Electrons flow from low potential to high potential in DC, i.e. from negative terminal to positive terminal. But how do they flow in AC, as the polarity changes every 10ms for 50Hz?
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when we use a DC battery into a circuit the flow of electrons in wire creates a clear picture as they flow from negative terminal to positive terminal. under AC(Alternating current) we know how energy flows and what does it mean by $I_{rms}$ or average current etc. But what about the pattern in which electrons move?
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This question stems from notations on schematics and seemingly conflicting information I'm seeing. I suspect that I'm seeing different vernacular for the same concepts--but I'm in a place where no one's ever told me that an "elevator" is a "lift". Then again, I may have the concept completely off, and need to be schooled so I don't blow up my workshop. :)  With DC: Batteries have +/- terminals. Most schematics I see show the circuit with voltage in and ground. I've heard that most schematics don't trace the return path to the negative terminal because that's understood and it doesn't need to clutter everything up. I've also heard that on a DC circuit, ground is SYNONYMOUS with the negative terminal. On schematics, I've seen V-in and ground, I've also seen V-in, ground, and a separate trace connecting to the negative terminal.   Then, we move to AC. There's a hot wire (positive), a neutral wire, and ground. I ~assume~ that in an AC circuit, positive correlates to positive, neutral to negative, and ground to ground. Transformers will correlate the +/- when changing DC.  What are the facts and what are the myths? How can I tell if I need to ground something to earth vs. "ground" to the negative terminal? When do I ground to the chassis of my device? Are there standard conventions in a schematic that would indicate GROUNDED ground vs. return-to-the-source ground? Or is that something you know from experience and doing an analysis of the circuit? Is it generally safe to assume I can connect ground to negative? Or are there cases in which that would be a Very Bad Thing, and how do I identify those cases?  Just trying to wrap my head around +/-/ground in AC vs. DC, and how that voltage is used...
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This is a question regarding the physics behind the observation.  I have guessed the answer to the question, but I may be wrong, so I want to wait for the responses before posting it.  Some major considerations:   A battery's positive terminal does have a positive potential. ie, a test positive charge will repel it and a test negative charge will attract it. Vice versa for negative terminal. From the paper below (Section 1.2.1), it seems abundantly clear that the battery will have positive and negative potential on respective terminals. Given 'point 1', above, connecting the positive terminal of battery A to negative terminal of battery B will lead to current flow in the conductor. If the potential difference only comes into effect when the circuit is closed using the terminals of the same battery (due to cell chemistry), then does that mean that 'point 1' will not be valid after this experiment (connecting positive terminal of A to negative terminal of B)? If it is only a tiny quantity of current that flows, then will it ever stop? And what determines this quantity? (Equation would be nice)   Related paper:
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For conversion of AC into pulsating DC we used Rectifiers circuits. But I read somewhere (from a well-known site) that the circuit does not work in ideal condition if I used my first circuit. The second circuit is the correct circuit as per their view ideally in ideal condition.  As I used the first circuit many times but on breadboard only so it adds the resistance of wires so might be it works.But I do not understand the reason behind for that extra resistance actually.  Please share your knowledge regarding this. It will be a great help in clearing my concepts on this topic.
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Sorry, i am novice in electronics, trying to understand basic fundamental of electronics.  1. How does capacitor conduct current when two plates aren't physically connected.
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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?
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Suppose we have the "basic" stuff like a battery 2 piece of wire and a bulb. Battery has a potential difference. But from where electrons flow to make the bulb light? from wire or from battery or from both? also if electrons flow from battery and they go through the wire (conductor) then why in insulators this doesnt happen? insulators dont give electrons but why they dont let electrons flow?
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At the risk of being blasted for cross-posting, I am posting this question again here in the Physics department. It is a duplicate of  in the EE forum. However, I would like to get some answers from more of a Physicist perspective and less of an Engineer. I don't like the fact I have to post it twice either, but I don't know how else to get the right audience to answer it here on StackExchange.  We use them every day and those in the know fully understand the functional characteristics of BJT transistors. There are documents and links galore explaining the operational math. There are even tons of nice videos that explain current theories of how they physically work. ( Most of the latter given by folks that speak "Tele-marketer English" for some reason. )  However, I have to admit, even after 40+ years, a lot of it I have to accept at face value, since the descriptions of how the collector junction fits into the equation is always a bit hand-wavy.   Anyhow, that aside, there is one facet I really just don't get. It seems to defy the laws of physics, Kirchhoff's Laws et.al.  I'm talking about your standard saturated common emitter circuit.  It is known, and we accept, that when saturated, the collector voltage will be less than the base voltage. We obviously use that to our advantage in circuits and chose parts to give us as low a Vce-Sat as possible for a particular load current.     All fine and dandy, till you look at the quintessential mode of a typical NPN transistor...    How the heck can the collector be a lower voltage than the base in that sandwich?   Even if you add in some back-EMF type voltage in there to account for it, the collector current would be going the wrong way through the base-collector junction..    The EE group have some answers that present theories that, to the most part, explain the phenomenon, but some of that theory seems rather conflicting.
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Drift velocity (explained to me as how fast the electrons are moving) is really slow. My book says the electrons move at around 10 mm/ s.  If electrons move so slowly how do circuits work so fast? If you make a basic circuit with just a light bulb, the bulb lights up almost immediately after you connect the wire to the positive terminal.
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For historical reasons, I understand that a proton's charge was assigned the label of "positive", the electron's charge was assigned the label of negative, and, crucially, equations of current in E &amp; M are written with the notion that current is the flow of positive charge despite the rarity of actual flows of protons in real systems (which I understand are mostly biological systems).   If the possibility existed to instantaneously and at no cost, retcon these associations such that protons have a negative charge, electrons have a positive charge, and current remains the flow of positive charge, would there be any beneficial or detrimental outcomes?  For instance, would there be:   new interpretations of physical laws that would provide different or better insights into any relevant subfield of Physics significant simplifications to notation in any field of Physics or Engineering significant increases in understanding in the fields of electricity, electronics, electrical engineering, radiation/radiative transport, etc. a noticeably easier learning curve for any subfield of physics dealing with the concept of charge   Additionally, is there a permutation of positive charge, negative charge, and flow of current definition that would improve upon the one I suggested?
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Consider a Photoelectric effect experiment apparatus as shown in Figure 11.1    The variation of photocurrent with the voltage applied across $\mathrm{A}$ and $\mathrm{C}$ is as shown in Figure 11.3.    My question –     Why is the current non-zero even if the voltage is zero?    Consider a situation in which the voltage across the plates is zero. Photons of sufficient frequency strike the outermost surface of the cathode, emitter plate $\mathrm{C}$. Electrons become free from the attraction of nucleus as their net energy becomes positive.  Why I think that there should be no net current?  As soon as photons strike the outermost surface of the cathode(assumed to be $1$ to $10$ atomic diameters thick), electrons become free. Now, as the electrons have kinetic energy, they'll come out of that outermost surface. Now, the electrons(with kinetic energy) should have an equal probability of going in every direction of the surface. So, around half of the electrons should go in the right direction towards the anode and strike it and half should go to the left, trying to penetrate the inner surfaces of the cathode plate.   Now, the electrons which will go to the right will face some field due to the space charge present in the tube but nevertheless, some electrons will definitely reach the anode plate. After they strike with the anode plate, they'll face some resistance but still, some electrons will reach the low resistance copper wires connected to the anode plate and milliammeter.  Now, consider the electrons which go towards left. They'll definitely face resistance from cathode plate but some will still reach the copper wire.  Now, as there is no potential difference applied between $\mathrm{A}$ and $\mathrm{C}$, there is no need of commutator and voltmeter. So, the ammeter is connected in series with the anode plates and cathode plates with copper wires.  Now, if there is steady state current in the circuit, electrons emitted to any one of the sides should dominate over the other. But the number of electrons emitted on both the sides is same, and all the electrons should face the same resistance as the current(electrons) should flow in a closed loop. But if the number and velocity of electrons coming from both the sides are same, there should be no net current. But there is. Why?  Edit –  From the answers of anna v and mmainville, it can be seen that the emission of electrons is dependent on the angle of emission of light.  But still, the question remains. If the electrons emitted towards anode have enough energy to complete the circuit and again reach the holes in cathode despite the resistance in between, the electrons emitted to the left should have enough energy to come to anode plate via copper wires and again get ejected through anode plate by the virtue of their kinetic energy and reach the cathode plate. In this case too, the current should be zero which is not true. Why?
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Electricity takes the path of least resistance!   Is this statement correct?    If so, why is it the case?  If there are two paths available, and one, for example, has a resistor, why would the current run through the other path only, and not both?
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My question is basically what exactly is electricity? I've simply been told before that it's a flow of electrons, but this seems too basic and doesn't show that electricity is instant.  What I mean is turning a switch has no delay between that and a light coming on. Is it really instantaneous? Or is it just so fast that we don't notice it?
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I'm trying to purchase several switches for a DC system with moderately low current (a few amps at 12V). When browsing for switches in parts catalogs (e.g.  and ) there are always separate voltage ratings for AC and DC applications. I understand that AC switches are less expensive because the switching time does not need to be as fast and arcing is less of a concern. There is only one current rating listed and I am not sure if it applies to AC switching, DC switching, or both.  Does this current rating apply to both the rated AC and DC voltages? Is there a good rule-of-thumb for converting between AC and DC voltage/current ratings? What is the highest current I could reasonably expect from a DC switch before moving to a relay?
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OK, very basic question here.  I read lots of books, searched quite a bit, and every description I read seemed to talk about the flow of electrons and right away go too deep in theory for me to grasp the basic principle of their use.  I understand a resistor limits the "flow", so that an LED doesn't blow up for example. But I fail to understand exactly what a resistor does to current and voltage...  Do resistors affect both current and voltage? In what manner?
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A man asked me this question in suburban train when I told him that I am an electrical engineer. Of course I replied him that current is the flow of electrons and in order for the current to flow there should be a potential difference between them. But the man didn't know what electron or potential difference was either.
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If voltage is a potential difference, describing how much energy will be lost per unit charge as charge carriers pass through the resistor, why would the rate at which charge flows (i.e. current, $=dq/dt$) affect the amount of energy lost per unit charge?  I suppose I am asking why a quasi-"extensive" property of the system (current) would affect a quasi-"intensive" property of the system (voltage drop across the resistor).  Apologies if the answer is obvious; I was having trouble understanding why the rate of charge flow would factor into voltage drop whilst trying to rationalize Ohm's law for myself.