krugertown
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a 20cm long hollow nichrome tube of inner diameter 2.8mm, outer diameter 3.0mm is connected to a 3.0V battery. What is the current in the tube?
I think this is a I = V/R problem. So i used R=([tex]\rho[/tex]*L)/A. Where L is the length of the wire and A is the outer area minus the inner area and rho is the resistivity of nichrome.
A =.003^2*[tex]\pi[/tex] - .0028^2*[tex]\pi[/tex] = 3.64*10^-6
R = ((1.5*10^-6) * (.02))/3.64*10^-6 = .0823
I = 3/.0823 = 36.44Amps
However the answer is 9.1A apparently. Where have i gone wrong?
Thanks!
btw if the greek symbols appear superscript they are not supposed to be I can't quite figure that part out
I think this is a I = V/R problem. So i used R=([tex]\rho[/tex]*L)/A. Where L is the length of the wire and A is the outer area minus the inner area and rho is the resistivity of nichrome.
A =.003^2*[tex]\pi[/tex] - .0028^2*[tex]\pi[/tex] = 3.64*10^-6
R = ((1.5*10^-6) * (.02))/3.64*10^-6 = .0823
I = 3/.0823 = 36.44Amps
However the answer is 9.1A apparently. Where have i gone wrong?
Thanks!
btw if the greek symbols appear superscript they are not supposed to be I can't quite figure that part out