1. D.C. component of the waveform shown below is—
   
   
   

1. 1. 0.5 V

   
   
   

   2. 1.5 V

   
   
   

   3. 2.5 V

   
   
   

   4. 3.5 V

   
   
   

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1. View Hint View Answer Discuss in Forum

   Here equation of f (t) = mt + 4
   

   f (t) =		–	1	t + 4
   			2

   
   where, m = slope of the given waveform
   

   Now, D.C. component =	1			f(t) dt
   	T		T

   
   

   =	1		2	– (t/2) + 4 dt
   	2		0

   
   
   

   =	1			–t2	+ 4t		2
   	2			4			0

   
   

   =	1			–4	+ 4 × 2
   	2			4

   
   = 3.5 V

   Correct Option: D

   Here equation of f (t) = mt + 4
   

   f (t) =		–	1	t + 4
   			2

   
   where, m = slope of the given waveform
   

   Now, D.C. component =	1			f(t) dt
   	T		T

   
   

   =	1		2	– (t/2) + 4 dt
   	2		0

   
   
   

   =	1			–t2	+ 4t		2
   	2			4			0

   
   

   =	1			–4	+ 4 × 2
   	2			4

   
   = 3.5 V

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2. D.C. component of the waveform shown below is—
   
   
   

1. 1. 0

   
   
   

   2. 2.5 mA

   
   
   

   3. 5 mA

   
   
   

   4. 10 mA

   
   
   

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1. View Hint View Answer Discuss in Forum

   D.C. component =	1			f(t) dt
   	T		T

   
   
   

   =	1		4	f(t) dt
   	4		0

   
   

   =	1			2	10 dt +		4	0 dt
   	4			0			2

   
   

   =	20	= 5mA
   	4

   Correct Option: C

   D.C. component =	1			f(t) dt
   	T		T

   
   
   

   =	1		4	f(t) dt
   	4		0

   
   

   =	1			2	10 dt +		4	0 dt
   	4			0			2

   
   

   =	20	= 5mA
   	4

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3. D.C. component of the waveform shown below is—
   
   
   

1. 1. 2

   
   
   

   2. 1

   
   
   

   3. 0.5

   
   
   

   4. 0

   
   
   

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1. View Hint View Answer Discuss in Forum

   Average value = D.C. component of wave
   

   =	1			f(t) dt =	1		2	f(t) dt
   	T		T		2		0

   
   
   

   =	1			1	(2) dt +		2	(–2) dt
   	2			0			1

   
   

   =	1	[(2t)10 + (–2t)21]
   	2

   
   

   =	1	[2 – 4 + 2] = 0
   	2

   Correct Option: D

   Average value = D.C. component of wave
   

   =	1			f(t) dt =	1		2	f(t) dt
   	T		T		2		0

   
   
   

   =	1			1	(2) dt +		2	(–2) dt
   	2			0			1

   
   

   =	1	[(2t)10 + (–2t)21]
   	2

   
   

   =	1	[2 – 4 + 2] = 0
   	2

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4. The 2-port network of fig. (1) has Y-parameter .The network is excited as shown in fig. (2). If I_X = I_Y, the current I drawn from the source would be—
   
   
   

1. 1. V (Y₁₁ + Y₁₂ + Y₂₁ + Y₂₂)

   
   
   

   2. V (Y₁₁ + Y₂₂)

   
   
   

   3. V (Y₁₁)

   
   
   

   4. V (Y₁₁ + Y₂₂ + Y₁₂ – Y₂₁)

   
   
   

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1. View Hint View Answer Discuss in Forum

   From the figure (1)
   I₁ = V₁ Y₁₁ + V₂ Y₁₂ (i)
   I₂ = V₁ Y₂₁ + V₂ Y₂₂ (ii)
   As from the figure (2) it is clear that terminal 2 and terminal 1 are at the same potential i.e.
   V₁ = V₂ = V
   and I = I₁ + I₂
   or I = V (Y₁₁ + Y₁₂) + V (Y₂₁ + Y₂₂)
   or I = V (Y₁₁ + Y₁₂ + Y₂₁ + Y₂₂)

   Correct Option: A

   From the figure (1)
   I₁ = V₁ Y₁₁ + V₂ Y₁₂ (i)
   I₂ = V₁ Y₂₁ + V₂ Y₂₂ (ii)
   As from the figure (2) it is clear that terminal 2 and terminal 1 are at the same potential i.e.
   V₁ = V₂ = V
   and I = I₁ + I₂
   or I = V (Y₁₁ + Y₁₂) + V (Y₂₁ + Y₂₂)
   or I = V (Y₁₁ + Y₁₂ + Y₂₁ + Y₂₂)

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5. The time constant of the network shown in the figure is—
   
   
   

1. 1. 2 RC

   
   
   

   2. RC

   
   
   

   3. 	CR
      	4

   
   
   

   4. 	CR
      	2

      
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   τ = R _eq. C _eq C_eq = C + C = 2C
   

   so, time constant Req =	2R . 2R	= R
   	2R + 2R

   
   τ = R. 2C = 2RC

   Correct Option: A

   τ = R _eq. C _eq C_eq = C + C = 2C
   

   so, time constant Req =	2R . 2R	= R
   	2R + 2R

   
   τ = R. 2C = 2RC

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