In an electrical cable,there is a single wire | vedclass

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(A) The resistance of a wire is given by $R = \rho \frac{l}{A} = \rho \frac{l}{\pi r^2}$.Initially,for the single wire: $R = 5 \, \Omega$ and $r_1 = 9

Vedclass · Accepted Answer

$7.5$

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(A) The resistance of a wire is given by $R = ho \frac{l}{A} = ho \frac{l}{\pi r^2}$.Initially,for the single wire: $R = 5 \, \Omega$ and $r_1 = 9 \, mm$.So,$5 = ho \frac{l}{\pi (9 	imes 10^{-3})^2} \implies ho l = 5 \pi (81 	imes 10^{-6}) = 405 \pi 	imes 10^{-6} \, \Omega \cdot m^2$.Finally,for the new cable consisting of $6$ wires in parallel,each with radius $r_2 = 3 \, mm$:The resistance of each individual small wire is $R' = ho \frac{l}{\pi r_2^2} = ho \frac{l}{\pi (3 	imes 10^{-3})^2} = \frac{ho l}{\pi (9 	imes 10^{-6})}$.Substituting $ho l$ from the initial condition: $R' = \frac{405 \pi 	imes 10^{-6}}{\pi (9 	imes 10^{-6})} = \frac{405}{9} = 45 \, \Omega$.Since there are $6$ such wires connected in parallel,the equivalent resistance $R_{eq}$ is given by $\frac{1}{R_{eq}} = \frac{1}{R'} + \frac{1}{R'} + \dots + \frac{1}{R'} = \frac{6}{R'}$.Therefore,$R_{eq} = \frac{R'}{6} = \frac{45}{6} = 7.5 \, \Omega$.

In an electrical cable,there is a single wire of copper with a radius of $9 \, mm$. Its resistance is $5 \, \Omega$. The cable is replaced by $6$ different insulated copper wires,each with a radius of $3 \, mm$. Now,the total resistance of the cable will be ............... $\Omega$.

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