electric circuits - Potential drop in a resistor

What causes a potential drop in a resistor or load? Why a does wire with neglible resistance have the same potential across it?

Answer

There are many ways to solve it.

Let's start with Ohm's law, $V=iR$. This law may also be formulated in a more insightful, microscopic form, especially in the simplest microscopic version of Ohm's law (due to Gustav Kirchhoff):

$$ \vec J = \sigma \vec E $$

Here, $\vec J$ is the current density, $\sigma$ is the conductivity of the material, and $\vec E$ is the electric field, i.e. the potential (voltage) drop per unit distance (including the direction of the decrease).

These arguments can be visualized so easily as follows. Imagine the circuit. The switch is open. This means, the charges could not flow through the circuit and complete the circuit. This is due to the gap in between the switch. So the applied energy is not converted to current even though the resistance is still there. Hence voltage will be there, where you applied it.

Now, in the second case, there is no resistance. This means the supplied energy (the voltage) is completely utilized by the charges to flow through the circuit. There is nothing left so as to be detected by any external circuit (a voltmeter probably). Hence one cannot see any voltage difference between the conductors. You can see energy somewhere if it is there. But if it's being utilized, then you will not see it there, but it's there in some other form - here it is the kinetic energy of the charges.

The energy concept here belongs to phonons. Actually, the electrons scatter off phonons in the resistor and the electron kinetic energy gets converted to energy of the phonons. The energy that goes into the phonons gets distributed around the resistor and ends up as heat.