Question

A wire of $25\ \Omega$ resistance is cut into n pieces of equal length. If these pieces are connected in parallel, the equivalent resistance is $1\ \Omega$, then n is:

• A. 3
• B. 6
• C. 8
• D. 5
• E. 4

Hint:

Connecting $n$ pieces of a wire in parallel reduces resistance by a factor of $n^2$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This problem involves understanding how the resistance of a wire depends on its length and how resistors combine in parallel.
Step 2: Key Formula or Approach:
1. Resistance and Length: The resistance of a uniform wire is directly proportional to its length (\( R \propto L \)). If a wire of resistance \( R_{total} \) is cut into n equal pieces, the resistance of each piece (\( R_{piece} \)) will be: \[ R_{piece} = \frac{R_{total}}{n} \] 2. Parallel Combination: When n identical resistors, each with resistance \( R_{piece} \), are connected in parallel, their equivalent resistance (\( R_{eq} \)) is: \[ R_{eq} = \frac{R_{piece}}{n} \] Step 3: Detailed Explanation:
We are given: - Total resistance of the original wire, \( R_{total} = 25 \, \Omega \) - The wire is cut into n pieces. - Equivalent resistance of these pieces in parallel, \( R_{eq} = 1 \, \Omega \) First, find the resistance of one of the small pieces (\( R_{piece} \)) in terms of n: \[ R_{piece} = \frac{R_{total}}{n} = \frac{25}{n} \] Now, use the formula for the parallel combination of these n identical pieces: \[ R_{eq} = \frac{R_{piece}}{n} \] Substitute the expression for \( R_{piece} \) into this equation: \[ R_{eq} = \frac{(25/n)}{n} = \frac{25}{n^2} \] We are given that \( R_{eq} = 1 \, \Omega \). So, we can set up the equation to solve for n: \[ 1 = \frac{25}{n^2} \] \[ n^2 = 25 \] Take the square root (n must be positive): \[ n = \sqrt{25} = 5 \] Step 4: Final Answer:
The value of n is 5.