1. Match the switch arrangements on the top row to the steady-state V-I characteristics on the lower row. The steady state operating points are shown by large black dots.
   
   
   

1. 1. A - I , B - II , C - III , D - IV

   
   
   

   2. A - II , B - IV , C - I , D - III

   
   
   

   3. A - IV , B - III , C - I , D - II

   
   
   

   4. A - IV , B - III , C - II , D - I

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   Negative voltage, current = 0

   Correct Option: C

   
   Negative voltage, current = 0

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2. In the single phase voltage controller circuit shown in the figure given below, for what range of triggering angle (α), the output voltage (v₀) is not controllable?
   
   

1. 1. 0° < α < 45°

   
   
   

   2. 45° < α < 135°

   
   
   

   3. 90° < α < 180°

   
   
   

   4. 135° < α < 180°

   
   
   

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1. View Hint View Answer Discuss in Forum

   R + jX_L = (50 + j 50) Ω
   

   ∴ tan φ =	ω L	=	XL	=	50	= 1
   	R		R		50

   
   ⇒ φ = 45°
   Hence firing angle α must be greater than 45° Thus for 0 < α < 45°, V_O is uncontrollable.
   

   Correct Option: A

   R + jX_L = (50 + j 50) Ω
   

   ∴ tan φ =	ω L	=	XL	=	50	= 1
   	R		R		50

   
   ⇒ φ = 45°
   Hence firing angle α must be greater than 45° Thus for 0 < α < 45°, V_O is uncontrollable.
   

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3. The equivalent circuits of a diode, during forward biased and reverse biased conditions, are shown in the figure given below.
   
   If such a diodes is used in clipper circuit of figure given above, the output voltage (v₀) of the circuit will be

1. 1. 
      

   
   
   

   2. 
      

   
   
   

   3. 
      

   
   
   

   4. 

   
   
   

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1. View Hint View Answer Discuss in Forum

   During +ve cycle of input voltage, V_i,
   When V_i > 5.7 volts
   Diode becomes forward-biased and the circuit will be as shown here
   
   R_f is the diode forward resistance.
   Output voltage, V₀ = 5.7 + R_fI
   During – ve cycle of input voltage V_i, Diode D is reversed-biased and cut-off.
   The circuit will be as shown below.
   
   

   V0 = 10 sinωt .	10 kΩ	= 5 sinωt
   	10 kΩ + 10 kΩ

   
   

   Correct Option: A

   During +ve cycle of input voltage, V_i,
   When V_i > 5.7 volts
   Diode becomes forward-biased and the circuit will be as shown here
   
   R_f is the diode forward resistance.
   Output voltage, V₀ = 5.7 + R_fI
   During – ve cycle of input voltage V_i, Diode D is reversed-biased and cut-off.
   The circuit will be as shown below.
   
   

   V0 = 10 sinωt .	10 kΩ	= 5 sinωt
   	10 kΩ + 10 kΩ

   
   

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4. A 3-phase Voltage Source Inverter is operated in 180° conduction mode. Which one of the following statements is true?
   

1. 1. Both pole-voltage and line-voltage will have 3^rd harmonic components
      

   
   
   

   2. Pole-voltage will have 3^rd harmonic component but line-voltage will be free from 3^rd harmonic

   
   
   

   3. Line-voltage will have 3^rd harmonic component but pole-voltage will be free from 3^rd harmonic

   
   
   

   4. Both pole-voltage and line-voltage will be free from 3^rd harmonic components

   
   
   

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1. View Hint View Answer Discuss in Forum

   Line voltage,
   
   

   For n = 3, VL = 0. as cos		nπ		= 0
   		6

   
   Pole voltage,
   
   For n = 3, V_P ≠ 0.
   Hence it is not free from third hormonics.

   Correct Option: B

   Line voltage,
   
   

   For n = 3, VL = 0. as cos		nπ		= 0
   		6

   
   Pole voltage,
   
   For n = 3, V_P ≠ 0.
   Hence it is not free from third hormonics.

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5. In the chopper circuit shown, the main thyristor (T_M) is operated at a duty ratio of 0.8 which is
   

   much larger the commutation interval. If the maximum allowable reapplied	dv	on TM
   	dt

   
   is 50 V/μs, what should be the theoretical minimum value of C₁? Assume current ripple through L₀ to be negligible.
   
   

1. 1. 0.2 μF

   
   
   

   2. 0.02 μF

   
   
   

   3. 2 μF

   
   
   

   4. 20 μF
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   I0 =	100	= 12.5 amp
   	0.8

   
   Duty ratio of T_M = 0.8
   ∴ Current through M, I_M = 12.5 × 0.8 = 10A
   

   Now IM = C	dVM
   	dt

   
   

   ∴ 10 = C ×	50
   	10-6

   
   

   or C =	50	× 10-6 = 0.2 μF
   	10

   
   

   Correct Option: A

   I0 =	100	= 12.5 amp
   	0.8

   
   Duty ratio of T_M = 0.8
   ∴ Current through M, I_M = 12.5 × 0.8 = 10A
   

   Now IM = C	dVM
   	dt

   
   

   ∴ 10 = C ×	50
   	10-6

   
   

   or C =	50	× 10-6 = 0.2 μF
   	10

   
   

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