24P1Q27

27 A

First calculate the current in the transmission cables.

$\displaystyle \begin{aligned}(P&=VI) \\ 100.00\times {{10}^{6}}&=400\times {{10}^{3}}I \\ I&=250\text{ A} \end{aligned}$

Total resistance of the cables, $\displaystyle r=2\times 100\times 0.030=6.0\text{ }\Omega\text{ }$

Power lost in the cables $\displaystyle ={{I}^{2}}r={{250}^{2}}\times 6.0=375000\text{ W}$

Power input to the sub-station $\displaystyle =100.00\times {{10}^{6}}-375000=99.625\text{ MW}$

Uncategorized

One thought on “24P1Q27”

Pingback: A-Level 2024 – xmPhysics

Leave a comment Cancel reply