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:
- Total resistance of the wire: \( 20 \, \Omega \)
- The wire is divided into 10 equal parts.

Step 1: Resistance of each part
The resistance of each part is:

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

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

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

Step 3: Total number of parallel combinations
Since there are 10 parts and they are paired in groups of 2, the total number of parallel combinations is:

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

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

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

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