Correct Answer - Option 2 : 401 A and 231.5 A

Given:

V_{L} = 1000 V (Line Voltage)

Motor output power = 500 kW

Power factor, cos ɸ = 0.8 (Lagging)

Efficiency, η = 0.9

Motor input power, \({P_{in}} = \frac{{500}}{{0.9}} = 555.55\;kW\)

Line current, \({I_L} = \frac{{{P_{in}}}}{{\sqrt 3 {V_L}\cos \phi }} = \frac{{555.55 \times {{10}^3}}}{{\sqrt 3 \times 1000 \times 0 \cdot 8}} = 401\;A\)

As the motor is delta connected, current in each phase of the motor is

\({I_P} = \frac{{{I_L}}}{{\sqrt 3 }} = \frac{{401}}{{\sqrt 3 }} = 231.5\;A\)