Inverting & Non-Inverting Amplifier Circuit Questions and Answers

This set of Electric Circuits Multiple Choice Questions & Answers (MCQs) focuses on “The Inverting and Non-Inverting Amplifier Circuit”.

1. The opamp in the Inverting circuit is in __________
a) Linear region
b) Saturation
c) Cut-off region
d) Non-linear region
View Answer

Answer: a
Explanation: We assume that the opamp is in linear region.

2. In an Inverting Amplifier circuit, the output voltage v_o is expressed as a function of ____________
a) Input current
b) Output current
c) Source voltage
d) Source current
View Answer

Answer: c
Explanation: The goal of an inverting circuit is to express output voltage v_o as a function of source voltage vs.

3. The other name for Gain is ____________
a) Scaling factor
b) Output
c) Amplifying factor
d) Scaling level
View Answer

Answer: a
Explanation: The gain is also known as scaling factor and it is the ratio of R_f/R_s in case of an Inverting amplifying circuit.

4. If V_CC = 12V and vs=1mV, then R_f/R_s is _____________
a) >12000
b) <12000
c) 12000
d) 1
View Answer

Answer: b
Explanation: R_f/R_s ≤ │V_CC/vs│.

5. In the expression v_o= -Av_n, A is called ______________
a) Closed loop gain
b) Closed loop fault
c) Open loop fault
d) Open loop gain
View Answer

Answer: d
Explanation: A is called open loop gain.

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6. The circuits of an inverting and Non-Inverting amplifying comprises of __________ and _______ number of resistors.
a) 3, 2
b) 2, 3
c) 2, 2
d) 3, 3
View Answer

Answer: b
Explanation: Inverting amplifying circuit- R_s, R_f.
Non-Inverting amplifying circuit – R_s, R_f, R_g.

7. The condition for a Non-inverting amplifying circuit to operate in linear region operation _____________
a) (R_s+R_f)/R_s < │V_CC/v_g│
b) (R_s+R_f)/R_s ≠ │V_CC/v_g│
c) (R_s+R_f)/R_s > │V_CC/v_g│
d) (R_s+R_f)/R_s = │V_CC/v_g│
View Answer

Answer: a
Explanation: Assume that opamp is ideal. The condition for the linear region operation in a Non-inverting amplifying circuit is (R_s+R_f)/R_s <│VCC/vg│.

8. If R_s= 3Ω, Rf= 6Ω then the relation between v_o and v_g in case of a Non-Inverting amplifying circuit.
a) v_o= 9v_g
b) v_o= 6v_g
c) v_o= 3v_g
d) v_o= v_g
View Answer

Answer: c
Explanation: v_o= ((R_s+R_f)/R_s) *v_g.

9. If R_s= 5Ω, R_f= 25Ω and -2.5V ≤ v_g ≤ 2.5V. What are the smallest power supply voltages that could be applied and still have opamp in linear region?
a) ±9V
b) ±2.5V
c) ±6V
d) ±15V
View Answer

Answer: d
Explanation: vo= ((R_s+R_f)/R_s) *v_g. By substituting the values, we have v_o=6v_g.
v_o=6(-2.5) = -15
v_o=6(2.5) =15.

10. If an inverting amplifying circuit has a gain of 10 and ±15V power supplies are used. The values of input for which opamp would be in the linear region?
a) ±1.25
b) ±1.5V
c) ±2.25
d) ±0.5
View Answer

Answer: b
Explanation: Gain= R_f/R_s= 10 and v_o= (-R_f/R_s)*v_s.
→ v_o= -10v_s and given -12V≤ v_o ≤ 12V.
→ -15= -10v_s. So, v_s= 1.5V
→ 15=-10v_s. So, v_s=-1.5V.

11. If the gain of an inverting amplifying circuit is 13 and ±22V power supplies are used. What range of input values allows the opamp to be in linear region?
a) ±1.69
b) ±1.35V
c) ±2.28
d) ±0.5
View Answer

Answer: a
Explanation: Gain= Rf/Rs= 13 and v_o= (-R_f/R_s)*v_s.
→ v_o= -13v_s and given -22V≤ v_o ≤ 22V.
→ -22= -13v_s. So, v_s=1.692 V
→ 22=-13v_s. So, v_s=-1.692V.

12. The input applied to an Inverting amplifier is ______________
a) Equal to output
b) Equal to Inverted output
c) Not equal to output
d) Output is equal to input
View Answer

Answer: b
Explanation: The name itself indicates it is an Inverting amplifier. So, the input applied is inverted and is given as output. Suppose the input applied is sinusoidal then, the output is

Inverting amplifier is inverted & is given as output in the given graph

13. In R₁=10kΩ, R_f=100kΩ, v₁=1V. A load of 25kΩ is connected to the output terminal. Calculate i₁ and v_o.

a) 0.5mA, 10V
b) 0.1mA, 10V
c) 0.1mA, -10V
d) 0.5mA, -10V
View Answer