Would a heating element have a very high resistance, or a very low resistance? (All comments in this post are based around the fact that the voltage is the same for each situation) I would have thought that a higher resistance would have resulted in more heat loss, but I've been taught that the higer the current, the more energy is lost to heat. Therefore, a lower resistance would release more heat. Which one is right? Thanks for any help.
The energy the heating element puts off has to come from somewhere, doesn't it? You probably have a formula in power output as a function of current and voltage in your textbook, as well as a formula for current as a function of voltage. Those should answer your question. In any case, imagine electrons trying to get through the wire. If the resistance is very low, they can scoot right through there as if it were greased- not losing much energy! If the resistance is very high there's a lot of "friction"- the electrons lose energy. Where do you think that energy goes? (That's the reverse of my first question!)
Yes, I know that P=VI, and that V=IR. Let's assume that the voltage is constant, say 12V. In this case, if the resistance was 1 Ohm, then the current would be 12A. Therefore, the total power would be 144W. If the resistance was 12 Ohms, then the current would be 1A, making the total power is 12W. So, according to this, a lower resistance would emit more heat? But that doesn't seem to make sense to me, and contradicts your second statmemt, HallsofIvy. Is there a problem with my calculations, or is it just that a lower resistance will emit more heat? Thanks for any help.
P= VI but that is NOT the heat. For one thing, heat is energy, not power. The heat a resistor produces is the energy loss given by I^{2} R. If you assume a fixed voltage and resistance, then I= V/R so the heat produced is given by (V/R)^{2}(R)= V^{2}/R. By golly it is true! LOWER resistance will produce a higher current and that is more important than the resistance! Lower resistance will produce more heat. Now that is surprising.
Yeah, I thought that was surprising. I did the calculations, and found that, and then came and posted on the forum to confirm. Wouldn't that mean that normal wires, with really low resistance have the most heat output? And wouldn't that be a waste of energy? I'm confused.
In the Heating element example you are assuming that the wire is the only resistance in the circuit. and therefore all the voltage is felt across it. With house wiring, the wire delivers electricity to a load that has a much greater resistance than the wires. Therefore the vast majority of the voltage is dropped across the load, and very little across the wires. The lower the resistance of the wires, the smaller the voltage droped by them, and the less heating they experience for any given load. This brings up a "reallife" point. In your example, you used a constant voltage applied across the heating element to compare what happens when you change resistances. In reality, if you use the same volatge source, this won't happen. Any real voltage source has an internal resistance associated with it, and can be treated like an ideal voltage source in series with a resistance. Now if you change resistances, the following happens. If you use a very large resistance for the element, it drops the majority of the voltage, but limits the current and you get little heating in the element. If you use a very small resistance, most of the voltage is dropped across the internal resistance of the battery, and you get little voltage across the element and you get little heating. If you plot the heat (wattage usage) of the element vs the resistance of the element, you will find that the maximum occurs when the Element's resistance equals that of the internal resistance of the source. This is also the reason for impedance matching in electronic circuits; The maximum signal strength is transfered when the impedance of output and input match.
There is yet another factor. Consider 2 equal diameter wires with the same resistance but of different materials, say copper and nichrome, since the resistivity of copper is something like 50 times less then that of nichrome the copper wire will be much longer then the nichorme. Now each is the same resistance so each will suffer the same I^{2}R loss, but since the copper wire is ~50 times the length of the nichrome wire it will be able to dissapate 50 times the heat at any given temperature (this is assuming equal emissitivity). Which wire will be hotter? The shorter nichrome wire of course. So you need to look at more then just the resistance to evaluate the temperature of a conducting wire.
BTW: It is exactly the high I^{2}R losses that gave us as AC power net rather then Edisons DC distribution system. High voltage AC does not see a wire the same way DC voltages do.
Ok, here is the question that I was given that all of this theory relates to: An electric blanket is connected to a 240V power supply (240V in Australia!) and contains two resistive elements, each with a resistance of R. THe switch connected to the blanket has three settings, high, medium, and low. The switch changes the wiring, so that for one setting the resistors are connected in parallel, for another setting they are connected in series, and for the final setting only one element is connected in the circuit. Which circuit setup corresponds to which heating setting? (Sorry, I don't think that was very well worded... oh well.) I found that the lowest resistance is parallel, the medium resistance is one element, and the highest resistance is series. Does this mean that High=Parallel, Medium=1 Element, and Low=Series? Thanks for the help everyone!
Not really surprising. Air is considered a good insulator (high resistance). Do we see an excessive amount of heating in the air surrounding conductors as opposed to conductors hooked to a heating element?
We have AC instead of DC because of the convenience in stepping up and down the voltages with transformers which cannot work with DC. AC sees resistance no differently than DC. HOWEVER, AC sees something that DC does not. It is called reactance and is quantified in ohms the same as resistance. But they are not the same.
It is easier than this Hi there, You can look at this from a far more simple point of view. Yes the source has an internal resistance and that if more current is drawn from that source more voltage is dropped and power consumed by that internal resistance before even leaving the source to do work on elements within the circuit; and no power is not heat, but none of this is needed to answer your question. The simple relationship between current, voltage and resistance is voltage is the product of the resistance and the current through that resistance. If you halve the resistance you double the current for the same voltage, i.e. V = IR, R(new) = R(old)/2 V = I(new)*R(new) = I(new)*R(old)/2 I(new) = 2*V/R(old) and V = I(old)R(old) = I(new)R(new), so I(new) = 2*I(old)R(old)/R(old) = 2*I(old) So indeed halving the resistance doubles the current flow. Power is related to current and resistance by the product of the resistance and the square of the current, so the power consumed by some resistance is: P(old) = I^2R and now with our new values of resistance and current we get P(new) = (2I)^2*R/2 = 4I^2/2R = 2I^2/R = 2*P(old) So halving the resistance doubled the current and the power taken from the source by that resistance. Look at it like this. If the resistance is increased from some small value to an eventual open the current starts off large and eventually drops to zero when the resistance has reached an open. No current flow means no energy transfer and hence no power is delivered to the resistance. Now if instead the resistance is started off large and decreased towards zero Ohms the current increases proportionally. This means more charges with a kinetic energy are being pushed through the resistance per unit time, more energy transfer in a given amount of time means more power is delivered. I'm not arguing with any one reply here, I just want to make this easier for the original poster of the question. While the source does have an internal resistance that takes voltage before being outputted and drops it within itself, also consuming power in the process, and we also have to assume that the source can deliver an increased current from decreasing a load resistance, and no power is not heat; heat is energy and power is energy taken/ used/ converted/ stored per unit time, but you want to know simply why a lower resistance needs more power than a greater one. It is because the greater the resistance the more likely collisions will occur with the charge flow transferring some of the kinetic energy from an electron and generating heat in the process, but far less often because the amount of charge that is flowing is perhaps half of what it was. When the resistance is decreased, more charges can flow per unit time and therefore transfer more energy into the resistance than before. It is less likely that collisions will occur with a lower resistance for the same charge flowing through it, but given the greater amount of charges that make it through the lower resistance, the collisions that do occur happen more often because there are so many more of them. In colliding with the resistive material they transfer some of their kinetic energy into heat within the resistance. So, in summary, the reason a lower resistance takes more energy from the source per unit time (power) is because more charge is allowed to flow when the resistance is lowered therefore transferring more energy from the source to the resistive load. This is the "visual" way of looking at it. Personally I just go with the mathematics which shows that the power is increased using a smaller resistance with the same applied voltage. Which is what I showed above. I hope this helps. Many Smiles, Craig
No Not exactly because the "wires" you speak of are used to transfer energy to some electronic device for useful electrical work to be done on it, the input resistance of these devices are typically far larger than what the wire resistance is. Therefore more of the supplied voltage is dropped across the load attached to the ends of the wires leaving less dropped across the wires themselves. So they dissipate very little power in delivering energy to some device. If by chance the device you connect power to also has a very low resistance, say about what the wires have, then the wires will take half the voltage and drop it across themselves and the other half across the device; but with such a small resistance overall with this circuit there's a good chance the current draw will be too large and the supply will not be able to deliver that much current. If it comes from a wall outlet you'll trip a breaker, if you are using a small power supply you'll load it down until it reaches a voltage that is equal to the current it is outputting times the resistance attached to it; which may be far smaller than what it was set at before being connected to the circuit. That's why almost all loads have a large input resistance when attached to some supply. Ideally the supply has no resistance as well as the wires, but of course they both do have some small resistance. The smaller the better in this case because this allows more of the energy to be delivered to the load and not transferred into thermal energy by the resistance of the wires. So no, you want nice low resistance wires delivering your energy to the load as less energy is lost to those wires being so low in resistance compared to what is delivered to a nice high resistance load or device. Craig
Hello my friends: This particular topic can be quite confusing, and Craig has laid out the math above and explained it accurately. I would like to help demonstrate these principles using a simple analogy I just came up with: Make a fist with one of your hands: Your fist represents R (Resistance), the tighter you squeeze your fist, the greater the resistance and vice versa. Make a fist with your other hand but with your pointer finger straight out: Your finger represents the electrons. Experiment 1: Tighten your fist as much as possible (Comfortably): This represents a high-resistance, or a resistor with a high ohm value. Take your finger from your other hand and try to push it in without loosening your fist. It should be very hard, but do not push harder. Because you have much resistance it is not possible for your finger (the electrons) to push through as fast (lower current) which means friction is minimal (low heat transfer). However, if you were to take your finger and push harder (raising the voltage) you will find that the flow (current) increases and thus the friction and heat increases too. Experiment 2: Loosen your fist but not all the way: This represents a low-resistance, or a resistor with a lower ohm value. Take your finger from your other hand and try to push it in without loosening your fist. Now, you will notice the finger is able to slide through easier, which means more electrons can squeeze through and thus more friction (more heat emitted). Conclusion: Heat generated by resistance (or friction) is proportional to the resistance and the current (Joules 1st law), and the current is proportional to the resistance and voltage (Ohms Law). In fact, if you do the above hand experiment correctly you can actually feel the heat generated by the friction and see first hand (no pun intended) how current and resistance are proportional. Quick Tips: - Electrical resistance, current and voltage behave much like your hands do in the above experiments. - At fixed resistances, current only increases when the voltage increases. Question: So do heating elements have high resistances or low resistances? Answer: It is all proportional to the application. If you only used 5 Ohms of resistance on 120V, you are using 24 amps of power, or in watts that is 2,880. For example, a common hair dryer using 1200 watts is using a heating element of 12 ohms and draws 10 amps. Notice how proportional everything is? While higher resistances create more friction, that friction can only occur with flow (current) and current cannot flow as fast with larger resistances. To calculate Q (Heat Transfer in Joules) we use the formula: Joules 1st Law: Q [tex]\alpha[/tex] I²R (Heat generated is proportional to current squared multiplied by the resistance.) To calculate voltages, currents and resistances we use the formula: Ohm’s Law: V=IR, I=V/R, R=V/I To calculate P (Power in watts): P = VI Notice how Joules and Watts are equal? 1 Watt = 1 Joule/second Example: Electrical extension cords heat up when a heavy load is connected to them because the wire within the extension cord has some resistance and while these small resistances may not be much when current flows are high, these small resistances create friction (heat). To overcome these losses in the utility electrical distribution systems the voltage is raised significantly, thus reducing current flow and less energy is lost in the form of heat. As a seperate but important note, the lower current also reduces the conductor (wire) size. Bottom line, heat generated from resistance is dependant on current flow. Lower resistances would allow more current, more friction and thus more heat. If there is no resistance, no heat would be generated because there is no opposition to current flow. -Fred ------------------------------------------------------------------------------ NOTE: Merged Experiment 3 with Experiment 1 - Thank you for the advice Craig!
Hi there, I do like your analogy, the only thing I might add, and I am not criticizing you in anyway, is that Experiment number 3 and number 1 are word for word the same. Except at the very end of experiment 3 where you use another analogy that would be the same as a voltage increase. I agree completely with the statement that increasing the voltage basically means increasing the pressure applied upon the electrons forcing more through the higher resistance material therefore converting more useful electrical energy to heat energy. You could have put that statement at the end of experiment 1, since it and experiment 3 are the same with the exception of that last statement, but other than just pointing out a redundancy in what you wrote, I like your analogy and hopefully the OP will also, as it is he/she that really needs to grasp an understanding of these concepts. So kudos to you, as I tried explaining it, but did so more in a mathematical way, and often people get a better understanding of something when they can relate it back to something else they do understand. The tightening or loosening of the fist idea to represent a greater or lower resistance is clever, as is using your finger as the current flow and the force at which you push your finger into your fist as the voltage potential applied; I like it. Craig
Hi Craig -- Thank you. I too prefer the mathematical approach but remember when I was learning this stuff how confusing it was. I hope people do use the math and just use this idea to as you say "grasp" the concept. I agree 100% and removed experiment 3 and merged it in to experiment 1. Let me know what you think. Thank you. -Fred
Yeah it was a long time ago when I was in college (as a student anyway) and I can remember some of these things being confusing. I always found that, and I do not know why, analyzing something from a mathematical point of view always gave me insight into the problem as well. Most people were simply bewildered by the math and needed some kind of analogy to something they knew to get insight into what they were trying to understand; again for me the mathematics provided that insight for me, but I have always loved mathematics. I have a professor friend of mine, we teach in two completely different styles. I like to use mathematics whenever I can to explain something and he almost always uses analogies to explain things. When it actually comes time to use a function to predict an outcome before it occurs, his analogies often allow him to look at the problem differently and model it in a way I wouldn't have thought of that is usually easier and yields the same results. He'll do whatever he can to avoid using complex calculus, he hasn't liked it since we were undergraduate students about 20 years ago; but somehow he has always found a way to get an answer to a question, most often just using Algebra when I would have immediately used a differential equation and solved for it that way. So his students seem to like him more than my students like me. We bicker back and forth about the importance of a solid mathematical understanding, or at least being able to model the problem mathematically and correctly, and his use of analogies to solve some complex problems. I do envy his ability to come up with analogies that work so very well in getting students to understand certain problems, but I also think he is doing the students a disservice in not teaching them that not every problem has an analogy that can be used to help one solve a problem and sometimes you just have to go with your mathematics and hope it leads you to the correct result. He also teaches many of the undergraduate courses we offer so he can get away with using analogies and Algebra for most of the problems; even when sometimes his Algebraic equations start to look even more complicated than using a simple line integral or what have you; but he just hates complex calculus and will not use it in his courses. Anyway, I am just babbling now, I looked back at your modified reply; I think it looks great. You hit on every point, and again the analogies you used I thought were clever. Many Smiles, Craig
It's funny you mention this, and a good point too. I am a big fan of using math but for some reason my mind starts first with thinking about the overall picture and then from there, I work to get the math. I would prefer your way, math first, because it is more scientific and can save a lot of time. Perhaps I can re-train my mind. Thank you for the help. -Fred
Hello again, Wow, I have just the opposite wish, I wish I were more like you and my friend in that I had a "feeling" about what it is I am trying to solve and then would apply the math to it. Instead I just look at everything that should be accounted for, write my mathematical model, solve the equations and then it's like, now I see what's going on, but if I didn't do the math first to solve the problem I would have no real insight into the problem, unless it is an easy problem. Anybody for the most part can learn how to properly apply a function to a problem to find a solution, but only some people have a feeling for what they wish to find before they even begin trying to model it mathematically. That's a great capability I wish I had. It has been a pleasure conversing with you. Hope to see your replies around, there are too few people at this forum that are actually helpful, I think you may be one of the helpful people. Take care, Craig