81. The value of i (t) < 0 is given by the relation for the circuit shown below—
    
    
    

1. 1. [1 – (– 2.5)] e ^{– 1 / 0.1} – .25

   
   
   

   2. [1 + (– 2.5)] e ^{– 1 / 0.1} + .25

   
   
   

   3. [1 + 2.5] e ^{– 1 / 100} + .25

   
   
   

   4. 1 u (– t)

   
   
   

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1. View Hint View Answer Discuss in Forum

   For t < 0, 50 u (t) → not exist, so the circuit becomes like
   
   

   i L (0–) = i L (0+) = 2 ×	10	= 1 amp.
   	10 + 10

   
   so, i (t) = 1 u (– t)

   Correct Option: D

   For t < 0, 50 u (t) → not exist, so the circuit becomes like
   
   

   i L (0–) = i L (0+) = 2 ×	10	= 1 amp.
   	10 + 10

   
   so, i (t) = 1 u (– t)

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82. Switch S is in position (1) for a long time at t = 0 comes to position (2). Find i _L (t)—
    
    
    
    

1. 1. i (t) = (1.25 – 0.25) e ^{– t / 5 × 103} + .25

   
   
   

   2. i (t) = (1.25 – 0.25) e ^{– t / 5 × 10–3} + .25

   
   
   

   3. i (t) = (.25 – 1.25) e ^{– t / 0.5 × 10–3} + .5

   
   
   

   4. i (t) = (.25 – .25) e ^{– t / 0.5 × 10–3} + .5

   
   
   

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1. View Hint View Answer Discuss in Forum

   It may be noted that in this problem differential approach becomes complex. So we will apply here general approach.
   We know that
   i (t)=[i (0) – i (∞)] e ^{–t / τ} + i (∞) .....(i)
   

   here i (0–) = i (0+) =	50	= 1.25 amp.
   	40

   
   

   i (∞) =	50	= .25
   	40

   
   

   τ =		L	=	20 × 10–3	= .5 × 10–3
   		R		40

   
   so on putting these values in equation (i), we get
   i _L (t) = (1.25 – 0.25) e ^{–1 5 × 10–3} + .25

   Correct Option: B

   It may be noted that in this problem differential approach becomes complex. So we will apply here general approach.
   We know that
   i (t)=[i (0) – i (∞)] e ^{–t / τ} + i (∞) .....(i)
   

   here i (0–) = i (0+) =	50	= 1.25 amp.
   	40

   
   

   i (∞) =	50	= .25
   	40

   
   

   τ =		L	=	20 × 10–3	= .5 × 10–3
   		R		40

   
   so on putting these values in equation (i), we get
   i _L (t) = (1.25 – 0.25) e ^{–1 5 × 10–3} + .25

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83. The Laplace transform of the function f (t) = e^{t2 – 4t} will be—

1. 1. 		1	+	e–4s	e– t
      		s2		2s

   
   
   

   2. 	e–4s2
      	s3

   
   
   

   3. 	e–4s
      	s3

   
   
   

   4. does not exist

   
   
   

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1. View Hint View Answer Discuss in Forum

   The Laplace transform is given by the relation for any function f (t).
   

   L {f(t)} =		∞	f (t) e–st dt
   		0

   
   Provided that the integral must be exist within the limit.
   But here,
   

   L {f(t)} =		∞	et2 – 4t. e–st dt
   		0

   Correct Option: D

   The Laplace transform is given by the relation for any function f (t).
   

   L {f(t)} =		∞	f (t) e–st dt
   		0

   
   Provided that the integral must be exist within the limit.
   But here,
   

   L {f(t)} =		∞	et2 – 4t. e–st dt
   		0

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84. The Laplace transform of the given function—
    

    f (t) =		t, t ≥ 4 sec
    		0, Otherwise	will be—

1. 1. 	e–4s
      	s2

   
   
   

   2. 	e–4(s + 1)
      	s2

   
   
   

   3. 	e–4s
      	s

   
   
   

   4. e–4s		t, t ≥ 4 sec
      		4s + 1
      		s2

      
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   

   Given f (t) =		t, t ≥ 4 sec
   		0, otherwise

   
   f (t) = t u (t – 4)
   or f (t)= (t – 4 + 4) u (t – 4)
   or f (t)= (t – 4) u (t – 4) + 4 u (t – 4)
   

   or f (s) =		1	e– 4s +	4	e– 4s
   		s2		5

   
   
   

   or f (s) = e– 4s		1	+	4
   		s2		s

   
   

   or f (s) = e– 4s		4s + 1
   		s2

   Correct Option: D

   
   

   Given f (t) =		t, t ≥ 4 sec
   		0, otherwise

   
   f (t) = t u (t – 4)
   or f (t)= (t – 4 + 4) u (t – 4)
   or f (t)= (t – 4) u (t – 4) + 4 u (t – 4)
   

   or f (s) =		1	e– 4s +	4	e– 4s
   		s2		5

   
   
   

   or f (s) = e– 4s		1	+	4
   		s2		s

   
   

   or f (s) = e– 4s		4s + 1
   		s2

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85. The initial voltage on the capacitor is V_c (0^–) = 2 V, I (s) = u (t) cos t find V (t) for t > 0—
    
    
    

1. 1. V (t) =		1	cos (t – 45º) +	3	e– t
      		2		2

   
   
   

   2. V (t) = 2 cos (t – 45º) +	3	e+ t
      	2

   
   
   

   3. V (t) = cos (t – 45º) +	1	e– 2t
      	2

   
   
   

   4. V (t) = cos (t + 45º) +	3	e– 2t
      	2

   
   
   

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1. View Hint View Answer Discuss in Forum

   Here Zeq =	1 .( 1 / Jω)	, ω = 1 on comparing with I (s) = u (t) cos
   	1 + (1 / Jω)

   
   
   

   or Z eq =	1
   	1 + J

   
   Since initial voltage on the capacitor is V(0^–) = 2 V,
   

   so, V (t) J (s), Zeq = cos t.	1	+ A e– st
   	1 + J

   
   

   s=	1	cos (t– 45°) + A e– st given that at t = 0. V (0) = 2 V
   	√2

   
   

   so, 2 =	1	2 cos (0 – 45º) + A e–5.0
   	√2

   
   

   2 =	1	√2 + A
   	√2

   
   

   or A =	3
   	2

   
   

   S =		1	=	1	= 1
   		RC		1 x 1

   
   

   V (t) = cos (t – 45°) +	3	e– t
   	2

   Correct Option: A

   Here Zeq =	1 .( 1 / Jω)	, ω = 1 on comparing with I (s) = u (t) cos
   	1 + (1 / Jω)

   
   
   

   or Z eq =	1
   	1 + J

   
   Since initial voltage on the capacitor is V(0^–) = 2 V,
   

   so, V (t) J (s), Zeq = cos t.	1	+ A e– st
   	1 + J

   
   

   s=	1	cos (t– 45°) + A e– st given that at t = 0. V (0) = 2 V
   	√2

   
   

   so, 2 =	1	2 cos (0 – 45º) + A e–5.0
   	√2

   
   

   2 =	1	√2 + A
   	√2

   
   

   or A =	3
   	2

   
   

   S =		1	=	1	= 1
   		RC		1 x 1

   
   

   V (t) = cos (t – 45°) +	3	e– t
   	2

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