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 \({I_{sc}} = \frac{{400}}{{300}} \times 30 = 120\;A\)
\(\frac{{{I_{st}}with\;Y{\rm{\Delta }}\;starter}}{{{I_{st}}with\;DOL}} = \frac{1}{3}\)
\({I_{st}}with\;Y{\rm{\Delta }}\;starter = \frac{{120}}{3} = 40\;A\)
\( \Rightarrow \frac{{{I_{st}}}}{{{I_{fl}}}} = \frac{{40}}{{21}} = 1.9\)