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

Answer 1

The correct option is B

$\frac{2{\sigma }_1{\sigma }_2}{{\sigma }_1+{\sigma }_2}$

Step 1: Dependence of resistance on conductivity

1. Resistance depends on the length of the wire, the cross-sectional area of the wire, and the conductivity of the wire.
2. The equation for the resistance can be written as follows:

$R=\frac{L}{\sigma A}$

Here, $R$ is the resistance, $L$ is the length of the wire, $\sigma$ is the conductivity of the wire, and $A$ is the cross-sectional area of the wire.

Step 2: Resistances connected in series

1. When two or more resistances are connected in series, the equivalent resistance can be calculated by adding the individual resistances.
2. Combining the equations, the effective conductivity of the wire can be calculated as follows:

$$\begin{gathered} R_{\mathrm{equivalent}}=R_1+R_2 \\ \frac{\left(L+L\right)}{{\sigma }_{\mathrm{effective}}\times A}=\frac{L}{{\sigma }_{1}A}+\frac{L}{{\sigma }_{2}A} \\ \frac{2}{{\sigma }_{\mathrm{effective}}}=\frac{1}{{\sigma }_1}+\frac{1}{{\sigma }_2} \\ {\sigma }_{\mathrm{effective}}=\frac{2{\sigma }_1{\sigma }_2}{{\sigma }_1+{\sigma }_2} \end{gathered}$$

Hence, the effective conductivity of the wire is $\frac{2{\sigma }_1{\sigma }_2}{{\sigma }_1+{\sigma }_2}$.

Therefore, option (B) is correct.