Physics School Project

[holtadit]

I am doing a small project for physics this year on the variation of current with voltage of a metallic conductor.

Now all i want to know is the following:

How to calculate the current flowing through a 0.1m long wire of Silver which has a cross sectional area of 0.001 m2. Hence what is the resistance.

The same as above for Gold, Copper, Lead and Aluminium.

Please include the answers with relevant equations. Im doing IGCSEs by the way.

[wassup]

I don't think u can calculate this because u need resistivity for this, then only u can find the resistance n then d current.
The first formula will be R=pl/A (Resistivity*Length/Area), then I=V/R. But u don't know the resistivity(p).

[holtadit]

http://en.wikipedia.org/wiki/Resistivity
 here is the required resistivity values u need. Plz help me!!!!!

[astarmathsandphysics]

I will answer when i get home

[holtadit]

By the way astar when i used the formulas given i got the following answer for current 15094339.62 amps. It looks wrong... so whats your advice ??

Please do include the units for each value.

[Ghost Of Highbury]

i am extremely curious to know how did u find out the cross-sectional area of the wire..(it is so thin!)

ohk,,

plz reply with the values u took

for

length of the wire - ?
  and

cross-section area of the silver wire - ?

------------------

also the voltage u used .

[holtadit]

i am extremely curious to know how did u find out the cross-sectional area of the wire..(it is so thin!)

ohk,,

plz reply with the values u took

for

length of the wire - ?
  and

cross-section area of the silver wire - ?

I made up the values for cross sectional area and length....

If u need to change around any of the values voltage, area or length to make it more logical then please do so.

My teacher didnt give us any values nor did we do an actual experiment, yet im supposed to do the project. So i made the values up....

[Ghost Of Highbury]

I made up the values for cross sectional area and length....

If u need to change around any of the values voltage, area or length to make it more logical then please do so.

hope ur estimation for the area is close..

and yea...dont make up values for length...its not cumbersome to measure it with a meter rule...

yea...so what values did u take for the current value u posted recently i.e 15094339.62

[holtadit]

the ones i posted in the first post.... once again none of these values have been tried out experimentally so please modify if required

[astarmathsandphysics]

Definitely wrong. Did you change to metres

[holtadit]

okay i did the following

I multiplied (1.59*10^-8) by 0.1m which is my made up length. I got (1.59*10^-9) which I divided by 0.001m^2 which is my cross-sectional area.

I got 1.59*10^-6 i.e. RESISTANCE

Using V/R=I to calculate current (24V battery) i did 24/(1.59*10^-6) to get 15094339.62 amps for current.
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So using the area of wire = 0.001 m2 (or any other value to make things more logical)
            the resistivity of silver = 1.59*10^-8 (as on wikipedia)
            the length of wire = 0.1 m (or any other value to make things more logical)   
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Please calculate the resistance of the wire (using the parameter given) and hence the current flowing through it if 24 V supply is provided.

[holtadit]

I guess no one is interested in my above posts anymore 

[astarmathsandphysics]

.001m2 for a wire is massive. Like 3cm by 3cm. I think you should have divided by 1000^2. That only puts you out by a factor of 1000 though. There is another mistake. I will look when i get home.

[holtadit]

I think you should have divided by 1000^2

Maybe that was a typo by you but if i divide by 1000^2 i get 1.509433962*10^16 amps 

[astarmathsandphysics]

Assume wire is 1mm^2 =10^-06m^2
            the resistivity of silver = 1.59*10^-8 (as on wikipedia)
            the length of wire = 0.1 m (or any other value to make things more logical)   
R=pl/A=1.59^10-8 *0.1/10^-6=1.59*10^-3 ohms
I=V/R=24/1.59*10^-3 =15000 Amps