266. Calculate the value of ν₀:
     
     
     

1. 1. 6 V

   
   
   

   2. – 6 V

   
   
   

   3. –10 V

   
   
   

   4. 10 V

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: C

   NA

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267. Calculate the value of Aνd =	V0
     	(V1 – V2)

     
     
     

1. 1. 8

   
   
   

   2. – 6

   
   
   

   3. 6

   
   
   

   4. – 8

   
   
   

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1. View Hint View Answer Discuss in Forum

   V_A = V₁
   V_B = V₂
   due to virtual ground. Let us assume that current I is flowing in the loop given below:
   

   =	VA – VB
   	2 kΩ

   
   By applying KVL in the loop.
   V_o – 6I – 2 I – 4 I = 0
   or V_o = 12 I
   

   Vo = 12.	(VA – VB)
   	2

   
   

   or Vo =	12	. (V1 – V2)
   	2

   
   

   or	Vo	= 6 = Avd
   	V1 – V2

   
   Hence alternative (C) is the correct choice.

   
   

   Correct Option: C

   V_A = V₁
   V_B = V₂
   due to virtual ground. Let us assume that current I is flowing in the loop given below:
   

   =	VA – VB
   	2 kΩ

   
   By applying KVL in the loop.
   V_o – 6I – 2 I – 4 I = 0
   or V_o = 12 I
   

   Vo = 12.	(VA – VB)
   	2

   
   

   or Vo =	12	. (V1 – V2)
   	2

   
   

   or	Vo	= 6 = Avd
   	V1 – V2

   
   Hence alternative (C) is the correct choice.

   
   

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268. The approximate value of input impedance of a common emitter amplifier with emitter resistance Re is given by:

1. 1. h_ie + A_l R_e

   
   
   

   2. h_ie + (1 + h_fe) R_e

   
   
   

   3. h_ie

   
   
   

   4. (1 + h_fe) R_e

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: B

   NA

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269. Calculate V₀
     
     
     

1. 1. – 7.5 V

   
   
   

   2. 7.5 V

   
   
   

   3. 8 V

   
   
   

   4. – 8 V

   
   
   

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1. View Hint View Answer Discuss in Forum

   V_A = 2.5 V (due to virtual ground) By using potential divider method
   

   VA = Vo.	4
   	4 + 8

   
   

   or 2.5 = Vo.	1
   	3

   
   or V_o = 7.5

   
   

   Correct Option: B

   V_A = 2.5 V (due to virtual ground) By using potential divider method
   

   VA = Vo.	4
   	4 + 8

   
   

   or 2.5 = Vo.	1
   	3

   
   or V_o = 7.5

   
   

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270. Find V₀
     
     
     

1. 1. 4 V

   
   
   

   2. – 4 V

   
   
   

   3. 5 V

   
   
   

   4. – 5 V

   
   
   

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1. View Hint View Answer Discuss in Forum

   V– = 5 V (due to virtual ground) apply KCL at node A
   

   0.1 × 10–3 =	Vo – VA
   	10 kΩ

   
   or V_o = V_A – 0.1 × 10^–3 × 10 × 10³
   V_o = 5 – 1 = 4 V

   
   

   Correct Option: A

   V– = 5 V (due to virtual ground) apply KCL at node A
   

   0.1 × 10–3 =	Vo – VA
   	10 kΩ

   
   or V_o = V_A – 0.1 × 10^–3 × 10 × 10³
   V_o = 5 – 1 = 4 V

   
   

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