Question

Two wires of equal length, one of aluminium and the other of copper have the same resistance. Which of the two wires is lighter? Hence explain why aluminium wires are preferred for overhead power cables. $({\rho }_{Al}=2.63\times {10}^{-8}\Omega m,{\rho }_{Cu}=1.72\times {10}^{-8}\Omega m$, Relative density of Al = 2.7, of Cu = 8.9.)

Answer 1

$\text{Given:}$

For aluminium: $d_1=2.7,{\rho }_{Al}=2.63\times {10}^{-8}\Omega m$, mass $=m_1$, Area $=A_1$

For copper: $d_2=8.9,{\rho }_{Cu}=1.72\times {10}^{-8}\Omega m$, mass $=m_2$, Area $=A_2$

$\text{Step 1: Resistance of two wires}$

As length of both wires is same, say $\ell$.

For aluminium, $R_1={\rho }_{Al}\frac{\ell }{A_1}$ ...........$\left(1\right)$

For copper, $R_2={\rho }_{Cu}\frac{\ell }{A_2}$ ...........$\left(2\right)$

As given, $R_1=R_2$

From Equations $\left(1\right)$ and $\left(2\right)$

$\Rightarrow \frac{A_1}{A_2}=\frac{{\rho }_{Al}}{{\rho }_{Cu}}$ ..........$\left(3\right)$

$\text{Step 2: Mass of two wires}$

For aluminium: $m_1=A_1\ell d_1$ ..........$\left(4\right)$

For copper: $m_2=A_2\ell d_2$ ..........$\left(5\right)$

$\text{Step 3: Ratio of masses of wires}$

Equation $\left(4\right)÷\left(5\right)$

$\frac{m_1}{m_2}=\frac{A_1}{A_2}\frac{d_1}{d_2}$

Put the value of $\frac{A_1}{A_2}$ from eq. $\left(3\right)$

$\frac{m_1}{m_2}=\frac{{\rho }_{Al}}{{\rho }_{Cu}}\frac{d_1}{d_2}$ ..........$\left(6\right)$

$\text{Step 4: Substituting all the values in eq. (6)}$

$\frac{m_1}{m_2}=\frac{2.63\times {10}^{-8}}{1.72\times {10}^{-8}}\times \frac{2.7}{8.9}$

$\frac{m_1}{m_2}=0.4638$ ...........$\left(7\right)$

$\text{Step 5:}$

From equation $\left(7\right)$ we can say that mass of aluminum wire($m_1$), is less than mass of copper wire($m_2$).

Hence aluminum is lighter than copper.

Since aluminum is lighter, it is preferred for overhead power cables over copper.