226. The Y-parameters of a 2-port network are:
     

     [Y] =		5		3		S
     		1		2

     
     A resistor of I ohm is connected across as shown below. The new Y-parameter would be—
     
     
     

1. 1. 		6		4		S
      		2		3

   
   
   

   2. 		6		2		S
      		0		3

   
   
   

   3. 		5		4		S
      		2		2

   
   
   

   4. 		4		4		S
      		2		1

   
   
   

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1. View Hint View Answer Discuss in Forum

   Y-parameter of 1 Ω resistor network are
   

   = det		1		–1
   		–1		1

   
   
   

   New Y-parameter =		5	3		+		1	–1
   		1	2				–1	1

   
   

   =		6		2
   		0		3

   
   Hence alternative (B) is the correct choice.

   Correct Option: B

   Y-parameter of 1 Ω resistor network are
   

   = det		1		–1
   		–1		1

   
   
   

   New Y-parameter =		5	3		+		1	–1
   		1	2				–1	1

   
   

   =		6		2
   		0		3

   
   Hence alternative (B) is the correct choice.

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227. For the 2-port as given below [Ya] =		2		0		mS
     		0		10

     
     

     The value	V0	is—
     	Vs

     
     
     

1. 1. 3

      3.2

   
   
   

   2. 1

      16

   
   
   

   3. 2

      33

   
   
   

   4. 1

      17

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given Ya =		2		0		mS.
   		0		10

   
   Its Z-parameter equivalent can be given as
   

   Z =	1		0
   	2 × 10–3		1
   	0		10 × 10–3

   
   

   =		500		0
   		0		100

   
   The equivalent Z-parameter at terminals AA′ and BB′
   

   Zeq =		500	0		+		100	100
   		0	100				100	100

   
   

   =		600		100
   		100		200

   
   or V₁ = 600 I₁ + 100 I₂ (A)
   V₂ = 100 I₁ + 200 I₂ (B)
   V_S = 60 I₁ + V₁ = 660 I₁ + 100 I₂ (C)
   V₂ = V₀ = – 300 I₂ (D)
   and V₂ = 100 I₁ + 200 I₂ (E)
   from (D) and (E)
   – 300 I₂ = 100 I₁ + 200 I₂
   or – 500 I₂ = 100 I₁
   or – 5 I₂ = I₁
   
   From equation (C)
   V_S = 660. (– 5 I₂) + 100 I₂
   or V_S = – 3300 + 100 I₂ = – 3200 I₂
   

   Now,	V0	=	–300 I2	=	3
   	Vs		–3200		32

   
   Hence alternative (A) is the correct choice.

   Correct Option: A

   Given Ya =		2		0		mS.
   		0		10

   
   Its Z-parameter equivalent can be given as
   

   Z =	1		0
   	2 × 10–3		1
   	0		10 × 10–3

   
   

   =		500		0
   		0		100

   
   The equivalent Z-parameter at terminals AA′ and BB′
   

   Zeq =		500	0		+		100	100
   		0	100				100	100

   
   

   =		600		100
   		100		200

   
   or V₁ = 600 I₁ + 100 I₂ (A)
   V₂ = 100 I₁ + 200 I₂ (B)
   V_S = 60 I₁ + V₁ = 660 I₁ + 100 I₂ (C)
   V₂ = V₀ = – 300 I₂ (D)
   and V₂ = 100 I₁ + 200 I₂ (E)
   from (D) and (E)
   – 300 I₂ = 100 I₁ + 200 I₂
   or – 500 I₂ = 100 I₁
   or – 5 I₂ = I₁
   
   From equation (C)
   V_S = 660. (– 5 I₂) + 100 I₂
   or V_S = – 3300 + 100 I₂ = – 3200 I₂
   

   Now,	V0	=	–300 I2	=	3
   	Vs		–3200		32

   
   Hence alternative (A) is the correct choice.

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228. The T-parameters of a 2-port network are
     

     [T] =		2		1		.
     		1		1

     
     If such two 2-port network are cascaded, the Z-parameter for the cascaded network is—

1. 1. 

   
   
   

   2. 

   
   
   

   3. 

   
   
   

   4. 

   
   
   

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1. View Hint View Answer Discuss in Forum

   First the cascaded T-parameters of a network
   

   Teq =		2	1				2	1
   		1	1				1	1

   
   

   =		5		3
   		3		2

   
   

   =		A´		B´
   		C´		D´

   
   the basic equation of transmission parameter
   V₁ = A′ V₂ – B′ I₂ (A)
   I₁ = C′ V₂ – D′ I₂ (B)
   or V₁ = 5 V₂ – 3 I₂ (C)
   I₁ = 3 V₂ – 2 I₂ (D)
   In order to calculate Z-parameter of this cascaded network, we should make the equations in the form of V₁ in terms of I₁ and I₂ and V₂ is terms of I₁ and I₂ i.e.
   

   V1 =	5	I1 +	1	I2 (E)
   	3		3

   
   

   V2 =	I1	+	3	I2 (F)
   	3		2

   
   on comparing equation (E) and (F) with standard equation of Z-parameters i.e.
   V₁ = Z₁₁ I₁ + Z₁₂ I₂
   V₂ = Z₂₁ I₁ + Z₂₂ I₂,
   we get
   

   Z11 =	5	; Z12 =	1
   	3		3

   
   

   Z21 =	1	; Z22 =	2
   	3		3

   
   Hence alternative (C) is the correct choice.

   Correct Option: C

   First the cascaded T-parameters of a network
   

   Teq =		2	1				2	1
   		1	1				1	1

   
   

   =		5		3
   		3		2

   
   

   =		A´		B´
   		C´		D´

   
   the basic equation of transmission parameter
   V₁ = A′ V₂ – B′ I₂ (A)
   I₁ = C′ V₂ – D′ I₂ (B)
   or V₁ = 5 V₂ – 3 I₂ (C)
   I₁ = 3 V₂ – 2 I₂ (D)
   In order to calculate Z-parameter of this cascaded network, we should make the equations in the form of V₁ in terms of I₁ and I₂ and V₂ is terms of I₁ and I₂ i.e.
   

   V1 =	5	I1 +	1	I2 (E)
   	3		3

   
   

   V2 =	I1	+	3	I2 (F)
   	3		2

   
   on comparing equation (E) and (F) with standard equation of Z-parameters i.e.
   V₁ = Z₁₁ I₁ + Z₁₂ I₂
   V₂ = Z₂₁ I₁ + Z₂₂ I₂,
   we get
   

   Z11 =	5	; Z12 =	1
   	3		3

   
   

   Z21 =	1	; Z22 =	2
   	3		3

   
   Hence alternative (C) is the correct choice.

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229. The equivalent circuit across a-b will be—
     
     
     

1. 1. 

   
   
   

   2. 

   
   
   

   3. 

   
   
   

   4. 

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   The above circuit can be redrawn as
   

   Correct Option: A

   
   The above circuit can be redrawn as
   

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230. For the circuit shown below
     
     
     

1. 1. Power absorbed by the dependent source is 80W

   
   
   

   2. Power delivered by the dependent source is 80W

   
   
   

   3. Power absorbed by the Independent source is 80W

   
   
   

   4. Power delivered by the Independent source is 80W

   
   
   

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1. View Hint View Answer Discuss in Forum

   The given circuit
   
   Let the current supplied by the Independent voltage source is I.
   KVL in the loop
   

   20– 5 I– 5		I +	V1		= 0
   			5

   
   

   or 20 – 10 I – 5 ×	V1	= 0
   	5

   
   

   or I =	20 – V1
   	10

   
   

   or I =	20 – 20	= 0 (since, V1 = 20V given)
   	10

   
   Thus there is no power delivered or absorbed by independent source.
   Now, power delivered by the dependent source
   

   =		V1		2	5 watt
   		5

   
   

   =		20		2	× 5 = 16 × 5 = 80 W
   		5

   Correct Option: B

   The given circuit
   
   Let the current supplied by the Independent voltage source is I.
   KVL in the loop
   

   20– 5 I– 5		I +	V1		= 0
   			5

   
   

   or 20 – 10 I – 5 ×	V1	= 0
   	5

   
   

   or I =	20 – V1
   	10

   
   

   or I =	20 – 20	= 0 (since, V1 = 20V given)
   	10

   
   Thus there is no power delivered or absorbed by independent source.
   Now, power delivered by the dependent source
   

   =		V1		2	5 watt
   		5

   
   

   =		20		2	× 5 = 16 × 5 = 80 W
   		5

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