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.

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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?

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

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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

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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 ?.

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