A 3 phase, 60 hz, 138 kv transmission line is 250 miles long and delivers power to a 50 mw load at 132 kv and at 0.92 la...

$$\begin{gathered} Given: \\ r=0.175ohm/mile=43.75ohm \\ L=0.302mH/mile \\ {X}_L=jWL=j28.46ohm \\ C=0.0245\mu f/mile \\ Y=j2.30\times {10}^{-3} \\ \cos \varphi =0.92lag \\ P=50Mw \\ f=60HZ \\ {V}_R=132Kv \end{gathered}$$ 

$$\begin{gathered} P=\sqrt{3}{V}_R{I}_R\cos \varphi \\ {I}_R=\frac{50\times {10}^6}{\sqrt{3}\times 132\times {10}^3\times 0.92} \\ {I}_R=237.7A \\ {I}_Y=V_R\times \frac{Y}{2} \\ {I}_Y=j87.64A \\ Applying KCL \\ {I}_z=I_R+I_Y \\ {I}_z=218.75\angle -1.45°A \end{gathered}$$

$$\begin{gathered} Now Consider \mathrm{\pi } \mathrm{model}: \\ {\mathrm{V}}_{\mathrm{s}}={\mathrm{V}}_{\mathrm{R}}+I_{z}Z \\ {\mathrm{V}}_{\mathrm{s}}=\frac{132}{\sqrt{3}}\times {10}^3+218.75\angle -1.45\left(43.75+\mathrm{j}28.46\right) \\ {\mathrm{V}}_{\mathrm{s}}=76.210\mathrm{K}+9.570\mathrm{K}+\mathrm{j}6.223K-\mathrm{j}0.2419K+0.157\mathrm{K} \\ {\mathrm{V}}_{\mathrm{s}}=85.937+\mathrm{j}5.981 \\ {\mathbf{V}}_{\mathbf{s}}\mathbf{=}\mathbf{86}\mathbf{.}\mathbf{14}\mathbf{\angle }\mathbf{3}\mathbf{.}\mathbf{98}\mathbf{°}\mathbf{KV}\mathbf{\left(}\mathbf{Answer}\mathbf{\right)} \end{gathered}$$