What is ohms law (was Ic=C x dv/dt)

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Ratch

Mar 10, 2013
1,098
Most materials are, or can be considered to be ohmic at a single point.

The resistance of a non-ohmic material has to measured at each value of current. The resistance of a ohmic material only has to be measured once.

Ratch

ElectronicPotatoe

Dec 17, 2016
24
J and E define the resistivity and therefore the conductivity. The following is from Wikipedia.

View attachment 33478

Ratch

I read the Wikipedia article. That does not mean that is the definition of conductivity...

It could be just a constant that Ohm used for his law... And that's what it is. If what you were saying made sense, the same could be said about resistance:

"V=IR is the definition of resistance."

And that's wrong. Constants are used a lot in science to formulate equations. That doesn't mean the equations are the definition of the constant.

Ratch

Mar 10, 2013
1,098
I read the Wikipedia article. That does not mean that is the definition of conductivity...
If V/R is the definition of resistance, then its reciprocal is also the definition of conductivity. Conductivity is just another way of representing a property of a material.

It could be just a constant that Ohm used for his law... And that's what it is. If what you were saying made sense, the same could be said about resistance:

What constant would that be? Resistance is not a constant. It varies with the kind of material and the shape of the material.

"V=IR is the definition of resistance."

And that's wrong. Constants are used a lot in science to formulate equations. That doesn't mean the equations are the definition of the constant.

Wikipedia is correct on this subject. Constants don't vary. Resistance does vary, so it is not a constant. Equations are used extensively in definitions. For instance, speed is defined as distance traveled divided by the time.

Ratch

Laplace

Apr 4, 2010
1,252
Reactance only applies to sinusoidal steady state. Transient behavior requires differential equations. Linear differential equations like those found in most circuits are not too bad.

Note that the original posting that started this discussion (and which seems to have been lost from this thread) described a problem that was in the sinusoidal steady state, and therefore required only the use of reactance and Ohm's Law (because reactance is measured in ohms) for solution. Only one of a perverse nature would lead the OP along to the unnecessary terrain of transient behavior and differential equations.

Good thing we don't live in Columbus's time, or you would believe the Earth is flat. Similarly, in Galileo's time. a lot of folks believed the Earth was the center of the universe.

There is a well-respected contemporary author whose thesis is, "The World is Flat".

That is TTT. Charge flows, current is already defined as charge flow. Current flow literally means "charge flow flow". The current into the terminal has the same value as the current leaving the opposite terminal. That is because the same amount of charge accumulating on the capacitor plate per unit time is the same as the amount of charge leaving on the opposite plate per unit time. That does not mean that current exists through the capacitor. If it did, the capacitor would be a resistor.

Current flows. Learn the language or be labeled a stubborn curmudgeon. Also, if current did not flow through the capacitor dielectric then there would be a discontinuity in the current, and that cannot happen. That is why Maxwell's equations include a "displacement current" proportional to the rate of change of the electric field in the dielectric. Current flow is not caused only by the movement of charge but also by a changing electric field.

No it doesn't. There are two circuits involved. One circuit accumulates the charge, the other depletes the charge. The two circuits are isolated from each other by the capacitor's dielectric. The charges that enter a capacitor are not the same charges that leave.

The two "circuits" are not isolated but rather connected by the displacement current. It can also be said that the charges (electrons) that enter a wire are not the same charges that exit the wire. Ever hear of 'drift velocity' of electrons in a conductor? So tracking the identity of individual charges in a capacitor is a bogus concept.

No it doesn't. Circuitwise yes, but on the capacitor itself, no. Kirchoff's law does not apply to a capacitor, because a capacitor stores and dispenses charges.

A capacitor does not store charge. A fully charged capacitor contains no more electric charge than an uncharged capacitor.

Mesh analysis works because the capacitor stores and dispenses charge by equal amounts per unit time. This fools a lot of folks into thinking that the current exists through the capacitor. The mathematics of mesh analysis does not know the difference, but those who understand how a capacitor works know what is really happening.

Mesh Analysis and Nodal Analysis are unconcerned with what may happen inside a connected device. A resistor, a capacitor, or an inductor are just two-terminal devices having a simple mathematical description of the voltage-current behavior at the two terminals. Nothing else matters.

ElectronicPotatoe

Dec 17, 2016
24
If V/R is the definition of resistance, then its reciprocal is also the definition of conductivity. Conductivity is just another way of representing a property of a material.

That doesn't mean J=σE is not Ohm's law, and is instead, the definition of conductivity.

Ratch

Mar 10, 2013
1,098
Note that the original posting that started this discussion (and which seems to have been lost from this thread) described a problem that was in the sinusoidal steady state, and therefore required only the use of reactance and Ohm's Law (because reactance is measured in ohms) for solution. Only one of a perverse nature would lead the OP along to the unnecessary terrain of transient behavior and differential equations.

Look at the OP's first post again. The title of the thread was I = C dv/dt, and he asked how to obtain dv/dt. That is a differential equation question, not a steady state condition.

There is a well-respected contemporary author whose thesis is, "The World is Flat".

Which is easily disproven by any picture taken of the Earth from space.

[Current flows. Learn the language or be labeled a stubborn curmudgeon.

Charge flows, current already means flow. Don't have to say it twice. Don't use technical slang when precision is valued.

Also, if current did not flow through the capacitor dielectric then there would be a discontinuity in the current, and that cannot happen.

You are TTT again. If current existed through the capacitor, it would be leaky. Capacitors that are leaky are discarded. If a capacitor did not store a charge on one of its plates and dispense an equal charge on its opposite plate, how could it store energy? How could it exhibit reactance and an exponential curve for the current when a constant voltage is applied?

That is why Maxwell's equations include a "displacement current" proportional to the rate of change of the electric field in the dielectric. Current flow is not caused only by the movement of charge but also by a changing electric field.

I am quoting from a book called Introduction to Electrodynamics, Third Edition, by David J. Griffiths.
"Maxwell called his extra term the displacement current:

It's a misleading name, since it has nothing to do with current, except that it adds to J in Ampere's law."

Do you know what you are talking about?

The two "circuits" are not isolated but rather connected by the displacement current.

You are way out in left field. The dielectric is impermeable to charge transport. The only inside connection between the plates of a capacitor is an electric field which appears when the plates have unequal charges.

It can also be said that the charges (electrons) that enter a wire are not the same charges that exit the wire. Ever hear of 'drift velocity' of electrons in a conductor? So tracking the identity of individual charges in a capacitor is a bogus concept.

Let me make it clearer. The charges next to one side of the dielectric will never cross to the opposite side of the dielectric.

A capacitor does not store charge. A fully charged capacitor contains no more electric charge than an uncharged capacitor.

A capacitor does not store net charge. However, one plate does store a charge and, the opposite plate depletes its charge by the same amount, for a net change of zero.

Mesh Analysis and Nodal Analysis are unconcerned with what may happen inside a connected device. A resistor, a capacitor, or an inductor are just two-terminal devices having a simple mathematical description of the voltage-current behavior at the two terminals.

Isn't that what I said in the last post? The mathematics of mesh analysis do not discern that a capacitor accepts charge into one terminal and releases an equal amount of charge at the opposite terminal.

Nothing else matters.

That depends on how deep you want to go into understanding what is happening.

Ratch

Ratch

Mar 10, 2013
1,098
That doesn't mean J=σE is not Ohm's law, and is instead, the definition of conductivity.

Look at the second paragraph of the quote from Haliday & Resnick found in post #8 of this thread. They keep saying that Ohm's law is a definition of a property of a material, specifically linearity. You want it to be a method of calculation.

Ratch

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,508
Look at the OP's first post again. The title of the thread was I = C dv/dt, and he asked how to obtain dv/dt. That is a differential equation question, not a steady state condition.

We're not in Kansas any more. No need to limit yourself to the original request.

ElectronicPotatoe

Dec 17, 2016
24
You want it to be a method of calculation.

No, sir, I think you made that up. I even quoted the part where it said it is a property. It has a consequence, which is the equation. That doesn't mean that equation is the definition of conductivity (which is what you said). It's just Ohm's law expressed in therms of an equation.

"More specifically, Ohm’s law states the following: For many materials (including most metals), the ratio of the current density to the electric field is a constant s that is independent of the electric field producing the current"

7th edition, Vol 2, page 756.

Ratch

Mar 10, 2013
1,098
No, sir, I think you made that up. I even quoted the part where it said it is a property. It has a consequence, which is the equation. That doesn't mean that equation is the definition of conductivity (which is what you said). It's just Ohm's law expressed in therms of an equation.

An equation by itself cannot express linearity. Ohm's law requires linearity. The equation has to be analyzed to determine if it is linear. Therefore, an equation by itself cannot express Ohm's law.

If resistivity can be determined by an equation, then so can conductivity, which is the reciprocal of resistivity.

Ratch

ElectronicPotatoe

Dec 17, 2016
24
An equation by itself cannot express linearity. Ohm's law requires linearity. The equation has to be analyzed to determine if it is linear. Therefore, an equation by itself cannot express Ohm's law.

it is a property. It has a consequence, which is the equation.

You are just summarizing what I´ve been saying...

Ratch

Mar 10, 2013
1,098
You are just summarizing what I´ve been saying...

Fine, we agree then. For the material to follow Ohm's law, the resistance of a material must not change if measured using different currents. Or, the resistivity of a material must not change if measured using different current densities.

Ratch

Laplace

Apr 4, 2010
1,252
I am quoting from a book called Introduction to Electrodynamics, Third Edition, by David J. Griffiths.
"Maxwell called his extra term the displacement current:

It's a misleading name, since it has nothing to do with current, except that it adds to J in Ampere's law."

Do you know what you are talking about?

You should refer to these class notes for a course in Applied Electromagnetics.

http://whites.sdsmt.edu/classes/ee382/notes/382Lecture5.pdf

Then you may be able to understand the following diagram showing displacement current flowing through the dielectric of a capacitor.

Ratch

Mar 10, 2013
1,098
You should refer to these class notes for a course in Applied Electromagnetics.

http://whites.sdsmt.edu/classes/ee382/notes/382Lecture5.pdf

Then you may be able to understand the following diagram showing displacement current flowing through the dielectric of a capacitor.

View attachment 33545

I read through the notes in the attachment. No where does it say that conduction current exists through the capacitor. Some sources I have come across call the displacement current a "fictitious" current. It has the units of current, but it is not real current. I see no conflict between the notes and what I averred.

The best explanation I found so far is from Wikipedia, as quoted below. They claim that a magnetic field exists between the capacitor plates equivalent to what the conduction current would generate. This equivalent magnetic field, when taken into consideration, corrects a discrepancy of Ampere's law. But again, this does not mean that charges pass through the dielectric.

Ratch

Laplace

Apr 4, 2010
1,252
I read through the notes in the attachment. No where does it say that conduction current exists through the capacitor.

Correct. That is because displacement current is not conduction current.

Some sources I have come across call the displacement current a "fictitious" current. It has the units of current, but it is not real current. I see no conflict between the notes and what I averred.

But what you aver is not the the whole picture. What you aver leads to a discontinuity in the current flow. What you aver is what Maxwell found necessary to correct.

The best explanation I found so far is from Wikipedia, as quoted below. They claim that a magnetic field exists between the capacitor plates equivalent to what the conduction current would generate.

That is because the displacement current is a real current that generates a real magnetic field. In an ideal capacitor the magnitude of the displacement current is equal to the magnitude of the conduction current. That is why "a magnetic field exists between the capacitor plates equivalent to what the conduction current would generate".

This equivalent magnetic field, when taken into consideration, corrects a discrepancy of Ampere's law. But again, this does not mean that charges pass through the dielectric.

This 'discrepancy' is a fundamental property that cannot be ignored. Current flow is not due simply to the movement of charge (conduction current), but is also caused by a time variance in the electric field (displacement current).

Nevertheless, this can be ignored for the purposes of circuit analysis which focuses only on the external terminal behavior of the capacitor. Nothing else matters.

Ratch

Mar 10, 2013
1,098
Correct. That is because displacement current is not conduction current.

Therefore, charge does not pass through the dielectric.

But what you aver is not the the whole picture. What you aver leads to a discontinuity in the current flow. What you aver is what Maxwell found necessary to correct.

The cap inputs current and outputs current both at the same time and at the same value. You could say that there is a discontinuity at each plate of the cap, but no net discontinuity exists. Max added the magnetic field term to "fix" Ampere's law. He should not have called that term "displacement current". That was a misnomer. Many authorities agree on that point.

That is because the displacement current is a real current that generates a real magnetic field. In an ideal capacitor the magnitude of the displacement current is equal to the magnitude of the conduction current. That is why "a magnetic field exists between the capacitor plates equivalent to what the conduction current would generate".

The mag field is not caused by any "fictional" charge moving through the dielectric.

This 'discrepancy' is a fundamental property that cannot be ignored. Current flow is not due simply to the movement of charge (conduction current), but is also caused by a time variance in the electric field (displacement current).

Why not just say current instead of current flow? The time variance of the electric field causes a mag field to form, but does not generate a current. You can show the mag field is equivalent to what the same value of conduction current would generate, but that fictional current does not exist, and the corrective term should not be called current.

Nevertheless, this can be ignored for the purposes of circuit analysis which focuses only on the external terminal behavior of the capacitor. Nothing else matters.

It is ignored because the net discontinuity in a cap does not exist..

Ratch

CDRIVE

Hauling 10' pipe on a Trek Shift3
May 8, 2012
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Chris

Laplace

Apr 4, 2010
1,252
Therefore, charge does not pass through the dielectric.

You definitively state the obvious. Hmmmm.
The cap inputs current and outputs current both at the same time and at the same value. You could say that there is a discontinuity at each plate of the cap, but no net discontinuity exists. Max added the magnetic field term to "fix" Ampere's law. He should not have called that term "displacement current". That was a misnomer. Many authorities agree on that point.

When anyone relies on the "Many Authorities Agree" defense, you know they have lost the argument because what that really means is, "While many authorities agree, most authorities do not subscribe to the crackpot proposition being purveyed." Meanwhile, Maxwell has survived the test of time.
The mag field is not caused by any "fictional" charge moving through the dielectric.

That statement is based on an obvious misconception. Displacement current is not caused by the flow of charge. Only conduction current is caused by the flow of charge.
Why not just say current instead of current flow?

Why not just say "current flow" like everyone else? C'mon, you know you want to!
The time variance of the electric field causes a mag field to form, but does not generate a current. You can show the mag field is equivalent to what the same value of conduction current would generate, but that fictional current does not exist, and the corrective term should not be called current.

Maxwell does not agree with you. He called it 'displacement current'. However, when ALL authorities agree that Maxwell is wrong, then you should post the announcement.
It is ignored because the net discontinuity in a cap does not exist..

Quite irrelevant! It is ignored because in circuit analysis we don't use a 'capacitor', rather we use a mathematical model of an ideal capacitor terminal characteristics. The fundamental passive components for circuit analysis are the resistor, capacitor, and inductor; each one given as a two terminal device with a mathematical definition relating the current flowing through the device terminals to the voltage existing between the device terminals. Current flows through the device model by definition; you can't argue with a definition.

Ratch

Mar 10, 2013
1,098
You definitively state the obvious. Hmmmm.

Then we agree that charge does not pass through a dielectric.

When anyone relies on the "Many Authorities Agree" defense, you know they have lost the argument because what that really means is, "While many authorities agree, most authorities do not subscribe to the crackpot proposition being purveyed." Meanwhile, Maxwell has survived the test of time.

When the explanation is well stated and proved, then it is not a crackpot proposal anymore. I do not know anyone with a full knowledge of electrodynamics who thinks that the fictional current is really a current. Max's correction to his law has stood the test of time, but not the name he gave the term.

That statement is based on an obvious misconception. Displacement current is not caused by the flow of charge. Only conduction current is caused by the flow of charge.

That is a contradiction. Current by definition is a flow of charge. The only kind of current that exists is conduction current. Fictional or displacement current means current that does not exist by definition. Max should have given his correctional term some other name to avoid a misnomer.

Why not just say "current flow" like everyone else? C'mon, you know you want to!

Nope, I am not attracted to ridiculous redundancies like "charge flow flow".

Maxwell does not agree with you. He called it 'displacement current'. However, when ALL authorities agree that Maxwell is wrong, then you should post the announcement.

The Max has been proven wrong on his name selection for the additional correction term he came up with.

The following snippet is from David Griffiths book, Introduction to Electrodynamics.

So when ALL authorities agree we did it, then we can say conclusively that a manned Moon landing was achieved? Some folks still think it was studio staged.

Quite irrelevant! It is ignored because in circuit analysis we don't use a 'capacitor', rather we use a mathematical model of an ideal capacitor terminal characteristics.

You would not be able to use that math model if a net discontinuity existed.

The fundamental passive components for circuit analysis are the resistor, capacitor, and inductor; each one given as a two terminal device with a mathematical definition relating the current flowing through the device terminals to the voltage existing between the device terminals. Current flows through the device model by definition; you can't argue with a definition.
View attachment 34129

The definition could not be used if a net discontinuity existed.

Ratch

Laplace

Apr 4, 2010
1,252
The definition could not be used if a net discontinuity existed.
So it becomes patently obvious that circuit analysts should model current flowing through a capacitor without being assaulted by curmudgeons claiming that Maxwell was wrong. Case closed.

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