I was just wondering that since resistance is required for a potential difference to build up, what happens when we have a perfectly conducting wire, 0 resistance? Will current still flow through the medium? Or am I getting something wrong?
Electronic – Will electricity flow in a perfectly conductive medium
conductorsphysicsresistors
Related Solutions
A wire is just a very small resistor. The limiting case is a superconductor. So, ask yourself what happens in these circuits:
simulate this circuit – Schematic created using CircuitLab
You know that the current in a resistor is given by Ohm's law:
$$ I = \frac{E}{R} $$
So as \$R \to 0\$, then \$I \to \infty\$. When \$R=0\$, then you are dividing by zero and the universe explodes. Fortunately, all wires and things we use as voltage sources (batteries, lab power supplies, wall warts...) have some (very small) resistance, so this doesn't happen in practice.
To know what the use of resistance is, consider what would happen if we had none. Voltage is the result of current that wants to flow, but can't. If there were no resistance, then all the current in the universe could flow, and pretty soon, all the electrical energy in the universe would be depleted, leaving you with no voltage anywhere, and no way to do any electrical work.
As long as we are considering what happens as the resistance gets smaller, we might as well consider what happens as it gets bigger:
That is, as the resistance gets higher, less current flows. As \$R \to \infty\$, \$I \to 0\$. When \$R = \infty\$, you have an open circuit, and no current flows. This is just the case of a battery sitting on your desk. There's also no work being done in this case, because although the voltage is exerting a force on the charge in the circuit, it can't move it.
Keep in mind that the wire is a metal and thus already has a great deal of free electrons. The battery is not a source of electrons, it's a pump to move what is already there.
Still, you can view the system as a very small capacitor, with the wires being the plates. They would be relatively far apart and have very little surface area so the capacitance would be very very small. So, yes, I think there would be electron transfer, but it would be negligible and it would only last an instant.
Best Answer
Let's assume you connect a 1.5V battery across a perfectly conducting wire(0 ohm resistance). Then, yes, current would flow across the wire. But since the resistance of the wire is 0, there would be no potential difference/voltage drop across the wire.
It is true that potential difference is required to drive current. In any circuit, the potential difference is maintained by the power source/battery. For example, let's say a 10v battery is supplying power to a circuit. The battery maintains a potential difference of 10v across the terminals of the circuit. This 10v is divided across various components of the circuit depending on how much work is being done to push current across that component. If we were to consider a simple voltage divider as shown below,
The battery maintains voltage "V" across R1 and R2. This voltage "V" gets divided into "V1" across "R1" and "V2" across "R2".(V=V1+V2) If R1>R2 then V1>V2 or vice versa. As resistance increases, more work needs to be done to push current through the component hence more voltage appears across that component. So, if the wire is perfectly conducting then NO WORK NEEDS TO BE DONE TO PUSH CURRENT THROUGH THE WIRE. Hence, maximum current would flow. If you were to connect this perfectly conducting wire directly across a battery, you would be short circuiting that battery because ideally infinite current would flow through the wire.
Long story short, resistance doesn't generate voltage/potential difference. Power source/battery does. A part of the potential difference generated by the battery appears across a component whenever work needs to be done(when there is resistance). If there is no resistance/no work needs to be done, current would still flow as long as battery is connected.