Sigmarus Posted April 25, 2017 Share Posted April 25, 2017 ^Please answer it in the most basic but clearest way(Im not the best in science).Thanks alot! Link to comment Share on other sites More sharing options...
dimreepr Posted April 25, 2017 Share Posted April 25, 2017 (edited) ^Please answer it in the most basic but clearest way(Im not the best in science).Thanks alot! The fundamental answer is friction. Edited April 25, 2017 by dimreepr Link to comment Share on other sites More sharing options...
Sigmarus Posted April 25, 2017 Author Share Posted April 25, 2017 Hmm but what does friction have to do to affect the electrons flowing? Link to comment Share on other sites More sharing options...
dimreepr Posted April 25, 2017 Share Posted April 25, 2017 (edited) Resistance. Edited April 25, 2017 by dimreepr Link to comment Share on other sites More sharing options...
Sigmarus Posted April 25, 2017 Author Share Posted April 25, 2017 Mm probably a bit in depth?I can't really understand it with one word :/,Thanks though:) Link to comment Share on other sites More sharing options...
dimreepr Posted April 25, 2017 Share Posted April 25, 2017 If you want to stop something (resistance), it takes work "(it converts work to heat)." and we all get hot when we work. Link to comment Share on other sites More sharing options...
Externet Posted April 25, 2017 Share Posted April 25, 2017 Would it be good to explain it as a pipe where water flows freely (as electrons flow in a big wire) and inserting a resistive to flow piece of rag into the pipe, where water has to find crevices and pores in order to flow turbulently (as electrons would move in a thinner/resistive wire) ? Or several rows of people getting into stadium gates (non-resistive wire) and all the rows traffic restricted to a single open gate (resistive wire), causing friction between people . That turbulence/friction is the heat producer, were collisions to obstacles, among water molecules, electrons or people arms scraping against a wall in the restricted flow. Link to comment Share on other sites More sharing options...
Bender Posted April 25, 2017 Share Posted April 25, 2017 (edited) Typically, it is the reverse, since the voltage is constant. A resistive heater has a very low resistance. The reason is that the current is larger if the resistance is lower. P=U^2/R But if there are different wires and components in a circuit, most heat is indeed produced in the highest resistance, because the current is constant throughout the circuit. P=I^2 R Edited April 25, 2017 by Bender 1 Link to comment Share on other sites More sharing options...
StringJunky Posted April 25, 2017 Share Posted April 25, 2017 (edited) ^Please answer it in the most basic but clearest way(Im not the best in science).Thanks alot! The more electrons you try to draw through a particular diameter of wire, in any given unit of time, the hotter the wire will get because there's more collisions occurring with the ions in metal. The number of collisions increases with narrower wires and, also, longer wires than shorter wires because the incidence of collisions still increases These collisions cause electrons to be temporarily excited to a higher energy state. When they drop back down again to a lower energy state they release a photon. That photon is your heat. Think of resistance as the degree of obstruction to the free flow of the electrons. Edited April 25, 2017 by StringJunky Link to comment Share on other sites More sharing options...
keyestothefuture Posted April 25, 2017 Share Posted April 25, 2017 Heat is movement at that level so as the wire heats up, the electrons that pass along it move more. As this movement is random (it occurs in all directions), electrons tend to bump into each other and the atoms of the wire more often. This disturbs the flow of electrons. Just think of electrons flow along cold wires as "smooth" and along hot wires as "rough". Hope this helps. Link to comment Share on other sites More sharing options...
studiot Posted April 25, 2017 Share Posted April 25, 2017 The more electrons you try to draw through a particular diameter of wire, in any given unit of time, the hotter the wire will get because there's more collisions occurring with the ions in metal. The number of collisions increases with narrower wires and, also, longer wires than shorter wires because the incidence of collisions still increases These collisions cause electrons to be temporarily excited to a higher energy state. When they drop back down again to a lower energy state they release a photon. That photon is your heat. Think of resistance as the degree of obstruction to the free flow of the electrons. Photons or phonons? Link to comment Share on other sites More sharing options...
StringJunky Posted April 25, 2017 Share Posted April 25, 2017 (edited) Photons or phonons? I'm on about the photons that are released into the environment in sentence. Not savvy on phonons but that's something to do with lattice vibration, isn't it; they are part of what's going on inside the wire? We've got to bear in mind the OP knows very little about this, so a complete description is not going to be possible in a few posts and talking about exotic phenomena is not going to help them. Edited April 25, 2017 by StringJunky Link to comment Share on other sites More sharing options...
studiot Posted April 25, 2017 Share Posted April 25, 2017 I don't know many wires that emit photons directly into the environment from electrical action. The resistance is internal to the wire. I am assuming you mean that due to the increased heating effect of current, as resistance increases the black body radiation from the wire will increase? But photons internal to the wire? Link to comment Share on other sites More sharing options...
StringJunky Posted April 25, 2017 Share Posted April 25, 2017 I don't know many wires that emit photons directly into the environment from electrical action. The resistance is internal to the wire. I am assuming you mean that due to the increased heating effect of current, as resistance increases the black body radiation from the wire will increase? But photons internal to the wire? You've never seen a heated coil? Link to comment Share on other sites More sharing options...
studiot Posted April 25, 2017 Share Posted April 25, 2017 You've never seen a heated coil? Yes of course, but the light given off from say a light bulb is generated by increased agitation of the tungsten lattice. Not directly by the (passage of) electric current or by the electric current itself. You would see the same light if you heated the wire with a blowtorch and passed no current whatsoever. That is what I meant by indirectly increasing the black body radiation (and its frequency) Link to comment Share on other sites More sharing options...
StringJunky Posted April 26, 2017 Share Posted April 26, 2017 Yes of course, but the light given off from say a light bulb is generated by increased agitation of the tungsten lattice. Not directly by the (passage of) electric current or by the electric current itself. You would see the same light if you heated the wire with a blowtorch and passed no current whatsoever. That is what I meant by indirectly increasing the black body radiation (and its frequency) OK. I'm just trying to give a bit more detail than has already being given. I'll leave it to you because you know much more than I do on this. Link to comment Share on other sites More sharing options...
studiot Posted April 26, 2017 Share Posted April 26, 2017 ^Please answer it in the most basic but clearest way(Im not the best in science).Thanks alot! I note you are seeking a basic explanation so There are two ways to approach this. The electrical engineer's approach which is offered at elementary level, even in Physics. Bender has already begun this. This is empirical (based on observation and experience) and simply accepts there is a property we call resistance that connects more basic electrical properties by Ohms Law. The physicists approach. However physicists do not (usually) work in terms of resistance. They study conductance, which is the reciprocal of resistance. This allows them to study why wires and other things conduct at all. So if we take Ohms Law V = IR and the observation that the power dissipated equals the product of the voltage and the current, P=IV we can generate two new equations. [math]P = IV = I(IR) = {I^2}R[/math] [math]P = IV = \left( {\frac{V}{R}} \right)V = \frac{{{V^2}}}{R}[/math] Now the power dissipated ends up as heat, so this tells us where the heat in your question comes from. The first equation tells us that the power dissipated is directly proportional to the square of the current and to the resistance. So if we maintain the current at a constant level and increase the resistance, more heat will be generated. However, and this is very important, Ohms Law tells us that you cannot specify all three of resistance, current and voltage. Once you have specified two of them the third is fixed (by Ohms law). So if we specify the current, and change the resistance, the voltage must vary to suit (goes up). But, as Bender pointed out, most electrical supplies (batteries, the mains and so on) have a fixed voltage. So if the voltage is fixed and we increase the resistance the second equation shows an inverse proportionality between power and resistance. So the power goes down if fix the votlage and increase the resistance, but the current varies goes down to suit. OK so this is all empirical. But why does things conduct at all and what is resistance? Well this question is not studied until university/college level physics and even today is not fully understood. There are several theories including the free electron theory and the band theory. Here is a non mathematical roundup. It is an observed fact that different materials conduct electricity to differing degrees. We distinguish three main categories Conductors Semiconductors Non Conductors In the free electron theory differning materials are able to release or free some (a very small proportion) of their electrons. These electrons are free to move about within the solid and become responsible for the ability of the material to conduct. Of course they leave behind much more massive atoms or ions, firmly locked in the lattice of the material. It is a good job these are not also free to move about or the material would fall apart! These free electrons 'drift' around in aimless or random fashion, moved on by thermal agitation. So there is no net movement in any direction. If we then impress a voltage difference the electrons are pushed in a preferred direction by this voltage. The greater the voltage, the more electrons are moved. A net movement of charge (electrons) constitutes a current so we have Yes you guessed it Ohms Law! The net current is proportional to the voltage. Now in a perfect crystal these free electrons are completely free to move. This is due to the perfectly regular nature of a perfect crystal exactly matching the quantum requirements for movement. Perfect crystals have zero resistance. But no crystal is perfect, certainly not the crystals making up a wire. There are several different imperfections, but all imperfections offer some resistance to free electron movement in their immediate environment. Imperfections include Impurities The fact that the wire is made of lots of small crystals joined together at random orientations so the moving electrons keep coming up against crystal boundaries. Thermal vibration distorting the perfect regularity of the lattice as the fixed ions/atoms vibrate about their mean positions. Does this help? 1 Link to comment Share on other sites More sharing options...
Sriman Dutta Posted April 26, 2017 Share Posted April 26, 2017 Resistance of a conductor is a measure of obstruction to the path of electric current. It is mainly caused due to repulsion between moving electrons due to potential difference between the ends of the conductor and the electrons of the atoms of the conductor. The heat is generated mainly due to these collisions. [math] Q= I^2 Rt [/math] where Q is heat, I is current, R is resistance and t is time. Link to comment Share on other sites More sharing options...
Itoero Posted April 26, 2017 Share Posted April 26, 2017 A thick copper wire has lower resistance than an otherwise identical thin copper wire. The resistance is also proportional to the length. A long copper wire has higher resistance than an otherwise-identical short copper wire. Resistance is a measure of how much resistance there is for free electrons moving through the solid. The energy that moves the electrons forward is converted to heat because of friction. A light bulb is a nice example. A wire filament is heated to such a high temperature (has high resistance) that it glows with visible light. When filaments are thinner, have higher resistance, they will heat up more and give brighter light. Link to comment Share on other sites More sharing options...
Bender Posted April 26, 2017 Share Posted April 26, 2017 (edited) A light bulb is a nice example. A wire filament is heated to such a high temperature (has high resistance) that it glows with visible light. When filaments are thinner, have higher resistance, they will heat up more and give brighter light. No, see above. The lower the resistance of the lamp, the higher its wattage. Two light bulbs in series will, together, only consume half of the energy of a single light bulb. They will give less than half the light, because they become even less efficient. (In some countries, such light bulbs are no longer sold, because they wast 95% of their power to heat.) Edited April 26, 2017 by Bender Link to comment Share on other sites More sharing options...
studiot Posted April 26, 2017 Share Posted April 26, 2017 Resistance of a conductor is a measure of obstruction to the path of electric current. It is mainly caused due to repulsion between moving electrons due to potential difference between the ends of the conductor and the electrons of the atoms of the conductor. The heat is generated mainly due to these collisions. [math] Q= I^2 Rt [/math] where Q is heat, I is current, R is resistance and t is time. Surely not because that would imply that the resistance of the conductor would disappear if you took the voltage away. The resistance of a material has nothing to do with the voltage applied to it in general. (Yes I know there are such things as VDRs) Link to comment Share on other sites More sharing options...
Itoero Posted April 26, 2017 Share Posted April 26, 2017 No, see above. The lower the resistance of the lamp, the higher its wattage. Two light bulbs in series will, together, only consume half of the energy of a single light bulb. They will give less than half the light, because they become even less efficient. (In some countries, such light bulbs are no longer sold, because they wast 95% of their power to heat.) Ok, but I'm not talking about the wattage. I'm pointing out that the diameter of the filament influences the resistance, which influences brightness. Link to comment Share on other sites More sharing options...
Bender Posted April 26, 2017 Share Posted April 26, 2017 Ok, but I'm not talking about the wattage. I'm pointing out that the diameter of the filament influences the resistance, which influences brightness. How is the brightness not related to the wattage? Link to comment Share on other sites More sharing options...
Itoero Posted April 26, 2017 Share Posted April 26, 2017 How is the brightness not related to the wattage?It is. So the lower the resistance, the higher the wattage... But resistance causes brightness/heat. Then you have higher wattage with lower brightness/heat? Link to comment Share on other sites More sharing options...
Bender Posted April 26, 2017 Share Posted April 26, 2017 It is. So the lower the resistance, the higher the wattage... But resistance causes brightness/heat. Then you have higher wattage with lower brightness/heat? Remember: the voltage over a light bulb remains constant at 220 V (or 110 V in other countries). It is the current that causes the brightness, and a higher resistance means a lower current. See also Studiot's breakdown. If you have two light bulbs of the same "colour", the wire is at the same temperature and the efficiency is equal. If the efficiency is equal, and the wattage increases, the brightness also increases. (and obviously the 95% heat also increases) Link to comment Share on other sites More sharing options...
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