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  • A wire of resistance R, length l and area of cross-section A, is cut into two parts, having their lengths in the ratio 1 : 2. The shorter wire is now stretched till its length becomes equal to that of the longer wire. If they are now connected in paral

A wire of resistance R, length l and area of cross-section A, is cut into two parts, having their lengths in the ratio 1 : 2. The shorter wire is now stretched till its length becomes equal to that of the longer wire. If they are now connected in parallel, find the net resistance of the combination.

 

 

Answers (1)

Given, Resistance = R

length = l

area of crosssection = A

If it cut in two pieces in the ratio 1 :2  then, 

length of shorter wire = \frac{l}{3}\;

length of longer wire = \frac{2l}{3}\;

    We know, R=\frac{\rho l}{A}\; implies R\propto \frac{l}{A} for the same materials

 R_{s}=\frac{R}{3}

 R_{l}=\frac{2R}{3}

If the shorter wire is stretched then to 2l area become half that is A/2, 

                \frac{R_{s}}{R'_{s}}=\frac{\frac{l'}{A}}{\frac{2l'}{A/2}}

                \frac{R_{s}}{R'_{s}}=\frac{1}{4} , R_{s}'=4R_{s}=\frac{4R}{3}

If it is connected in parallel,

            \frac{1}{R_{eitecfiv}}=\frac{3}{4R}+\frac{3}{2R}

                                =\frac{9}{4R}

therefore the required net resistance

=\frac{4R}{9} 

Posted by

Safeer PP

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