321. A 220 V, DC shunt motor is operating at a speed of 1440 rpm. The armature resistance is 1.0 Ω and armature current is 10 A. If the excitation of the machine is reduced by 10%, the extra resistance to be put in the armature circuit to maintain the same speed and torque will be
     

1. 1. 1.79 Ω

   
   
   

   2. 2.1 Ω

   
   
   

   3. 3.1 Ω

   
   
   

   4. 18.9 Ω

   
   
   

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1. View Hint View Answer Discuss in Forum

   I_a1 = 10
   Now flux is decreased by 10%, so φ₂ = 0.9 φ₁
   Torque is constant, & T = k_T φ I_a
   Then, I_a1 φ₁ = I _a2 φ₂
   

   ⇒ Ia2 =	10	= 11.11 A
   	0.9

   
   

   Now , N ∝	Eb
   	φ

   
   

   ⇒	N1	=	Eb1	×	φ2
   	N2		Eb2		φ1

   
   

   =	220 - Ia1 r1	×	0.9φ1
   	220 - Ia2 (r1 + R)		φ1

   
   

   ∴ 1 =	210 × 0.9
   	220 - 11.11 (1 + R)

   
   ⇒ 1 + R = 2.79
   ⇒ R = 1.79 Ω
   
   

   Correct Option: A

   I_a1 = 10
   Now flux is decreased by 10%, so φ₂ = 0.9 φ₁
   Torque is constant, & T = k_T φ I_a
   Then, I_a1 φ₁ = I _a2 φ₂
   

   ⇒ Ia2 =	10	= 11.11 A
   	0.9

   
   

   Now , N ∝	Eb
   	φ

   
   

   ⇒	N1	=	Eb1	×	φ2
   	N2		Eb2		φ1

   
   

   =	220 - Ia1 r1	×	0.9φ1
   	220 - Ia2 (r1 + R)		φ1

   
   

   ∴ 1 =	210 × 0.9
   	220 - 11.11 (1 + R)

   
   ⇒ 1 + R = 2.79
   ⇒ R = 1.79 Ω
   
   

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322. A single-phase air core transformer, fed from a rated sinusoidal supply, is operating at no load. The steady state magnetizing current drawn by the transformer from the supply will have the waveform
     
     
     

1. 1. 

   
   
   

   2. 

   
   
   

   3. 
      

   
   
   

   4. 

   
   
   

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1. View Hint View Answer Discuss in Forum

   It is an air core transformer. So, there is no saturation effect.

   Correct Option: C

   It is an air core transformer. So, there is no saturation effect.

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323. A three-phase, salient pole synchronous motor is connected to an infinite bus. It is operated at no load at normal excitation. The field excitation of the motor is first reduced to zero and then increased in the reverse direction gradually. Then the armature current
     

1. 1. increases continuously

   
   
   

   2. first increases and then decreases steeply

   
   
   

   3. first decreases and then increases steeply

   
   
   

   4. remains constant

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: B

   NA

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324. A 4-point starter is used to start and control the speed of a
     

1. 1. dc shunt motor with armature resistance control

   
   
   

   2. dc shunt motor with field weakening control

   
   
   

   3. dc series motor

   
   
   

   4. dc compound motor

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: A

   NA

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325. The locked rotor current in a 3-phase, st ar connected 15 kW, 4-pole, 230 V, 50 Hz induction motor at rated conditions is 50 A. Neglecting losses and magnetizing current, the approximate locked rotor line current drawn when the motor is connected to a 236 V, 57 Hz supply is
     

1. 1. 58.5 A

   
   
   

   2. 45.0 A

   
   
   

   3. 42.7 A

   
   
   

   4. 55.6 A

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   For the rotor at blocked position, slip, s = 1.
   As magnetizing current is negligible, therefore
   

   IBR =	VBR
   	√(r1 + r2)² + (x1 + x2)²

   
   

   =	VBR
   	√R² × x²

   
   R = r₁ + r₂ , x = x₁ + x₂ and, P_BR = I_BR² .R
   

   ⇒	15 × 103	= R
   	3 × 502

   
   ⇒ R = 2 Ω
   

   then , 50 =	230 / √3
   	√R² × x²

   
   

   =	230 / √3
   	√4 + x²

   
   ⇒ x = 1.74 Ω . at f = 50 Hz
   At f = 57 Hz, x '
   

   = 1.74 ×	57	= 1.98 Ω
   	50

   
   hence, at f = 57 Hz, V_BR = 236 V,
   

   I' =	236 / √3
   	√R² + x'²

   
   

   I' =	236 / √3	≅ 46 Amp.
   	√2² + 1.98²

   
   

   Correct Option: B

   
   For the rotor at blocked position, slip, s = 1.
   As magnetizing current is negligible, therefore
   

   IBR =	VBR
   	√(r1 + r2)² + (x1 + x2)²

   
   

   =	VBR
   	√R² × x²

   
   R = r₁ + r₂ , x = x₁ + x₂ and, P_BR = I_BR² .R
   

   ⇒	15 × 103	= R
   	3 × 502

   
   ⇒ R = 2 Ω
   

   then , 50 =	230 / √3
   	√R² × x²

   
   

   =	230 / √3
   	√4 + x²

   
   ⇒ x = 1.74 Ω . at f = 50 Hz
   At f = 57 Hz, x '
   

   = 1.74 ×	57	= 1.98 Ω
   	50

   
   hence, at f = 57 Hz, V_BR = 236 V,
   

   I' =	236 / √3
   	√R² + x'²

   
   

   I' =	236 / √3	≅ 46 Amp.
   	√2² + 1.98²

   
   

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