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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

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 is
 

Option: 1

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


Option: 2

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


Option: 3

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


Option: 4

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


Answers (1)

best_answer

Resistance of each part = \mathrm{\frac{R}{2 n}}
For ' n ' such part connected in series, equivalent resistances, say

\mathrm{R_1=n=\left[\frac{R}{2 n}\right]=\frac{R}{2} }

Similarly, equivalent resistance, say \mathrm{R_2}, for another set of n identical, respectively, in parallel resistances would be

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

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

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

Posted by

himanshu.meshram

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