A wire of length 'l' and resistance 100Ω is divided into 10 equal parts. The first 5 parts are connected in series ​

Question

A wire of length 'l' and resistance \(100\Omega\) 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: 

(1) \(26\Omega\) 

(2) \(52\Omega\) 

(3) \(55\Omega \)

(4) \(60\Omega\) 

Answer 1

Correct option is : (2) \(52\Omega\)

Wire resistance = \(100\Omega\)

Divide into 10 equal parts 

so each part resistance r = \( \frac{100}{10} + 10\Omega\) 

 

\(Req. = 5(10) + \frac{10}{5}\)

\(=52 \Omega\)