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We always assign a voltage drop to a resistor when a current goes through it in a closed circuit. I was wondering if this voltage drop is due to dissipated heat from the resistor or is there another reason?

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marked as duplicate by Ignacio Vazquez-Abrams, PeterJ, Daniel Grillo, Matt Young, JRE Jan 25 at 16:04

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

up vote 11 down vote accepted

The heat dissipation and volt drop are related, but I would not describe the dissipation as the cause of the drop. As electrons pass through a resistance, they lose energy as they interact with electrons in the conducting material. As energy is given up to the material, it gains thermal energy so its temperature rises. The moving electrons lose potential energy and hence there is a drop in voltage. This is similar to a gas passing through a narrow pipe, losing pressure and causing heating by friction.

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"The moving electrons lose kinetic energy and hence there is a drop in voltage" If the electrons have lost kinetic energy, they're moving slower. If the electrons after the resistor are moving faster than the electrons before it, then electrons must be piling up inside the resistor, so the resistor must be gaining a negative charge. Only, it isn't. – David Richerby Jan 22 at 4:52
@DavidRicherby You seem to be assuming that the distance between the moving electrons is the same on both sides. If however the distance between the electrons changed as their speed changed, no pileup would be happening. – kasperd Jan 22 at 10:25
@kasperd If the distance between the electrons decreases, then there are more electrons per volume in areas where the electrons are moving slower, so those areas acquire a negative charge. Only, they don't. – David Richerby Jan 22 at 16:21
@DavidRicherby You have convinced me that it is probably a valid assumption to assume that the distance between the moving electrons is the same on both sides. – kasperd Jan 22 at 16:27
@DavidRicherby to give a full answer requires going into some quantum physics stuff, that is beyond me, but I have a good grasp of the basics. I will try to give a clearer explanation but be aware that it is an approximation to reality that aids understanding. In a conductor, where are free electrons which are not bound to the cores of atoms. We will ignore holes here. Electrons are charged and repel each other. At the negative supply terminal there will be a higher density of electrons with greater repulsive force. At the positive terminal the density will have reduced, velocity increased ... – user1582568 Jan 22 at 17:02

You can get a voltage drop across a capacitor and current can be flowing but no power will be dissipated so no, your question....

I was wondering if this voltage drop is due to dissipated heat from the resistor

Is incorrect.

Power is a by-product (for want of a better word) of current flow and voltage i.e. power = volts x amps = heat dissipation but, this doesn't happen in a "reactive component" such as a capacitor or inductor.

Voltage is really hard to explain but can be defined as: -

The voltage between two points is equal to the work done per unit of charge against a static electric field to move the charge between two points

This isn't an intuitive answer and I struggle with it a lot but hopefully someone other than me will give a really good explanation of what voltage is.

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The voltage drop across a capacitor is different and there is another equation for power across the capacitor that depends on capacitance and voltage and it has nothing to do with current. – Maryann Ethan Jan 25 at 22:15
@MaryE - there is no such thing as "power across the capacitor" so I'm confused by your comment. Also you say "he voltage drop across a capacitor is different" and I ask different to what? – Andy aka Jan 26 at 8:18
You are right. Capacitor is a passive element and power is equal to voltage time . I was confusing myself! – Maryann Ethan Jan 26 at 10:13

The voltage drop times the current is the electrical power being dumped into the resistor. This causes the resistor to heat up, not the other way around. Heating a resistor won't cause current thru it or voltage across it.

The voltage drop is what resistors do when current flows thru them. One way to think about that is it's simply that way by definition of what a resistor is:

  V = A * Ω

where V is Volts, A Amps, and Ω Ohms.

Another way to think about it is that the voltage is the force required to squeeze the current thru the resistor. Higher resistances resist current more, so require a larger force (more voltage) to make the same current pass thru.

To use the water analogy, a resistor is like a constriction in a pipe. More flow thru the pipe means more pressure across the constriction. Conversely, more pressure across the constriction means more flow thru it.

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What is voltage?

If we drop right down to basic physics, we find that both Charge and Energy are quantities with a Conservation Law. These Laws hold in both Classical and Quantum physics, in Newtonian and General Relativity. So we can be safe in taking these as having some sort of fundamental existence.

Voltage OTOH is not mentioned at all. Voltage only appears as a defined quantity that's handy to work with, as the potential energy of an electrical field. Voltage is defined as the change of energy associated with the movement of a charge (to within a scaling factor and dimension depending on what units we are using for energy and charge and whether it's per charge or absolute).

So when we push some charge (a current flowing for some time) through a resistor, see a voltage across it, and see energy released as heat from the resistor, it's not even appropriate to ask whether the heat causes the voltage or vice versa, the voltage is just a definition of what is happening with the charge movement.

If we have a conductor through which no energy is associated with the movement of charge, then there is no voltage drop across it (@Andy), and it's called a superconductor.

An analogy is height, potential energy for a gravitational field, change of which is the energy associated with moving a mass. A superconductor is like an air table, where the mass can slide sideways without a change of potential energy. Letting it drop against a frictional restraint generates heat in the 'frictor'.

The definition of voltage and the gravity analogy works for storage of energy in capacitors, inductors, height and velocity as well, but let's keep it simple for the moment with just finite or zero resistance.

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Somehow my head is spinning ;) – Rev1.0 Jan 22 at 14:04
This answers the question. – Rob Jan 25 at 14:06

I'll just leave it here as an illustration to the other answers: enter image description here

(The image is taken from here. Some attribute it to Eberhard Sengpiel).

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Nice illustration, but you really should give it proper attribution. – Olin Lathrop Jan 21 at 22:53
@OlinLathrop you are absolutely right. – Eugene Sh. Jan 21 at 22:59
Hmmm I wonder what it looks like for a super-conductor LOL – Andy aka Jan 21 at 23:07
I think, the illustration gives a rough impression on the current-determining properties of a resistor but says nothing about the voltage drop caused by a current. – LvW Jan 22 at 10:58
ha ha ...nice, but you have to interpret every pixel of this illustration...it is already posted some time ago on forum – GR Tech Jan 22 at 12:55

You need to see this at an atomic level. Current flows due to the flow of electrons (in the direction opposite to its flow). Now electrons flow due to presence of valence electrons in the valence band. i.e they don't actually flow, but jump from one copper atom valence band to the next. So consider two scenarios

  1. A copper wire across a battery(short circuit), and the potential difference across the battery terminals be 10 electrons(not the usual volt notation, simplifying the definition of voltage here). So what happens is, these 10 electrons rush into the copper wire, onto the valence band of first 10 atoms, and push the previous one's to the next 10, this follows til they come out from the other end.(for current flow, the electron that entered is not the one that comes out first). this short produces a large current(10 electrons worth large).

  2. add a resistor to the battery and wire, now what happens is that when reaching the resistor, those 10 electrons do not have enough conductive atoms to jump onto. assuming 5 electrons get "trapped" due to lack of conductive electrons. these 5 electrons cause the voltage drop across the resistor.

The heat dissipated is due to these trapped electrons jumping and colliding with the walls of the resistor.

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