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A piece of conducting wire of resistance R is cut into 2n equal parts. Half the parts are connected in series to form a bundle and remaining half in parallel to form another bundle. These bundles are then connected to give the maximum resistance. The maximum resistance of the combination isA. `R/2(1+1/n^2)`B. `R/2 (1+n^2)`C. `R/(2(1+n^2))`D. `R (n+1/n)` |
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Answer» Correct Answer - A a. Resistance of each part `=R//2n`. For n such parts connected in series, equivalent resistances, say `R_1 = n = [R/(2n)] = R/2` Similarly, equivalent resistance, say `R_2`, for another set of n identical, respectively , in parallel resistance would be `1/n (R/(2n)) = R/(2n^2)` For getting maximum of `R_1 and R_2`, the resistances should be connected in series and hence, `R_(eq) = R_1 + R_2 = R/2 (1+1/n^2)`. |
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