1. Assuming ideal elements in the circuit shown below, the voltage V_ab will be—
   
   
   
   

1. 1. – 3V
      

   
   
   

   2. 0V
      

   
   
   

   3. 3V
      

   
   
   

   4. 5V

   
   
   

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1. View Hint View Answer Discuss in Forum

   The given circuit:
   
   Applying KVL in the loop, we get,
   V_ab – 2 × 1 + 5 = 0
   or
   V_ab = 2 – 5
   = – 3V.

   Correct Option: A

   The given circuit:
   
   Applying KVL in the loop, we get,
   V_ab – 2 × 1 + 5 = 0
   or
   V_ab = 2 – 5
   = – 3V.

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2. In figure, the value of the source voltage is—
   
   
   
   

1. 1. 12V
      

   
   
   

   2. 24V
      

   
   
   

   3. 50V
      

   
   
   

   4. 44V

   
   
   

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1. View Hint View Answer Discuss in Forum

   The given figure
   
   Applying KCL at node C, we get
   

   =	VC - 0	= 1 + 2
   	10

   
   or
   V_C = 30V
   and
   

   =	Vs - VC	= 2
   	10

   
   or
   V_S = 20 + V_c = 20 + 30
   = 50V
   Hence alternative (C) is the correct choice.

   Correct Option: C

   The given figure
   
   Applying KCL at node C, we get
   

   =	VC - 0	= 1 + 2
   	10

   
   or
   V_C = 30V
   and
   

   =	Vs - VC	= 2
   	10

   
   or
   V_S = 20 + V_c = 20 + 30
   = 50V
   Hence alternative (C) is the correct choice.

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3. In figure, the value of R is—
   
   
   
   

1. 1. 10Ω
      

   
   
   

   2. 18Ω
      

   
   
   

   3. 24Ω
      

   
   
   

   4. 12Ω

   
   
   

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1. View Hint View Answer Discuss in Forum

   The given circuit:
   
   Applying KCL at node C,
   

   =	VC - VA	+	VC - VB	+	VC	= 0
   	14		1		2

   
   or
   

   Vc		1	+	1	+	1		=	VA	+	VB	=	100	+	40
   		14		1		2			14		1		14		1

   
   or
   

   Vc		1 + 14 + 7		=	100 + 40 × 14
   		14			14

   
   or
   VC = 30V
   Now, current in 14Ω resistor,
   

   I14Ω =	100 - VC	=	100 - 30
   	14		14

   
   = 5A
   So,
   I_R = 10 – 5 = 5A
   and
   

   R =	VA - VB	=	100 - 40
   	IR		5

   
   or
   R = 60/5 ≈ 12Ω
   Hence alternative (D) is most correct choice.

   Correct Option: A

   The given circuit:
   
   Applying KCL at node C,
   

   =	VC - VA	+	VC - VB	+	VC	= 0
   	14		1		2

   
   or
   

   Vc		1	+	1	+	1		=	VA	+	VB	=	100	+	40
   		14		1		2			14		1		14		1

   
   or
   

   Vc		1 + 14 + 7		=	100 + 40 × 14
   		14			14

   
   or
   VC = 30V
   Now, current in 14Ω resistor,
   

   I14Ω =	100 - VC	=	100 - 30
   	14		14

   
   = 5A
   So,
   I_R = 10 – 5 = 5A
   and
   

   R =	VA - VB	=	100 - 40
   	IR		5

   
   or
   R = 60/5 ≈ 12Ω
   Hence alternative (D) is most correct choice.

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4. The voltage V for the network shown below is always equal to—
   
   
   
   

1. 1. 9V
      

   
   
   

   2. 5V
      

   
   
   

   3. 1V
      

   
   
   

   4. None of these

   
   
   

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1. View Hint View Answer Discuss in Forum

   The given circuit:
   
   From above figure
   V – 4 – 5 = 0
   or
   V = 9 V
   Hence alternative (A) is the correct choice.

   Correct Option: A

   The given circuit:
   
   From above figure
   V – 4 – 5 = 0
   or
   V = 9 V
   Hence alternative (A) is the correct choice.

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5. In the circuit shown in the figure, the value of the current i will be given by—
   
   
   
   

1. 1. 0·5A
      

   
   
   

   2. 1·25A
      

   
   
   

   3. 1·75A
      

   
   
   

   4. 2·5A

   
   
   

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1. View Hint View Answer Discuss in Forum

   The given circuit:
   
   

   Va = 5 ×	1	=	5	V
   	1 + 1		2

   
   

   Vb = 4Vab ×	1	=	Vab
   	1 + 3		4

   
   V_ab = V_a – V_b = 5/2 – V_ab
   or
   2V_ab = 5/2
   or
   V_ab = 5/4 V
   or
   V_ab = 1.25 V
   and
   

   i =	4 × Vab	=	4 × 1.25	= 1.25A
   	4		4

   
   Hence alternative (B) is the correct choice.

   Correct Option: A

   The given circuit:
   
   

   Va = 5 ×	1	=	5	V
   	1 + 1		2

   
   

   Vb = 4Vab ×	1	=	Vab
   	1 + 3		4

   
   V_ab = V_a – V_b = 5/2 – V_ab
   or
   2V_ab = 5/2
   or
   V_ab = 5/4 V
   or
   V_ab = 1.25 V
   and
   

   i =	4 × Vab	=	4 × 1.25	= 1.25A
   	4		4

   
   Hence alternative (B) is the correct choice.

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