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Physics School Project

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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:

--- Quote ---I think you should have divided by 1000^2
--- End quote ---

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

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