Question

(c18p22) Two cylindrical bars, each with diameter of 2.10 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 iron (resistivity 9.71e-8 ohm*m) and is 0.135 m long. What is the resistance between the ends of the welded bar at 20 oC ?

Answer 1

Sol:

Given:
diameter (D) =0.021 m
Radius (r) =0.0105 m
T= 20°C

We have resistance
R=ρL/A.......................................(1)
Where
ρ is the resistivity of the metals
ρ (iron) = 9.71*10^-8 Ωm
ρ (copper) = 1.72*10^-8 Ωm

L is the length of each bar
Li=0.135 m and Lc=0.35 m

A is the area of each bar and it is taken as a cross section
A= πr^2

So from equation(1),we have

R(iron) =9.71*10^-7*(0.135 )/(3.14 * 0.0105^2)

R (iron) = 3.785*10^-5 Ω

R(cu) =1.72*10^-8*(.350 )/(3.14*0.0105^2)

R (cu) = 1.738*10^-5 Ω

Since the two bars are next to each other this is considered a SERIES thus you add the resistance together

R = R(iron) + R(cu)
= 3.785*10^-5 +1.738*10^-5
=5.523*10^-5 Ω

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