61. Switch S in position (a) for a long time and moves to be at t = 0. The value of V_C and dVC / dt at t = 0⁺—
    
    
    

1. 1. 12 V, – 8 V

   
   
   

   2. 12 V, 8 V

   
   
   

   3. 4 V, – 8 V

   
   
   

   4. 4 V, 8 V

   
   
   

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1. View Hint View Answer Discuss in Forum

   or		IC	=	- 4		dVC	=	Ic	=	- 4	= - 8 V
   		C		1 / 2		dt		C		1 / 2

   Correct Option: A

   or		IC	=	- 4		dVC	=	Ic	=	- 4	= - 8 V
   		C		1 / 2		dt		C		1 / 2

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62. A series R-L circuit is initially released. A step voltage is applied to the circuit. If is the time constant of the circuit, the voltage across R and L will be same at time t equal to—

1. 1. log_e 2

   
   
   

   2. loge	1
      	2

   
   
   

   3. 1	loge 2

   
   
   

   4. 		1	loge	1
      				2

   
   
   

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1. View Hint View Answer Discuss in Forum

   V_L = e ^{– R / L x (t)}
   
   V_R = 1 – e ^{– R / L x (t)}
   (V_R + V_L = 1)
   According to the given condition
   V_L = V_R
   e ^{– R / L x (t)} = 1 – e ^{– R / L x (t)}
   or 2 e e ^{– R / L x (t)} = 1
   or e e ^{– R / L x (t)} = 1 / 2
   or ^{– R / L x (t)} = log_e 1 / 2
   

   or – t =	L	(loge 1 – loge 2)
   	R

   
   

   or t =	L	loge 2
   	R

   
   or t = log_e 2

   Correct Option: A

   V_L = e ^{– R / L x (t)}
   
   V_R = 1 – e ^{– R / L x (t)}
   (V_R + V_L = 1)
   According to the given condition
   V_L = V_R
   e ^{– R / L x (t)} = 1 – e ^{– R / L x (t)}
   or 2 e e ^{– R / L x (t)} = 1
   or e e ^{– R / L x (t)} = 1 / 2
   or ^{– R / L x (t)} = log_e 1 / 2
   

   or – t =	L	(loge 1 – loge 2)
   	R

   
   

   or t =	L	loge 2
   	R

   
   or t = log_e 2

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63. For the circuit shown below current to through the inductor is given by the relation—
    
    
    

1. 1. 5 amp.

   
   
   

   2. (1.1 – 1) + 1

   
   
   

   3. (1.1 – 1) + 1

   
   
   

   4. None of these

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given circuit
   
   

   i L (0–) = i L (0+) =	15	where R eq = 25 || (10 + 20)
   	Req

   
   

   		25 × 30	=	25 × 30	Ω
   		25 + 30		55

   
   

   or i L (0–) =	15 × 55	= 1.1 amp.
   	25 × 30

   
   Calculation for i _L (∞):
   Circuit becomes here the current through the inductor
   I_L (∞)=I₁ + I₂
   

   I1 =	VA – 0
   	20

   
   

   =		15 × 10 / 10 + 10 (- 0)	=	150	=	3
   		20		20 x 20		8

   
   
   

   and I2 =		15	=	3
   		25		8

   
   

   IL =		3	+	3	=	24 + 15	=	39	1 amp.
   		5		8		40		40

   
   As we know that the general equation for the current
   
   i (t) = [i (0_–) – i (∞)] e ^–t + i (∞) .....(i)
   Calculation for the time constant ():
   

   =		L	=	1 / 2
   		Req		[(10 || 100 + 20) || 25]

   
   

   =		1	=	1
   		2(25 / 2)		25

   
   Now, on putting these values in equation (i), we get.
   i (t) = [1.1 – 1] e ^{– t / 1/25} + 1

   Correct Option: D

   Given circuit
   
   

   i L (0–) = i L (0+) =	15	where R eq = 25 || (10 + 20)
   	Req

   
   

   		25 × 30	=	25 × 30	Ω
   		25 + 30		55

   
   

   or i L (0–) =	15 × 55	= 1.1 amp.
   	25 × 30

   
   Calculation for i _L (∞):
   Circuit becomes here the current through the inductor
   I_L (∞)=I₁ + I₂
   

   I1 =	VA – 0
   	20

   
   

   =		15 × 10 / 10 + 10 (- 0)	=	150	=	3
   		20		20 x 20		8

   
   
   

   and I2 =		15	=	3
   		25		8

   
   

   IL =		3	+	3	=	24 + 15	=	39	1 amp.
   		5		8		40		40

   
   As we know that the general equation for the current
   
   i (t) = [i (0_–) – i (∞)] e ^–t + i (∞) .....(i)
   Calculation for the time constant ():
   

   =		L	=	1 / 2
   		Req		[(10 || 100 + 20) || 25]

   
   

   =		1	=	1
   		2(25 / 2)		25

   
   Now, on putting these values in equation (i), we get.
   i (t) = [1.1 – 1] e ^{– t / 1/25} + 1

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64. Which one statement is correct for the circuit given below?
    
    
    

1. 1. Output in stable

   
   
   

   2. Output in unstable

   
   
   

   3. Cannot be calculated from the given data

   
   
   

   4. None of these

   
   
   

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1. View Hint View Answer Discuss in Forum

   The output of the given circuit
   

   V0 =		1	∫ i dt =	1	∫ 2 dt =	2
   		C		C		C

   
   
   Which is unstable, because it directly proportional to the time that's why the output of the given circuit unstable in nature.

   Correct Option: B

   The output of the given circuit
   

   V0 =		1	∫ i dt =	1	∫ 2 dt =	2
   		C		C		C

   
   
   Which is unstable, because it directly proportional to the time that's why the output of the given circuit unstable in nature.

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65. Given i_L (0⁺) = 0, V_C (0^–) = 0 in the steady state V_C equals to—
    
    
    

1. 1. 200 V

   
   
   

   2. 100 V

   
   
   

   3. 0 V

   
   
   

   4. – 100 V

   
   
   

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1. View Hint View Answer Discuss in Forum

   Apply KVL to the loop
   
   100 = L di(t) dt + 1 C ∫ i (t) dt
   taking L.T. to both side, we get
   
   = 100 V

   Correct Option: B

   Apply KVL to the loop
   
   100 = L di(t) dt + 1 C ∫ i (t) dt
   taking L.T. to both side, we get
   
   = 100 V

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