91. The common mode voltage is completely alternated at the out put of the differential amplifier shown in the figure, if
    
    

1. 1. R₁ + R₂ = R₃ + R₄

   
   
   

   2. R₁ – R₂ = R₃ – R₄

   
   
   

   3. 	R1	=	R3
      	R2		R4

   
   
   

   4. 	R2	=	R3
      	R1		R4

      
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: B

   NA

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92. In an non-inverting OP-AMP summer is shown in the figure, output voltage V₀ will be
    
    

1. 1. 		3		sin (100t)
      		2

      
      

   
   
   

   2. 3 sin (100t)

   
   
   

   3. sin (100 t)

   
   
   

   4. 2 sin (100 t)

   
   
   

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1. View Hint View Answer Discuss in Forum

   V0 = -2		1 +	2R		+ (2 + sin 100 t)		1 +	2R
   			R					R

   
   = – 6 + 6 + 3 sin 100 t = 3 sin 100 t

   Correct Option: B

   V0 = -2		1 +	2R		+ (2 + sin 100 t)		1 +	2R
   			R					R

   
   = – 6 + 6 + 3 sin 100 t = 3 sin 100 t

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93. Magnitude of the gain in the inverting op-amp amplifier circuit shown in the figure be x with switch S₁ open. When switch S₁ is closed, then magnitude of gain becomes
    
    

1. 1. 	x
      	2

      
      

   
   
   

   2. – x
      

   
   
   

   3. 2x

   
   
   

   4. – 2x

   
   
   

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1. View Hint View Answer Discuss in Forum

   	Vo	= A =	-2R	= -2 = x
   	Vi		R

   
   When switch S₁ is closed,
   

   A =	-R	= -1 =	x
   	R		2

   Correct Option: A

   	Vo	= A =	-2R	= -2 = x
   	Vi		R

   
   When switch S₁ is closed,
   

   A =	-R	= -1 =	x
   	R		2

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94. If I .C. amplifier shown in the figure has an input volt age V_i = 0.1 cos (2π × 100t) volt s, t hen waveform of the output voltage V₀ is
    
    
    

1. 1. 
      

   
   
   

   2. 
      

   
   
   

   3. 
      

   
   
   

   4. 
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: D

   NA

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95. In the given circuit, V_in = 10 mV dc maximum, maximum possible output offset volt age V_oo caused by the input voltage V_io with respect to ground is
    
    

1. 1. 60 mV dc 0

   
   
   

   2. 110 mV dc

   
   
   

   3. 130 mV dc

   
   
   

   4. 150 mV dc

   
   
   

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1. View Hint View Answer Discuss in Forum

   Applying KVL at the inverting node, we have
   

   	Vin - 0	+	Vin - V0	= 0
   	R1		Rf

   
   

   ⇒ V0 =		1	+	1		RF = 110 mV
   		R1		RF

   
   

   Correct Option: B

   Applying KVL at the inverting node, we have
   

   	Vin - 0	+	Vin - V0	= 0
   	R1		Rf

   
   

   ⇒ V0 =		1	+	1		RF = 110 mV
   		R1		RF

   
   

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