The starting current of a three-phase induction motor is equal to three times the full-load current. If the full-load slip is 2%, then the starting torque as a percentage of full-load torque is

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BPSC AE Paper 6 (Electrical) 25 Mar 2022 Official Paper
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  1. 18% of full-load torque
  2. 6% of full-load torque
  3. 36% of full-load torque
  4. None of the above

Answer (Detailed Solution Below)

Option 1 : 18% of full-load torque
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संकल्पना:

प्रारंभिक बलाघूर्ण और पूर्ण भार बलाघूर्ण के बीच अनुपात को निम्न द्वारा ज्ञात किया गया है:

\(\frac{{{T}_{st}}}{{{T}_{fl}}}={{s}_{fl}}\times {{\left( \frac{{{I}_{st}}}{{{I}_{fl}}} \right)}^{2}}\)

जहाँ Tst प्रारंभिक बलाघूर्ण है।

Tfl पूर्ण भार बलाघूर्ण है।

sfl पूर्ण भार पर सर्पी है।

Ist प्रारंभिक धारा है।

Ifl रेटेड धारा है।

गणना:

3-ϕ वाले प्रेरण मोटर की प्रारंभिक धारा रेटेड धारा का 3 गुना होती है अर्थात्

Ist = 3 Ifl

पूर्ण भार सर्पी (sfl) = 2 %

\(\frac{{{T}_{st}}}{{{T}_{fl}}}={0.02}\times {{\left( \frac{{3}}{{1}} \right)}^{2}}\)

= 18 %

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