86. A OP-AMP has an offset voltage of 1m V and is ideal in all other respects. If this op-amp is used in the circuit shown in the figure, then output voltage will be (select nearest value)
    
    

1. 1. 1 mV

   
   
   

   2. 1 V

   
   
   

   3. ± 1 V

   
   
   

   4. 0 V

   
   
   

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1. View Hint View Answer Discuss in Forum

   For the given op-amp, we have
   

   Vo = Vos × gain =	1 × 10-3 × 106	= 1 V
   	103

   
   It can be either positive or negative i.e. ± 1V. voltage follower

   Correct Option: C

   For the given op-amp, we have
   

   Vo = Vos × gain =	1 × 10-3 × 106	= 1 V
   	103

   
   It can be either positive or negative i.e. ± 1V. voltage follower

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87. An op-amp is open-loop gain of 10⁵ and open-loop upper cut-off frequency of 10 Hz. If this OP-AMP is connected as an amplifier with a closed-loop gain at 100, then new uper cut-off frequency will be
    

1. 1. 10 Hz

   
   
   

   2. 100 Hz

   
   
   

   3. 10 kHz

   
   
   

   4. 100 kHz

   
   
   

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1. View Hint View Answer Discuss in Forum

   Upper cut-off frequency at closed loop gain
   = f_o × (1 + βA) = 10 × 100 = 10 kHz
   

   Correct Option: C

   Upper cut-off frequency at closed loop gain
   = f_o × (1 + βA) = 10 × 100 = 10 kHz
   

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88. Expression for the output voltage V_o in terms of the input voltage V₁ and V₂ in the circuit shown in the figure, assuming operational amplifier to be ideal is V_o = A₁ V₁ + A₂ V₂ Values of A₁ and A₂ would be respectively
    
    

1. 1. 9 and – 10

   
   
   

   2. 9.9 and – 10

   
   
   

   3. – 9 and 10

   
   
   

   4. – 9.9 and 10

   
   
   

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1. View Hint View Answer Discuss in Forum

   VB = V1 -	V1	× 11 = 0.9 V1
   	110

   
   V_A = V_B = 0.9 V₁
   

   Again	V0 - VA	=	VA - V2
   	100 K		10 K

   
   

   ⇒	V0 - 0.9 V1	=	0.9 V1 - V2
   	100		10

   
   ⇒ V₀ = 9.9 V₁ – 10 V₂
   ∴ A₁ = 9.9 and A₂ = – 10

   Correct Option: B

   VB = V1 -	V1	× 11 = 0.9 V1
   	110

   
   V_A = V_B = 0.9 V₁
   

   Again	V0 - VA	=	VA - V2
   	100 K		10 K

   
   

   ⇒	V0 - 0.9 V1	=	0.9 V1 - V2
   	100		10

   
   ⇒ V₀ = 9.9 V₁ – 10 V₂
   ∴ A₁ = 9.9 and A₂ = – 10

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89. In the differentiating circuit given in the figure, function of R₁ is to
    
    

1. 1. enable circuit to approach ideal differentiator

   
   
   

   2. maintain high input impedance

   
   
   

   3. eliminate high frequency noise spikes

   
   
   

   4. prevent oscillations at high frequency

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: B

   NA

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90. The circuit connection of OP-AMP given in the Fig. (a) and Fig. (b) represent
    
    

1. 1. logarithmic amplifiers for both Fig. (a) and (b)

   
   
   

   2. detectors for both Fig. (a) and (b)

   
   
   

   3. detectors for Fig. (a) and logarithmic amplifier for figure (b)

   
   
   

   4. logarithmic amplifier for Fig. (a) and detector for Fig. (b)

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: D

   NA

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