Question

Two cylindrical bars, each with diameter of 2.30 cm, are welded together end-to-end. One of the original bars is copper (resistivity 1.72e-8 ohm*m) and is 0.350 m long. The other bar is tungsten (resistivity 5.65e-8 ohm*m) and is 0.155 m long. What is the resistance between the ends of the welded bar at 20 ^oC ?

Answer 1

You are going to use the equation
R=pL/A
p is the resistivity of the metals
p (tungsten) = 5.65*10^-8 ohm*m
p (copper) = 1.72*10^-8 ohm*m
L is the length of each bar
A is the area of each bar and it is taken as a cross section
The equation for the area will be A= (3.14)r²
They give you the diameter so you have to divide by 2 to get the radius

R(tungsten) =(5.65*10^-8) * (.155 m)/ (3.14*0.0115²)
R(copper) = 1.72*10^-8 * (.350 m)/ (3.14 * .0115²)
R (tungsten) = 2.108*10^-5 ohm
R (copper) = 1.449*10^-5 ohm
Since the two bars are next to each other this is considered a series thus you add the resistance together
R = R(tungsten) + R(copper) = 3.557*10^-5 ohms