(A) The resistance of the block of aluminum is given by:
where
is the aluminum resistivity
L is the length of the cylinder
A is the cross-sectional area of the cylinder
We already know the aluminum resistivity () and the length of the cylinder (L=20 m), so we must find the cross-sectional area A, which is given by the difference between the area of the larger cylinder and the area of the radial hole:
where and (assuming that the 20 mm removed radially refers to the radius of the hole).
Therefore, the cross-sectional area is
Substituting into the initial formula of the resistance, we find:
(B) The resistivity of the tungsten is , so a cylinder of tungsten of the same size would have a resistance of
The behaviour of the resistance versus temperature is given by:
where is a coefficient that for aluminum is equal to , R0 is the resistance of the piece of aluminum we found at point (A), and R is the resistance of the tungsten. Re-arranging the formula and substituting, we find
So, the temperature must increase by 259.3 degrees.
(C) The power dissipated is given by:
where I=30 A is the current. Substituting the numbers into the formula, we find