Correct Answer - Option 1 : 1.9

Given power output = 10 kW

\({\eta _{fl}} = 0.86\)

Input power, \( = \frac{{10}}{{0.86}} = 11.628\;kW\)

\({P_m} = \sqrt 3 {V_L}{I_{fl}}\cos \phi \)

\({I_{fl}} = \frac{{11.629 \times {{10}^3}}}{{\sqrt 3 \times 400 \times 0.8}} = 20.97\;A\)

Short circuit current at 100 V is 30 A

Short circuit at rated voltage is \({I_{sc}} = \frac{{400}}{{300}} \times 30 = 120\;A\)

\(\frac{{{I_{st}}with\;Y{\rm{\Delta }}\;starter}}{{{I_{st}}with\;DOL}} = \frac{1}{3}\)

\({I_{st}}with\;Y{\rm{\Delta }}\;starter = \frac{{120}}{3} = 40\;A\)

\( \Rightarrow \frac{{{I_{st}}}}{{{I_{fl}}}} = \frac{{40}}{{21}} = 1.9\)