301. For the circuit shown below gain is A_ν = ν₀/ν_i = – 10. The value of R is:
     
     
     

1. 1. 600 kΩ

   
   
   

   2. 450 kΩ

   
   
   

   3. 4.5 MΩ

   
   
   

   4. 6 MΩ

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given

   V0	= – 10
   Vi

   
   KCL at node A
   

   VA – Vi	+	VA – VB	= 0 ...........(1)
   100		R

   
   KCL at node B
   

   VB – VA	+	VB	+	VB – V0	= 0 ...........(2)
   R		100		100

   
   on manipulating
   V_A = 0. (due to virtual ground)
   V₀ = – 10 V_i
   in equation (1) and (2) we get
   

   	0 – Vi	+	0 – VB	= 0
   	100		R

   
   

   or Vi = – VB.	100	.............(3)
   	R

   
   

   VB – 0	+	VB	+	VB	=	V0
   R		100		100		100

   
   

   or VB		1	+	1	+	1		=	–10Vi
   		R		100		100			100

   
   

   or VB

   1

   +

   2

   =

   –16

   ·

   –VB

   100

   R

   100

   100

   R

   
   

   or	1	+	2	=	10
   	R		100		R

   
   

   or	10	–	1	=	2
   	R		R		100

   
   

   or	10R – R	=	2
   	R2		100

   
   

   or	9	=	2
   	R		100

   
   or R = 450 kΩ

   
   

   Correct Option: B

   Given

   V0	= – 10
   Vi

   
   KCL at node A
   

   VA – Vi	+	VA – VB	= 0 ...........(1)
   100		R

   
   KCL at node B
   

   VB – VA	+	VB	+	VB – V0	= 0 ...........(2)
   R		100		100

   
   on manipulating
   V_A = 0. (due to virtual ground)
   V₀ = – 10 V_i
   in equation (1) and (2) we get
   

   	0 – Vi	+	0 – VB	= 0
   	100		R

   
   

   or Vi = – VB.	100	.............(3)
   	R

   
   

   VB – 0	+	VB	+	VB	=	V0
   R		100		100		100

   
   

   or VB		1	+	1	+	1		=	–10Vi
   		R		100		100			100

   
   

   or VB

   1

   +

   2

   =

   –16

   ·

   –VB

   100

   R

   100

   100

   R

   
   

   or	1	+	2	=	10
   	R		100		R

   
   

   or	10	–	1	=	2
   	R		R		100

   
   

   or	10R – R	=	2
   	R2		100

   
   

   or	9	=	2
   	R		100

   
   or R = 450 kΩ

   
   

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302. In circuit shown below, the input voltage νi is 0.2 V. The output voltage ν0 is:
     
     
     

1. 1. 6 V

   
   
   

   2. – 6 V

   
   
   

   3. 8 V

   
   
   

   4. – 8 V

   
   
   

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1. View Hint View Answer Discuss in Forum

   From figure.
   

   VA = –	50	. Vi
   	10

   
   or V_A = – 5 V_i
   

   and V0 = –	150	(VA)
   	25

   
   V₀ = – 6. (– 5 V_i)
   V₀ = 30 V_i
   V₀ = 30 × 0.2
   V₀ = 6 V.

   
   

   Correct Option: A

   From figure.
   

   VA = –	50	. Vi
   	10

   
   or V_A = – 5 V_i
   

   and V0 = –	150	(VA)
   	25

   
   V₀ = – 6. (– 5 V_i)
   V₀ = 30 V_i
   V₀ = 30 × 0.2
   V₀ = 6 V.

   
   

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303. A circuit using an op-amp is shown below. It has:
     
     
     

1. 1. voltage series feedback

   
   
   

   2. voltage shunt feedback

   
   
   

   3. current series feedback

   
   
   

   4. current shunt feedback

   
   
   

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1. View Hint View Answer Discuss in Forum

   I_i = I_f
   

   If =	–V0
   	Rf

   
   Since feedback current is proportional to the output voltage in Node. Hence circuit is voltage shunt feedback.

   
   

   Correct Option: B

   I_i = I_f
   

   If =	–V0
   	Rf

   
   Since feedback current is proportional to the output voltage in Node. Hence circuit is voltage shunt feedback.

   
   

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304. The input to the circuit shown below is ν_i = 2 sin ωt mV. The current i₀ is:
     
     
     

1. 1. – 2 sin ωt µA

   
   
   

   2. – 7 sin ωt µA

   
   
   

   3. – 5 sin ωt µA

   
   
   

   4. 0

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given that, V_i = 2 sin ωt mV
   

   V0 = –	10	. Vi → (1)
   	1

   
   KCL at node B
   

   	V0	+	V0 – VA	= i0
   	4		10

   
   V_A = 0
   (There4; due to virtual ground)
   

   or	V0	+	V0	= i0
   	4		10

   
   

   or	–10Vi	+	10Vi	= i0
   	4		10

   
   or i₀ = – 3.5 V_i
   or i₀ = – 3.5.2 sin ωt µA
   or i₀ = – 7 sin ωt µA

   
   

   Correct Option: B

   Given that, V_i = 2 sin ωt mV
   

   V0 = –	10	. Vi → (1)
   	1

   
   KCL at node B
   

   	V0	+	V0 – VA	= i0
   	4		10

   
   V_A = 0
   (There4; due to virtual ground)
   

   or	V0	+	V0	= i0
   	4		10

   
   

   or	–10Vi	+	10Vi	= i0
   	4		10

   
   or i₀ = – 3.5 V_i
   or i₀ = – 3.5.2 sin ωt µA
   or i₀ = – 7 sin ωt µA

   
   

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305. Calculate Av for the circuit given below:
     
     
     

1. 1. – 10

   
   
   

   2. 10

   
   
   

   3. 13.46

   
   
   

   4. – 13.46

   
   
   

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1. View Hint View Answer Discuss in Forum

   Since non-inverting terminal is at ground potential. Thus due to virtual ground V_A = 0 V. It means there will be no current in 60 kΩ.
   

   so, Av =	V0	= –	400	= – 10
   	Vi		40

   
   

   Correct Option: A

   Since non-inverting terminal is at ground potential. Thus due to virtual ground V_A = 0 V. It means there will be no current in 60 kΩ.
   

   so, Av =	V0	= –	400	= – 10
   	Vi		40

   
   

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