Explanation

(a) [image:3_image173.emf]
$V_{AB} = I.R_{AB} = \frac{I.\rho.L_{AB}}{A_{1}} = \frac{I.\rho\left( \frac{L}{2} \right)}{\pi(2r)^{2}} = \frac{I\ \rho.\left( \frac{L}{2} \right)}{\pi 4r^{2}}$
$V_{AB} = \frac{I\ \rho.L}{8\pi r^{2}}$
$V_{BC} = I.R_{BC} = \frac{I.\rho.L}{A_{2}}$
$= \frac{I.\rho.\frac{L}{2}}{\pi(r^{2})} = \frac{I.\rho.L}{2\pi r^{2}} \Rightarrow \frac{V_{AB}}{V_{BC}} = \frac{\frac{I.\rho.L}{8\pi r^{2}}}{\frac{I.\rho.L}{2\pi r^{2}}} = \frac{2}{8} = \frac{1}{4}$
$V_{AB} = \frac{V_{BC}}{4}$
Now for power loss
$P_{AB} = V_{A.B}.I$
$P_{BC} = V_{BC}.I$
$\frac{P_{AB}}{P_{BC}} = \frac{V_{AB}}{V_{BC}} = \frac{1}{4} \Rightarrow V_{AB} = \frac{P_{BC}}{4}$