# Electric Circuits Questions and Answers – The Inverting and Non-Inverting Amplifier Circuit

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

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

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

Explanation: The goal of an inverting circuit is to express output voltage vo 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

Explanation: The gain is also known as scaling factor and it is the ratio of Rf/Rs in case of an Inverting amplifying circuit.

4. If VCC = 12V and vs=1mV, then Rf/Rs is _____________
a) >12000
b) <12000
c) 12000
d) 1

Explanation: Rf/Rs ≤ │VCC/vs│.

5. In the expression vo= -Avn, A is called ______________
a) Closed loop gain
b) Closed loop fault
c) Open loop fault
d) Open loop gain

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

Explanation: Inverting amplifying circuit- Rs, Rf.
Non-Inverting amplifying circuit – Rs, Rf, Rg.

7. The condition for a Non-inverting amplifying circuit to operate in linear region operation _____________
a) (Rs+Rf)/Rs < │VCC/vg
b) (Rs+Rf)/Rs ≠ │VCC/vg
c) (Rs+Rf)/Rs > │VCC/vg
d) (Rs+Rf)/Rs = │VCC/vg

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

8. If Rs= 3Ω, Rf= 6Ω then the relation between vo and vg in case of a Non-Inverting amplifying circuit.
a) vo= 9vg
b) vo= 6vg
c) vo= 3vg
d) vo= vg

Explanation: vo= ((Rs+Rf)/Rs) *vg.

9. If Rs= 5Ω, Rf= 25Ω and -2.5V ≤ vg ≤ 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

Explanation: vo= ((Rs+Rf)/Rs) *vg. By substituting the values, we have vo=6vg.
vo=6(-2.5) = -15
vo=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

Explanation: Gain= Rf/Rs= 10 and vo= (-Rf/Rs)*vs.
→ vo= -10vs and given -12V≤ vo ≤ 12V.
→ -15= -10vs. So, vs= 1.5V
→ 15=-10vs. So, vs=-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

Explanation: Gain= Rf/Rs= 13 and vo= (-Rf/Rs)*vs.
→ vo= -13vs and given -22V≤ vo ≤ 22V.
→ -22= -13vs. So, vs=1.692 V
→ 22=-13vs. So, vs=-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

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

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

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

Explanation: i1= v1/R1 = 1V/10kΩ = 0.1mA
V0= -(Rf/R1)*v1 = -(100kΩ/10kΩ)*1V = -10V.

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