Why is the potential in conducting wire of negligible resistance is same as the potential of the terminal of the battery to which it is connected? I am a high school student and I am very confused "why is the potential in conducting wire of negligible resistance is same as the potential of the terminal of the battery to which it is connected?" I mean say we have a conducting wire of negligible resistance which is connects a terminal of a resistor to a terminal of a battery, the my teacher told me that because of the difference in potential of terminal of battery and circuit there would be an electric field established inside the circuit which would lead to flow of charges(current) but if the same current is passing through each component of the circuit then how the potential of conducting wire changes and becomes the same as the terminal of battery whereas across resistor it changes according the value of resistance? you can apply ohm's law its valid for both resistor and wire but I want to understand it logically, we know that potential can only change if there is change in configuration of charges? but this configuration doesn't seems to get change inside the wire as well as in the resistor because net charge on them is always 0 as same current is entering and leaving then how potential of different component changes?
 A: It's all about surface charges.
When the wire is connected to the battery and the circuit is open, the charge will distribute itself on the surface of the wire in order to make the electric field inside zero, and the surface equipotential. So, same potential as the battery terminal.
When the circuit is closed on a resistor, the charge will distribute itself on the surface of the wire in order to create an electric field directed along the wire and compliant with Ohm's law in the conductor (j = sigma E, with sigma very small and ideally zero in a perfect conductor). The voltage drop on a good conductor will be negligible and you will see the same potential as the battery at the resistor's terminals. Charge will also accumulate at the interfaces between the wires and the resistive material (and also on the surface of the resistor) producing an electric field inside that is much stronger and that will be responsible for the voltage drop across the resistor.
See also this Is the electric field in a wire constant? for the refecences.
A: Say you have a battery ($V$), a resistor ($R$) in a closed circuit, connected together by a cable with a resistance $r$ ($r << R$).
Then the voltage between both ends of the resistor is
$$ v = RI = R V /(r + R) =  V (1 + r / R) \approx  V (1 -r/R)$$
