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

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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.
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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.

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

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.

Sanfoundry Global Education & Learning Series – Electric Circuits.

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