A cable consists of a copper wire of radius 9 | vedclass

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(A) The resistance of a wire is given by $R = ho \frac{l}{A} = ho \frac{l}{\pi r^2}$.For the original wire: $R_1 = ho \frac{l}{\pi (9 	imes 10^

Vedclass · Accepted Answer

$7.5$

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(A) The resistance of a wire is given by $R = ho \frac{l}{A} = ho \frac{l}{\pi r^2}$.For the original wire: $R_1 = ho \frac{l}{\pi (9 	imes 10^{-3})^2} = 5 \, \Omega$.For a single new wire of radius $r_2 = 3 \, mm$: $R_2 = ho \frac{l}{\pi (3 	imes 10^{-3})^2}$.Comparing $R_1$ and $R_2$: $\frac{R_2}{R_1} = \frac{(9 	imes 10^{-3})^2}{(3 	imes 10^{-3})^2} = (3)^2 = 9$.So,$R_2 = 9 	imes R_1 = 9 	imes 5 = 45 \, \Omega$.Since there are $6$ such wires connected in parallel,the equivalent resistance $R_{eq}$ is given by $\frac{1}{R_{eq}} = \frac{1}{R_2} + \frac{1}{R_2} + \dots + \frac{1}{R_2} = \frac{6}{R_2}$.Therefore,$R_{eq} = \frac{R_2}{6} = \frac{45}{6} = 7.5 \, \Omega$.

$A$ cable consists of a copper wire of radius $9 \, mm$ having a resistance of $5 \, \Omega$. If this cable is replaced by $6$ copper wires each of radius $3 \, mm$,what will be the new resistance of the cable in $\Omega$?

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