A wire of length 'l' and resistance 100 Ω 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:

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Accepted Answer

(4) 52 Ω. Each part is (100)/(10)=10 Ω. Five in series: 5×10=50 Ω. Five in parallel: (10)/5=2 Ω. The two groups in series: 50+2=52 Ω.

A wire of length 'l' and resistance 100 Ω 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:

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Why the other options are wrong