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

The Correct Option is B

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

Answer 2

Calculating Total Resistance

Step 1: Calculate the Resistance of Each Segment

The total resistance of the wire is 100Ω. Dividing it into 10 equal parts:

R_segment = 100Ω / 10 = 10Ω (resistance of each part).

Step 2: Calculate the Resistance of the First 5 Parts in Series

When resistors are connected in series, the total resistance is the sum of individual resistances:

R_series = 5 × R_segment = 5 × 10Ω = 50Ω

Step 3: Calculate the Resistance of the Next 5 Parts in Parallel

For resistors in parallel, the total resistance is given by:

1 / R_parallel = 1 / R_segment + 1 / R_segment + ... (5 times) 

Substituting R_segment = 10Ω:

1 / R_parallel = 5 × (1 / 10) = 1 / 2

R_parallel = 2Ω

Step 4: Combine the Series and Parallel Resistances

The total resistance of the final combination is the sum of R_series and R_parallel:

R_total = R_series + R_parallel = 50Ω + 2Ω = 52Ω