231. Voltage regulation of a large transformer is mainly influenced by

1. 1. no-load current and load power factor

   
   
   

   2. winding resistances and load power factor

   
   
   

   3. leakage fluxes and load power factor

   
   
   

   4. winding resistances and core loss

   
   
   

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   % Regulation =	I(Re2cosθ ± Xe2sinθ)
   	V2

   
   cos θ is power factor

   Correct Option: B

   % Regulation =	I(Re2cosθ ± Xe2sinθ)
   	V2

   
   cos θ is power factor

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232. The given figure represents a transformer with two windings 1 and 2 wound on the core as shown. By applying a voltage V₁ across winding 1, a voltage V₂ is induced across winding 2. In an idealised condition V₂ would lag V₁ by
     

1. 1. zero degrees

   
   
   

   2. 90 degrees

   
   
   

   3. 180 degrees

   
   
   

   4. 270 degrees

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   Figure shows that two fluxes produced by the windings are opposing each other, this inducing voltages having phase difference of 180 degrees.

   Correct Option: C

   
   Figure shows that two fluxes produced by the windings are opposing each other, this inducing voltages having phase difference of 180 degrees.

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233. Two single-phase transformer rated 1000 kVA and 500 kVA have per unit leakage impedance of (0.02 + j0.06) and (0.025 + j0.08) respectively. Then, the largest kVA load delivered by the parallel combination of these two transformers without overloading any one is

1. 1. 1825.3 kVA

   
   
   

   2. 1377.44 kVA

   
   
   

   3. 1500 kVA

   
   
   

   4. 3202.74 kVA

   
   
   

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1. View Hint View Answer Discuss in Forum

   Choosing base of 1000 kVA,
   Z_e1(1000 kVA) = (0.02 + j 0.06) pu
   

   Ze2(500 kVA) =	1000	. (0.025 + j 0.08) p.u. = 0.05 + 0.16 j
   	500

   
   

   Now, S1 =	Ze2	. S
   	Ze1 + Ze2

   
   

   ⇒ S = S1 .	Ze1 + Ze2
   	Ze2

   
   

   = 1000 .	0.07 + j0.22
   	0.05 + j0.16

   
   = 1377.44 kVA,
   

   and S2 =	Ze1	. S1
   	Ze1 + Ze2

   
   

   ∴ S = 500	0.07 + j0.022	= 1825.3 kVA
   	0.02 + j0.06

   
   As, Z_e1 < Z_e2, transformer 1 reaches its rated kVA first,
   ∴ largest kVA load = 1377.44 k VA

   Correct Option: B

   Choosing base of 1000 kVA,
   Z_e1(1000 kVA) = (0.02 + j 0.06) pu
   

   Ze2(500 kVA) =	1000	. (0.025 + j 0.08) p.u. = 0.05 + 0.16 j
   	500

   
   

   Now, S1 =	Ze2	. S
   	Ze1 + Ze2

   
   

   ⇒ S = S1 .	Ze1 + Ze2
   	Ze2

   
   

   = 1000 .	0.07 + j0.22
   	0.05 + j0.16

   
   = 1377.44 kVA,
   

   and S2 =	Ze1	. S1
   	Ze1 + Ze2

   
   

   ∴ S = 500	0.07 + j0.022	= 1825.3 kVA
   	0.02 + j0.06

   
   As, Z_e1 < Z_e2, transformer 1 reaches its rated kVA first,
   ∴ largest kVA load = 1377.44 k VA

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234. A 2400/240 V, 200 kVA single-phase transformer has a core loss of 1.8 kW at rated voltage. Its equivalent resistance is 1.1 percent. Then, the transformer efficiency at 0.9 pf and on full-load is

1. 1. 97.57%

   
   
   

   2. 98.05%

   
   
   

   3. 97.82%

   
   
   

   4. 96.56%

   
   
   

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1. View Hint View Answer Discuss in Forum

   Pcu-loss	= re(p.u.)
   kVA rated

   
   ⇒ P_cu-loss = 0.011 × 200 kVA = 2.2 kW
   

   ∴ η =	200 × 0.9
   	200 × 0.9 + 2.2 + 1.8

   
   

   =	180	= 97.2 %
   	184

   Correct Option: C

   Pcu-loss	= re(p.u.)
   kVA rated

   
   ⇒ P_cu-loss = 0.011 × 200 kVA = 2.2 kW
   

   ∴ η =	200 × 0.9
   	200 × 0.9 + 2.2 + 1.8

   
   

   =	180	= 97.2 %
   	184

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235. The hysteresis eddy current losses of a single phase transformer working on 200 V, 50 Hz supply are P_h and P_e respectively. The per cent age decreases in these, when operated on a 160 V, 40 Hz supply are

1. 1. 32, 36

   
   
   

   2. 20, 36

   
   
   

   3. 25, 20

   
   
   

   4. 40, 80

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: B

   NA

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