Question

Two wires of same length and thickness having specific resistances 6 \(\Omega\) cm and 3 \(\Omega\) cm respectively are connected in parallel. The effective resistivity is \(\rho \Omega\) cm. The value of \(\rho\), to the nearest integer, is __________.

Hint:

The harmonic mean of resistivities is used here because the wires are identical in dimensions.

Answer 1

Correct Answer: 4

Step 1: Relate resistance to resistivity: \(R = \frac{\rho l}{A}\). For wires in parallel, \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\).
Step 2: Since the wires are combined in parallel, the effective length remains \(l\), but the total area becomes \(2A\). \[ \frac{2A}{\rho_{eff} l} = \frac{A}{\rho_1 l} + \frac{A}{\rho_2 l} \implies \frac{2}{\rho_{eff}} = \frac{1}{\rho_1} + \frac{1}{\rho_2} \]
Step 3: Calculate \(\rho_{eff}\). \[ \frac{2}{\rho_{eff}} = \frac{1}{6} + \frac{1}{3} = \frac{1+2}{6} = \frac{3}{6} = \frac{1}{2} \] \[ \rho_{eff} = 4 \ \Omega \text{ cm} \]