Đáp án:
$\begin{align}
& {{P}_{hp}}=27000\text{W} \\
& \frac{{{N}_{1}}}{{{N}_{2}}}=\frac{1}{5} \\
\end{align}$
Giải thích các bước giải:
$L=100km;P=3MW;U=100kV;1km\Leftrightarrow 0,3\Omega $
a) công suất hao phí:
${{P}_{hp}}=\dfrac{R.{{P}^{2}}}{{{U}^{2}}}=100.0,3.\dfrac{{{({{3.10}^{6}})}^{2}}}{{{({{100.10}^{3}})}^{2}}}=27000W$
b)
$\begin{align}
& {{P}_{hp}}\downarrow 25lan\Rightarrow \dfrac{{{P}_{hp1}}}{{{P}_{hp2}}}=25=\dfrac{U_{2}^{2}}{U_{1}^{2}} \\
& \Rightarrow \dfrac{{{U}_{2}}}{{{U}_{1}}}=5=\dfrac{{{N}_{2}}}{{{N}_{1}}} \\
\end{align}$
$\dfrac{{{N}_{1}}}{{{N}_{2}}}=\frac{1}{5}$