Question

A wire of length and resistance \(100\) is divided into 10 equal parts. The first \(5\) parts are connected in series while the next \(5\) parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is:

• A. \(26\Omega\)
• B. \(52\Omega\)
• C. \(26\Omega\)
• D. \(26\Omega\)

Answer 1

The Correct Option is B

The resistance of each wire segment is calculated as $R_1 = \frac{100}{10} = 10 \, \Omega$.

Series connection of the first 5 segments:
     $R_s = 5 \cdot 10 = 50 \, \Omega$.
Parallel connection of the next 5 segments:
     $R_p = \frac{10}{5} = 2 \, \Omega$.

Total resistance calculation:
     $R_total = R_s + R_p = 50 + 2 = 52 \, \Omega$.