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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:
Each piece = 100/10 = 10 Ω; Series of 5 = 50 Ω; Parallel of 5 = 10/5 = 2 Ω -> Total = 50 + 2 = 52 Ω!
Since resistance of a uniform wire is directly proportional to its length (R = ρl/A), dividing the wire into 10 equal parts divides its total resistance equally among the 10 segments:
The first 5 parts are connected end-to-end in series. For series combination, individual resistances simply add up:
The next 5 parts are connected in parallel across common terminals. For 5 identical resistors in parallel:
The problem states that these two combinations are now connected in series with each other. Therefore, the total equivalent resistance is:
Diagnose every choice: Know exactly why the right option works and why distractors fail.
Calculation error; occurs if parallel branch is miscalculated as 5 Ω.
Incorrect equivalent resistance.
Half of the correct answer; happens if student divides the final result by 2.
Exact solution: 50 Ω (series) + 2 Ω (parallel) = 52 Ω.
Students frequently write R_parallel = n × r instead of r/n, leading to 50 + 50 = 100 Ω, or misplace decimal factors.
"Resistance is proportional to length; resistors in series add linearly, while identical resistors in parallel divide by the number of branches."
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