An electric cable of copper has just one wire of radius 9 mm. Its resistance is 14 $\Omega$. If this single copper wire of the cable is replaced by seven identical well insulated copper wires each of radius 3 mm connected in parallel, then the new resistance of the combination will be :

<p>Initially, copper wire radius (r₁) = 9 mm</p> <p>Resistance (R) = 14 $\Omega$</p> <p>We know, $R = {{\rho L} \over A} = {{\rho L} \over {\pi r_1^2}} = 14$</p> <p>Now this copper wire is replaced by 7 parallel copper wire of resistance R₁.</p> <p>$\therefore$ Equivalent resistance of 7 parallel copper wire,</p> <p>$${1 \over {{R_{eq}}}} = {1 \over {{R_1}}} + {1 \over {{R_1}}} + {1 \over {{R_1}}} + {1 \over {{R_1}}} + {1 \over {{R_1}}} + {1 \over {{R_1}}} + {1 \over {{R_1}}}$$</p> <p>$\Rightarrow {1 \over {{R_{eq}}}} = {7 \over {{R_1}}}$</p> <p>$\Rightarrow {R_{eq}} = {{{R_1}} \over 7}$</p> <p>$= {1 \over 7} \times {{\rho L} \over {\pi r_2^2}}$</p> <p>$$ = {1 \over 7} \times {{\rho L} \over {\pi {{\left( {{{{r_1}} \over 3}} \right)}^2}}}$$ [as ${r_2} = {{{r_1}} \over 3}$ ; ${r_1} = 9$ m and ${r_2} = 3$]</p> <p>$= {1 \over 7} \times {{\rho L} \over {\pi r_1^2}} \times 9$</p> <p>$= {1 \over 7} \times 14 \times 9$</p> <p>$= 18\,\Omega$</p>