16. Given the maximum efficiency of a transformer of a 500 kVA, 3300/500V, 50 Hz single-phase transformer is 97% and occurs at 3/4 of full-load, unity power factor. If the impedance is 10%, the regulation at full-load, power factor 0.8 lagging will be _____ % .
    

1. 1. 8.25%

   
   
   

   2. 17.52%

   
   
   

   3. 7.52%

   
   
   

   4. 10.52%

   
   
   

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   η = 0.97 =		500 ×	3	× 1
   			4
   	500 ×	3	+ Pcore + Pcu ×	9
   		4		16

   
   

   ⇒ Pcore + Pcu ×	9	= 11.60
   	16

   
   ⇒ P_cu = 10.31 kW (at full-load)
   

   ∴ re (p.u.) =	Pcu	=	10.31	= 0.02062
   	kVArated		500

   
   x_e (p.u.) = √Z_e² - r_e²
   = √0.1² - (0.02062)² = 0.0978
   ∴ Regulation = r_e cos θ + x_e sin θ
   = 0.02062 × 0.8 + 0.0978 × 0.6 = 7.52%
   

   Correct Option: C

   η = 0.97 =		500 ×	3	× 1
   			4
   	500 ×	3	+ Pcore + Pcu ×	9
   		4		16

   
   

   ⇒ Pcore + Pcu ×	9	= 11.60
   	16

   
   ⇒ P_cu = 10.31 kW (at full-load)
   

   ∴ re (p.u.) =	Pcu	=	10.31	= 0.02062
   	kVArated		500

   
   x_e (p.u.) = √Z_e² - r_e²
   = √0.1² - (0.02062)² = 0.0978
   ∴ Regulation = r_e cos θ + x_e sin θ
   = 0.02062 × 0.8 + 0.0978 × 0.6 = 7.52%
   

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17. A single-phase transformer when supplied from 220V, 50 Hz has eddy current loss of 50 W. If transformer is connected to a voltage of 330 V, 50 Hz, then eddy current loss will be _______ W .

1. 1. 11.25 W

   
   
   

   2. 112.5 W

   
   
   

   3. 21.05 W

   
   
   

   4. 1125 W

   
   
   

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1. View Hint View Answer Discuss in Forum

   P_e = k_e f² B_m² , and V = 4.44 f B_m AN
   

   ∴	Pe1	=		f1 Bm1		2	=		V1		2
   	Pe2			f2 Bm2					V2

   
   

   ⇒ Pe2 = Pe1.		V1		2	= 50.		330		2	= 112.5 watts
   		V2					220

   Correct Option: B

   P_e = k_e f² B_m² , and V = 4.44 f B_m AN
   

   ∴	Pe1	=		f1 Bm1		2	=		V1		2
   	Pe2			f2 Bm2					V2

   
   

   ⇒ Pe2 = Pe1.		V1		2	= 50.		330		2	= 112.5 watts
   		V2					220

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18. In a single-phase transformer, magnitude of leakage reactance is twice that of resistance of both primary and secondary. With secondary short-circuited, the input power factor is _______ .

1. 1. 	2
      	√5

   
   
   

   2. 	1
      	√3

   
   
   

   3. 	5
      	√2

   
   
   

   4. 	3
      	√4

   
   
   

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   NA

   Correct Option: A

   NA

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19. A 100 k VA, 2400 / 240V, 50 Hz single phase transformer has an exciting current of 0.64 A and a core loss of 700 watts, when its high-voltage side is energised at rated voltage and frequency. If load current is 40 A at 0.8 power factor of on the LV side then the primary current will be _______ A .

1. 1. 4.58 A

   
   
   

   2. 45 A

   
   
   

   3. 45.8 A

   
   
   

   4. None of these

   
   
   

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1. View Hint View Answer Discuss in Forum

   Load current, I_L (to h.v. side) = 4 ∠ – 37° amp.
   

   IC =	PC	=	700	= 0.292 A
   	V		2400

   
   I_φ = √I_e² - I_C²
   = √0.64² - 0.292² = 0.569
   Primary current, I₁ = I_C - j I_φ + I_L
   = 0.292 – j0.569 + 3.19 – 2.40j
   = 3.48 – 2.96 j = 4.58 ∠ – 40.3°
   

   Correct Option: A

   Load current, I_L (to h.v. side) = 4 ∠ – 37° amp.
   

   IC =	PC	=	700	= 0.292 A
   	V		2400

   
   I_φ = √I_e² - I_C²
   = √0.64² - 0.292² = 0.569
   Primary current, I₁ = I_C - j I_φ + I_L
   = 0.292 – j0.569 + 3.19 – 2.40j
   = 3.48 – 2.96 j = 4.58 ∠ – 40.3°
   

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20. For a 1.15 kVA, 460/230 V transformer connected to a standard single phase 50 Hz supply, the no load current is likely to be _______ A .

1. 1. 0.1 A

   
   
   

   2. 0.01 A

   
   
   

   3. 1 A

   
   
   

   4. 10 A

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: A

   NA

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