261. An amplifier has an input power of 2 mW. The power gain of amplifier has 60 dB. The output power will be:

1. 1. 6 µW

   
   
   

   2. 120 µW

   
   
   

   3. 2 mW

   
   
   

   4. 2 W

   
   
   

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1. View Hint View Answer Discuss in Forum

   Power gain = 20 log10		Po
   		Pi

   
   

   or 60 = 20 log10		Po
   		Pi

   
   

   or 3 = log10		Po
   		Pi

   
   

   or	Po	= 103
   	Pi

   
   or P_o = P_i × 10³ = 2 × 10^–3 × 10³ = 2 W

   Correct Option: D

   Power gain = 20 log10		Po
   		Pi

   
   

   or 60 = 20 log10		Po
   		Pi

   
   

   or 3 = log10		Po
   		Pi

   
   

   or	Po	= 103
   	Pi

   
   or P_o = P_i × 10³ = 2 × 10^–3 × 10³ = 2 W

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262. In figure below assume that the diode and op-amps are ideal. For an input V_in = sin ωt, the output voltage V₀ is:
     
     
     

1. 1. full wave rectified with a peak value of 1

   
   
   

   2. full wave rectified with a peak value of – 1

   
   
   

   3. half wave rectified with a peak value of – 1

   
   
   

   4. half wave rectified with a peak value of ± 1

   
   
   

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1. View Hint View Answer Discuss in Forum

   The op-amp (1) circuit represent precision half wave rectifier conducting for negative cycle at the inverting input, and op-amp (2) is a unity follower.
   Hence alternative (C) is the correct choice.

   Correct Option: C

   The op-amp (1) circuit represent precision half wave rectifier conducting for negative cycle at the inverting input, and op-amp (2) is a unity follower.
   Hence alternative (C) is the correct choice.

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263. Figure shown a Schmitt trigger circuit op-amp. The output V₀ is limited to + 10 V and – 5 V connecting suitably chosen Zener diodes at the output. The lower and upper trigger voltage are respectively:
     
     
     

1. 1. – 1 V and 0.5 V

   
   
   

   2. – 0.5 V and 1.0 V

   
   
   

   3. – 1 V and – 0.5 V

   
   
   

   4. 0.5 V and 1.0 V

   
   
   

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1. View Hint View Answer Discuss in Forum

   VUT = – (– Vsat).	R2	= – (– 5).	1	= 0.5 V
   	dR		10

   
   

   VLT = – Vsat.	R2	= – 10.	1	= – 1 V
   	dR		10

   Correct Option: A

   VUT = – (– Vsat).	R2	= – (– 5).	1	= 0.5 V
   	dR		10

   
   

   VLT = – Vsat.	R2	= – 10.	1	= – 1 V
   	dR		10

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264. For the circuit given below V₀:
     
     
     

1. 1. V₁

   
   
   

   2. – V₁

   
   
   

   3. V1

      2

   
   
   

   4. –V1

      2

   
   
   

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1. View Hint View Answer Discuss in Forum

   Vo = – Vi.		R
   		R

   
   or V_o = – V_i

   
   

   Correct Option: B

   Vo = – Vi.		R
   		R

   
   or V_o = – V_i

   
   

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265. The	Output voltage	for the circuit shown below is:
     	Input voltage

     
     
     

1. 1. + 1 when S open, 1 +	R2	when S closed
      	R1

   
   
   

   2. R4	–	R2	when S open, + 1 when S closed
      R3		R1

   
   
   

   3. – 1 when S open, 1 +	R3	when S closed
      	R4

   
   
   

   4. + 1 when S open,	–R2	when S closed
      	R1

   
   
   

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1. View Hint View Answer Discuss in Forum

   When switch S is open. Circuit becomes
   i₂ = 0 becomes op-amp draws negligible current.
   V_d = 0, since V_A = V_B = V_i
   

   i1 =	Vi – VA	= 0 (Vi = VA)
   	R1

   
   

   At node A, i1 =	VA – Vo	= 0 or	Vo	= 1
   	R1 + R2		Vi

   
   When S is closed, it functions as inverting amplifier
   

   with gain = –	R2
   	R1

   
   Hence alternative (D) is the correct choice.

   
   

   Correct Option: D

   When switch S is open. Circuit becomes
   i₂ = 0 becomes op-amp draws negligible current.
   V_d = 0, since V_A = V_B = V_i
   

   i1 =	Vi – VA	= 0 (Vi = VA)
   	R1

   
   

   At node A, i1 =	VA – Vo	= 0 or	Vo	= 1
   	R1 + R2		Vi

   
   When S is closed, it functions as inverting amplifier
   

   with gain = –	R2
   	R1

   
   Hence alternative (D) is the correct choice.

   
   

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