Question

Two wires of equal diameters having resistivities ${\rho }_1$ and ${\rho }_2$ and lengths $x_1$ and $x_2$ respectively are joined in series. The equivalent resistivity of the combination is

• A. $\frac{{\rho }_1{x}_1+{\rho }_2{x}_2}{x_1+x_2}$
• B. $\frac{{\rho }_1{x}_2-{\rho }_2{x}_1}{x_1-x_2}$
• C. $\frac{{\rho }_1{x}_2+{\rho }_2{x}_1}{x_1+x_2}$
• D. $\frac{{\rho }_1{x}_1-{\rho }_2{x}_2}{x_1-x_2}$

Answer 1

The correct option is A $\frac{{\rho }_1{x}_1+{\rho }_2{x}_2}{x_1+x_2}$

Given $d_1=d_2$

$\Rightarrow R_1=\frac{{\rho }_1{L}_1}{A}=\frac{{\rho }_1{x}_1}{A}$

$R_2=\frac{{\rho }_2{L}_2}{A}=\frac{{\rho }_2{x}_2}{A}$

$(R_1+R_2)=\frac{{\rho }_1{x}_1+{\rho }_2{x}_2}{A}$

$\frac{{\rho }_{eq}\times (x1+x_2)}{A}=\frac{{\rho }_1{x}_1+{\rho }_2{x}_2}{A}$

${\rho }_{eq}=\frac{{\rho }_1{x}_1+{\rho }_2{x}_2}{x_1+x_2}$