# Why is my idea of voltage drop wrong?

when battery is connected to 2 ends of a wire, potential difference is created, this causes electric field causing electrons to move, the field gives the electron electric potential and the thus potential decreases leading to an increase in kinetic energy of the electron, when a resistor is placed and the electron moves within the resistor then its kinetic energy is lowered due to frequent collision and energy is lost as heat, is this loss of energy in heat called the voltage drop ? is this the V in ohm's law ? [it surely can't be the change in electric potential for different positions of the electron as it would always be constant, as it depends only on position]

I have realised that a resistor provides a difficult path for electrons to move, so the battery needs to do more work to move the electron across the resistor, this sudden increase in work done causes the potential difference measured by the resistor. Is this the correct takeback from all your answers ?

• @AntonBert The energy comes from the electromagnetic field around the circuit. In a resistor, whatever is carrying the current (often electrons, sometimes something else) couples that to heat. The current carriers gain tiny amounts of energy from the field and immediately lose that to heat. The details are complicated, messy, and add nothing to your understanding of electric circuits. Best to ignore them at this level. Ohm never heard of electrons, but he figured this out anyway. Commented Jul 3 at 14:22
• Some minor comments (v10). (1) The intermix of ALL CAPS, bold, italic, (parentheses), and [brackets], combined with unconventional capitalization and unconventional spacing around punctuation, make this question unnecessarily hard to read. As a rule, it's better to edit text until it is readable with a minimum of decoration. (2) If you have found a possibly-relevant related question, please link to it in addition to describing it. If the explanation at a related question is confusing, elaborate about which parts make sense and which don't, so that we can focus our answers.
– rob
Commented Jul 3 at 14:26
• @JohnDoty That seems like an answer, rather than a comment.
– rob
Commented Jul 3 at 14:26
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– Buzz
Commented Jul 3 at 20:38

Voltage drop means voltage loss, typically in the context of electrical current flowing through a series of resistors or a piece of resistive wire. Let's say we have a loop of resistive wire leading from the positive power supply terminal to the negative power supply terminal. The total resistance is R and the source voltage is V. That voltage is the effort variable which acts to perform work by pushing electrons through the wire against the resistance of the wire, establishing a current I which is the flow variable. (Effort)x(flow) = (power) which equals I^2R.

The kinetic energy of the electrons themselves does not enter the picture. It is the voltage at each point along the length of the loop which steadily decreases.

• thanks you helped me get a good understanding of voltage Commented Jul 3 at 14:56
• @AntonBert If this answer helped you, please consider voting it up (which you haven't). Commented Jul 3 at 20:25
• I am sorry but my reputation crossed 15 overnight(its 6 am now) Commented Jul 4 at 0:36
• i have upvoted it Commented Jul 4 at 16:23

To simplify matters, assume that the mobile charged particles under consideration are positive and ignore the thermal motion of the charged particles.

I am going to compare the motion of a charged particle with a gravitational analogy.

Imagine a heavy ball at the top of a hill.
Imagine a positive charge in the wire where the wire is connected to the positive terminal of a battery.*

Relative to the bottom of the hill the heavy ball has some gravitational potential energy per unit mass, gravitational potential.
Relative to the negative terminal of the battery where the other end of the wire is connected the positive charge has some electric potential energy per unit charge, electrical potential.

As the ball moves down the hill its gravitational potential decreases $$\dots$$
As the charge moves towards the negative terminal of the battery its electrical potential decreases, ie there is a potential/voltage drop.

$$\dots$$ and the kinetic energy of the ball increases.
$$\dots$$ and the kinetic energy of the charged particle increases.

The hill is populated with a lot of small trees.
The wire has within in it ions in fixed mean positions within a lattice.

When the ball hits a tree it momentarily stops and then starts moving again from rest as it continues down the hill.
When the charged particle hits a lattice ion it momentarily stops and then starts moving again from rest as it continues along the wire.

The net effect is that as the ball rolls down the hill its gravitational potential [energy] decreases and that lost gravitational potential energy has been converted into deformation of the tree, sound, and heat.
The net effect is that as the charge moves along the wire its electric potential [energy] decreases, and that lost electric potential energy has been converted into an increase in the vibrational kinetic energy of the ions in the lattice, the motion of the charged particles has caused a heating effect.

• isn't gravitational potential is converted to kinetic energy which is converted to heat etc... is a one- way process right. My thinking is that if there is a forest or not, in the end gravitational potential decreases due to height from the ground. So moving across a forest dosen't lower potential , it just converts kinetic to heat. Kinetic energy is converted to heat This is my problem, then why are we calling loss of KE 'voltage drop' ? Commented Jul 3 at 9:24
• elelctric potential isn't lost due to resistance it is lost due to change i position in wire right ? Commented Jul 3 at 9:40
• As the ball moves down the hill it loses gravitational potential energy and it is converted into kinetic energy. I introduced the tree to make my gravitational model more like the electrical one in which electrons collide with lattice ions. Commented Jul 3 at 9:51
• electric potential isn't lost due to resistance it is lost due to change i position in wire right ? As the charge moves along the wire the electric potential changes. Commented Jul 3 at 9:52
• i really like the way you explained this, but why do we say 'voltage' drop , in a conductor too kinetic energy of electrons is reduced due to collisions right ? Commented Jul 3 at 9:53

AlphaPhoenix made a series of videos on this topic that are just exemplary. You should begin by watching this, and then this.

so voltage decreases along the length of the conductor, on connecting a resistance, kinetic energy of electron is reduced

You are extremely wrong on this, even though it is also extremely common for students to have this as their first few misconceptions. In the water model understanding of electricity, as AlphaPhoenix does a good job explaining, a battery just tries to keep a voltage difference = water level height difference between its two ends, and it does that by pumping electrons from one end to the other.

Whatever it is you are doing analysis in those simple circuits, are always done at steady state. The transients have already gone. That is, however long it takes for the circuit to settle down into the steady state, is so fast, that you need to watch how much effort AlphaPhoenix put into the 2nd video to even measure it.

In steady state operation, everywhere along a simple loop circuit, the currents are the same. That means that the electrons have the exact same average KE everywhere along the circuit, in the conducting wires, and in the resistor, everywhere the same. The electrons have to be passing through every point along the loop at the same average speed. There is no "kinetic energy of electron is reduced".

The water model, with the water level height difference, is why we say that "voltage drop". Just watch the 1st video, he does a good job explaining.

• From what I remember, alphaphoenix assumes that water is compressible to justify the propagation of the perturbations along the pipes. I don't recall the need for compressibility to be a requisite to explain how pressure changes along an hydraulic circuit. Commented Jul 2 at 4:40
• i thought about the decrease in KE due to external resistance, this can be significant right ? Commented Jul 3 at 5:10
• I just told you that there CANNOT be a decrease in KE, external resistance or not. "external" resistance would not even make sense. Even if the electrons collided with the ions and lost some KE, that KE is immediately replenished by the PE, by the electric force that the voltage drop must equal to, that gives it back. Commented Jul 3 at 6:05
• i am starting to see your point now, but if it is not the KE then what causes the voltage drop ? Commented Jul 3 at 14:48
• from neils nelson's answer, if voltage is the effort, then i guess since a resistor offers a path with more collisions we can say that more effort is required to move through it, so the battery should make more effort to make the electron travel is this incraese in effort is called voltage drop ? Commented Jul 3 at 14:53

First of all drift velocity of electron is negligible i.e. $$10^{-4} \,\text{m}/\text{s}$$. So the changes in it are not considered to be effective. Whereas, the voltage drop is actually energy drop in reality if we look simple equation of voltage ($$V=E/R$$). $$E$$ represents the electric field across the circuit. Resistors just convert this energy present in electric field into some other form such as light or heat. For your question, if we don't provide steady voltage, then if you consider it's resistance at B or at last point of wire, the potential drop will be maximum as battery is disconnected and that much finite electric energy is converted to heat.

Indeed, the charge carriers are accelerated because of the potential difference and hence the electric field. Potential energy is converted to kinetic energy. Due to scattering they reach an equilibrium velocity, the drift velocity. Thus the excess kinetic energy is converted into heat. This conversion is not the cause but the ultimate result of the potential energy loss.

The present theory, notably the Poynting vector, gives quite a different picture. This point is discussed in Feynman's Lectures II, Ch 27.5, Feynman's Lectures also have a sound track where you can hear Feynman's actual spoken words and where he calls this picture 'intuitively screwy'.