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

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

To solve this problem, we first need to calculate the resistance of each segment of the wire. The total resistance of the wire is \(100\Omega\) and it is divided into 10 equal parts. Therefore, the resistance of each part is:

\(R_{\text{each}} = \frac{100\Omega}{10} = 10\Omega\)

Next, consider the connection of the first 5 parts in series. The equivalent resistance for a series combination is the sum of individual resistances. Thus,

\(R_{\text{series1}} = 5 \times 10\Omega = 50\Omega\)

Now, consider the parallel connection of the next 5 parts. The equivalent resistance \(R_{\text{parallel}}\) for parallel combination of identical resistors is calculated using:

\(\frac{1}{R_{\text{parallel}}} = \frac{1}{10\Omega} + \frac{1}{10\Omega} + \frac{1}{10\Omega} + \frac{1}{10\Omega} + \frac{1}{10\Omega} = \frac{5}{10\Omega} = \frac{1}{2\Omega}\)

Solving for \(R_{\text{parallel}}\), we have:

\(R_{\text{parallel}} = 2\Omega\)

The series combination of \(R_{\text{series1}}\) and \(R_{\text{parallel}}\) gives the final resistance:

\(R_{\text{final}} = R_{\text{series1}} + R_{\text{parallel}} = 50\Omega + 2\Omega = 52\Omega\)

Therefore, the resistance of the final combination is \(52\Omega\).