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. $26\Omega$
• C. $26\Omega$
• D. $26\Omega$

Answer 1

Answer: (B)

$26\Omega$

Answer 2

Each part of the wire has resistance $R_1 = \frac{100}{10} = 10 \, \Omega$.
- The first 5 parts connected in series:
$R_s = 5 \cdot 10 = 50 \, \Omega$.
- The next 5 parts connected in parallel:
$R_p = \frac{10}{5} = 2 \, \Omega$.
- Total resistance:
$R_total = R_s + R_p = 50 + 2 = 52 \, \Omega$.