JEE Main & Advanced Sample Paper JEE Main - Mock Test - 23

  • question_answer
    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 resistance of the combination is

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

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

    C) \[\frac{R}{2(1+{{n}^{2}})}\]  

    D)        \[\left( n+\frac{1}{n} \right)\]

    Correct Answer: A

    Solution :

    [a] Resistance of each part \[=\frac{R}{2n}\] For 'n' such parts connected in series, equivalent resistances, say \[{{R}_{I}}=n\left[ \frac{R}{2n} \right]=\frac{R}{2}.\]Similarly, equivalent resistance say \[{{R}_{2}}\] for another set of n identical respectively in parallel would be \[\frac{1}{n}\left( \frac{R}{2n} \right)=\frac{R}{2{{n}^{2}}}.\] For getting maximum of \[{{R}_{1}}\]& \[{{R}_{2}},\] they should be connected in series & hence, \[{{R}_{eq}}={{R}_{1}}+{{R}_{2}}=\frac{R}{2}\left( 1+\frac{1}{{{n}^{2}}} \right)\]

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