71. When two identical stage with upper cut off at ωh are cascaded, the overall 3 dB point is at:

1. 1. ω_h

   
   
   

   2. 2 ω_h

   
   
   

   3. 0.5 ω_h

   
   
   

   4. 0.64 ω_h

   
   
   

---

1. View Hint View Answer Discuss in Forum

   ω_h* = ω_h √2_1/n – 1
   = ω_h √2_1/2 – 1 (since n = 2)
   = ω_h √0·414
   = 0.64 ω_h

   Correct Option: D

   ω_h* = ω_h √2_1/n – 1
   = ω_h √2_1/2 – 1 (since n = 2)
   = ω_h √0·414
   = 0.64 ω_h

---

72. The gain bandwidth product of a two stage CE amplifier is:

1. 1. the same as that of one stage

   
   
   

   2. greater than that of one stage

   
   
   

   3. less than that of one stage

   
   
   

   4. the product of the two gain bandwidth products of each stage

   
   
   

---

1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: B

   NA

---

---

73. A CE amplifier has a resistor R_F connected between collector and base. R_F = 40 K, R_C (collector load resistance) = 4 K, Given h_fe = 50, r_π = 1 K. The output resistance R₀ is given by:

1. 1. 40 K

   
   
   

   2. 20 K

   
   
   

   3. 4 K

   
   
   

   4. 0.66 K

   
   
   

---

1. View Hint View Answer Discuss in Forum

   R0 = R′L = Rc || RF =	4 × 40	=	160	= 3.64 k
   	4 + 40		44

   
   output resistance after feedback is given by (since RF is the feedback resistance connected between the collector and base)
   

   Rof =	R0
   	1 + Aβ

   
   (∴ due to voltage shunt feedback)
   

   Aβ =	RC	. β =	4	. 50 =	50
   	RF + AC		40 + 4		11

   
   

   or 1 + Aβ = 1 +	50	=	61
   	11		11

   
   

   Now, Rof =	R0	=	3·64	= 0.66 kΩ
   	1 + Aβ		61/11

   
   Hence alternative (D) is the correct choice.

   Correct Option: D

   R0 = R′L = Rc || RF =	4 × 40	=	160	= 3.64 k
   	4 + 40		44

   
   output resistance after feedback is given by (since RF is the feedback resistance connected between the collector and base)
   

   Rof =	R0
   	1 + Aβ

   
   (∴ due to voltage shunt feedback)
   

   Aβ =	RC	. β =	4	. 50 =	50
   	RF + AC		40 + 4		11

   
   

   or 1 + Aβ = 1 +	50	=	61
   	11		11

   
   

   Now, Rof =	R0	=	3·64	= 0.66 kΩ
   	1 + Aβ		61/11

   
   Hence alternative (D) is the correct choice.

---

74. A CE amplifier has an unbypassed R_E Given RC (collector load) is 10 k ohms, R_E = 1 k ohms, r_π = 500 ohms, h_fe = 100. The voltage gain | A_v | is approximated by:

1. 1. 2000

   
   
   

   2. 1000

   
   
   

   3. 20

   
   
   

   4. 10

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Given that R_c = 10 kΩ
   R_E = 1 kΩ
   r_π = 500 Ω
   h_fe = β = 100
   | A_V |=?
   

   | AV | =	β·Rc
   	rπ + (1 + β)RE

   
   

   =	100 × 10 × 103	≈ 10
   	500 + (1 + 100)103

   Correct Option: D

   Given that R_c = 10 kΩ
   R_E = 1 kΩ
   r_π = 500 Ω
   h_fe = β = 100
   | A_V |=?
   

   | AV | =	β·Rc
   	rπ + (1 + β)RE

   
   

   =	100 × 10 × 103	≈ 10
   	500 + (1 + 100)103

---

---

75. A transistor amplifier has voltage gain of 100 and R_i = 1 K. Negative voltage feedback with β = 1/5 is applied using unbypassed R_E. The R_i with feedback is:

1. 1. 1	k ohms
      21

   
   
   

   2. 21 k ohms

   
   
   

   3. 20 k ohms

   
   
   

   4. 50 ohms

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Since the given amplifier uses voltage series feedback. Therefore after feedback input resistance increases.
   R_if = R_i (1 + Aβ)
   

   or Rif = 1 k		1 + 100 ×	1
   			5

   
   or R_if = 1 k (21)
   or R_if = 21 k ohms.

   Correct Option: B

   Since the given amplifier uses voltage series feedback. Therefore after feedback input resistance increases.
   R_if = R_i (1 + Aβ)
   

   or Rif = 1 k		1 + 100 ×	1
   			5

   
   or R_if = 1 k (21)
   or R_if = 21 k ohms.

---