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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 the remaining half in parallel to form another bundle. These bundles are then connected to give the maximum resistance. The maximum resistance of the combination is

Option: 1

   \frac{R}{2}\left(1+\frac{1}{n^2}\right)


Option: 2

\frac{R}{2}\left(1+n^2\right)


Option: 3

\frac{R}{2\left(1+n^2\right)}


Option: 4

R\left(n+\frac{1}{n}\right)


Answers (1)

best_answer

\text { Resistance of each part }=\frac{R}{2 n}

 For ' n ' such part connected in series, equivalent resistances, say

R_1=n=\left[\frac{R}{2 n}\right]=\frac{R}{2}
Similarly, equivalent resistance, say R_2 , for another set of  n identical, respectively, in parallel resistances would be

R_2=\frac{1}{n}\left(\frac{R}{2 n}\right)=\frac{R}{2 n^2}

For getting maximum of R_1 \ and \ R_2, the resistances should be connected in series and hence,

R_{e q}=R_1+R_2=\frac{R}{2}\left(1+\frac{1}{n^2}\right)

Posted by

Ritika Jonwal

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