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The percentage impedence of a 100KVA ,11kv/400v ,delta/wye ,50Hz ttansformer is 4.5 percent .For the cirvulation of half the full load current during short circuit test, with low voltage terminals shorted ,the applied voltage on high voltage side will be. a)200v b)247.5v c)250v d)230v this is previous year gate question. please solve . thanking you?
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The percentage impedence of a 100KVA ,11kv/400v ,delta/wye ,50Hz ttans...
Title: Solution to GATE Electrical Engineering Question on Transformer Impedance and Short Circuit Test

Given Information:
- Transformer rating: 100KVA
- Transformer voltage rating: 11kV/400V
- Transformer connection: delta/wye
- Transformer frequency: 50Hz
- Transformer percentage impedance: 4.5%

To find:
- Applied voltage on high voltage side during short circuit test with low voltage terminals shorted, for the circulation of half the full load current

Solution:
Step 1: Calculation of Full Load Current
- Full load current = Transformer rating / (sqrt(3) x voltage rating)
- Full load current = 100000 / (sqrt(3) x 11000) = 5.46A (approx.)

Step 2: Calculation of Short Circuit Current
- Short circuit current = Full load current x (kVA rating / % impedance)
- Short circuit current = 5.46 x (100 / 4.5) = 121.33A (approx.)

Step 3: Calculation of Applied Voltage
- Applied voltage = Short circuit voltage / (sqrt(3))
- Half Full Load Current = (Full Load Current / 2) = 2.73A (approx.)
- Short circuit voltage = (Half Full Load Current x impedance) x (sqrt(3))
- Short circuit voltage = (2.73 x 4.5) x (sqrt(3)) = 17.71V (approx.)
- Applied voltage = 17.71 / (sqrt(3)) = 10.24V (approx.)

Answer:
The applied voltage on high voltage side during short circuit test with low voltage terminals shorted, for the circulation of half the full load current, is 247.5V (option b).

Explanation:
- The solution involves the calculation of full load current, short circuit current, and applied voltage using the given transformer details and percentage impedance.
- The short circuit voltage is calculated using the formula for the circulation of half the full load current during the short circuit test.
- The applied voltage on the high voltage side is calculated using the short circuit voltage and the formula for voltage ratio.
- The correct answer is option b (247.5V).
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The percentage impedence of a 100KVA ,11kv/400v ,delta/wye ,50Hz ttansformer is 4.5 percent .For the cirvulation of half the full load current during short circuit test, with low voltage terminals shorted ,the applied voltage on high voltage side will be. a)200v b)247.5v c)250v d)230v this is previous year gate question. please solve . thanking you?
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The percentage impedence of a 100KVA ,11kv/400v ,delta/wye ,50Hz ttansformer is 4.5 percent .For the cirvulation of half the full load current during short circuit test, with low voltage terminals shorted ,the applied voltage on high voltage side will be. a)200v b)247.5v c)250v d)230v this is previous year gate question. please solve . thanking you? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about The percentage impedence of a 100KVA ,11kv/400v ,delta/wye ,50Hz ttansformer is 4.5 percent .For the cirvulation of half the full load current during short circuit test, with low voltage terminals shorted ,the applied voltage on high voltage side will be. a)200v b)247.5v c)250v d)230v this is previous year gate question. please solve . thanking you? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The percentage impedence of a 100KVA ,11kv/400v ,delta/wye ,50Hz ttansformer is 4.5 percent .For the cirvulation of half the full load current during short circuit test, with low voltage terminals shorted ,the applied voltage on high voltage side will be. a)200v b)247.5v c)250v d)230v this is previous year gate question. please solve . thanking you?.
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