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The ratio of starting to full load current for a 10kW, 400 V, 3 phase induction motor with star delta starter, given the full load efficiency as 0.86, the full load pf is 0.8 and short circuit current is 30 A at 100 V is
1. 1.9
2. 1.8
3. 2.4
4. 3.2

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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 \({I_{sc}} = \frac{{400}}{{300}} \times 30 = 120\;A\)

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

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

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

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