71. A voltage source V_AB = 4 sinωt is applied to the terminals A and B of the circuit shown in the given figure. The diodes are assumed to be ideal. The impedance by the circuit across the terminals A and B is _________ kΩ .
    
    

1. 1. 10

   
   
   

   2. 20

   
   
   

   3. 45

   
   
   

   4. 66

   
   
   

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1. View Hint View Answer Discuss in Forum

   Diode D₁ conducts through 10 kΩ on the extreme right and diode D₂ is blocked. Diode D₂ conducts through 10 kΩ in the middle branch and diode D₁ is blocked. Thus source always sees a resistance of 10 kΩ.

   Correct Option: A

   Diode D₁ conducts through 10 kΩ on the extreme right and diode D₂ is blocked. Diode D₂ conducts through 10 kΩ in the middle branch and diode D₁ is blocked. Thus source always sees a resistance of 10 kΩ.

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72. In the circuit shown in the given figure, transistor is in the active mode. V_BE = 0.7 volt. The best approximation without knowing h_fe for V_C as shown in the figure is _________ V
    

1. 1. 3.4

   
   
   

   2. 4

   
   
   

   3. 4.3

   
   
   

   4. None of the above

   
   
   

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1. View Hint View Answer Discuss in Forum

   IC ≃ IE =	(4 - 0.7)	volt = 1 mA
   	3.3 kΩ

   
   ∴ V_C = 8 – 3.7 × 1 = 4.3 volts .
   

   Correct Option: C

   IC ≃ IE =	(4 - 0.7)	volt = 1 mA
   	3.3 kΩ

   
   ∴ V_C = 8 – 3.7 × 1 = 4.3 volts .
   

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73. The centre tap full-wave single-phase rectifier circuit uses 2 diodes as shown in the given figure. The rms voltage across each diode is _________ V
    

1. 1. 295.3

   
   
   

   2. 395.3

   
   
   

   3. 91

   
   
   

   4. 315

   
   
   

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1. View Hint View Answer Discuss in Forum

   PIV across each diode = 2 × 280 √2 volts.
   When diode is conducting (during half cycle) voltage across the diode is zero.
   Hence rms value of the voltage across the diode
   

   =	PIV	= 280 √2 = 395.3 V
   	2

   Correct Option: B

   PIV across each diode = 2 × 280 √2 volts.
   When diode is conducting (during half cycle) voltage across the diode is zero.
   Hence rms value of the voltage across the diode
   

   =	PIV	= 280 √2 = 395.3 V
   	2

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74. For the oscillator circuit shown in the given figure, expression for the time period of oscillation can be given by (where t = RC)
    
    

1. 1. τ ln 3

   
   
   

   2. 2 τ ln 3

   
   
   

   3. τ ln 2

   
   
   

   4. 2 τ ln 2

   
   
   

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1. View Hint View Answer Discuss in Forum

   T = 2RC ln	1 + β
   	1 - β

   
   

   Here , β =	1	=	R
   	2		R + R

   
   ∴ T = 2 RC ln 3 = 2 τ ln 3

   Correct Option: B

   T = 2RC ln	1 + β
   	1 - β

   
   

   Here , β =	1	=	R
   	2		R + R

   
   ∴ T = 2 RC ln 3 = 2 τ ln 3

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75. The circuit shown in the figure
    
    

1. 1. is an oscillating circuit and its output is a square wave

   
   
   

   2. is one whose output remains stable in ‘1’ state

   
   
   

   3. is one having output semains stable in ‘0’ state

   
   
   

   4. having a single pulse of 3 times propagation delay.

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: A

   NA

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