306. A 400 V, 50 Hz, 4 pole, 1400 rpm, star connected squirrel cage induction motor has the following parameters referred to the stator
     R_r' = 1.0 Ω , X_s = X_r' = 1.5 Ω
     Neglect stator resistance and core and rotational losses of the motor.
     The motor is controlled from a 3-phase voltage source inverter with constant V/f control. The stator line-to-line voltage (rms) and frequency to obtain the maximum torque at starting will be
     

1. 1. 20.6 V, 2.7 Hz

   
   
   

   2. 133.3 V, 16.7 Hz

   
   
   

   3. 266.6 V, 33.3 Hz

   
   
   

   4. 323.3 V, 40.3 Hz

   
   
   

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1. View Hint View Answer Discuss in Forum

   For maximum torque, slip,
   

   sm =	R'r	(Neglecting stator resistance)
   	Xsm + X'rm

   
   For starting torque, s_m = 1
   X_sm + X '_rm = R_r ' (∴ R'_r = 1 Ω)
   ∴ 2π f_m L_s 2π f_m + L_r ' = 1
   

   ⇒ fm =	1
   	2π(Ls + Lr)

   
   where, m indicates value of reactance corresponding to maximum torque frequency and f_m is frequency at maximum torque.
   

   But Ls =	Xs	=	1.5
   	2π × 50		2π × 50

   
   

   and L'r =	1.5
   	2π × 50

   
   Putting values of L_s and L'_r, we get
   

   fm =	1				=	50	= 16.7 Hz
   	1.5	+	1.5			3
   	50		50

   
   

   For constant	v	,
   	f

   
   

   	v1	=	400	= 8
   	f1		50

   
   
   ⇒ v₂ = f₂ × 8 = 16.7 × 8 = 133.3 volts
   Hence answer is (133.3V, 16.7 Hz).
   

   Correct Option: B

   For maximum torque, slip,
   

   sm =	R'r	(Neglecting stator resistance)
   	Xsm + X'rm

   
   For starting torque, s_m = 1
   X_sm + X '_rm = R_r ' (∴ R'_r = 1 Ω)
   ∴ 2π f_m L_s 2π f_m + L_r ' = 1
   

   ⇒ fm =	1
   	2π(Ls + Lr)

   
   where, m indicates value of reactance corresponding to maximum torque frequency and f_m is frequency at maximum torque.
   

   But Ls =	Xs	=	1.5
   	2π × 50		2π × 50

   
   

   and L'r =	1.5
   	2π × 50

   
   Putting values of L_s and L'_r, we get
   

   fm =	1				=	50	= 16.7 Hz
   	1.5	+	1.5			3
   	50		50

   
   

   For constant	v	,
   	f

   
   

   	v1	=	400	= 8
   	f1		50

   
   
   ⇒ v₂ = f₂ × 8 = 16.7 × 8 = 133.3 volts
   Hence answer is (133.3V, 16.7 Hz).
   

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307. A 400 V, 50 Hz, 30 hp,three-phase induction motor is drawing 50 A current at 0.8 power factor lagging. The stator and rotor copper losses are 1200 W. The air-gap power of the motor will be
     

1. 1. 23.06 kW

   
   
   

   2. 24.11 kW

   
   
   

   3. 25.01 kW

   
   
   

   4. 26.21 kW

   
   
   

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1. View Hint View Answer Discuss in Forum

   Motor input power in stator = √3 VI cos φ
   = √3 400 × 50 × 0.8 = 27.71 kW
   ∴ Air gap power = Motor input to stator – Stator copper loss – Coreloss
   = 27.71 – 1.5 – 1.2 = 25.01 kW

   Correct Option: C

   Motor input power in stator = √3 VI cos φ
   = √3 400 × 50 × 0.8 = 27.71 kW
   ∴ Air gap power = Motor input to stator – Stator copper loss – Coreloss
   = 27.71 – 1.5 – 1.2 = 25.01 kW

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308. A 230 V, 50 Hz, 4-pole, single-phase induction motor is rotating in the clockwise (forward) direction at a speed of 1425 -rpm. If the rotor resistance at standstill is 7.8 Ω. then the effective rotor resistance in the backward branch of the equivalent circuit will be
     

1. 1. 2 Ω

   
   
   

   2. 4 Ω

   
   
   

   3. 78 Ω

   
   
   

   4. 156 Ω

   
   
   

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1. View Hint View Answer Discuss in Forum

   Synchronous speed,
   

   Ns =	120 × f	=	120 × 50	= 1500 rpm
   	P		4

   
   

   ∴ Slip , s =	1500 - 1425	= 0.05
   	1500

   
   ∴ Resistance of backward branch,
   

   rb =	r2	=	7.8	= 4 Ω
   	2 - s		2 - 0.05

   Correct Option: B

   Synchronous speed,
   

   Ns =	120 × f	=	120 × 50	= 1500 rpm
   	P		4

   
   

   ∴ Slip , s =	1500 - 1425	= 0.05
   	1500

   
   ∴ Resistance of backward branch,
   

   rb =	r2	=	7.8	= 4 Ω
   	2 - s		2 - 0.05

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309. In a stepper motor, the detent torque means
     

1. 1. minimum of the static torque with the phase winding excited.

   
   
   

   2. maximum of the static torque with the phase winding excited.

   
   
   

   3. minimum of the static torque with the phase winding unexcited.

   
   
   

   4. maximum of the static torque with the phase winding unexcited.

   
   
   

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1. View Hint View Answer Discuss in Forum

   Minimum of static torque with the phase winding unexcited.
   

   Correct Option: C

   Minimum of static torque with the phase winding unexcited.
   

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310. Three single-phase transformers are connected to form a 3-phase transformer bank. The transformers are connected in the following manner.
     
     The transformer connection will be represented by
     

1. 1. Y d0

   
   
   

   2. Y d1

   
   
   

   3. Y d6

   
   
   

   4. Y d11

   
   
   

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1. View Hint View Answer Discuss in Forum

   Phase displacement = – 30°
   Phasor
   
   

   Correct Option: B

   Phase displacement = – 30°
   Phasor
   
   

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