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 Subject: Resistors and Temperature Category: Science Asked by: ciee-ga List Price: \$15.00 Posted: 19 Nov 2004 10:32 PST Expires: 19 Dec 2004 10:32 PST Question ID: 431150
 ```When the temperature is increased to about say 50-60C what will happen to the copper resistor? I'm using a 9V Energizer battery with copperwire as the resistor. What would happen to the volt and amp if the copper resistor is hot? Note: The battery temperature stays the same.``` Clarification of Question by ciee-ga on 20 Nov 2004 08:19 PST ```Ok i guess its my fault, what i meant to ask is that what will happen to the battery volt and amp or power when the temperature in the resistor rises? Thats the question that i wanted to answered. Ok the resistivity increases when the temperature increases, but what happen to the volt, amp, or power? Increased or decrease?```
 Subject: Re: Resistors and Temperature Answered By: hedgie-ga on 20 Nov 2004 09:06 PST Rated:
 ```Hello Ciee hf gave you lot of good numbers in the comment- but perhaps too much too soon. Let's start with basics - so you get a feel for what to expect. Than you can verify by measurement if it actually works that way. Let's start with the Ohm Law - described here: http://www.allaboutcircuits.com/vol_1/chpt_2/2.html Ignore the power stuff for now. You can get some real resistors in the Radioshack they look like this: http://www.allaboutcircuits.com/vol_1/chpt_2/4.html Notice the power-rating http://www.allaboutcircuits.com/vol_1/chpt_2/3.html Now what is this good for: Before you coonect anything, ALLWAYS estimate the current I which will flow - like this I= 9 / ( R0+ Ra + R1 + R2 .. ) Here is a picture of hooking up the ammeter http://www.doctronics.co.uk/meter.htm Ra is Resistance of ammeter - it is almost zero R0 is Internal resistance of battery (I would guess some 5 to 10 Ohms) Lets use 5 Ohms (not shown on the fig) Let's say - as example: You pick R1=10 R2=5 Ohms Your I will be about 9/20 = about .5 A and thats OK for battery R1 will dissipate powet R1 *I *I = 10/4 and its power rating (see link above) should be at least 2.5W - so it does not heat too much -- and so on. If you go too much over over 1A battery will heat too much -- not good. If you connect Ammeter directly to battery (R1=R2=0) You are shorting it I=9/5 ~ 2A is too much, same if you just add copper wire .. You should always estimate the current - so nothing overheats and burns. Try to measure R0 of your battery (indirectly). And do not hesitate to ask if things are not clear after you look through the references. Then - if you start with fresh (cool wire) - set current to some .5A (using R1) and measure current, you will see it slowly creeps down, as the resistors heat up. Voltage will go down, slowly, as battery is discharging - and as resitances increasing and you should be able to differentiate between the two effects by letting the resistors to cool again .. and also by measuring the voltage of the battery directly, WITH NO LOAD http://www.ehow.com/how_16767_voltmeter.html Read the warnings on this site ! Only try all this on battery powered circuits - do not use or measure AC (household voltage). Resistance of Voltmeter is very large (more then 1000 Ohms) so you can connect it directly to the battery (unlike the ammeter) and load (current) will be almost zero. They have some basic books in Radioshack about all this - or try library - some theory is useful. Happy experimenting Hedgie``` Request for Answer Clarification by ciee-ga on 20 Nov 2004 10:16 PST ```Do i need to have a resistors like those you showed on the page of radio shack? I thought my copper wire already acted as a resistor? And what you are saying is that volt will go down if the resistors heats up right? Please explain, Thank you.``` Request for Answer Clarification by ciee-ga on 20 Nov 2004 10:46 PST ```Do i need to have a resistors like those you showed on the page of radio shack? I thought my copper wire already acted as a resistor? And what you are saying is that volt will go down if the resistors heats up right? Please explain, Thank you. Let say that the battery that i'm using never wear out(imagine) will the volt go down when the temperature increases? I asked this question because i'm getting different answer, people say the volt goes up when the temperature rises thats why it shorten out the battery.``` Clarification of Answer by hedgie-ga on 20 Nov 2004 18:10 PST ```1) re: I thought my copper wire already acted as a resistor? Resistance of the copper wire is almost zero. With just a wire accros the terminals your are creating a 'short' SEARCH TERM : short circuit (e.g.): http://www.exploratorium.edu/snacks/short_circuit.html The resistance is fraction of an Ohm - unless the wire is very long and thin and made from something less conductive then copper. Resistance of a wire can be calculated by expression given here http://en.wikipedia.org/wiki/Electrical_resistance You need something in the circuit which will limit your current . It does not have to look exactly like a resistor. A flashlight bulb will act as a resistor - but same reasoning applies: If you take a bulb from single cell (1.5 V flashlight) it will burn when connected to 9V. Its resistance is too low, current gets too high and bulb's power rating not sufficient to handle the heat generated -- so it burns. 2) voltage changes Let me re-phrase what I said about the voltage. There are several things going on. First, let's differentate between the voltage across the terminals and emf. The voltmeter measures the potential difference (voltage) across R not emf. That is shoen here: http://www.physchem.co.za/Current Electricity/Heating.htm#Heating As the batter is discharching the proportion of its internal voltage that gets through the internal resistance to appear at the load gets smaller, so the battery's ability to deliver power to the load decreases. http://en.wikipedia.org/wiki/Battery_(electricity) So there is basic, long term trend - the voltage across the terminals will go down with time as battery is slowly 'going dead' The very small effect, but mesurable, will be this: as load decreases, emf and internal resistance being same, voltage accross the terminal increease a bit - that should be measurable as your resistor heats up. It will small, -- like .1V typically - and will go away as resistor cools again. that folows from the Ohms law for your circuit. Complex things hapen when battery gets warm because the internal chemistry changes. You can get a short-term 'boost' . Also battery 'recovers' if left in a warm place overnight. Warm place means up tp 30C -- do not 'cook it' it may be dangerous !!: http://is.med.ohio-state.edu/policies/battery.htm Please do look at the link and then feel free to ask for more clarification if needed. Hedgie``` Clarification of Answer by hedgie-ga on 20 Nov 2004 18:20 PST ```This link in the above clarification http://www.physchem.co.za/Current Electricity/Heating.htm#Heating has a blank space in it (which it should not have) - what that happens it is not enough to click on it. One gas to paste it into the location filed of the browser. You may want read the section which says: The emf of a battery: Here is an example of a setup http://www.iscienceproject.com/labs/finished_labs/6364_ohmslaw.html They recommend even higher resistance: They must be at least 500 ? apiece. Good choices might be 1000 ? and something like 4700 ?.```
 ciee-ga rated this answer: and gave an additional tip of: \$2.00 ```Great answer, reply very quickly, did not repeat what everyone else did. over all great job. Satisfied with the annswer```

 ```In general, the resistivity of conductors increases with increasing temperature. The opposite is true for insulators. There is no, simple universal rule that quantatively predicts how resistance varies as a function of temperature in materials, though. See http://www.matsci.ucdavis.edu/MatSciLT/EMS-172L/Files/Resistivity.pdf. The resistivity (p) of a material can be used to calculate the resistance (R) of a particular piece of a material (e.g., a wire, or sheet) if you know the dimensions of the piece: R = p*L/A, where L is the length of the piece measured in the direction of current flow, and A is the cross-sectional area of the piece measured perpendicular to the current direction. (Obviously, this gets more complicated if the piece's dimensions are not uniform when measured in this manner.) In the case of many metals, including copper, the resistivity varies almost linearly with temperature (at least from room temperature to hundreds of degrees K), and are usually expressed as: p(T) = p0*(1+ a*(T-T0)), where p0 is the resistivity at some reference temperature, T0 usually near room temperature), and p(T) is the resistivity at the temperature of interest, T. For copper, a ~= 0.004 (for temperatures near 25C), which means that for every 1 degree C increase in temperature, the resistivity increases by ~0.4%. In your example, if you heat the resistor from 25 C to 55 C, the 30C increase in temperature would increase the resistivity (and resistance) by 30*0.4% = 12%. (ASIDE: In reality, even for a material of a given composition (e.g., pure copper), the actual values for a and p0 depend on the grain structure and defect densities of the material, which, in turn, depend on how the material has been "handled" (i.e., it's thermal and mechanical history). Annealed copper has slightly different electrical properties than cold-worked or drawn copper wire.) If we assume that your battery acts as a perfect constant-voltage source, then if the resistance of an resistor connected in series with the battery increases, the current passing through the circuit will decrease according to Ohm's Law: I(T) = V/R(T), where I(T) and R(T) are the current and resistance, which are now both functions of temperature. As R(T) goes up and V is held constant, I(T) must go down. The power dissipated by the resistor also varies as a function of temperature according to: P(T) = V^2/R(T) See. http://hypertextbook.com/physics/electricity/resistance/ (requires a free Adobe plugin) http://spiff.rit.edu/classes/phys273/manual/temp_coeff.html```
 ```Hi. Continuing from the expired question.... You are using a piece of copper wire as a resistor? The comment by hfshaw details copper resistance with temp but the load which will be imposed upon the battery will be much too extreme. One foot of copper wire, even if really thin, will still be a very small resistance. It won?t then be the wire which is limiting the current, but the internal resistance of the battery and connections (and ammeter if connected). If you want to see the effect of battery resistance, touch your copper wire across the terminals of a car battery -- it?ll explode! When you load the little 9 volt battery with what amounts to almost a short circuit, the terminal voltage will drop to a small fraction of a volt. If you are monitoring the voltage across the terminals when shorting with a copper wire, the voltage will fluctuate due to connection resistance and the battery getting distressed. So your question about what to record because of the fluctuations is not a meaningful test. As I said, aim for a maximum current of 100mA when checking out terminal voltage. Even that though utilises little 9 volters very poorly. Rechargeable (most) cells are much better in this respect because their internal resistance is much lower but let?s stick to alkaline, preferably single cell 1.5 volt ones for cost. Any battery which you have connected the copper wire across is not now new therefore unrepresentative. If you measure the voltage of a new battery you?ll find it is about 9.6 volts. Connect up a suitable load and you?ll see that it immediately drops a bit -- that?s because of the internal resistance. If you monitor the voltage over time, you?ll notice that it initially drops quite rapidly then flattens out and drops very slowly. Eventually it?ll start dropping very rapidly again, at which point the battery will be more or less exhausted. You could take two new batteries and chill one for a few hours in a fridge and heat the other to 60 degrees (a bit too hot to handle) and you should find a slight difference in voltage. If the two batteries are then each connected to the same sized load you?ll find that the hot one will be slightly higher voltage, again mainly because of the internal resistance which will be lower for the hot one. The greater the load, the greater will be the difference. Under reasonable loading, self heating of the battery will be insignificant but if you want to confirm the effect of internal resistance, short out a new battery with your copper wire for a few minutes and you will find that it?ll get warm. Take care though because some batteries can burst with the heat and gas internally generated. If you can get hold of some wire wound resistors -- 100 ohm and greater you?ll find that they too get warm. But the wire which they are wound with is formulated for low change with temperature. Never zero of course, but perfectly adequate for most purposes. I?d send you some but I think there is the Atlantic ocean between us. Does this all make sense? You?ve got P=V*A but I?ll give you others just in case. A= V/R V=A*R R=V/A Obviously those are just rearrangements. But starting from P=V*A and substituting from the previous formulas, P=(A squared) * R P=(V squared) / R It takes a while to get to grips with how current changes with voltage and resistance but stick in there. In most instances, R won?t change significantly and can be taken as constant. Light bulbs do, but they glow white hot. Can you read the resistor colour code -- very useful. If you were setting up a display at the fair using colour coded resistors (not usual for over a couple of watts or wirewound), a colour code chart might look good. Do you know how to make up different resistor values by connecting resistors in series and parallel? Do you want to know? I probably haven?t covered all your questions adequately so just drop a line again if needs be. BTW, thanks for the ?thankyou?. Best```
 ```Since this has been taken up by a ?researcher? do you want more info from me. There are some points made which you may find a little confusing. For instance, you?ll hardly notice any change in current as wirewound resistors heat up unless they get very hot. As I said before, the current must be kept low from a wee 9 volt battery -- say 100mA max -- or the results will not be valid. Your copper wire is indeed a resistor but the resistance is far too low. If your battery never wears out, the voltage will increase slightly with temperature, both off load and on. You mustn?t think in terms of the amp rating of batteries. That?s not it works for small ones. Best```