21. The rms voltage measured across on admittance (G + jB) is V. The reactive power for the element is—

1. 1. V² B

   
   
   

   2. – V² B

   
   
   

   3. V² G² + B²

   
   
   

   4. V² (G + jB)

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: A

   NA

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22. If L {f (t)} =
    	s2 + 2

    , then the value of lim t→∞ f (t)—

1. 1. cannot be determined

   
   
   

   2. Zero

   
   
   

   3. 1

   
   
   

   4. ∞

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given, L {f (t)} =
   	s2 + 2

   
   or f (t) = sin t.
   as t → ∞, f (t) = sin t goes through 1 to 0
   and – 1 to 0 to infinite times so the value of Lim t→∞ f(t) cannot be determined.

   Correct Option: A

   Given, L {f (t)} =
   	s2 + 2

   
   or f (t) = sin t.
   as t → ∞, f (t) = sin t goes through 1 to 0
   and – 1 to 0 to infinite times so the value of Lim t→∞ f(t) cannot be determined.

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23. 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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24. 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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25. The rms value of the current in the a.c. circuit as shown below in the figure will be—
    
    
    

1. 1. 2 A

   
   
   

   2. 4 A

   
   
   

   3. 5 A

   
   
   

   4. 8 A

   
   
   

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1. View Hint View Answer Discuss in Forum

   Since, 10² = V² R_rms + V²L_rms
   100 = V² R_rms + 36
   V² R_rms = 100 – 36 = 64
   V² R_rms = 64 = 8
   

   i =		V2 Rrms	=	8	= 4 A
   		R		2

   Correct Option: B

   Since, 10² = V² R_rms + V²L_rms
   100 = V² R_rms + 36
   V² R_rms = 100 – 36 = 64
   V² R_rms = 64 = 8
   

   i =		V2 Rrms	=	8	= 4 A
   		R		2

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