JEE Mains · Physics · STD 12 - 3. current electricity
A wire of resistance \(20 \Omega\) is divided into \(10\) equal parts. A combination of two parts are connected in parallel and so on. Now resulting pairs of parallel combination are connected in series. The equivalent resistance of final combination is _______\(\Omega\).
- A \(5\)
- B \(7\)
- C \(8\)
- D \(10\)
Answer & Solution
Correct Answer
(A) \(5\)
Step-by-step Solution
Detailed explanation
Each part has resistance \(=2 \Omega\) \(2\) parts are connected in parallel so, \(R=1 \Omega\) Now, there will be 5 parts each of resistance \(1 \Omega\), they are connected in series. \(\mathrm{R}_{\mathrm{xq}}=5 \mathrm{R}, \mathrm{R}_{\mathrm{sq}}=5 \Omega\)
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