-
Categories
-
Platforms
-
Content
A capacitor that's fully charged has Infinite resistance.
Presume that there are no capacitors, calculate the voltages at the various Nodes. Those Voltages will be the same after "All the Capacitors Have Fully Charged".
need a bit of assistance on this problem if you guys wouldn't mind. See attachment
Ah, but the wording was "fully charged".The wording of the question is not correct.
Instead of "fully charged"
Ah, but the wording was "fully charged".
I agree with my limited knowledge that all responses here are correct!".
The poor OP had a question. The only way to answer that is by using the questions format!.
If not he/she gets more confused. There is a reason for explaining physics, however, there is also a good reason to keep within the format of the question too.
After all, how do we know that the teacher hasn't given the correct information but used analogy to keep it simple? Or hundreds of examples so they all understood in their own way!.
I (wrongly) use fully charged and discharged every day!. But it's easier for people to understand immediately. So I can only imagine a school wants the pupils to get it quickly! Then the student/pupil who initiates further questions has obviously seen the light and gets that there is more to it!
Which is why schools don't teach physics first. I bet two out of ten get the physics explanation. But an analogy or 6, all students get it in their own way!.
Now, this is where I get shot down by the OP. He says, "I have studied physics for five years. There is nothing to charge a capacitor! They are energized. And then continues to explain the physics of an inductor!!.
Wow, did I really type all that?...
Martin
I don't, or never would disagree witth you!.
But some people simply cannot understand the physics! I am one of them!
To feed children in a class with physics of a capicitor is obviously good! But the majority wont be head strong to understand the physics! Hence an analogy works! Then the explanation as to why can commence.
I wrote a whole lot more here. In fact a full A4 page lol. I think the moderators saw my weakness and did their job!!.
Martin
I don't, or never would disagree witth you!.
But some people simply cannot understand the physics! I am one of them!
To feed children in a class with physics of a capicitor is obviously good! But the majority wont be head strong to understand the physics! Hence an analogy works! Then the explanation as to why can commence.
I wrote a whole lot more here. In fact a full A4 page lol. I think the moderators saw my weakness and did their job!!.
Martin
My physics textbook, "College Physics" by Sears & Zemansky, has this to say: The net charge on the capacitor as a whole is, of course, zero, and "the charge on a capacitor" is understood to mean the charge on either conductor without regard to sign.1 Capacitors are not ever "charged" (with what?). They are "energized" to a certain energy level (joules or voltage). Saying capacitors are charged is using technical slang that does not literally describe what is happening.
Once a capacitor has been fully charged, what remains except leakage current through the dielectric? The dielectric insulator may not have infinite resistance, but it is pretty close. If the voltage changes in a circuit with a fully charged capacitor, then one must wait an additional six time constants to regain the infinite resistance.2), A capacitor does not block current by presenting an infinite resistance to a voltage. It blocks current by presenting a counter voltage. This method of current blockage is not the same as an open circuit or infinite resistance. A capacitor supports charge flow (current) if the voltage changes, whereas no current exists in an open circuit.
My physics textbook, "College Physics" by Sears & Zemansky, has this to say: The net charge on the capacitor as a whole is, of course, zero, and "the charge on a capacitor" is understood to mean the charge on either conductor without regard to sign..
It does not seem too difficult to understand what is meant by physicists charging a capacitor. What are capacitors charged with? Obviously with protons and electrons, the ultimate natural unit of charge.
One conductor will have an excess of protons while the other conductor has an excess of electrons. In electronics the term "fully charged" means that the capacitor has been charging in the circuit for a time in excess of six RC time constants. Really, that's what it means.
Once a capacitor has been fully charged, what remains except leakage current through the dielectric?
The dielectric insulator may not have infinite resistance, but it is pretty close. If the voltage changes in a circuit with a fully charged capacitor, then one must wait an additional six time constants to regain the infinite resistance.
Laplace,
There is a reason this web site is called "Electronicspoint" and not "Physicspoint".
My 2 cents:
Leave those "Sears & Zemansky"and other Physics text books on the shelf ,
and move on to the Electronics ones...
The classic used to be "Basic Circuit Theory" by Desoer and Kuh.
It is an engineering grade book and thus University grade math is needed.
But,
it will take your knowledge and understanding several order of magnitudes forward
Webster's New World College Dictionary gives the relevant definition of slang: the specialized vocabulary and idioms of those in the same work: now usually called shoptalk, jargon. I don't seem to have any problem with electronics professionals using specialized technical jargon.That is a slang definition.
So we should change our terminology to state how many kilowatt-hours of energy are stored in a charged capacitor? How would that be at all useful? The capacitor is a charge storage device; equal and opposite amounts of charge are stored on each plate. It would make more sense to state the amount of charge stored, but even that would not be very useful because we have no way to measure the amount of charge stored. However, the amount of charge is proportional to the voltage of the charged capacitor, and we can measure the voltage. This is the reason for using the convention that a capacitor is charged to a particular voltage -- it is useful. So there is the question - Why would your proposal result in something that is even more useful? Show us how it would play out in solving circuit problems to specify capacitors energized in kilowatt-hours.A capacitor is a energy storage device, so its status should be defined with respect to the amount of energy it contains. That's "fully logical". You could say a capacitor is charged with energy, but then you may as well say it is "energized".
I seem to have missed the point of this comment. Sorry???Protons and electrons carry the smallest amount of discrete charge. MKS units of charge are coulombs.
My statement is true based on the highest possible authority. "Electronics All-in-One For Dummies" by Doug Lowe states: "In theory, the capacitor will never be fully charged because with the passing of each RC time constant the capacitor picks up only a percentage of the remaining available charge. … For all practical purposes, you can consider the capacitor fully charged after five time constants have elapsed."Who says? Why not 12 time constants. Or infinity?
Even so, it is the protons that provide the positive charge on the positive plate of the capacitor.By the way, only electrons move in a conductor, not protons. Protons stay with their atoms, which form the ionic core of the material. Protons are not charge carriers in a conductor like electrons are. Therefore, the number of protons on each plate of a capacitor always remain the same, even if the distribution of electrons varies.
That seems strange. The capacitor is in a circuit with a voltage source charging the capacitor. What role does the voltage source have in causing the voltage across the plates? Does the voltage source cause the electric field? If there were no capacitor connected, but there are two wires connected to the voltage source (open circuit), is there an electric field between the wires?An electric field, which causes a voltage across the plates.
But this is a failure to understand. My comment only applied to the fully charged condition of the capacitor, when current flow has ceased and the circuit equations are no longer relevant. If there are two nodes in a circuit with a voltage difference between them yet no current flows at either node, how is that not the definition of an open circuit?A fully energized cap is not the same as a open circuit, even if its dielectric if perfect. The circuit equations differ between the two, and the circuit performance will be different. Treating caps as an open circuit is a mathematical artifice used for calculating the final steady voltage or current. This is applicable to only a single particular circuit status.
Webster's New World College Dictionary gives the relevant definition of slang: the specialized vocabulary and idioms of those in the same work: now usually called shoptalk, jargon. I don't seem to have any problem with electronics professionals using specialized technical jargon.
So we should change our terminology to state how many kilowatt-hours of energy are stored in a charged capacitor? How would that be at all useful? The capacitor is a charge storage device; equal and opposite amounts of charge are stored on each plate. It would make more sense to state the amount of charge stored, but even that would not be very useful because we have no way to measure the amount of charge stored. However, the amount of charge is proportional to the voltage of the charged capacitor, and we can measure the voltage. This is the reason for using the convention that a capacitor is charged to a particular voltage -- it is useful. So there is the question - Why would your proposal result in something that is even more useful? Show us how it would play out in solving circuit problems to specify capacitors energized in kilowatt-hours.
I seem to have missed the point of this comment. Sorry???
My statement is true based on the highest possible authority. "Electronics All-in-One For Dummies" by Doug Lowe states: "In theory, the capacitor will never be fully charged because with the passing of each RC time constant the capacitor picks up only a percentage of the remaining available charge. … For all practical purposes, you can consider the capacitor fully charged after five time constants have elapsed."
Also, "Introductory Circuit Analysis, 3rd Ed" by Robert L Boylestad states (pg. 218) "For all practical purposes, therefore, a capacitor will charge to its final voltage in five time constants."
I know these references state 5τ while I said 6τ, but I believe in being really sure.
Even so, it is the protons that provide the positive charge on the positive plate of the capacitor.
That seems strange. The capacitor is in a circuit with a voltage source charging the capacitor. What role does the voltage source have in causing the voltage across the plates?
Does the voltage source cause the electric field?
If there were no capacitor connected, but there are two wires connected to the voltage source (open circuit), is there an electric field between the wires?
But this is a failure to understand. My comment only applied to the fully charged condition of the capacitor, when current flow has ceased and the circuit equations are no longer relevant. If there are two nodes in a circuit with a voltage difference between them yet no current flows at either node, how is that not the definition of an open circuit?
Capacitors that are "fully charged" (whatever that means) in a steady-state DC circuit are not conducting current. They can therefore be removed from the steady-state analysis as @Colin Mitchell stated in Post #2. Follow his instructions to find the potentials across all the resistors and hence the potentials of the capacitors connected in parallel with those resistances.need a bit of assistance on this problem if you guys wouldn't mind. See attachment
That was my point Hop, he didn't 'think' anything. Just following a homework paper. It is not the fault of the pupil to show the 'actual' paper involved.I will forgive you if you want to think of this operation as "charging" the capacitor.
Yeah, lazy editing strikes again. This type of phrasing is common in physics problem texts. It is used as a clue here to mean the capacitors are not part of the steady-state analysis. Once the student realizes that, the "problem" devolves to finding the voltages developed across three resistors connected in series. I have no idea what the student is expected to learn from all this... maybe that some "problems" are "tricky" to solve?My point was, they used 'fully charged' for that paper!.