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A uniform conductor of resistance R is cut into 20 equal pieces. Half of them are joined in series, and the remaining half of them are connected in parallel. If the two combinations are joined in parallel, then what would be the effective resistance of this new combination?

Asked in GSEB Board March 2020 (old course) · Symmetric and complex networks

• (1) (101R)/(200)
• (2) R/(202)
• (3) (201R)/(200)
• (4) (200R)/(101)

Step-by-step solution

Why the other options are wrong

• (1) (101R)/(200) = R/2 + R/(200) joins the two groups in series. The question joins them in parallel.
• (3) No single slip gives (201R)/(200); it equals R + R/(200), as if all twenty pieces were in series with the parallel group added on. Sanity check: a parallel combination is smaller than its smallest branch, R/(200).
• (4) No single slip gives (200R)/(101); it is the series sum (101R)/(200) turned upside down with R left on top. Sanity check: the answer must lie below R/(200), while this is almost 2R.

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