Suppose a ball is falling from a high point and we stop it midway using a plank.

enter image description here enter image description here enter image description here

Now the plank only affected the KE and total energy of the ball but not its PE.

The same way, resistance is a physical obstruction to the path of an electron which slows it down and is losing its kinetic energy but its potential energy should remain unaffected.

Now voltage is the measurement of the potential energy at a point.

Then why do we say there is a voltage drop across a resistor when the potential should be the same at both points (across the resistor)?

  • 3
    $\begingroup$ Why have you drawn a physical analogue of a capacitor then written as if you had drawn one of a resistor? $\endgroup$ Commented Apr 11, 2020 at 18:07
  • $\begingroup$ Keep also in mind that distinguishing between kinetic and potential energy of an electron in a crystal is a risky business. $\endgroup$
    – fraxinus
    Commented Apr 11, 2020 at 19:20

4 Answers 4


Your analogy is faulty.

Think instead of the falling ball as having already reached a constant terminal velocity because of air drag when its gravitational potential energy is 10 J. Then since it’s velocity is constant on the way down there is no change in its kinetic energy. Its loss of gravitational potential energy equals the heat dissipated in the air resistance.

The constant velocity ball is analogous to the constant velocity electron (constant current). The loss of gravitational potential energy per unit massis analogous to the loss of electrical potential energy per unit charge (voltage) if it were a coulomb of electrons. And the air resistance analogous to electrical resistance.

The gravity analogy I prefer is a block sliding at constant velocity down an incline plane with friction.

Hope this helps.

  • $\begingroup$ @bob_d, okay now let the ball fall in a vacuum or as in your preferred analogy a block sliding with no friction (0 electrical resistance), now according to ohms law, when R=0, V =0, which means no potential drop but as given in your analogy the block is still losing potential energy(getting converted into kinetic energy). $\endgroup$
    – Pranav K
    Commented Apr 11, 2020 at 11:33
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    $\begingroup$ @PranavK The short answer is Ohm's law simply does not apply to vacuums and superconductors (conductors with zero resistance). In fact Ohms law isn't really a law, but the linear relationship that may exist between voltage, current, and finite resistance. This has been brought up time and time again on this exchange. See the following: physics.stackexchange.com/questions/62664/…. The air resistance and incline plane resistance analogy applies to Ohms law where Ohms law is applicable.. $\endgroup$
    – Bob D
    Commented Apr 11, 2020 at 12:35

Let's assume that across a resistor, potential drop is zero when electric current passes through it. This is equivalent to stating (as you said) potential is same across it. Therefore there is NO electric field available to accelerate the electrons once they have collided with the kernels in the conductor. So either the electrons move in random direction after collision (determined by the collision itself) or it undergoes an inelastic collision with the kernel (it stops, which doesn't actually happens). This means that there is no net movement of electrons, i.e. no electric current through the conductor, contrary to what assumed. It is true that obstructions to the electrons(kernels) don't directly affect the potential energy of the elecrons. But for electric current to flow, it is needed to accelerate the electons again, which consumes it's potential energy. Of course, for a conductor in which there is no net current flowing, potential is constant all over it, as you stated.


Imagine one of those marble drop game made with a lot of tiny obstacles. If you drop the marble gravity will make it accelerate downwards until it will hit an obstacle. After hitting the obstacle it will have a different velocity, but The gravitational field will draw it downwards again until it hits the next obstacle and so on ... In this path the gravitational potential energy of the marble will decrease because it is moving on average from an higher to a lower point.

In a similar way electrons flow in a resistor under the effect of the electric potential. If the potential were the same at the extremities of the resistors there would not be any movement of the charges.


I think your analogy is wrong. For an experiment, let's take this

1) Potential difference is height difference (analogy)

2) Electrons/current matter - ball

Let's take a circuit. enter image description here

Consider the loop AB, there is 0 resistance. By ohm's law (formulated much before we came to know that electricity is particulate in nature) the potential difference is 0.

So, there is no change in height when the ball is moving from A to B.

But, between B and C, as you have a resistor, the potential difference is non-zero, therefore, there is a drop in "height" and the ball falls. Once it has crossed the resistor, the potential difference between Cand D is once again 0. So, the ball doesn't lose kinetic energy.

In your assumption, you have the ball falling even when it is moving through a conducting wire, where the potential difference is zero. While moving between two points of same potential, the electron in your analogy is losing potential energy.

So, the analogy should look more like this than like the one you gave

enter image description here


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