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Topic: Series and parallel combination of resistors· Current Electricity

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 com…

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. 60 \(\Omega\)

2. 26 \(\Omega\)

3. 52 \(\Omega\)

4. 55 \(\Omega\)

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Answer: 3. 52 \(\Omega\)

Key idea: Each of the 10 equal parts has resistance \(\frac{100}{10}=10\ \Omega\).

  1. Five parts in series: \(5\times10=50\ \Omega\).
  2. Five parts in parallel: \(\frac{10}{5}=2\ \Omega\).
  3. The two groups in series: \(50+2=52\ \Omega\).

Answer: 52 \(\Omega\), option 3.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report
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