Contradiction between ΔV in circuits' wires and Kirchoff's voltage law ?

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A simple electric circuit (such as one composed of a battery with Emf V and resistor with resistance R) is in an electrodynamic state since the battery's potential difference creates an electric field in the circuit's wires, which in turn moves charges around. So potential difference in wires is non-zero, otherwise charges won't move at all.

The definition of Kirchoff's voltage law is that the sum of voltages in a closed loop should equal to zero.

How come when we use this formula during circuit analysis, we consider the potential difference in wires to be zero when in fact it is not due to the presence of an electric field in the wires ? Like for the same example of the battery and the resistance, we write directly: V-RI=0 ?
 

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  • #2
I like Serena
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Hi ezadam, welcome to PF :smile:

Obviously the voltage difference over a wire is not zero, because indeed no current would flow.

Note that a wire has an internal resistance.

However, the internal resistance of a wire is usually much smaller than any resistors in a circuit.
So the voltage drop over the wire is insignificantly small compared to the voltage drop over a resistor.

In regular electronics, wires are modelled as having zero resistance, meaning the voltage drop is zero. This is however only to make the mathematical model easier.
 
  • #3
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By Ohm's Law,E=ρj so if ρ is zero no electric field can exist inside the wires even in electrodynamic state.Can somebody tell me whether the field appearing in this expression is the net or applied field.It seems fairly correct to assume that an external field can be applied to an ideal conductor.

I am guessing it is another unrealistic situation coming out of faulty assumptions.An ideal conductor means an average relaxation time τ of infinity.How is that even supposed to be possible?Now going the same way we derived an expression for drift velocity we can see for τ to be infinite either E is zero or drift velocity is infinite.

You remember the expression for electric field for an infinite charged plate.Try applying Kirchoff's law to that.You will see how confusing assumptions can get.
 
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Thanks I like Serena :biggrin: !!! Before I was into Openstudy.com but their forum community cannot be compared at all to the one here at PF. I am sure I'll be learning a lot from this place :smile:

I understand your explanations. Even assuming the existence of an internal resistance r for the wires and taking it into account in Kirchoff's voltage law, the result won't be much different from the idealized mathematical model because of the resistance's negligible value.

aim1732, neither do I know about that. Do current-carrying wires create external fields ? And if yes, does that field influence the internal field of other wires ? I had randomly come across something that explained that using the concept of "ring of charges", but I couldn't really understand and I forgot about the source.
 
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Thank you ezadam!
I appreciate to be appreciated! :smile:
 
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Yeah this is a good question, but you must remember that no potential difference does not mean charges cannot move across - it only means they will not change their energy by doing so.

If your wire has zero resistance (like a superconductor) then there is no potential difference across a segment of wire (under circuit analysis assumptions). But if there is a potential difference at any other points in the circuit (across elements) then charges will move since potential is a state thing (by definition) and does not care about path.
 
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Do current-carrying wires create external fields ? And if yes, does that field influence the internal field of other wires ?
External fields can not be created by the sources.By definition they come from outside.In this case the battery is the one that creates the field.My question is whether the charges that move under the influence of the external field create a counter-field just like in electrostats.That will ensure that they do not absorb energy from the battery just as Curl said.
 
  • #8
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Do current-carrying wires create external fields ? And if yes, does that field influence the internal field of other wires ?
sure....

do you know the "right hand rule" convention??

diagram of such an electric field here:

http://en.wikipedia.org/wiki/Right-hand_rule


does field influence INTERNAL field.....no... generally electrons repel and move to or near the skin of a conductor, so the field inside a conductor is usually zero.....

but the field does have a lot of influence....that's how transformerswork, for example... power from one wire is induced in an adjacent wire....

"My question is whether the charges that move under the influence of the external field create a counter-field just like in electrostats."

only when the circuit is opened....then charges stop moving.....things become static....
 
  • #9
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Naty1, by internal and external I mean applied and produced.Moving charges inside a conductor can not influence the external field.
 
  • #10
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sure....

do you know the "right hand rule" convention??

diagram of such an electric field here:

http://en.wikipedia.org/wiki/Right-hand_rule


does field influence INTERNAL field.....no... generally electrons repel and move to or near the skin of a conductor, so the field inside a conductor is usually zero.....

but the field does have a lot of influence....that's how transformerswork, for example... power from one wire is induced in an adjacent wire....

"My question is whether the charges that move under the influence of the external field create a counter-field just like in electrostats."

only when the circuit is opened....then charges stop moving.....things become static....
Naty1, are you sure that you are not talking about the magnetic field created around the wire ? I think that's where the right-hand rule applies. I am talking about a possible external electric field created by a wire.
You are right about the fact that external fields do not affect the total internal in a neutral conductor as demonstrated by Gauss's Law (charges inside are zero so field is necessarily zero). But the question about whether or not an external electric field is created by current-carrying wires remains ...
 
  • #11
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demonstrated by Gauss's Law (charges inside are zero so field is necessarily zero).
Faulty assumption!It merely implies integral of E.dS is zero.Dow can you suggest that field inside a wire is zero?How are the charges supposed to drift then?
There is a small charge density on the surface of the wire that produces an external field.But I don't know anything much about it.It is supposed to be negligible.
 
  • #12
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Dow can you suggest that field inside a wire is zero?How are the charges supposed to drift then?
That's why I was talking about a neutral conductor. In neutral conductors, free electrons rearrange themselves in a way that cancels the external electric field, thus total field inside a conductor is 0.

About current-carrying conductors, I don't know as I had specified ...
 

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