Question

A wire of resistance 20 \(\Omega\) is divided into 10 equal parts. A combination of two parts are connected in parallel and so on. Now resulting pairs of parallel combination are connected in series. The equivalent resistance of final combination is _______ \(\Omega\).

Answer 1

Correct Answer: 5

Given:
- Wire's total resistance: \( 20 \, \Omega \)
- Wire divided into 10 equal segments.

Step 1: Resistance per segment
The resistance of each segment is calculated as:

\[ R_{\text{segment}} = \frac{\text{Total resistance}}{\text{Number of segments}} = \frac{20 \, \Omega}{10} = 2 \, \Omega. \]

Step 2: Parallel connection of two segments
Each pair of segments is connected in parallel. The equivalent resistance of a parallel connection is:

\[ R_{\text{parallel\_pair}} = \frac{R_{\text{segment}}}{2} = \frac{2 \, \Omega}{2} = 1 \, \Omega. \]

Step 3: Total number of parallel pairs
With 10 segments paired in groups of 2, the total number of parallel pairs is:

\[ \text{Number of parallel pairs} = \frac{10}{2} = 5. \]

Step 4: Series connection of parallel pairs
The 5 parallel pairs (each with \( 1 \, \Omega \)) are connected in series. The equivalent resistance of a series connection is:

\[ R_{\text{eq}} = 5 \times R_{\text{parallel\_pair}} = 5 \times 1 \, \Omega = 5 \, \Omega. \]

Therefore, the final equivalent resistance is \( R_{\text{eq}} = 5 \, \Omega \).