8 \) (lagging), and load voltage \( V_L = 125 \, \text{kV} \).
- Apparent power \( S = \frac{P}{\text{PF}} = \frac{50 \times 10^6}{0.8} = 62.5 \, \text{MVA} \).
- Load current \( I_L = \frac{S}{\sqrt{3} \times V_L} = \frac{62.5 \times 10^6}{\sqrt{3} \times 125
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