311. Distributed winding and short chording employed in AC machines will result in
     

1. 1. increase in emf and reduction in harmonics

   
   
   

   2. reduction in emf and increase in harmonics

   
   
   

   3. increase in both emf and harmonics

   
   
   

   4. reduction in both emf and harmonics

   
   
   

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1. View Hint View Answer Discuss in Forum

   Due to distributed winding and short chording, both emf and harmonics are reduced.
   

   Correct Option: D

   Due to distributed winding and short chording, both emf and harmonics are reduced.
   

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312. Figure given below shows the extended view of a 2 pole dc machine with 10 armature conductors. Normal brush positions are shown by A and B, placed at the interpolar axis. If the brushes are now shifted, in the direction of rotation, to A' and B' as shown, the voltage waveform V_A'B' will resemble
     

1. 1. 
      

   
   
   

   2. 
      

   
   
   

   3. 
      

   
   
   

   4. 
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   

   Correct Option: A

   

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313. A 220V, 50 Hz, single-phase induction motor has the following connection diagram and winding orientations shown. MM' is the axis of the main stator winding (M₁M₂) and AA' is that of the auxiliar y winding (A₁A₂). Directions of the winding axes indicate direction of flux when currents in the windings are in the directions
     shown. Parameters of each winding are indicated. When switch S is closed, the motor
     
     

1. 1. rotates clockwise

   
   
   

   2. rotates anticlockwise

   
   
   

   3. does not rotate

   
   
   

   4. rotates momentarily and comes to a halt

   
   
   

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1. View Hint View Answer Discuss in Forum

   Z_AUX = r _a + jxL_a
   

   = 1 + j(2π × 50) ×	10
   	π

   
   = 1 + j 1000 ∼ 1000 ∠90°
   
   Z_m = r_m + jx_m
   

   = 0.1 + j (2 π × 50 ) ×	0.1
   	π

   
   = 0.1 + j 10 = ∠ 90°
   

   IAUX =	V	= Ia ∠ -90°
   	ZAω0

   
   

   IM =	V	= IM ∠ -90°
   	ZM

   
   ∴ Starting torque = k I_a I_m sin (I_a, I _m)
   = k' sin (θ) = 0
   So motor does not rotate at starting. Obviously 1φ motor is not self starting.

   Correct Option: C

   Z_AUX = r _a + jxL_a
   

   = 1 + j(2π × 50) ×	10
   	π

   
   = 1 + j 1000 ∼ 1000 ∠90°
   
   Z_m = r_m + jx_m
   

   = 0.1 + j (2 π × 50 ) ×	0.1
   	π

   
   = 0.1 + j 10 = ∠ 90°
   

   IAUX =	V	= Ia ∠ -90°
   	ZAω0

   
   

   IM =	V	= IM ∠ -90°
   	ZM

   
   ∴ Starting torque = k I_a I_m sin (I_a, I _m)
   = k' sin (θ) = 0
   So motor does not rotate at starting. Obviously 1φ motor is not self starting.

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314. A 3-phase squirrel cage induction motor supplied from a balanced 3-phase source drives a mechanical load. The torque-speed characteristics of the motor (solid curve) and of the load (dotted curve) are shown in the figure given below. Of the two equilibrium points A and B, which of the following options correctly describes the stability of A and B?
     
     

1. 1. A is stable B is unstable

   
   
   

   2. A is unstable B is stable

   
   
   

   3. Both are stable

   
   
   

   4. Both are unstable

   
   
   

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1. View Hint View Answer Discuss in Forum

   Since at point B,	dTm	<	dTL	Ω
   	dt		dt

   
   
   

   Since angle φ1 between	∂T2	and	∂T	at A is larger than φ2 at B,
   	∂s		∂s

   
   therefore, any fluctution of load at A would cause much speed variations with respect to the same at B.
   

   Correct Option: B

   Since at point B,	dTm	<	dTL	Ω
   	dt		dt

   
   
   

   Since angle φ1 between	∂T2	and	∂T	at A is larger than φ2 at B,
   	∂s		∂s

   
   therefore, any fluctution of load at A would cause much speed variations with respect to the same at B.
   

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315. The single phase, 50Hz, iron core transformer in the circuit has both the vertical arms of cross sectional area 20 cm² and both the horizontal arms of cross sectional area 10 cm². If the two windings shown were wound instead on opposite horizontal arms, the mutual inductance will
     
     

1. 1. double

   
   
   

   2. remain same

   
   
   

   3. be halved

   
   
   

   4. become one quarter

   
   
   

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1. View Hint View Answer Discuss in Forum

   M ∝ √A₁ A₂
   But A₁ = A₂ = A
   ∴ M ∝ A
   

   ∴ M2 =	A2	. M1 =	10	M1 =	M2
   	A1		20		2

   
   Since M = √L₁ L₂
   

   where L1 =	N1 φi	and φi = B1 A1
   	Li

   
   

   ∴ M ∝ √A1 A2 × L1 =	N1 B1 A1
   	Li

   
   Alternately
   L = μ₀ nA
   ∴ M = k √L₁ L₂
   
   

   Correct Option: C

   M ∝ √A₁ A₂
   But A₁ = A₂ = A
   ∴ M ∝ A
   

   ∴ M2 =	A2	. M1 =	10	M1 =	M2
   	A1		20		2

   
   Since M = √L₁ L₂
   

   where L1 =	N1 φi	and φi = B1 A1
   	Li

   
   

   ∴ M ∝ √A1 A2 × L1 =	N1 B1 A1
   	Li

   
   Alternately
   L = μ₀ nA
   ∴ M = k √L₁ L₂
   
   

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