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  1. Imagine a circuit consisting of just a battery and conducting wires which have zero dissipativity so that there is no loss of energy( Zero resistance). If the wires are connected to both terminals the current will flow in the direction of lower potential. Can we say that there has been a voltage drop in the current in going from positive terminal to negative terminal? I asked this question because voltage drop happens across resistor and I don't understand why it should occur only there. Because I think, considering a positive test charge its electric potential should decrease as it travel from higher potential to lower potential whether any dissipation occurs or not.Does voltage drop refers to loss in total energy of electron i.e. because in this case KE + Electric potential will be zero and the total energy of electrons remains unchanged and thus we can says that voltage drop is zero.
  2. Is voltage drop(assumed here to mean loss in electric potential) equal to work output under normal conditions. I think no - because electric potential difference gets converted into kinetic energy of electrons out of which only some gets converted into heat or is it the case that in
    a resistor entire kinetic energy gets converted into another energy such as heat?
  3. Is electric field inside circuit constant? If yes how? Will not the shape of circuit matter?
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marked as duplicate by Qmechanic Dec 8 '15 at 10:43

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    $\begingroup$ Sincerely, I don't understand the question. The expression "voltage drop" refers to the tension due to the circulation of a current in a resistive cable. The cable is seen as a medium to feed a distant charge with voltage : since a voltage appears at the cables terminals, the effective voltage at the charge is lower. That is the voltage drop. The use of this expression in other cases seems inappropriate to me. Finaly, I would not consider local physical values (electric field) and integrated values (current, 'voltage drop') at the same time when studying electric circuits (global per se). $\endgroup$ – TZDZ Nov 5 '14 at 10:42
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Can we say that there has been a voltage drop in the current

First, it is true that electrons leaving a resistor have less potential energy than those entering the resistor.

Second, assuming an ideal conductor, the electrons that flow along the length of the conductor have the same potential energy. This must be the case since there can be no electric field present inside an ideal conductor.

Third, your thought experiment is flawed as John Rennie has pointed out. One cannot imagine both an ideal battery and ideal wire connected together as this is to imagine a contradiction.

The ideal battery maintains a constant voltage for any current and the ideal wire maintains a zero voltage for any current. If the battery is a 9V battery, the associated KVL equation is

$$9V = 0V$$

which is a contradiction.

A physical battery has an internal resistance and, thus, a maximum current for 0V across.

That is to say, if one places an (effectively) ideal wire across the battery terminals, the voltage across the battery terminals becomes 0V and the current is a finite (and perhaps very large) value.

The magnitude of this current is called the short-circuit current for the battery.

So, in your thought experiment, assuming a non-zero internal resistance for the battery and steady state operation, there would be zero volts across the ideal conductor and a non-zero, finite current through. The electrons flowing through the conductor would enter and leave the conductor with the same potential energy.

By the way, in almost any case, connecting the terminals of a battery together is dangerous due to the large short-circuit current. The battery will certainly get very hot with the possibility of explosion and/or fire high.

In other words, keep this experiment in the realm of your imagination.

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    $\begingroup$ As a young and curious kid I did short the terminals of an old car battery with a long piece of insulated wire and I can attest that it will catch fire. $\endgroup$ – Floris Nov 5 '14 at 12:51
  • $\begingroup$ I still don't understand the term "voltage drop". Here is what I think:Voltage means potential difference across says A and B. So voltage drop means lessening of PD across the same points.How is that possible since electric potential is only location dependent(if we keep electric field same which is ,I believe for very unclear reasons!, is same inside circuit )and so it wouldn't matter whether there was a resistor in between AB; the PD would remain the same.I am a newbie. Please explain me like a small child! $\endgroup$ – Viham G Nov 6 '14 at 14:22
  • $\begingroup$ @VihamG, voltage drop is the potential difference,not a lessening of potential difference. I almost never use the term voltage drop or voltage rise and you don't have to either. I simply refer to the potential difference as the voltage across, e.g., "the voltage across a resistor". $\endgroup$ – Alfred Centauri Nov 6 '14 at 16:27
  • $\begingroup$ I hope I will get it right by knowing this:Does electric potential remain constant inside wire before resistor,it then decreases across resistor and we have a different constant electric potential after resistor inside the wire? If yes, please edit your answer to give intuition why EP changes only across resistor because electric potential decreases as +ve charge moves away from +ve terminal to -ve terminal.If no,then confusion remains unsolved. $\endgroup$ – Viham G Nov 7 '14 at 9:01
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    $\begingroup$ @Viham, the electric field accelerates charge, i.e., a charge 'feels' a force due to an electric field. In the absence of resistance, charge does not require an electric field to sustain a current. This a analogous to a mass not requiring a force to sustain movement in the absence of friction. $\endgroup$ – Alfred Centauri Nov 8 '14 at 11:47
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All batteries have some internal resistance. If you connect the terminals with a wire of zero resistance than there is no voltage drop across the wire and the whole voltage drop occurs within the battery due to its internal resistance.

If you go farther and require that you battery has zero internal resistance as well, then when you connect the wire the current will be infinite. This is unphysical so it isn't possible to make any sensible calculations about it.

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