- 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.
- 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?
- Is electric field inside circuit constant? If yes how? Will not the shape of circuit matter?
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.
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.