fluid" were a gas, I would agree with you.  But a liquid is incompressible, so so no work is done.  It is like applying a large force against a immovable object.  No work is done in that case, either.  The energy transfer to compress/expand a gas is well known in stoichiometic chemistry as PV.  No change in volume, no energy exchanged.

A fluid can be a gas or a liquid.  You are confused between potential and kinetic energy: static pressure which has potential energy and a pressure differential which clearly can do work moving a compress-able or in-compress-able fluid through a pipe.

No, you did not explain clearly what you were talkiing about.  If you are moving a mass of gas or liquid through a pipeline, then yes, that takes energy.  If you are only compressing a gas, that takes energy, too.  If you try to just compress a liquid, then no work is done.  The confusion does not come from the definition of potiential or kinetic energy.  It comes from not knowing what you mean.

Ratch
Hopelessly Pedantic

[amyk]

Someone just registers and 15 of their 16 posts are in this thread, claiming everyone else is wrong and they're right. Is anyone else thinking what I'm thinking...? 

[helius]

The derivation of N/m² from N·m/m³ is an allowed manipulation because a dot product is a scalar quantity.

Aren't you doing the same thing as above?  You are saying that energy/volume, which are both scalar quantities with no direction, is equivalent to pressure, which is a vector quantity, and thus has a direction.

I don't think so. What is the direction of a balloon inflated to 50 psi?
The force vector is integrated over a surface, conceptually taking the dot product of the surface normal at each patch.
\$ \displaystyle \iint_{s} \mathbf F \cdot d \mathbf \Sigma \, = \iint_{s} ( \mathbf F \cdot \mathbf n ) \, d \Sigma \$
The result is scalar.

[Ratch]

The derivation of N/m² from N·m/m³ is an allowed manipulation because a dot product is a scalar quantity.

Aren't you doing the same thing as above?  You are saying that energy/volume, which are both scalar quantities with no direction, is equivalent to pressure, which is a vector quantity, and thus has a direction.

I don't think so. What is the direction of a balloon inflated to 50 psi?
The force vector is integrated over a surface, conceptually taking the dot product of the surface normal at each patch.
\$ \displaystyle \iint_{s} \mathbf F \cdot d \mathbf \Sigma \, = \iint_{s} ( \mathbf F \cdot \mathbf n ) \, d \Sigma \$
The result is scalar.

You are calculating the divergence of an enclosed surface using vector calculus.  The result will be 50 times the area of the balloon surface in square inches.  How does that turn energy/volume into force/area?

Ratch
Hopelessly Pedantic

[Ratch]

Someone just registers and 15 of their 16 posts are in this thread, claiming everyone else is wrong and they're right. Is anyone else thinking what I'm thinking...? 

They probably think I haven't been on this forum very long.  Isn't that the logical conclusion?  Thus far, only a very few folks have challenged my propositions.  After all, I don't submit something without giving a logical reason or proof.

Ratch
Hopelessly Pedantic

[onlooker]

How does that turn energy/volume into force/area

The discussion about this is all over internet. Just google it. say wiki or any .edu sites. For example,
from http://hyperphysics.phy-astr.gsu.edu/hbase/press.html

"Pressure in a fluid can be seen to be a measure of energy per unit volume by means of the definition of work."

Or from wikipedia,

"Energy per unit volume has the same physical units as pressure, and in many circumstances is a synonym"

[Ratch]

How does that turn energy/volume into force/area

The discussion about this is all over internet. Just google it. say wiki or any .edu sites. For example,
from http://hyperphysics.phy-astr.gsu.edu/hbase/press.html

"Pressure in a fluid can be seen to be a measure of energy per unit volume by means of the definition of work."

Yes, and an even more detailed answer at http://physics.stackexchange.com/questions/216342/what-is-pressure-energy.  Both those treatments of the topic involve fluid dynamics and kinematics with Bernoulli's equation thrown in.  I was hoping to find a derivation of energy volume density to force/area pressure for a static system, if there is one.  Will keep looking.

Ratch
Hopelessly Pedantic

[onlooker]

I was hoping to find a derivation of energy volume density to force/area pressure for a static system, if there is one

Just as in most physical systems, if not all, static case is just a special case or limiting case of the general/non-static system and concepts apply equally.

[Ratch]

I was hoping to find a derivation of energy volume density to force/area pressure for a static system, if there is one

Just as in most physical systems, if not all, static case is just a special case or limiting case of the general/non-static system and concepts apply equally.

That observation does not help much.  For example, if a have quantity of water in a closed container under pressure, what will be the energy/volume?  If the tank is part of a tube or pipeline where the water is flowing, then the kinetic and potential energy can be calculated.  But what is the energy/volume it when the tank is isolated by itself?  I can easily determine the pressure of the tank, and it is supposed to be the same as the energy/volume.  Or is it for static systems?

By the way, I checked a physics book and discovered that pressure is considered a scalar quantity and not a vector quantity like I first thought.  That is because the pressure is always considered at right angles to the surface, so no direction variation is permitted.

Ratch
Hopelessly Pedantic

[IanB]

I was hoping to find a derivation of energy volume density to force/area pressure for a static system, if there is one.

You have been shown it. It comes from the dimensional identity that force-per-unit-area is the same as energy-per-unit-volume. This occurs because work done on a system causes a change in energy of the system by the same amount, and work equals force times distance.

Thus (in SI units), if I move 1 m³ of an incompressible fluid through a pressure differential of 1 N/m², I do work on the fluid equal to 1 m³ x 1 N/m² = 1 Nm = 1 J. I have therefore done 1 J of work on 1 m³ of fluid. By moving the fluid volume through a pressure differential of 1 N/m² I have increased its potential to do work by 1 J/m³. The two statements are equivalent.

Note that I was careful to stipulate an (ideal) incompressible fluid above, since the thermodynamics of compressible fluids are more complex and the analogy with electricity breaks down.

[IanB]

By the way, I checked a physics book and discovered that pressure is considered a scalar quantity and not a vector quantity like I first thought.  That is because the pressure is always considered at right angles to the surface, so no direction variation is permitted.

That is correct. I'm glad you discovered that by your own research.

[Ratch]

I was hoping to find a derivation of energy volume density to force/area pressure for a static system, if there is one.

You have been shown it. It comes from the dimensional identity that force-per-unit-area is the same as energy-per-unit-volume. This occurs because work done on a system causes a change in energy of the system by the same amount, and work equals force times distance.

Thus (in SI units), if I move 1 m³ of an incompressible fluid through a pressure differential of 1 N/m², I do work on the fluid equal to 1 m³ x 1 N/m² = 1 Nm = 1 J. I have therefore done 1 J of work on 1 m³ of fluid. By moving the fluid volume through a pressure differential of 1 N/m² I have increased its potential to do work by 1 J/m³. The two statements are equivalent.

Note that I was careful to stipulate an (ideal) incompressible fluid above, since the thermodynamics of compressible fluids are more complex and the analogy with electricity breaks down.

As I mentioned in the above post. What energy does a tank of fluid have at particular volume and pressure when it is isolated, and not part of a moving fluid stream?

Ratch
Hopelessly Pedantic

[IanB]

As I mentioned in the above post. What energy does a tank of fluid have at particular volume and pressure when it is isolated, and not part of a moving fluid stream?

Energy is relative, not absolute, so the correct question to ask is "How much energy does it take to fill the tank with fluid?"

If, for example, the tank were at the top of a tower (a water tower), then the water in the tank would have potential energy according to the formula E = Mgh, where M is the mass of liquid, g is the gravitational acceleration and h is the elevation above ground. It would take that much energy to pump the water up into the tower (plus friction losses), and you could get that much energy back again (minus friction losses) by letting the water down.

[IanB]

As I mentioned in the above post. What energy does a tank of fluid have at particular volume and pressure when it is isolated, and not part of a moving fluid stream?

Let's suppose you are thinking of a closed metal tank completely full of water at atmospheric pressure, and now suppose you want to add a small additional volume of water into the tank. The work required to do this is equal to the volume of water added times the pressure difference, which initially is zero. But after you have pumped in a bit of water the pressure in the tank has gone up (let's say the walls are elastic and have stretched a bit). So the next bit of water you want to pump in will require some work as the pressure difference is no longer zero. The next bit of water after that will require more work, and the next more work still. The work to keep pumping in more water will keep increasing as the pressure in the tank goes up and up.

Hopefully you will begin to see the similarity here with pumping charge into a capacitor.

The energy you can get back out of the tank after you have pumped water into it is essentially equal to the work you did pumping water into it. The water itself doesn't have energy, but the tank plus water system has stored some energy, just like the capacitor plus charge system has stored some energy when you charge it up.

This, and the recognition that both pressure and voltage are scalar fields of potential, goes further towards explaining why voltage is "electrical pressure".

[Ratch]

As I mentioned in the above post. What energy does a tank of fluid have at particular volume and pressure when it is isolated, and not part of a moving fluid stream?

Energy is relative, not absolute, so the correct question to ask is "How much energy does it take to fill the tank with fluid?"

If, for example, the tank were at the top of a tower (a water tower), then the water in the tank would have potential energy according to the formula E = Mgh, where M is the mass of liquid, g is the gravitational acceleration and h is the elevation above ground. It would take that much energy to pump the water up into the tower (plus friction losses), and you could get that much energy back again (minus friction losses) by letting the water down.

Agreed, but I specified an isolated tank.  How much energy to pressurize an isolated tank of fluid to a particular pressure?

Ratch

[IanB]

Agreed, but I specified an isolated tank.  How much energy to pressurize an isolated tank of fluid to a particular pressure?

If the tank is isolated, how do you propose to change its pressure?

[Ratch]

As I mentioned in the above post. What energy does a tank of fluid have at particular volume and pressure when it is isolated, and not part of a moving fluid stream?

Let's suppose you are thinking of a closed metal tank completely full of water at atmospheric pressure, and now suppose you want to add a small additional volume of water into the tank. The work required to do this is equal to the volume of water added times the pressure difference, which initially is zero. But after you have pumped in a bit of water the pressure in the tank has gone up (let's say the walls are elastic and have stretched a bit). So the next bit of water you want to pump in will require some work as the pressure difference is no longer zero. The next bit of water after that will require more work, and the next more work still. The work to keep pumping in more water will keep increasing as the pressure in the tank goes up and up.

Hopefully you will begin to see the similarity here with pumping charge into a capacitor.

The energy you can get back out of the tank after you have pumped water into it is essentially equal to the work you did pumping water into it. The water itself doesn't have energy, but the tank plus water system has stored some energy, just like the capacitor plus charge system has stored some energy when you charge it up.

This, and the recognition that both pressure and voltage are scalar fields of potential, goes further towards explaining why voltage is "electrical pressure".

Allowing the walls of the tank to be elastic is a backdoor way of saying the liquid is compressible. One way or the other, the volume of the liquid has to change for energy to exchange.

You might call it electrical pressure for an analog, but the it does not make sense for a real electrical circuit.

By the way, you don't pump up a capacitor with charge. You pump it up with energy.  The net charge of a capacitor is the same at 0 volts, 100 volts, or 1000 volts.  The amount of charge on one plate is exactly the same as the charge removed on the opposite plate for a net change of zero.  So it is wrong to say a capacitor is "charged".  It is instead energized.  Same for batteries.

Ratch
Hopelessly Pedantic

[Ratch]

Agreed, but I specified an isolated tank.  How much energy to pressurize an isolated tank of fluid to a particular pressure?

If the tank is isolated, how do you propose to change its pressure?

With completely rigid tank walls and a incompressible liquid, you cannot.  With gas, one can release the gas or pump in more.

Ratch

[helius]

You are calculating the divergence of an enclosed surface using vector calculus.  The result will be 50 times the area of the balloon surface in square inches.  How does that turn energy/volume into force/area?

You normalize the surface S to 1 unit of area, which is where the m^-2 comes from. I was showing you that the pressure is a scalar, even though it is derived from a vector. The force really needs to be calculated as a tensor field in this case, and I'll admit my tensor calculus is not very strong. But I hope I don't need to argue that this can't change the units of the problem.

In practice you would skip the tensors (since the air is isothermic and isobaric) and calculate the pressure using the ideal gas law, which is completely scalar. In \$ PV = nRT \$, R has units of joules per kelvin. Which means that the product of pressure and volume is energy 

[Ratch]

You are calculating the divergence of an enclosed surface using vector calculus.  The result will be 50 times the area of the balloon surface in square inches.  How does that turn energy/volume into force/area?

You normalize the surface S to 1 unit of area, which is where the m^-2 comes from. I was showing you that the pressure is a scalar, even though it is derived from a vector. The force really needs to be calculated as a tensor field in this case, and I'll admit my tensor calculus is not very strong. But I hope I don't need to argue that this can't change the units of the problem.

In practice you would skip the tensors (since the air is isothermic and isobaric) and calculate the pressure using the ideal gas law, which is completely scalar. in \$ PV = nRT \$, R has units of joules per kelvin. Which means that the product of pressure and volume is energy 

I agree with you.  My physics book says that pressure force is defined to be only in the normal direction from the surface.  Therefore no directional variation is possible, which makes pressure a scalar value.

Ratch
Hopelessly Pedantic

[IanB]

Allowing the walls of the tank to be elastic is a backdoor way of saying the liquid is compressible. One way or the other, the volume of the liquid has to change for energy to exchange.

If you had a (hypothetical) perfectly incompressible fluid contained within a (hypothetical) inelastic vessel, then the pressure inside the vessel could be any value you want it to be. In such a case no energy change is required for the pressure to change.

You might call it electrical pressure for an analog, but the it does not make sense for a real electrical circuit.

Everything has matched exactly so far. For instance, if I take an insulated metal sphere and move it in an electric field from a position of lower potential (let's say 0 V) to a position of higher potential (let's say 300 kV), then I have increased the voltage on the sphere by 300 kV. How much additional energy is stored in the sphere after I have done this? (This is the analog of the incompressible fluid in the inelastic vessel.)

By the way, you don't pump up a capacitor with charge. You pump it up with energy.  The net charge of a capacitor is the same at 0 volts, 100 volts, or 1000 volts.  The amount of charge on one plate is exactly the same as the charge removed on the opposite plate for a net change of zero.  So it is wrong to say a capacitor is "charged".  It is instead energized.  Same for batteries.

When a conventional two plate capacitor is charged up, charge gets moved from one plate to the other, and it takes work to do this. A capacitor can then do work on an external circuit by allowing the accumulated charge to flow back to the lower energy state. So it is quite legitimate to say the capacitor has been "charged up".

If you had a single pole capacitor (a metal sphere in free space), then you would actually have to add charge to it (or remove charge from it) when wanting to change its electrical potential.

[Ratch]

Allowing the walls of the tank to be elastic is a backdoor way of saying the liquid is compressible. One way or the other, the volume of the liquid has to change for energy to exchange.

If you had a (hypothetical) perfectly incompressible fluid contained within a (hypothetical) inelastic vessel, then the pressure inside the vessel could be any value you want it to be. In such a case no energy change is required for the pressure to change.

The pressure could only be zero because no volume change or compression is allowed.

You might call it electrical pressure for an analog, but the it does not make sense for a real electrical circuit.

Everything has matched exactly so far. For instance, if I take an insulated metal sphere and move it in an electric field from a position of lower potential (let's say 0 V) to a position of higher potential (let's say 300 kV), then I have increased the voltage on the sphere by 300 kV. How much additional energy is stored in the sphere after I have done this? (This is the analog of the incompressible fluid in the inelastic vessel.)

As I said before, I am not into analogs too much.

By the way, you don't pump up a capacitor with charge. You pump it up with energy.  The net charge of a capacitor is the same at 0 volts, 100 volts, or 1000 volts.  The amount of charge on one plate is exactly the same as the charge removed on the opposite plate for a net change of zero.  So it is wrong to say a capacitor is "charged".  It is instead energized.  Same for batteries.

When a conventional two plate capacitor is charged up, charge gets moved from one plate to the other, and it takes work to do this. A capacitor can then do work on an external circuit by allowing the accumulated charge to flow back to the lower energy state. So it is quite legitimate to say the capacitor has been "charged up".

No, the charge is added to one plate and deleted from the other plate.  The effect may be the same as a charge shift, but the net change of the charge is still zero.  You could say one plate is charged up, but then it is also legitimate to say the other plate is charged down.  It is unambiguous to say the cap is charged with energy, so one might as well say it is energized.

If you had a single pole capacitor (a metal sphere in free space), then you would actually have to add charge to it (or remove charge from it) when wanting to change its electrical potential.

Google did not have much information on a single-pole capacitor.  What they did show was not a metal sphere in space.  Where is its dielectric?  What is its voltage reference?  Show me a link describing one.

Ratch
Hopelessly Pedantic

[rs20]

The pressure could only be zero because no volume change or compression is allowed.

Wrong, pressure is defined as Force/Area. There is no requirement for changes in volume for a force to be applied, particularly in the hypothetical inelastic regime that IanB is referencing.

No, the charge is added to one plate and deleted from the other plate.  The effect may be the same as a charge shift, but the net change of the charge is still zero.  You could say one plate is charged up, but then it is also legitimate to say the other plate is charged down.  It is unambiguous to say the cap is charged with energy, so one might as well say it is energized.

You can say a capacitor is storing a certain amount of charge, and you can say that a capacitor is storing a certain amount of energy. It is pointless to try and claim that one view is better than the other, because each is useful in different circumstances. I think about the charge stored in a capacitor far more often than the energy; and in particular, if one end of the capacitor is connected to circuit ground (which is a very common use-case), the "charge on the capacitor" becomes a blindingly obviously unambiguous and useful concept.

Google did not have much information on a single-pole capacitor.  What they did show was not a metal sphere in space.  Where is its dielectric?  What is its voltage reference?  Show me a link describing one.

Dielectric: space. Voltage reference: theoretical sphere of infinite radius. Link: https://en.wikipedia.org/wiki/Capacitance#Self-capacitance . For example, planet Earth could be considered a single-pole capacitor of value 710uF.

Btw, I encourage you use the "Preview" feature to make sure your quotes are matched up properly, makes it less confusing to read.

[IanB]

Google did not have much information on a single-pole capacitor.  What they did show was not a metal sphere in space.  Where is its dielectric?  What is its voltage reference?  Show me a link describing one.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capsph.html#c2

[Ratch]

The pressure could only be zero because no volume change or compression is allowed.

Wrong, pressure is defined as Force/Area. There is no requirement for changes in volume for a force to be applied, particularly in the hypothetical inelastic regime that IanB is referencing.

We are talking about an incompressible liquid and inelastic tank walls.  If those two entities cannot be expanded or compressed, then how can a pressure be applied?

No, the charge is added to one plate and deleted from the other plate.  The effect may be the same as a charge shift, but the net change of the charge is still zero.  You could say one plate is charged up, but then it is also legitimate to say the other plate is charged down.  It is unambiguous to say the cap is charged with energy, so one might as well say it is energized.

You can say a capacitor is storing a certain amount of charge, and you can say that a capacitor is storing a certain amount of energy. It is pointless to try and claim that one view is better than the other, because each is useful in different circumstances. I think about the charge stored in a capacitor far more often than the energy; and in particular, if one end of the capacitor is connected to circuit ground (which is a very common use-case), the "charge on the capacitor" becomes a blindingly obviously unambiguous and useful concept.

No, I made it very clear that a capacitor has the same net charge when it is energized at 0, 100 volts, or 1000 volts.  That is because the charge added to one plate is removed from the opposite plate for a net change of zero.  Only the energy stored in the capacitor changes when a voltage is applied.

Google did not have much information on a single-pole capacitor.  What they did show was not a metal sphere in space.  Where is its dielectric?  What is its voltage reference?  Show me a link describing one.

Dielectric: space. Voltage reference: theoretical sphere of infinite radius. Link: https://en.wikipedia.org/wiki/Capacitance#Self-capacitance . For example, planet Earth could be considered a single-pole capacitor of value 710uF.

The article mentions a Van de Graaff generator and the planet Earth. No explanation of what constitutes the plates and dielectric. Neither is a single metal sphere.

Ratch
Hopeless Pedantic

[Ratch]

Google did not have much information on a single-pole capacitor.  What they did show was not a metal sphere in space.  Where is its dielectric?  What is its voltage reference?  Show me a link describing one.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capsph.html#c2

OK, I looked at the link.  Too bad the reference the link gives is missing.  A single-pole capacitor (SPC) does not look too practical, because it would take a a static generator to charge up its one plate.  Why does it have to be a metal sphere.  Why not a big piece of sheet metal?  I will concede, however, that a SPC does "charge up".  Unlike a two-pole capacitor.

Ratch
Hopelessly Pedantic

[Stephan_T]

A single-pole capacitor (SPC) does not look too practical, [...]

You find them practically everywhere. All you need is isolation (e.g. pair of rubber soles will do). To charge them, it just takes some friction (e.g. walk over a carpet of a certain material will do) and you yourself can get charged substantially. This has the potential (pun intended) to hurt you.
But the EMF of such a charge can also interfere with your multimeter and oscilloscope. I usually use a PVC-Installation tube as test device to experiment with that effect.
Depending on the impedance of the measurement circuit, it can easily cause voltage readings of hundreds of mV for hundreds of milliseconds just exposed to the tiny tip of an otherwise shielded oscilloscope probe. If you use unshielded wires it gets much worse.

[rs20]

We are talking about an incompressible liquid and inelastic tank walls.  If those two entities cannot be expanded or compressed, then how can a pressure be applied?

[/quote]

Depending on your point of view and your level of trust in infinities and infinitesimals:
- Pressure can be applied by adding an infinitesimal amount of incompressible liquid. The point here is that the pressure in a tank with incompressible liquid and inelastic tank walls is undefined, not zero -- try taking limits as compressibility approaches infinity, you'll find that the pressure does not converge to zero.
- Pressure can be applied having an (inelastic but moveable) sliding piston in the side of the box. Pressure inside the box equals normal force on the piston / area of piston.

No, I made it very clear that a capacitor has the same net charge when it is energized at 0, 100 volts, or 1000 volts.  That is because the charge added to one plate is removed from the opposite plate for a net change of zero.  Only the energy stored in the capacitor changes when a voltage is applied.

I agree that the net charge on a capacitor remains unchanged (neglecting self-capacitance of the whole assembly, of course). However, the turn of phrase "charge up a capacitor" remains obviously meaningful and useful, even if it is only referring to a charge imbalance within the capacitor.

[Ratch]

A single-pole capacitor (SPC) does not look too practical, [...]

You find them practically everywhere. All you need is isolation (e.g. pair of rubber soles will do). To charge them, it just takes some friction (e.g. walk over a carpet of a certain material will do) and you yourself can get charged substantially. This has the potential (pun intended) to hurt you.
But the EMF of such a charge can also interfere with your multimeter and oscilloscope. I usually use a PVC-Installation tube as test device to experiment with that effect.
Depending on the impedance of the measurement circuit, it can easily cause voltage readings of hundreds of mV for hundreds of milliseconds just exposed to the tiny tip of an otherwise shielded oscilloscope probe. If you use unshielded wires it gets much worse.

I have been thinking about that.  Statically charged particles are a nuisance, but are they really miniature  capacitors?  Most are not hollow conducting shells, so their geometry is not such that their charge is separated from their electric field.  It is something to think about

Ratch
Hopelessly Pedantic

[Ratch]

We are talking about an incompressible liquid and inelastic tank walls.  If those two entities cannot be expanded or compressed, then how can a pressure be applied?

Depending on your point of view and your level of trust in infinities and infinitesimals:
- Pressure can be applied by adding an infinitesimal amount of incompressible liquid. The point here is that the pressure in a tank with incompressible liquid and inelastic tank walls is undefined, not zero -- try taking limits as compressibility approaches infinity, you'll find that the pressure does not converge to zero.
- Pressure can be applied having an (inelastic but moveable) sliding piston in the side of the box. Pressure inside the box equals normal force on the piston / area of piston.

I have to apologize.  I said "pressure", but I meant energy.  I guess I got stuck on automatic.  The piston will not be able to transfer any energy to the incompressible liquid, because it will not be able to move further, and no more energy can be exchanged.  This is similiar to a heavy force being applied to a immovable object.  No work is done.

No, I made it very clear that a capacitor has the same net charge when it is energized at 0, 100 volts, or 1000 volts.  That is because the charge added to one plate is removed from the opposite plate for a net change of zero.  Only the energy stored in the capacitor changes when a voltage is applied.

I agree that the net charge on a capacitor remains unchanged (neglecting self-capacitance of the whole assembly, of course). However, the turn of phrase "charge up a capacitor" remains obviously meaningful and useful, even if it is only referring to a charge imbalance within the capacitor.

Charging a cap or battery is an example of technical slang prevalent the tech world.  Another two examples are "current flow", and NASA referring to their astronauts as "walking" in space.  If their tether breaks, would they walk away?  They should say "external excursion".  Even though folks understand what the slang means, it is not accurately descriptive of what is really happening.  Understanding what action is being performed is not the same as being precise about what is happening.   

Ratch
Hopelessly Pedantic

[rs20]

I have to apologize.  I said "pressure", but I meant energy.  I guess I got stuck on automatic.  The piston will not be able to transfer any energy to the incompressible liquid, because it will not be able to move further, and no more energy can be exchanged.  This is similiar to a heavy force being applied to a immovable object.  No work is done.

Sure, Energy = F * d; even if F is huge, d = 0 --> Energy = 0.

Charging a cap or battery is an example of technical slang prevalent the tech world.  Another two examples are "current flow", and NASA referring to their astronauts as "walking" in space.  If their tether breaks, would they walk away?  They should say "external excursion".  Even though folks understand what the slang means, it is not accurately descriptive of what is really happening.  Understanding what action is being performed is not the same as being precise about what is happening.   

That is just:

Hopelessly Pedantic

But more seriously; yes, there are lots of instances of English being not literally true. It is a fool's errand to try and fix this though; and the attempt can be truly harmful and distracting when it derails a pedagogical discussion.

P.S./ You're not nesting quotes correctly which makes things confusing; perhaps ignore my earlier request please. I do appreciate the effort thus far

[Ratch]

P.S./ You're not nesting quotes correctly which makes things confusing; perhaps ignore my earlier request please. I do appreciate the effort thus far

I don't understand the problem.  Every quote in post #65 was labeled with the author.

Ratch
Hopelessly Pedantic

[Stephan_T]

I have been thinking about that.  Statically charged particles are a nuisance, but are they really miniature  capacitors?  Most are not hollow conducting shells, so their geometry is not such that their charge is separated from their electric field.  It is something to think about

Why do you say "particles"? That term usually refers to small objects. Do you consider your self to be a particle?
And how do you define the term "capacitor"? I prefer a pragmatic definition and only call objects that intentionally have a capacitance a capacitor. The nuisance is usually called "parasitic capacitance" and there is a lot of that around us.

[Ratch]

I have been thinking about that.  Statically charged particles are a nuisance, but are they really miniature  capacitors?  Most are not hollow conducting shells, so their geometry is not such that their charge is separated from their electric field.  It is something to think about

Why do you say "particles"? That term usually refers to small objects. Do you consider your self to be a particle?
And how do you define the term "capacitor"? I prefer a pragmatic definition and only call objects that intentionally have a capacitance a capacitor. The nuisance is usually called "parasitic capacitance" and there is a lot of that around us.

Because dust, lint, powders, and other small particles are attracted by static electricity.

No, I don't consider myself a particle.  I might, however, possess a small amount of capacitance.

A capacitor is an object that can store electrical energy in a electrostatic field.  By that definition, I surmise that small, statically charged particles are miniature capacitors.

I was thinking of unwanted static electricity as a nuisance, not the nuisance of unwanted capacitance.

Ratch
Hopelessly Pedantic