11. A 300 kVA, 3300/400 V, three-phase transformer has its primary and secondary windings connected in delta for a dummy load test, for circulating full-load current, the magnitude of the injected voltage in open delta of HV windings is 360 V. The leakage impedance in per -unit system is _______

1. 1. 364

   
   
   

   2. 0.0064

   
   
   

   3. 64

   
   
   

   4. 0.0364

   
   
   

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1. View Hint View Answer Discuss in Forum

   Voltage across each phase of h.v. winding
   

   =	360	= 120 V = VSC
   	3

   
   V_Ph = 3300 V ;
   

   IPh =	300 × 103	=	100	= 30.3 amp.
   	3 × 3300		3.3

   
   

   ZeH =	VSC	=	120	= 3.96 Ω
   	ISC		30.3

   
   

   Zbase =	Vbase	=	3300	= 108.9 Ω
   	Ibase		30.3

   
   

   ∴ ZeH (p.u.) =	3.96	= 0.0364 p.u.
   	108.9

   
   

   Correct Option: D

   Voltage across each phase of h.v. winding
   

   =	360	= 120 V = VSC
   	3

   
   V_Ph = 3300 V ;
   

   IPh =	300 × 103	=	100	= 30.3 amp.
   	3 × 3300		3.3

   
   

   ZeH =	VSC	=	120	= 3.96 Ω
   	ISC		30.3

   
   

   Zbase =	Vbase	=	3300	= 108.9 Ω
   	Ibase		30.3

   
   

   ∴ ZeH (p.u.) =	3.96	= 0.0364 p.u.
   	108.9

   
   

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12. A 220 V, 60 Hz single-phase transformer has hysteresis loss of 340 watts and eddy current loss of 120 watts. If the transformer is operated from, 230 V, 50 Hz supply mains, then total core-loss, assuming steinmetz’s constant equal to 1.6 will be _______ w .
    

1. 1. 119.3 watt

   
   
   

   2. 539.3 watt

   
   
   

   3. 51.3 watt

   
   
   

   4. None of the above

   
   
   

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1. View Hint View Answer Discuss in Forum

   	V1	=	f1 Bm1
   	V2		f2 Bm2

   
   

   ∴	220	=	60	.	Bm1
   	230		50		Bm2

   
   ⇒ B_m2 = 1.255 B_m1
   

   ∴ Ph2 = Ph1 ×	f2 Bxm2
   	f1 Bxm1

   
   

   = 340 ×	50	(1.255)1.6 = 408.0 watts
   	60

   
   

   Pe2 = Pe1 ×		f2 .Bm2		2
   		f1 .Bm1

   
   

   = 120 ×		50	× 1.255		2	= 131.3 watts
   		60

   
   Total loss= P_h2 + P_e2 = 408 + 131.3 = 539.3 watt.
   

   Correct Option: B

   	V1	=	f1 Bm1
   	V2		f2 Bm2

   
   

   ∴	220	=	60	.	Bm1
   	230		50		Bm2

   
   ⇒ B_m2 = 1.255 B_m1
   

   ∴ Ph2 = Ph1 ×	f2 Bxm2
   	f1 Bxm1

   
   

   = 340 ×	50	(1.255)1.6 = 408.0 watts
   	60

   
   

   Pe2 = Pe1 ×		f2 .Bm2		2
   		f1 .Bm1

   
   

   = 120 ×		50	× 1.255		2	= 131.3 watts
   		60

   
   Total loss= P_h2 + P_e2 = 408 + 131.3 = 539.3 watt.
   

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13. A 400 / 100 V, 5kVA, single-phase two winding transformer is converted into an autotransformer to supply 400 volts from a 500 V voltage source. When tested as a two winding transformer at rated load and 0.85 p.f. lagging, efficiency is 0.95, than its efficiency as auto-transformer at rated-load and 0.8 p.f. lagging will be _______ % .

1. 1. 98.95%

   
   
   

   2. 45%

   
   
   

   3. 38.42%

   
   
   

   4. 9.995%

   
   
   

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1. View Hint View Answer Discuss in Forum

   kVA rating of auto-transformer
   = 500 × 50 = 25 kVA
   As a two- winding transformer,
   
   

   η =	output	= 0.95
   	output + losses

   
   

   ∴	5 × 0.8	= 0.95
   	5 × 0.8 + losses

   
   ⇒  Losses = 210.52 watt
   Since losses remain const ant, hence for an auto-transformer,
   

   η =	25 × 0.8	= 98.95%
   	25 × 0.8 + 0.210

   
   

   Correct Option: A

   kVA rating of auto-transformer
   = 500 × 50 = 25 kVA
   As a two- winding transformer,
   
   

   η =	output	= 0.95
   	output + losses

   
   

   ∴	5 × 0.8	= 0.95
   	5 × 0.8 + losses

   
   ⇒  Losses = 210.52 watt
   Since losses remain const ant, hence for an auto-transformer,
   

   η =	25 × 0.8	= 98.95%
   	25 × 0.8 + 0.210

   
   

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