76. The period of the output waveform for the circuit of figure shown, when triggered by a negative pulse is
    
    

1. 1. 0.75 ms

   
   
   

   2. 0.825 ms

   
   
   

   3. 0.525 ms

   
   
   

   4. 1.5 ms

   
   
   

---

1. View Hint View Answer Discuss in Forum

   T = (1.1) R_AC
   = 1.1(7.5 × 10³) (0.1 × 10^{– 6}) = 0.825 ms.

   Correct Option: B

   T = (1.1) R_AC
   = 1.1(7.5 × 10³) (0.1 × 10^{– 6}) = 0.825 ms.

---

77. What is the hysterisis voltage of the submitt trigger shown if V_sat = ± 10V ?
    
    

1. 1. 1.90 volts

   
   
   

   2. 1.45 volts

   
   
   

   3. 1.00 volt

   
   
   

   4. 0.90 volt

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Hysterisis voltage,
   

   VH =	2 R2 V0	=	2 × 5 × 10	= 1 volt
   	R1 + R2		100

   
   

   Correct Option: C

   Hysterisis voltage,
   

   VH =	2 R2 V0	=	2 × 5 × 10	= 1 volt
   	R1 + R2		100

   
   

---

---

78. In the switcling circuit shown, RC time constant is much longer as compared to the time period of its input pulse signal. V_out of this circuit is of the form
    
    

1. 1. 
      

   
   
   

   2. 
      

   
   
   

   3. 
      

   
   
   

   4. 
      

   
   
   

---

1. View Hint View Answer Discuss in Forum

   During ‘+’ ve cycle, diode turns ON and V₀ = 0V. Between positive peaks, diode is OFF and V₀ = 2V_in
   
   

   Correct Option: B

   During ‘+’ ve cycle, diode turns ON and V₀ = 0V. Between positive peaks, diode is OFF and V₀ = 2V_in
   
   

---

79. The low frequency gain of the low pass filter shown in the given figure is
    
    

1. 1. 10 dB

   
   
   

   2. 20 dB

   
   
   

   3. 30 dB

   
   
   

   4. 40 dB

   
   
   

---

1. View Hint View Answer Discuss in Forum

   	V2		=	R	.	1
   	V1			R1		RCs + 1

   
   

   = 100	1	≈ 100
   	10-3 j + 1

   
   

   ∴ 20 log		V2		= 40 dB
   		V1

   Correct Option: D

   	V2		=	R	.	1
   	V1			R1		RCs + 1

   
   

   = 100	1	≈ 100
   	10-3 j + 1

   
   

   ∴ 20 log		V2		= 40 dB
   		V1

---

---

80. The transfer function of a second order LP filter shown in the given figure is
    

1. 1. 	1
      	R2C2s2 + 3RCs + 1

   
   
   

   2. 	RCs
      	R2C2s2 + 3RCs + 1

   
   
   

   3. 	R2C2s2 + 1
      	R2C2s2 + 3RCs + 1

   
   
   

   4. 	R2C2s2
      	R2C2s2 + 3RCs + 1

      
      

   
   
   

---

1. View Hint View Answer Discuss in Forum

   
   Using ∆ – Y conversion,
   

   Z1 =	R2Cs
   	2RCs + 1

   
   

   Z2 =	R
   	2RCs + 1

   
   

   Z3 =	R
   	2RCs + 1

   
   From the figure, using KCL, we have
   

   V0	=	1
   Vin		R2C2s2 + 3RCs + 1

   
   

   Correct Option: A

   
   Using ∆ – Y conversion,
   

   Z1 =	R2Cs
   	2RCs + 1

   
   

   Z2 =	R
   	2RCs + 1

   
   

   Z3 =	R
   	2RCs + 1

   
   From the figure, using KCL, we have
   

   V0	=	1
   Vin		R2C2s2 + 3RCs + 1

   
   

---