296. In transformers, which of the following statements is valid ?
     

1. 1. In an open circuit test, copper losses are obtained while in short circuit test, core losses are obtained

   
   
   

   2. In an open circuit test, current is drawn at high power factor

   
   
   

   3. In a short circuit test, current is drawn at zero power factor

   
   
   

   4. In an open circuit test, current is drawn at low power factor

   
   
   

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1. View Hint View Answer Discuss in Forum

   In SC test, I₁ (FLC) is drawn at angle θ₁ In OC test, only I_o (NLC) is drawn at angle θ₂
   
   θ₂ > θ₁
   So cos θ₂ < cos θ₁
   So current is drawn a low power factor.
   

   Correct Option: D

   In SC test, I₁ (FLC) is drawn at angle θ₁ In OC test, only I_o (NLC) is drawn at angle θ₂
   
   θ₂ > θ₁
   So cos θ₂ < cos θ₁
   So current is drawn a low power factor.
   

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297. A three-phase squirrel cage induction motor has a starting torque of 150% and a maximum torque of 300% with respect to rated torque at rated voltage and rated frequency. Neglect the stator resistance and rotational losses. The value of slip for maximum torque is
     

1. 1. 13.48%

   
   
   

   2. 16.24%

   
   
   

   3. 18.92%

   
   
   

   4. 26.79%

   
   
   

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1. View Hint View Answer Discuss in Forum

   Tst	=	Z
   Tm		1	+ sm
   		sm

   
   

   ⇒	1.5	=	sm
   	3		1 - sm2

   
   ⇒ s_m² – 4 s_m + 1 = 0
   

   ⇒ sm =	4 ± √16 - 4	=	1	(4 - 2 √3)
   	3		2

   
   = 0.268 = 26.8%.

   Correct Option: D

   Tst	=	Z
   Tm		1	+ sm
   		sm

   
   

   ⇒	1.5	=	sm
   	3		1 - sm2

   
   ⇒ s_m² – 4 s_m + 1 = 0
   

   ⇒ sm =	4 ± √16 - 4	=	1	(4 - 2 √3)
   	3		2

   
   = 0.268 = 26.8%.

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298. A single - phase 50 – kVA, 250V/ 500 V two winding transformer has an efficiency of 95% at full load, unity power factor. If it is reconfigured as a 500 V/ 750 V auto-transformer, its efficiency at its new rated load at unity power factor will be
     

1. 1. 95.752%

   
   
   

   2. 97.851%

   
   
   

   3. 98.276%

   
   
   

   4. 99.241%

   
   
   

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1. View Hint View Answer Discuss in Forum

   Efficiency , η = 0.95 =	50
   	50 + Losses

   
   ⇒ Losses = 50 – 47.5 = 2.5 kW
   

   Now kVA = 50 ×	1
   	1 - K

   
   

   K =	500	=	2
   	750		3

   
   

   = 50 ×	1	= 50
   	{1 - (2 / 3)}

   
   

   ∴ New efficiency =	150	× 100 = 98.36%
   	150 + 2.5

   
   

   Correct Option: C

   Efficiency , η = 0.95 =	50
   	50 + Losses

   
   ⇒ Losses = 50 – 47.5 = 2.5 kW
   

   Now kVA = 50 ×	1
   	1 - K

   
   

   K =	500	=	2
   	750		3

   
   

   = 50 ×	1	= 50
   	{1 - (2 / 3)}

   
   

   ∴ New efficiency =	150	× 100 = 98.36%
   	150 + 2.5

   
   

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299. A three-phase, three-stack, variable reluctance step motor has 20 poles on each rotor and stator stack. The step angle of this step motor is
     

1. 1. 3°

   
   
   

   2. 6°

   
   
   

   3. 9°

   
   
   

   4. 18°

   
   
   

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1. View Hint View Answer Discuss in Forum

   Step angle , α =	360°	= 6°
   	3 × 20

   Correct Option: B

   Step angle , α =	360°	= 6°
   	3 × 20

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300. A 100 k VA. 415 V (line). star -connect ed synchronous machine generates rated open circuit voltage of 415 V at a field current of 15 A. The short circuit armature current at a field current of 10 A is equal to the rated armature current. The per unit saturated synchronous reactance is
     

1. 1. 1.731

   
   
   

   2. 1.5

   
   
   

   3. 0.666

   
   
   

   4. 0.577

   
   
   

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1. View Hint View Answer Discuss in Forum

   Synchronous line impedance,
   

   =	open circuit line voltage
   	√3 × short circuit phase current

   
   table> = 415 = 1.722√3 × 100 × 100 √3 × 415
   

   Correct Option: A

   Synchronous line impedance,
   

   =	open circuit line voltage
   	√3 × short circuit phase current

   
   table> = 415 = 1.722√3 × 100 × 100 √3 × 415
   

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