66. In a class A series fed amplifier using a transistor, under conditions the maximum ac power delivered is 1 watt.
    The maximum transistor dissipation capability should be:

1. 1. 1 watt

   
   
   

   2. 2 watts

   
   
   

   3. 3 watts

   
   
   

   4. 4 watts

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given P_{o(ac) max} = 1 watt
   P_{D (max)} =?
   

   Again 0.5 = η =	Po(ac)
   	Pi(dc)

   
   

   or 0.5 =	Po(ac)
   	PD(max) + Po(ac)

   
   

   or 0.5 =	1
   	PD(max) + 1

   
   or 0.5 P_{D (max)} + 0.5 = 1
   or 0.5 P_{D (max)} = 0.5
   or P_{D (max)} = 1
   Hence alternative (A) is the correct choice.

   Correct Option: A

   Given P_{o(ac) max} = 1 watt
   P_{D (max)} =?
   

   Again 0.5 = η =	Po(ac)
   	Pi(dc)

   
   

   or 0.5 =	Po(ac)
   	PD(max) + Po(ac)

   
   

   or 0.5 =	1
   	PD(max) + 1

   
   or 0.5 P_{D (max)} + 0.5 = 1
   or 0.5 P_{D (max)} = 0.5
   or P_{D (max)} = 1
   Hence alternative (A) is the correct choice.

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67. In class A direct coupled (series fed) power amplifier, maximum dissipation capability of the transistor is 2.5 watts, when delivering maximum a.c. power in the load is:

1. 1. 5 watts

   
   
   

   2. 2.5 watts

   
   
   

   3. 1.25 watts

   
   
   

   4. 0.625 watts

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given P_{D Max} = 2.5 Watts
   P_oac =?
   We know maximum efficiency of direct coupled class-A amplifier is 50% i.e., η = 50%
   

   η = 50% = 0.5 =	Po(ac)
   	Pi(dc)

   
   ∴ P_{i (dc)} = P_{o (ac)} + P_D) P_{i (dc)}
   = P_{o (ac)} + 2.5
   

   or 0.5 =	Po(ac)
   	Po(ac) + 2·5

   
   or 0.5P_o(ac) + 1.25 = P_{o (ac)}
   or 0.5 P_{o (ac)} = 1.25
   

   or Po (ac) =	1·25	= 2.5 watts.
   	0·5

   
   where, P_{o (ac)} = output ac power to the load
   P_{i (dc)} = input dc power to the transistor

   Correct Option: B

   Given P_{D Max} = 2.5 Watts
   P_oac =?
   We know maximum efficiency of direct coupled class-A amplifier is 50% i.e., η = 50%
   

   η = 50% = 0.5 =	Po(ac)
   	Pi(dc)

   
   ∴ P_{i (dc)} = P_{o (ac)} + P_D) P_{i (dc)}
   = P_{o (ac)} + 2.5
   

   or 0.5 =	Po(ac)
   	Po(ac) + 2·5

   
   or 0.5P_o(ac) + 1.25 = P_{o (ac)}
   or 0.5 P_{o (ac)} = 1.25
   

   or Po (ac) =	1·25	= 2.5 watts.
   	0·5

   
   where, P_{o (ac)} = output ac power to the load
   P_{i (dc)} = input dc power to the transistor

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68. For the resonant circuit shown, w₀ = 10⁵, Q = 50, R = 400Ω. The value of C is given by:
    
    
    

1. 1. 250 pF

   
   
   

   2. .1000 pF

   
   
   

   3. 500 pF

   
   
   

   4. 1.25 µF

   
   
   

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1. View Hint View Answer Discuss in Forum

   Q =	ω0L
   	R

   
   

   50 =	105·L
   	400

   
   L = 0.2
   

   Also ω01 =	1
   	√LC

   
   

   or ω02 =	1	or C =	1	=	1
   	LC		ω02·L		(105)2 × 0·2

   
   = 500 pf

   
   

   Correct Option: C

   Q =	ω0L
   	R

   
   

   50 =	105·L
   	400

   
   L = 0.2
   

   Also ω01 =	1
   	√LC

   
   

   or ω02 =	1	or C =	1	=	1
   	LC		ω02·L		(105)2 × 0·2

   
   = 500 pf

   
   

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69. For the circuit in the figure the feedback is:
    
    
    

1. 1. current series

   
   
   

   2. current shunt

   
   
   

   3. voltage series

   
   
   

   4. voltage shunt

   
   
   

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1. View Hint View Answer Discuss in Forum

   In the given circuit current is feedback in the form of node.
   Hence the circuit represents current shunt feedback.

   Correct Option: B

   In the given circuit current is feedback in the form of node.
   Hence the circuit represents current shunt feedback.

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70. A three stage amplifier with identical stage with lower cut off frequency per stage = f₁ is given overall negative feedback. Depending on the gain, the system may oscillate at a low frequency f_c given by:

1. 1. f1	= 1
      fc

   
   
   

   2. f1	=	1
      fc		√3

   
   
   

   3. f1	= √3
      fc

   
   
   

   4. f1	=	1
      fc		√2

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: C

   NA

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