0
$\begingroup$

As far as I understand, the electric potential is the amount of energy that a third party agent has to spend to move a positive charge from infinite separation to a point. Thus, the electric potential due to a battery cell depends only on the position of the charge with respect to the cell's terminals. So it seems illogical to me that the potential difference across a resistor should be different from the potential difference across a piece of wire with no resistance. I understand that pushing an electron through a wire and through a resistor requires different amounts of energy but I struggle to understand why it has anything to do with the difference of the electric potential.

$\endgroup$
4

1 Answer 1

1
$\begingroup$

Your understanding of electric potential is correct. It is indeed the amount of work done by an external agent to move a positive charge from infinity to a point in the electric field.

Now, let's consider the case of a resistor and a wire. The key difference between them is the presence of resistance. Resistance is a property that opposes the flow of electrons. When electrons move through a resistor, they collide with the atoms of the resistor's material. These collisions convert some of the electrical energy into heat, which is why resistors get warm when a current flows through them.

This energy loss appears as a drop in electric potential or voltage across the resistor. This is often referred to as a "voltage drop". According to Ohm's law, the voltage drops across a resistor is given by V = IR where I is the current through the resistor and R is the resistance.

On the other hand, an ideal wire has zero resistance. Therefore, there are no energy losses due to collisions of electrons, and hence no voltage drop. The electric potential remains the same across the wire.

In summary, the difference in electric potential across a resistor and a wire arises due to the energy losses in the resistor, which are absent in an ideal wire.

$\endgroup$
8
  • $\begingroup$ Thank you for your answer! My problem is that if electric potential that you have to do against the electric field, then the work that is done against the resistor shouldn't affect the electric potential. Here's an analogy for what I'm trying to say: if we have a ball and we raise it to a height of one meter, it would have same potential energy, regardless of whether there was an obstacle in the way. Yes, it would take more work to raise it if there was an obstacle, but it wasn't due to the gravitational field, so it doesn't count. $\endgroup$ Jan 16 at 1:44
  • $\begingroup$ I understand your confusion, but your analogy is not quite accurate. The electric potential difference between two points in a circuit is not the same as the gravitational potential energy of a ball at a certain height. The electric potential difference is the energy change per unit charge between two points, while the gravitational potential energy is the energy stored in the ball due to its position in the gravitational field. $\endgroup$
    – Vish
    Jan 16 at 17:14
  • $\begingroup$ When you raise a ball to a height of one meter, you are doing work against the gravitational field, and that work is stored as potential energy in the ball. The amount of work you do depends only on the initial and final heights of the ball, not on the path you take or the obstacles you encounter. The gravitational field is conservative, meaning that the work done by or against it depends only on the initial and final positions, not on the path $\endgroup$
    – Vish
    Jan 16 at 17:14
  • $\begingroup$ However, when you move a charge through a circuit, you are not only doing work against the electric field, but also against the resistance of the circuit elements. The resistance is a measure of how much the circuit elements oppose the flow of current. The work you do against the resistance is not stored as potential energy, but dissipated as heat. The resistance is non-conservative, meaning that the work done by or against it depends on the path, not just the initial and final positions. $\endgroup$
    – Vish
    Jan 16 at 17:15
  • $\begingroup$ Therefore, when you move a charge from one point to another in a circuit, the electric potential difference between those points is equal to the work done by or against the electric field, plus the work done by or against the resistance. The electric potential difference is also called the voltage drop, because it represents the loss of electric potential energy as the charge moves through the circuit. $\endgroup$
    – Vish
    Jan 16 at 17:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.