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. $\frac{\sigma_1\sigma_2}{\sigma_1+\sigma_2}$
• C. $\frac{2\sigma_1\sigma_2}{\sigma_1+\sigma_2}$
• D. $\frac{\sigma_1+\sigma_2}{2\sigma_1\sigma_2}$

Answer 1

The Correct Option is C

To find the effective conductivity of two metal wires connected in series, we need to understand how conductivity and resistance work in series and parallel circuits. Conductivity (\sigma) is the reciprocal of resistivity (\rho), and the relationship between resistance (R), resistivity (\rho), length (L), and area (A) of a wire is given by:

R = \frac{\rho \cdot L}{A}

For the wires in series, the total resistance (R_{\text{total}}) is the sum of their resistances:

R_{\text{total}} = R_1 + R_2

Given that \rho_1 = \frac{1}{\sigma_1} and \rho_2 = \frac{1}{\sigma_2}, and using the formula above, the resistances R_1 and R_2 can be expressed as:

R_1 = \frac{1}{\sigma_1} \cdot \frac{L}{A}, R_2 = \frac{1}{\sigma_2} \cdot \frac{L}{A}

Combining these, the total resistance is:

R_{\text{total}} = \left(\frac{1}{\sigma_1} + \frac{1}{\sigma_2}\right) \cdot \frac{L}{A}

The effective conductivity \sigma_{\text{eff}} for resistances in series, is given by:

\sigma_{\text{eff}} = \frac{1}{R_{\text{total}}} \cdot \frac{A}{L}

Substituting the value of R_{\text{total}}, we have:

\sigma_{\text{eff}} = \frac{1}{\left(\frac{1}{\sigma_1} + \frac{1}{\sigma_2}\right)}

After simplifying, the effective conductivity is found to be:

\sigma_{\text{eff}} = \frac{2\sigma_1\sigma_2}{\sigma_1+\sigma_2}

Thus, the correct answer is:

• \frac{2\sigma_1\sigma_2}{\sigma_1+\sigma_2}