Question

What equal length of an iron wire and a copper-nickel alloy wire, each of 2 mm diameter connected parallel to give an equivalent resistance of 3 \(\Omega\) ? (Given resistivities of iron and copper-nickel alloy wire are 12 \(\mu\Omega\) cm and 51 \(\mu\Omega\) cm respectively)

• A. 110 m
• B. 97 m
• C. 90 m
• D. 82 m

Hint:

Convert all units to SI (meters and Ohm-meters) early. Resistivity given in \(\mu\Omega \cdot \text{cm}\) must be multiplied by \(10^{-8}\) to get \(\Omega\text{m}\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
When two resistors are in parallel, their equivalent resistance \(R_{eq}\) is given by \(1/R_{eq} = 1/R_1 + 1/R_2\). Each wire's resistance is determined by its resistivity, length, and cross-sectional area.
Step 2: Key Formula or Approach:
1. \(R = \frac{\rho L}{A}\)
2. \(\frac{1}{R_{eq}} = \frac{A}{\rho_1 L} + \frac{A}{\rho_2 L} = \frac{A}{L} \left( \frac{1}{\rho_1} + \frac{1}{\rho_2} \right)\)
Step 3: Detailed Explanation:
Given:
\(d = 2 \text{ mm} \implies r = 1 \text{ mm} = 10^{-3} \text{ m}\)
\(A = \pi r^2 = \pi \times 10^{-6} \text{ m}^2\)
\(\rho_1 = 12 \text{ } \mu\Omega \cdot \text{cm} = 12 \times 10^{-8} \text{ } \Omega\text{m}\)
\(\rho_2 = 51 \text{ } \mu\Omega \cdot \text{cm} = 51 \times 10^{-8} \text{ } \Omega\text{m}\)
\(R_{eq} = 3 \text{ } \Omega\)
Substitute into the parallel formula:
\[ \frac{1}{3} = \frac{\pi \times 10^{-6}}{L} \left( \frac{1}{12 \times 10^{-8}} + \frac{1}{51 \times 10^{-8}} \right) \]
\[ \frac{1}{3} = \frac{\pi \times 10^{-6} \times 10^8}{L} \left( \frac{1}{12} + \frac{1}{51} \right) = \frac{100\pi}{L} \left( \frac{51 + 12}{12 \times 51} \right) \]
\[ \frac{1}{3} = \frac{100\pi}{L} \left( \frac{63}{612} \right) \]
\[ L = 300\pi \times \frac{63}{612} \approx 942.48 \times 0.10294 \approx 97.02 \text{ m} \]
Step 4: Final Answer:
The required length of the wires is approximately 97 m.