271. In the circuit shown below | V₀ | = 1 V to a certain set of values of ω, R and C. | V₀ | will remain as 1 V even if:
     
     
     

1. 1. None of these

   
   
   

   2. ω is halved

   
   
   

   3. R is halved

   
   
   

   4. C is doubled

   
   
   

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1. View Hint View Answer Discuss in Forum

   The given circuit represent all pass filter.
   

   T.F. = H (j) =	1 + (RC)2	= 1
   	1 + (RC)2

   
   when, R and C are changed by the same magnitude the transfer function is unchanged.
   Hence alternative (A) is the correct choice.

   Correct Option: A

   The given circuit represent all pass filter.
   

   T.F. = H (j) =	1 + (RC)2	= 1
   	1 + (RC)2

   
   when, R and C are changed by the same magnitude the transfer function is unchanged.
   Hence alternative (A) is the correct choice.

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272. Each transistor in the darlington pair shown below has h_FE = 100. The overall h_FE of the composite transistor neglecting leakage current is:
     
     
     

1. 1. 10000

   
   
   

   2. 10001

   
   
   

   3. 10100

   
   
   

   4. 10200

   
   
   

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1. View Hint View Answer Discuss in Forum

   The overall h′_FE of the final stage is given by relation = 100² + 2 × 100 = 10000 + 200 = 10200
   (h′_FE = h²_FE + 2 h_FE)

   Correct Option: D

   The overall h′_FE of the final stage is given by relation = 100² + 2 × 100 = 10000 + 200 = 10200
   (h′_FE = h²_FE + 2 h_FE)

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273. If the CMRR of the Op-amp is 60 dB, then magnitude of the output voltage is:
     
     
     

1. 1. 1 mV

   
   
   

   2. 100 mV

   
   
   

   3. 200 mV

   
   
   

   4. 2 V

   
   
   

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1. View Hint View Answer Discuss in Forum

   V– =	R	. 2 = 1 V
   	R + R

   
   

   V+ =	R	. 2 = 1 V
   	R + R

   
   V_d = V₊ – V_– = 0
   given that
   CMRR = 60 dB
   

   60 = 20 log10		AdM
   		ACM

   
   = 20 log₁₀ Ad – 20 log₁₀ A_c
   60 = 0 – 20 log₁₀ A_c
   or 3 = – log₁₀ A_c
   or A_c = 10^{– 3} = 0.001
   

   or Vo = ACM.		V+ + V–
   		2

   
   

   = 0.001		1 + 1
   		2

   
   = 0.001 = 1 mV

   
   

   Correct Option: A

   V– =	R	. 2 = 1 V
   	R + R

   
   

   V+ =	R	. 2 = 1 V
   	R + R

   
   V_d = V₊ – V_– = 0
   given that
   CMRR = 60 dB
   

   60 = 20 log10		AdM
   		ACM

   
   = 20 log₁₀ Ad – 20 log₁₀ A_c
   60 = 0 – 20 log₁₀ A_c
   or 3 = – log₁₀ A_c
   or A_c = 10^{– 3} = 0.001
   

   or Vo = ACM.		V+ + V–
   		2

   
   

   = 0.001		1 + 1
   		2

   
   = 0.001 = 1 mV

   
   

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274. An op-amp has an open-loop gain of 10⁵ and an openloop upper cut off frequency of 10 Hz. If this op-amp is connected as an amplifier with a closed-loop gain of 100. Then the new upper cut off frequency is:

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

   Given that
   A_OL = 10⁵
   f_H = 10 Hz
   A_CL = 100
   

   ACL =	AOL
   	1 + AOL · β

   
   

   1 + AOL β =	105	= 103
   	100

   
   Since due to close loop, gain reduces while uppercut off frequency increases by a factor of (1 + A_OL β)
   fH^* = f_H (1 + A_OL β)
   = 10 (10³)
   = 10 kHz
   Hence alternative (C) is the correct choice.

   Correct Option: C

   Given that
   A_OL = 10⁵
   f_H = 10 Hz
   A_CL = 100
   

   ACL =	AOL
   	1 + AOL · β

   
   

   1 + AOL β =	105	= 103
   	100

   
   Since due to close loop, gain reduces while uppercut off frequency increases by a factor of (1 + A_OL β)
   fH^* = f_H (1 + A_OL β)
   = 10 (10³)
   = 10 kHz
   Hence alternative (C) is the correct choice.

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275. Calculate ν₀
     
     
     

1. 1. – 12 V

   
   
   

   2. 12 V

   
   
   

   3. – 18 V

   
   
   

   4. 18 V

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: A

   NA

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