I have been reading a lot about the subject, I've seen lots of videos (youtube, udemy) and even read some scholar papers about it and everyone explains the above in so many different ways. We all know what the scientific definition of what voltage is but I still don't know what effect it has on the electrons.  The water analogy which is used so much, seems to fail when we compare higher voltage vs lower voltage, because the higher the pressure of water in a tank is, the faster the water current will be once the tap is open, but the same does not apply to electrons. In fact, from what I've read, the velocity at which electrons move (drift velocity) is quite slow - small drift changes which propagate across the wire is what allow the fast delivery of electricity.  The other thing I have a problem is is how the voltage is represented in a circuit. I keep reading over and over again and seeing in the water analogy too, that voltage is what pushes the electrons, but if that's the case why is voltage lower at the negative terminal?   If I put a few springs in a box and apply pressure to close the box, some springs will contract and will probably fly away once I open the box, so using this analogy, is voltage the force necessary to keep the springs tight in the box?  The electrons don't really need a push to move across the hire, they have that desire, they are attracted to the positive side, how what is really the voltage role here?  Using another analogy, it seems that voltage is to electrons what fuel is to my car. At the end of a working day all I want to do is to drive home. I will always have the desire to reach home, but if I don't have enough fuel in the tank, I will never be able to, in the same manner, if there's no enough voltage the electrons will never reach the other end of the hire, even if that's all they want.  To me, it seems that voltage is simply the force that pushes electrons apart from each other. I also see in so many videos, electrons flowing from the beginning of the circuit to the end, when in fact the electrons barely more.  The more I read the more confused I become.   UPDATE 1  This is the best explanation I have read so far:  "A voltage appears whenever there is an imbalance of electrical charge (i.e. electrons). Since like charges repel and opposite charges attract, any collection of electrically charged particles creates some kind of force on each other. If there is an imbalance of negative to positive, a kind of "pressure" or "push" is formed. In conducting materials, electrons are free to flow through the material, as opposed to being fixed in atoms, and will therefore flow to the point of least "pressure"."
18243
Bit of a strange question, but what is it? My physics teacher said it was kind of like a "push" that pushes electrons around the circuit. Can I have a more complex explanation? Any help is much appreciated.
89292
I have difficulties with understanding high voltage transmission and transformers working process. Particularly with this expression: "the higher the transmission line voltage, the smaller the current".  I will try to explain:  The first case - we have usual line transmission, without transformers:   Here is all straightforward: we have circuit, with   R = 50 Ω;  V = 1000 v;  I = 20 A   Everything is clear and understandable.  Now, the second case - we have high voltage line transmission with transformers:   Two trasformers was added: step up and step down. Why it was done? I know the answer, it is wrote everywhere on the internet - it was done, to decrease current in transmission line, in order to reduce losses, caused by high current. Voltage have increased, but power should stay the same, therefore current should be decrease, the formula P = I*V, Joule-Lenz law... etc. It's all clear.  But! Now we are have potential difference 10 000 volt in our transmission line. Resistance stays the same. Apply Ohm's law and get:  \$I = \frac {10000v}{20 Ω} = 500 A\$  500 amperes instead of 20 amperes in the first scheme. I understand, that it can't be the truth, because then there would not high voltage lines existence. But where I made error?  Edit1:  I have found good discussion on this topic . They working on this question thoroughly. Look there, if you are interested in the detailed answer too (first three pages needed to read).  Edit2:  Can anyone explain this moment:    We are turn on generator (AC source)  It starts produce AC current This action affects step up transformer by the magnetic field on its primary coil and step up transformer start works too, generating current on its secondary coil. Then current goes to the high voltage line After high voltage line, it comes to the step down transformer Step down transformer starts work too And only now is the turn of ‘Load’.   What I think - ‘Load’ is passive entity, it is take what it is given. How it can dictate and regulate magnitude of current which pass through high voltage line? 'Load' just sits on its place and waits what current will be on the step down transformer and uses, what has come. It can't sucks current, how it want. Furthermore, even if ‘Load’ turned off, circuit will closed and will work. Right?  In the first case (without transformers), we are not having closed circuit, before ‘Load’ is turned on. But in the second case we have closed circuit without 'Load'.  Second question: how ‘Load’ can determines and affects to the current in the high voltage line, when it is passive entity (in my opinion)? Is the current in the high voltage line depends from the load after step down transformer?  Found answer to the second question . Very good article.
49990
Say you have a conductor, filled with free electrons. The nuclei have a weak pull on the valence electrons so they are moving around in the conductor.  But the electrons don't leave the solid. If you bring a positively charged object near the conductor, the electrons might move to the other side of the object, but won't leave it, almost as if there's a "wall" blocking it.   Why do surfaces act like barriers/walls for free electrons?  I have gotten the following explanations:  1) They are attracted by the nuclei so they are pulled back.  My problem with this: Conductors have a weak attraction force on the electrons, so why does it take an extremely large force to counteract the pull by the nuclei? And also what makes the boundary so special?  If the electron can move within the solid freely, what makes it so much harder to move outside the solid, since the force by the nuclei is the same?  2) Think of the electrons as being held by the nuclei by a string. If you pull on it, you might be able to move the electron away but you would need a stronger force to break the "string" of the nuclei.  My problem with this: This analogy doesn't work. Elastic strings follow Hooke's Law:  $$F = kr$$ where $k$ is a constant and $r$ is the distance. As the distance increases, the force pulling back increases.  Electromagnetic forces however follow Coloumb's Law:  $$F = \frac{kq_1q_2}{r^2}$$  As the distance increases, force pulling back decreases.  I am not looking for a mathematical proof. I am looking for an intuitive explanation. Something like an analogy would help a lot, but any other explanation you feel would be more helpful is welcome.  Also note that I know that you can get electrons to leave a conductor, but this requires a high voltage.  Related:
2043
Current is the amount of electrons passing through a wire. Can we say that voltage is the speed of those electrons?
27801
Let's say that we have a car travelling at a constant velocity (V) on a flat plane , let's assume that it's applied force or the force applied from its engine ( Fapp ) balances the frictional force ( Ffrict ) , hence the power due to Fapp is equal to Fapp×V , now let's assume that the same car now moves up an incline with the same engine power as previously stated and at a constant velocity ...but this time it will travel at a much lower constant velocity since the force which the engine exerts is much higher now in order to balance Fparalell and Ffrict , but what I don't understand is what caused this reduction in velocity , if the engine applies an equal force to that of the counteracting forces shouldn't that mean that there is no net force hence no change in velocity ,also if p=FV and if we now apply a greater force for the same velocity V why does the power increase , I know the answer can be simply explained using mathematics but my intuition says that power is the rate at which kinetic energy is transferred to a particle , so why would a greater force at the same constant velocity mean that you are transferring a greater amount of KE per unit time ?
83111
If we want to transmit electic current for a long distance, we must minimize a heat that releases because of the resistanse. We cannot make a cable wide because it is expensive and it will be massive. So, we must decrease a current because $Q = I^2R\Delta t$. A power that we want to transmit is constant, $P = UI$, so we must increase the voltage. Nevertheless, $Q = \frac{U^2}{R}\Delta t$, so if we increase the voltage, the heat will increase too. And for addition, $U = IR$, so if we decrease the current, the voltage will decrease too.  Could you, explaine in details where I have a mistake.  P.S. sorry for my English, I am not a native speaker.
8031
I know this question sounds silly, as if there was a potential difference a current would be created when the terminals are connected together and this would mean energy has come from somewhere.  The reason I ask this though is that from my understanding of the depletion region and built in potential of a diode it seems like if you connected a voltmeter across the whole diode it would show the value of the built in potential.  This is explained in the image below:    At first, electrons flow from the n type to the p type because there are a higher concentration in the n type, and holes do vise versa. This is called the diffusion current. The first electrons and holes to cross the pn boundary are the ones which are closest to it; these carriers recombine when they meet each other and are then no longer a carrier. This means there is a depletion region of no carriers near the pn boundary. because electrons have left the n type material, and holes have left the p type material, there is a surplus of positive and negative charge on the n and p side of the pn boundary respectively. This causes an electric field that opposes the diffusion current, and so no more electrons or holes cross the boundary and combine. In short, only the electrons and holes near the boundary combine, because after they have done that an electric field is formed that prevents any more carriers from crossing. The current due to this electric field is called drift current, and when in equilibrium this will equal the diffusion current. Because there is an electric field at the boundary (pointing from the positive charge to the negative charge) there is an associated voltage. This is called the built in potential.  If you sample the electric field at each point along the diode from left to right, you would start with 0 in the p region because there are an equal number of protons and electrons. As you approach the depletion region you would see a small electric field pointing back towards the p region, caused by acceptor impurities which now have an extra electron (due to recombination) and therefore now have a net negative charge. This electric field would increase in strength as you get closer to the boundary, and then die away as you get further away.  This electric field means there is a voltage, as shown in graph (d). The p side is at an arbitrary potential, and the n side is at a potential higher than this because there is an electric field between them. This means there is a potential difference across the depletion region; this is known as the built-in potential.  But why when I connect a volt-meter across the whole diode will I not see this built in potential?  Edit:  I've also asked this question on another  And they say that the reason is because the net potential difference across the diode once it has been connected to a wire or volt meter is zero, because a metal-semiconductor junction has an electric field between it, and the overall effect of the electric fields at the junctions of the m-p-n-m materials results in zero potential difference.  Regardless of the fact that voltage does not equal electrostatic potential difference, can you please confirm that there is also no net electrostatic potential difference across a diode connected to a wire or volt meter because of the m-s junctions? I'd like to get a physicists answer on this, to see if this is the truth or just a simplified explanation to make it easier to accept.
23555
We all know that the amount of current flowing b/w the plates is independent of the frequency. If I were to ask why doesn't it depend then you'd probably say that it depends on the number of electrons rather than the frequency or you might show me a current vs frequency graph.  But here is my theoretical argument. Electric current is defined as the amount of charge flowing through a point/space/wire per unit time. Let I be current due to photo electrons ejected by photons of some frequency. Now if I would increase the frequency, the kinetic energy of the ejected electron would increases, therefore it will move faster and therefore contribute more current (charge will through the same point many times per unit time) . Thereby, giving a net current greater than I.  What is wrong with my argument?  After thinking a lot, I came up with a possible reason. However fast the electrons may move, but when they hit the metal plate, they would lose their velocity and start moving in drift velocity. Now there is a problem with this argument. If this were to be true then I should expect to see a big chunk of extra electrons getting stuck at the anode.  If my reasoning was correct, then why can't I see a big charge buildup at the anode.
92973
I know that electric current is the flow of electrons but electrons have a very slow drift speed (about 2mm/s). How is it that electric current reaches its destination at almost the speed of light?
1365
I understand all the concepts of what voltage is using all the analogies but some things related to the drop of voltage across a circuit confuses me.   If I had a short circuit and attached a voltmeter I would get a potential difference reading of 0 volts. How is current then going through the wire if it is 'X' volts at any point in the wire? Let's say I have a DC 9V battery with a load in the middle. I read somewhere (probably mistaken here) that the voltage drop in this situation must be 9V or rather, that the sum of load resistances must be equal to voltage of the source. I mean, there is a variable amount of resistance to each load so at the base of the load I might have 9V and at the end of the load I might have 5V with a 4V potential difference.
109978
How is it possible that in scenario 1 resistor R1 converts 5V of energy to heat and in scenario 2 the same resistor R1 converts 2.5V of energy to heat?  The resistor R1 in both scenarios are exacly the same, I don't seem to grasp what happens on the physical level to the electrons how they "know" when to dump all their energy at once in resistor R1 in scenario 1 or to divide the energy they have proportionally over the resistors R1 and R2 in scenario 2.  I have read an explenation , but it's still not clear. How is it possible that the voltage is different at point A going through the same amount of resistance? I would expect a linear relation between voltage and resistance.  Could someone explain what physically happens that causes this?         &ndash; Schematic created using   Edit  Is my understanding right? According to   The resistance drops liniar along the path.  In scenario 1 at the start of R1 the resistance the current encounters is 100\$\Omega\$ and at the end of R1 it encounters 0\$\Omega\$.  In scenario 2 at the start of R1 the resistance the current encounters is 200\$\Omega\$ and at the end of R1 it encounters 100\$\Omega\$.  If this is true, why does point A in scenario 2 have more volt left if it has encountered a higher resistance of 200\$\Omega\$ at the beginning opposed to scenario 1 where it encountered 100\$\Omega\$ at the beginning?
3891
I recently came to know about the Conventional Current vs. Electron Flow issue. Doing some search I found that the reason for this is that Benjamin Franklin made a mistake when naming positive and negative charges. There is even this little comic about that   My question is, how can a naming convention be wrong? Perhaps I don't understand what is the actual mistake here, I don't know. But I don't see how a naming convention could be wrong or right. There is no right and wrong about that. It could've been any other name, like charge A and charge B.  I'll appreciate anyone can help me understand what is wrong in all this.
63750
I would like to know why does electrons flow through a wire connected to a battery as said in .It tells that "the electrons that are stripped from the carbon electrode is collected on the zinc can" and then tells     The electrons at the negative terminal want to go to positive terminal, they just need a way to get there. In our light bulb circuit, the way to get there is through the wire.   Why couldn't the electrons go to the positive terminal through the electrolyte(potassium hydroxide) instead of going through the wire?
61248
Imagine a battery with the positive terminal connected to ground.  The positive side has a high eletrical potential, ground obviously has a lower electric potential.  Electricity wants to seek a lower potential, and there's no significant resistance between the terminal and the ground.  Shouldn't electricity flow?  I know it doesn't but I don't understand why.  I know you have to close the loop, but why is this necessary?
90682
I am a little unclear on what  actually is and I am trying to find out what the physical cause for voltage is. Is it the speed of the electrons through the conductor or is it an excess of electrons on the supply side of the conductor. I somewhat understand electromagnetic induction in transformers but I can't seem to figure what the voltage is physically.
89382
In the  (semi-classical, but should still apply here I think), the conducting electrons are in a constant electric field, and, in between collisions with the lattice ions (that happen, on average, in a time $\tau$ since their last collision) in which their velocities are reset to an average of 0, they are accelerated and gain a momentum  $$p = qE\tau (=mv)$$, where $q$ is its charge and $E$ is the electric field from the applied voltage. This obviously means in between the collisions, the electron accelerates $a$, from $v_f = v_i + a\tau$.  Accelerating charges give off . So does a DC carrying wire give off radiation one can measure?  I suspect it does, but it's of an insanely small magnitude. A quick order of magnitude calculation with the following values gives me:  $$q \sim 10^{-19} \,{\rm C}, \\ E\sim 1\,{\rm V}/1\,{\rm cm}=100\,{\rm V/meter},\\ m\sim10^{-30}\,{\rm kg} \\ \Rightarrow a = qE/m \sim 10^{13}{\rm m/s}^2 \\ \Rightarrow P \sim q^2a^2/\epsilon_0c^3 \sim 10^{-28}\,{\rm W}$$  Which is obviously pretty small. I'm not sure what's the smallest power that can be measured, but I'm guessing by increasing the applied voltage one could increase the radiated power. But is the idea even fundamentally sound?
101859
We could imagine a simple electronic circuit composed by a power source and a resistor.   It is usual find descriptions as "The moving charged particles in an electric current are called charge carriers. In metals, one or more electrons from each atom are loosely bound to the atom, and can move freely about within the metal. These conduction electrons are the charge carriers in metal conductors." (wikipedia ). That made me imagine electrons going from power -, running by the wires, crossing the resistor and returning to power +, where they suffer a new kick to power -. All that at near the speed of light.  In this model, it is easy to define "electrical current intensity" as proportional to the number of electrons per second that cross one specific point of the circuit.  However, looking for the physics of resistivity, I can read the phrase "The actual drift velocity of electrons is very small, in the order of magnitude of a meter per hour." (wikipedia ). This is a complete different point of view, where electrons are more or less fixed around a fixed position (or moving very slowly), an electricity is a wave of "collisions" or "jumps" of electrons, like in a Newton's cradle (or "do the wave" as public in a football match). Obviously, we do not need severals hours from the switch on to the light on.  So, the question is, do electrons move around a circuit ? References to band theory are welcome.  A related question is: if electrons doesn't moves along the circuit, what is "electrical current intensity"?  Note: I known there are similar questions already posted, but answers of the ones I've read are not very detailed (could be I've not found the best ones). Please, be a few patient before "close vote".