Line resistance
$=$ length of two wire line $\times$ resistance per unit length
Line resistance $(R)=2 \times 15 \mathrm{~km} \times 0.5 \frac{\Omega}{\mathrm{km}}=15 \Omega$
Virtual a.c in the line, $P=E_v I_v$
$800 \times 10^3=4000 I_V$ or $I_{\mathrm{v}}=200 \mathrm{~A}$
(a) Line power loss $P_{\text {loss }}=I_v^2 R$
$$
=(200)^2 \times 15=600 \mathrm{~kW}
$$
(b) Assuming no power loss due to leakage, total power need to be supply by the power plant.
$$
\begin{aligned}
P_{\text {total }} &=P_{\text {loss }}+P_{\text {output }} \\
&=600 \mathrm{~kW}+800 \mathrm{~kW}=1400 \mathrm{~kW}
\end{aligned}
$$
(c) Potential drop in the line,
$$
V=I_{\mathrm{v}} R=200 \times 15=3000 \mathrm{~V}
$$
So, the voltage output of step-up transformer at the plant should be $4000+3000=7000 \mathrm{~V}$.
hence at the plant the step-up transformer should be $440-7000 \mathrm{~V}$