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Okay, even if the wire has zero resistance, then charges are being transferred in the direction of electric field, which means there must be a potential drop according to the distance travelled by the charges in the direction of electric field. But, in my book, as it is explained, it seems like the potential remains constant throughout the journey of the charge and that it abruptly changes when the charge reaches the other end of the battery. Well, that's just like saying that a body dropped from a height has an abrupt potential drop when it reaches the ground and has no potential drop along its journey if there is no air friction.

I think the potential drop still happens. The difference is only that in case of an ideal wire, the potential drop corresponds to the increase in kinetic energy of the electrons, because the electrons can accelerate forever in this case when there are no positive ions to collide to, while in case of a real wire, the potential drop mostly corresponds to the heat developed.

Qmechanic
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Dove
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3 Answers3

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Zero resistance for a conducting wire is an approximation, one that's so close to reality and so useful that it often gets treated as fact. Here's a table that lists AWG 22 copper wire (which is the smallest hole on my wire stripper, diameter 0.6 mm) as having a resistance of 53 milli-ohms per meter. So imagine a circuit like this:

      -------------------- one meter thin wire -------------------------
 1 V battery                                                       100 ohms
      -------------------- one meter thin wire -------------------------

If your "thin wires" behave like the thin wires in my table, then each of them is a "resistor" with roughly $0.050\,\Omega$ resistance, and the voltage drop across the resistor is closer to $0.999\rm\,V$ than to the $1.000\rm\,V$ across the battery. But most electronic components have manufacturing tolerances of about 1%, so that'd be a tricky difference to measure. It's safe to leave out when you're teaching circuits the first time.

Finite resistance in a long cable becomes important if you're sending a lot of current a long way, but for analyzing a circuit that fits in your hand you can usually neglect the transmission line.

rob
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  • So, even in an idea wire (By ideal I mean exactly ideal, a wire having no positive nuclei and no electron viscocity) potential drop happens accroding to the work done in the direction of electric field, it's just negligible and is significant if the wire is long, right? – Dove Feb 12 '17 at 05:12
  • In a superconducting wire, there isn't any electric field, no work is done within the wire, and your ideal behavior is exact. This is why an electric current can be sustained indefinitely in a superconducting loop of wire. In that case the "work done" that you have in mind happens at the superconducting/normal-conducting junction, and is usually called "contact resistance." – rob Feb 12 '17 at 05:18
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Apply Ohm's law $\Delta V = iR$ across the two points, you'll see, since $R = 0\implies\Delta V = 0$.

Explanation:

Consider the image:

enter image description here

The wire is of zero resistance. $EMF$ of battery is $E$ volts.

From ohm's law, current $I$ in the circuit is $I = \dfrac{E}{R}$.

Now you may apply $KVL$ to find potentials at A, B, C. When you calculate, you'll find out that $$V_A = V_c$$ This result shouldn't be surprising even though electrons have externally flowed from $C$ to $A$, because Electric field exists only inside the resistor!

That electric field has given the electrons some kinetic energy at point $C$, and they flow with the same velocity. There is no electric field inside the zero resistance wire.

Here is a nice visual example from this post on Physics.SE:

enter image description here

Community
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Max Payne
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  • That doesn't explain what I've written that charges are still transferred in the direction of electric fied even with zero resistance. BTW, at least in my book, Ohm's law is derived by using the concept of drift velocity. There is no concept of drift velocity if the ressistance is zero, i.e. if there are no positive nuclei in the wire to interfere. – Dove Feb 12 '17 at 05:00
  • @Dove What you said in the post was that wire was not having any resistance. But you didn't say that the circuit had zero resistance. In fact, ohm's law is only applicable for circuits with non-zero resistance, otherwise current $i \to \infty$ (acc. to ohm's law). Think of this as: when an agency works on a body or against a body, it gains or loses potential energy. But when this external agency is absent (resistance), there is no need for potential to change. – Max Payne Feb 12 '17 at 08:37
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Another way to look at it is if you connect across a potential with a zero ohm wire the potential difference gets reduced to zero. It would be impossible to maintain a potential under zero ohms.

PhysicsDave
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