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- Thread starter Bashyboy
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- #27

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That doesn't make any sense to me. By "total rate of charge transfer", do you mean "current"? If so, then current is inversely proportional to the resistance.The drift velocity is a consequence of the charge density and the total rate of charge transfer.

See this link (or any text book)

The resistance is not relevant.

In this article, the claim is made that drift velocity is proportional to the mean-free path d between collisions.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/ohmmic.html

The conductivity (the inverse of the resistance) is similarly proportional to the mean free path, and so we have:

[itex]v_d \propto \dfrac{1}{R}[/itex]

So what you said is, I believe, completely wrong.

- #28

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The problem I have is that after reading your post, one feels more confused than when he started, and yet feels embarrassed to ask clarifying questions, for fear of being ridiculed. That's just the impression I get.I think the OP can answer for himself, whether it is a question or a question based on a misconception. But, as we seem to be agreeing about the actual facts, what's the problem?

- #29

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stevendaryl, have you seen my message that I posted, just moments before you posted yours?

- #30

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I think that the answer is that there is negligible energy in the circuit in the form of kinetic energy of charges, because the drift velocity is so small, typically.stevendaryl, have you seen my message that I posted, just moments before you posted yours?

Conservation of energy would tell us that:

[itex]\sum_i P_i = \dfrac{dU}{dt}[/itex]

where [itex]P_i[/itex] is the power generated or consumed by component [itex]i[/itex], and [itex]U[/itex] is the energy in the circuit in the form of kinetic energy of charges. So since the kinetic energy of charges is negligibly changed by applying a voltage, we can approximate:

[itex]\dfrac{dU}{dt} = 0[/itex]

So we get:

[itex]\sum_i P_i = 0[/itex]

The power [itex]P_i[/itex] generated or consumed by component [itex]i[/itex] is just equal to [itex]I_i V_i[/itex], where [itex]I_i[/itex] is the current through component [itex]i[/itex], and [itex]V_i[/itex] is the voltage change from one end of the component to the other. So we have:

[itex]\sum_i I_i V_i = 0[/itex]

For a circuit without junctions that split the current, the current is the same for all components, so we would have:

[itex]\sum_i I_i V_i = I \sum_i V_i = 0[/itex]

[itex] \Rightarrow \sum_i I V_i = 0[/itex]

To extend the argument to the case of circuits with junctions is more work, but I assume something similar applies.

- #31

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Well current is indeed inversely proportional to resistance , voltage is directly proportional to current and

I think sophiecentaur is not necessarily wrong in this case as the amount of current is associated with the drift velocity , so as the current gets higher the velocity also increases and the heat dissipated by the conductor will also increase as heat is a property which arises from the kinetic energy of particles in a piece of matter as we know.

So basically it all ties up.

Now we had a debate with sophiecentaur on some earlier posts whether using electrons as describing electricity and current is useful for the inexperienced or layman and I do partly agree to him that for starters we shouldn't use quantum mechanical objects one of which is the electron.

Although this is a thing that must be left in the hands of the OP to decide, if he or she is comfortable and capable of understanding answers given by electrons not current and voltage then we should provide the answers in a way that the OP wants or has asked us to do so.If he or she hasn't said in how complicated terms the answers should be then we should ask which is actually what i have seen done here by many professional responders.

I think sophiecentaur is not necessarily wrong in this case as the amount of current is associated with the drift velocity , so as the current gets higher the velocity also increases and the heat dissipated by the conductor will also increase as heat is a property which arises from the kinetic energy of particles in a piece of matter as we know.

So basically it all ties up.

Now we had a debate with sophiecentaur on some earlier posts whether using electrons as describing electricity and current is useful for the inexperienced or layman and I do partly agree to him that for starters we shouldn't use quantum mechanical objects one of which is the electron.

Although this is a thing that must be left in the hands of the OP to decide, if he or she is comfortable and capable of understanding answers given by electrons not current and voltage then we should provide the answers in a way that the OP wants or has asked us to do so.If he or she hasn't said in how complicated terms the answers should be then we should ask which is actually what i have seen done here by many professional responders.

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- #32

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I apologize for the rudeness of that post.The problem I have is that after reading your post, one feels more confused than when he started, and yet feels embarrassed to ask clarifying questions, for fear of being ridiculed. That's just the impression I get.

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i don't think the post was rude I just think a little misunderstanding arose in this discussion

- #34

sophiecentaur

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That's nice to knowWell current is indeed inversely proportional to resistance , voltage is directly proportional to current and

I think sophiecentaur is not necessarily wrong in this case as the amount of current is associated with the drift velocity , so as the current gets higher the velocity also increases and the heat dissipated by the conductor will also increase as heat is a property which arises from the kinetic energy of particles in a piece of matter as we know.

So basically it all ties up.

Now we had a debate with sophiecentaur on some earlier posts whether using electrons as describing electricity and current is useful for the inexperienced or layman and I do partly agree to him that for starters we shouldn't use quantum mechanical objects one of which is the electron.

Although this is a thing that must be left in the hands of the OP to decide, if he or she is comfortable and capable of understanding answers given by electrons not current and voltage then we should provide the answers in a way that the OP wants or has asked us to do so.If he or she hasn't said in how complicated terms the answers should be then we should ask which is actually what i have seen done here by many professional responders.

I had the luxury, in my early education, of not being 'allowed' to try to get to grips with electricity by using electrons. It has been a real help all my life. Once I got on a degree course, it was obvious that they are not the way into the subject and I never experienced any conflict.

A few decades ago, someone decided that everything Scientific could be explained in terms of particles, when talking to young would-be Scientists. It got into School curricula and has been responsible for such a lot of confusion ever since. This is because the massive step between the observed phenomenon and the QM description (or 'quasi' QM description, rather) is too much to cope with in one go. When you get down to it, I think I'd actually prefer the water analogy for electricity than the totally half-arsed analogy that people get sold, involving electrons. These 'electrons' that are used in the model are nothing like the real thing. They have significant mass (like peas) and their motion (at almost light speed) is represented by animated dotted lines around a circuit in a movie. They just beg to be assigned momentum, Kinetic Energy and significant masses and they 'bump into' atoms on the way round a circuit, to represent resistance. Madness.

Anyone who claims to have a good explanation about any electrical phenomenon, based on that model, should just try to find it echoed in any reputable Science text book - and they consider why they can't find it. Time to re-consider that particular approach, I think.

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- #36

sophiecentaur

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All the battery needs to "know" is how much current is needed to maintain its voltage across its terminals. It hasn't a brain with which to 'know anything' but charges are produced (internally) at its terminals until the current it releases is limited by the PD that exists (externally) at its terminals. How long it takes for that voltage to appear will be determined by the step response of the circuit. This could be a few ns or ten days. This is where the idea of trying to explain the process in 'mechanical' terms becomes pointless.

Kirchoff does not claim to apply at switch-on so why try to reconcile what it says with the switch on situation or to prove him 'wrong', in some way?

There are always resistors in a circuit that Kirchoff II describes. Read it.

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Also the battery doesn't try to compensate for anything she just does her chemical reactions and current flows until the reactions are over and so is current.

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Sure the charge carrier/electron model is the more fundamental one. i.e. You can derive kirchoff's laws from the charge carrier model but not the other way around, but it's not a good choice for designing electric circuits.

However there is no reason not to discuss what's happening on the level of electrons if you are interested in that.

To understand why electrons behave the way they do you need to look at the electric fields. Let's say you have a battery and a wire with a high resistance connecting it's two terminals. The wire's diameter is equal everywhere.

Now the current will be equal everywhere in the wire and the electrons will move at an equal speed everywhere. Electrons in a wire do not behave like water molecules in a pipe. They do not simply bump into each other and thereby pushing each other along. They are always moved by an electric field. The current density in metal is ALWAYS equal to the electric field strength inside the metal times the conductivity.

That means the electric field inside the wire must be equal everywhere.

But that seems strange since the electric field of a battery should look like the magnetic field of a magnet. i.e. it should be stronger close to the battery and weaker further away. So how can it be equal over the entire length of the wire? There is only one possible solution. There is "static charge" on the surface of the wire that is distributed such that the superposition of the field of the battery and the field of the charges produce a field that is equal everywhere in the wire.

If you have a network of resistors those "static charges" will distribute themselves such that an equilibrium is reached i.e. all the currents are such that the "static charges" do not change anymore. That equilibrium is reached when kirchoff's laws are fulfilled.

The battery always provides the same amount of energy independantly of the resistance. Some of that energy may be released in the internal resistance of the battery though.It's as if the battery knows that there are resistors, and compensates for this by providing more energy than the current (or electron) would need if there were no resistors. Is this correct?

If we replace the wire in my example with one made from a metal with higher resistivity, the electrons will move through it at a slower speed but will experience more "friction". But in total the energy they receive from the battery and the energy they release into the wire will stay the same.

- #39

sophiecentaur

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That's a good way to look at it.But that seems strange since the electric field of a battery should look like the magnetic field of a magnet. i.e. it should be stronger close to the battery and weaker further away. So how can it be equal over the entire length of the wire? There is only one possible solution. There is "static charge" on the surface of the wire that is distributed such that the superposition of the field of the battery and the field of the charges produce a field that is equal everywhere in the wire.

If you have a network of resistors those "static charges" will distribute themselves such that an equilibrium is reached i.e. all the currents are such that the "static charges" do not change anymore. That equilibrium is reached when kirchoff's laws are fulfilled.

.

It's like a string of capacitors, connected between the wires, in parallel. The Electric field between the wires (in V/m) will depend upon the spacing (can vary wildly, over the circuit) and will be far greater than the fields within the wire. It's only when you get to the Resistor that the 'series' field becomes significant.

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