Question

In an aluminum (Al) bar of square cross section, a square hole is drilled and is filled with iron (Fe) as shown in the figure. The electrical resistivities of Al and Fe are $2.7\times {10}^{-8}\Omega$ m and $1.0\times {10}^{-7}\Omega$ m, respectively. The electrical resistance between the two faces P and Q of the composite bar is

• A. $\frac{2475}{64}\mu \Omega$
• B. $\frac{1875}{64}\mu \Omega$
• C. $\frac{1875}{49}\mu \Omega$
• D. $\frac{2475}{132}\mu \Omega$

Answer 1

Answer: (B)

$\frac{1875}{64}\mu \Omega$

$R_{Al}$ and $R_{Fe}$ are parallel to each other.
$R=\frac{\rho l}{A}$
Al:
$Area=7^2-2^2=45m{m}^2$
$\rho =2.7\times {10}^{-8}\Omega m$
$l=50mm$
Fe:
$Area=2^2=4m{m}^2$
$\rho =1.0\times {10}^{-7}\Omega m$
$l=50mm$
Substituting values:
$R_{Al}=30\mu \Omega$
$R_{Fe}=1250\mu \Omega$
$R_{total}=\frac{R_{Al}{R}_{Fe}}{R_{Al}+R_{Fe}}=\frac{30\times 1250}{30+1250}=\frac{1875}{64}\mu \Omega$