Question

The resistivity of Cu is $0.015\mu$ ohm - m at 300 K. Addition of each atomic % of Ni and Ag causes an increase in resistivity by 0.012 $\mu$ ohm - m and $0.16\mu$ ohm - m respectively. Compute resistivity of Cu - Ni - Ag alloy at 300 K when 0.25% Ni and 0.4 atomic % Ag is added to Cu

• A. $0.082\mu$ ohm - m
• B. $0.018\mu$ ohm - m
• C. $0.006\mu$ ohm - m
• D. none

Answer 1

The correct option is A $0.082\mu$ ohm - m

Given,
The resistivity of Cu at 300K is ${\rho }_{Cu}=0.015\mu \Omega -m$
While making alloy, addition of each atomic % of Ni and Ag causes an increase in resistivity by $0.012\mu \Omega -m$ and $0.16\mu \Omega -m$ respectively.
The increase in resistivity by addition of 25 % Ni is
${\rho }_{Ni}=0.012\times 0.25$
${\rho }_{Ni}=0.003\mu \Omega -m$
And the increase in resistivity by addition of 0.4% Ag is
${\rho }_{Ag}=0.16\times 0.4$
${\rho }_{Ag}=0.064\mu \Omega -m$
The resistivity of alloy is
${\rho }_{alloy}={\rho }_{Cu}+{\rho }_{Ni}+{\rho }_{Ag}$
${\rho }_{alloy}=0.015+0.003+0.064$
${\rho }_{alloy}=0.082\mu \Omega -m$