# What does Potential Difference between a load/resistor mean?

Guys this has been bugging me for ages and all the answers i found online were unsatisfactory.

My questions is, what does difference in potential or potential drop after current passes through a load/ resistor mean? Current is constant throughout the circuit with a resistor hence we cannot say that the electron loses kinetic energy after passing through the load. So what changed when an electron passes through a load that causes the load/resistor to heat up while the potential drops?

UPDATE: ok im sorry guys but i believe my question is rather vague. This is an updated version of my question

I wish to know where did the energy come from that results in a potential difference between a load. Is it kinetic energy of the electrons? But it is impossible since current is constant throughout the circuit

QUESTION Current is constant throughout the circuit with a resistor hence we cannot say that the electron loses kinetic energy after passing through the load.

SOLUTION Current throughout the circuit with a resistor is constant , no doubt about that. But to be fundamental, current in a circuit is set up by the electric field, not by electrons. For instance if there are 100 bulbs connected in series in an electric circuit, on switching on the cell, all the bulbs are lighted at the same time, clearly proving that electric field and not electrons set up the current. Again electrons have a very low drift velocity of a few $mm/s$ as they suffer a large number of collisions among themselves. And yes the electrons do lose kinetic energy and that is dissipated in the form of heat in the resistor. The external energy supply i.e. the cell in this case supplies the necessary energy to keep the current constant. That is the sole loss in energy in our domestic electrical circuits, i.e. to say Loss in K.E. of electrons = Heat in resistor. $$E_{cell} = Work_{electrons} + Energy _{dissipated}$$

QUESTION So what changed when an electron passes through a load that causes the load/resistor to heat up while the potential drops?

SOLUTION As the electron has to overcome some electrical resistance, it does some work which is dissipated in the form of heat. And potential difference being the driving force of the circuit, current flow is resisted in a resistor and as current prefers a resistance free medium for propagation, the driving force of the current lessens and across the resistor, there occurs what we call "a potential drop". It can also be verified by using Ohm's Law.

I explained this in layman's language. Hope it helps.

• Comments are not for extended discussion; this conversation has been moved to chat. Sep 20, 2015 at 6:54

To maintain a current you need to "push" the charge through any obstacles on the path.

If there is resistance against the current, then the "push" must be large enough to overcome this. The potential difference is this "push". Of course, as soon as the resistor has been passed, then a large "push" is no longer needed to make the current keep moving. Now what is needed is only a "push" to move through the remaining resistances of the path. So the potential difference drops and is lower.

See the picture in this other answer: Could someone intuitively explain to me Ohm's law? It makes it pretty intuitive.

If you are in doubt when considering electric circuits, think of it as a flow of water in pipes.

• The current is the flow of water,
• a resistance can be a narrow part of the pipe, and
• the potential difference (voltage) is the pressure (there is a potential difference, if the pressures are different at either side of some "resistance".)
• Im sorry steven but i still do not understand. Firstly, potential difference as i know is the ammount of electrical energy converted to other forms of energy and the "push" that you are referring to is electromotive force of the source. What i wish to know is what happens at the quantum level when an electron passes through a load/resistor that causes it's electrical potential to drop Sep 19, 2015 at 12:21