Question

Two metal wires of identical dimensions are connected in series. If ${\sigma }_1$ and ${\sigma }_2$ are the conductivities of the metal wires respectively, the effective conductivity of the combination is-

• A. $\frac{{\sigma }_1+{\sigma }_2}{2{\sigma }_1{\sigma }_2}$
• B. $\frac{2{\sigma }_1{\sigma }_2}{{\sigma }_1+{\sigma }_2}$
• C. $\frac{{\sigma }_1{\sigma }_2}{{\sigma }_1+{\sigma }_2}$
  
• D. $\frac{{\sigma }_1+{\sigma }_2}{{\sigma }_1{\sigma }_2}$
  

Answer 1

The correct option is B $\frac{2{\sigma }_1{\sigma }_2}{{\sigma }_1+{\sigma }_2}$

As we know Resistance of a wire is,

$R=\frac{\rho l}{A}=\frac{l}{\sigma A}[\because \sigma =\frac{1}{\rho }]$

$\Rightarrow R_1=\frac{l}{{\sigma }_{1}A}\text{and}{R}_2=\frac{l}{{\sigma }_{2}A}$

For series combination,

$R_s=R_1+R_2$

$\frac{2}{{\sigma }_{eff}A}=\frac{l}{{\sigma }_{1}A}+\frac{l}{{\sigma }_{2}A}$

$\frac{2}{{\sigma }_{eff}}=\frac{l}{{\sigma }_1}+\frac{l}{{\sigma }_2}$

$\frac{2}{{\sigma }_{eff}}=\frac{{\sigma }_1+{\sigma }_2}{{\sigma }_1{\sigma }_2}$

${\sigma }_{eff}$ =$\frac{2{\sigma }_1{\sigma }_2}{{\sigma }_1+{\sigma }_2}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.