36. In the given figure, the switch was closed for a long time before opening at t = 0. The voltage V_x at t = 0⁺ is
    
    

1. 1. 25 V

   
   
   

   2. 50 V

   
   
   

   3. – 50 V

   
   
   

   4. 0 V

   
   
   

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1. View Hint View Answer Discuss in Forum

   At t = 0⁺ , the circuit will be as shown below.
   
   I _L (0⁺) = 0
   Then V_i = – 50 V

   Correct Option: C

   At t = 0⁺ , the circuit will be as shown below.
   
   I _L (0⁺) = 0
   Then V_i = – 50 V

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37. In the circuit of given figure, assume that the diodes are ideal and meter is an average indicating ammeter. The ammeter will read
    
    

1. 1. 0.4 √2 A
      

   
   
   

   2. 0.4 A
      

   
   
   

   3. 0.8	A
      π

      
      

   
   
   

   4. 0.4	mA
      π

   
   
   

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1. View Hint View Answer Discuss in Forum

   The meter will read during positive half for halfwave rectifier.
   

   Vav =	Vm	=	4	=	0.4	mA.
   	πR		π10K		π

   
   

   Correct Option: D

   The meter will read during positive half for halfwave rectifier.
   

   Vav =	Vm	=	4	=	0.4	mA.
   	πR		π10K		π

   
   

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38. In the circuit shown, i(t) is a unit step current. The steady-state value of v(t) is
    
    

1. 1. 2 V
      

   
   
   

   2. 3 V

   
   
   

   3. 6 V

   
   
   

   4. 9 V

   
   
   

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1. View Hint View Answer Discuss in Forum

   At steady-state condition, capacitor is open and inductor is shorted, then
   

   current through 6 Ω = 1 ×	3	=	1	Ω
   	3 + 6		3

   
   

   and V(t) = L	di	+ 6i
   	dt

   
   

   = 0 +		6 ×	1		= 2 volts
   			3

   
   

   Correct Option: A

   At steady-state condition, capacitor is open and inductor is shorted, then
   

   current through 6 Ω = 1 ×	3	=	1	Ω
   	3 + 6		3

   
   

   and V(t) = L	di	+ 6i
   	dt

   
   

   = 0 +		6 ×	1		= 2 volts
   			3

   
   

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39. The system function H(s) =	1
    	s + 1

    
    For an input signal cos t, the steady state response is
    

1. 1. 1	cos		t -	π
      √2				4

      
      

   
   
   

   2. cos t
      

   
   
   

   3. cos		1 -	π
      			4

      
      

   
   
   

   4. 1	cos t
      √2

      
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   Response, C (s) = H (s). R (s)
   

   =	1	.	s
   	s + 1		s2 + 1

   
   

   C(t) = -	1	e-t +	1	cos		t -	π
   	2		√2				4

   
   At steady state e^{– t} → 0
   

   ∴ C(t) =	1	.cos		t -	π
   	√2				4

   
   

   Correct Option: A

   Response, C (s) = H (s). R (s)
   

   =	1	.	s
   	s + 1		s2 + 1

   
   

   C(t) = -	1	e-t +	1	cos		t -	π
   	2		√2				4

   
   At steady state e^{– t} → 0
   

   ∴ C(t) =	1	.cos		t -	π
   	√2				4

   
   

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40. The steady state in the circuit, shown in the given figure is reached with S open. S is closed at t = 0. The current I at t = 0⁺ is
    
    

1. 1. 1 A
      

   
   
   

   2. 2 A
      

   
   
   

   3. 3 A
      

   
   
   

   4. 4 A

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: B

   NA

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