- #1
Mythian
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This problem is about a cylinder, which its length is 'L', its radius 'a' and its resistivity 'p'.
We pierce a hole in the cylinder (this hole has a radius of a/2)
I have now to
a) prove that the resistance equation(between the ends) of this hollow cylinder is
R= 4/3 pL/(pi)a^2
b) prove that the total resistance of the same hollow cylinder filled with a new resistivity material p2 is:
R= 4/3 pL/(pi)a^2 X p2/p2+p
I tried to prove A but it gives me R=pL/pi(b^2 - a^2)...which is not what I want. I have NO idea where does the 4/3 from the equation of the question comes from. If someone could help me with a hint or something, it would be much appreciated.
We pierce a hole in the cylinder (this hole has a radius of a/2)
I have now to
a) prove that the resistance equation(between the ends) of this hollow cylinder is
R= 4/3 pL/(pi)a^2
b) prove that the total resistance of the same hollow cylinder filled with a new resistivity material p2 is:
R= 4/3 pL/(pi)a^2 X p2/p2+p
I tried to prove A but it gives me R=pL/pi(b^2 - a^2)...which is not what I want. I have NO idea where does the 4/3 from the equation of the question comes from. If someone could help me with a hint or something, it would be much appreciated.