Operational Amplifier Application MCQ [Free PDF] - Objective Question Answer for Operational Amplifier Application Quiz

21. An amplifier in which the output voltage is equal to an average input voltage?

A. Summing amplifier
B. Weighting amplifier
C. Scaling amplifier
D. Averaging amplifier

Answer: D

An averaging amplifier can be used as an averaging circuit, in which the output voltage is equal to the average of all the input voltages.

22. Find out the gain value by which each input of the averaging amplifier is amplified? ( Assume there are four inputs)

A. 0.5
B. 0.25
C. 1
D. 2

Answer: B

In an averaging amplifier, the gain by which each input is amplified must be equal to the lower number of inputs.

= > R_F /R = 1/n , where n = number of inputs

∴ R_F /R = 1/4 = 0.25 (Four inputs)

So, each input in the averaging amplifier must be amplified by 0.25.

23. 3v, 5v, and 7v are the three input voltages applied to the inverting input terminal of the averaging amplifier. Determine the output voltage?

A. -5v
B. -10v
C. -15v
D. -20v

Answer: A

The output voltage,

V_o = -[(V_a+V_b+V_c)/3]

= -[(3+5+7)/3] = -5v.

24. The following circuit represents an inverting scaling amplifier. Compute the value of R_oM and V_O?

A. V_O = -0.985v ; R_oM = 111.11Ω
B. V_O = -2.567v ; R_oM = 447.89Ω
C. V_O = -1.569v ; R_oM = 212.33Ω
D. V_O = -1.036v ; R_oM = 320.56Ω

Answer: D

V_O = – {[(R_F/R_a)×V_a]+[(R_F/R_b)×V_b] + [(R_F/R_c)×V_c]}

= – {[(10kΩ/1kΩ)×3.3mv] + [(10kΩ/1.25kΩ)×5mv] +

[(10kΩ/820Ω)×7.9mv]} = -1.036v.

R_oM = [R_a||R_b||R_c||R_F]

= [(R_a×R_b)/(R_a+ R_b)] || [(R_c×R_F)/( R_c+ R_F)] =

[(1kΩ×1.25kΩ)/(1kΩ+1.25kΩ)] || [(820Ω×10kΩ)/(820Ω+10kΩ)] =

555.55||757.85

= [(555.55 ×757.85)/(555.55+757.85)] = 320.56Ω.

25. Which type of amplifier has output voltage equal to the average of all input voltages?

A. Inverting averaging amplifier
B. Non-inverting averaging amplifier
C. Non-inverting summing amplifier
D. Inverting scaling amplifier

Answer: B

In a non-inverting averaging amplifier, the non-inverting input voltage is the average of all inputs, with a positive sign.

26. Expression for output voltage of non-inverting summing amplifier with five input voltage?

A. V_o = 5×( V_a + V_b+ V_c+ V_d+ V_e)
B. V_o = [1+( R_f/R₁)]× ( V_a + V_b+ V_c+ V_d+ V_e)
C. V_o = V_a + V_b+ V_c+ V_d+ V_e
D. V_o = ( V_a + V_b+ V_c+ V_d+ V_e) /5

Answer: C

The output voltage of the non-inverting summing amplifier is (1+ ( R_f / R₁ )) times the average of all input voltages in the circuit.

Since there are five input voltages = > (1+ ( R_f / R₁ )) = 5

Therefore, V_o = 5 × ( V_a + V_b+ V_c+ V_d+ V_e) /5

= > V_o = (V_a + V_b+ V_c+ V_d+ V_e).

27. Find the value of V₁ in the circuit shown below?

A. 4v
B. 2v
C. 3v
D. None of the mentioned

Answer: D

Using the superposition theorem the voltage V₁ at non-inverting terminal is

V₁ = V_a/4 + V_b/4+ V_c/4+ V_d/4

= [V_a + V_b+ V_c+ V_d] /4

= [4+(-3v)+6v+(-1v) ] /4 = 1.5v.

28. If the gain of a non-inverting averaging amplifier is one, determine the input voltages if the output voltage, if the output voltage is 3v?

A. V₁ = 6v ,V₂ = 3v and V₃ = 2v
B. V₁ = 9v ,V₂ = 5v and V₃ = -4v
C. V₁ = 8v ,V₂ = -6v and V₃ = 1v
D. V₁ = 7v ,V₂ = 4v and V₃ = -3v

Answer: D

As the output voltage = Average of all input voltage = sum of input voltage /3

∴ sum of input voltage = 3×3 = 9.

From the given option, the combination of input voltage 7v, 4v, and -3v gives the value 9v.

29. In the circuit shown, supply voltage = ±15v, V_a = +3v , V_b = -4v , V_c = +5v, R = R₁ = 1kΩ and R_F = 2kΩ. 741 op-amp has A = 2×10⁵ and R₁ = 10kΩ. Determine the output voltage internal resistance of the circuit?

A. V_o ≅3v , R_iF = 6.67MΩ
B. V_o ≅3v , R_iF = 7MΩ
C. V_o ≅3v , R_iF = 9.2MΩ
D. V_o ≅3v , R_iF = 3.5MΩ

Answer: A

The output voltage

V_o = [1 + (R_F/R₁)] × [ (V_a+V_b+V_c/3)]

= [1+(2kΩ/1kΩ)] ×[(3-4+5)/3] = 2.67 ≅ 3v.

Internal resistance of circuit, R_iF = R _i [A×R₁/ (R₁+ R_F)] =

100Ω×[(200000×1kΩ)/(1kΩ+2kΩ)]

= > R_iF = 6.67 MΩ.

30. Find the type of amplifier that cannot be constructed in differential configuration?

A. Summing amplifier
B. Scaling amplifier
C. Averaging amplifier
D. Subtractor

Answer: C

In differential op-amp configuration, an amplifier produces the sum or difference between two input terminals of the op-amp. So, averaging is not possible in this type of configuration.

31. Calculate the output voltage, when a voltage of 12mv is applied to the non-inverting terminal and 7mv is applied to inverting terminal of a subtractor.

A. 19mv
B. 5mv
C. 1.7mv
D. 8.4mv

Answer: B

Output voltage of a subtractor

V_o = V_{non-inverting terminal} – V_{inverting terminal}

= 12mv-7mv = 5mv.

32. How many additional sources are connected to each input terminal to obtain an eight-input summing amplifier?

A. Six
B. Three
C. Four
D. Eight

Answer: B

An eight-input summing amplifier can be constructed using a basic differential amplifier if six additional input sources are used by connecting three input sources to inverting and non-inverting input terminals through resistors.

33. Calculate the output voltage for the summing amplifier given below, where R = 2kΩ and R_L = 10kΩ.

A. 4v
B. 18v
C. 8v
D. None of the mentioned

Answer: A

The output voltage for summing amplifier is given

V_o = -V_a -V_b +V_c +V_d

= 3-4+6+5 = 4v.

34. The output voltage of a summing amplifier is equal to (assume the sum of input voltage as V_n )

A. V (non-inverting terminal)+ V_n (inverting terminal)
B. V_n (non-inverting terminal)+ (-V_n (inverting terminal)
C. -V_n (non-inverting terminal)+ (-V_n (inverting terminal)
D. -V_n (non-inverting terminal)+ V_n (inverting terminal)

Answer: B

The output voltage of the summing amplifier is equal to the sum of the input voltage applied to the non-inverting terminal plus the negative sum of the input voltage applied to the inverting terminal.

35. Strain gauge is an example of which device?

A. Transducer
B. Voltage follower
C. Integrator
D. Differentiator

Answer: A

A strain gauge is a device when subjected to pressure or force undergoes a change in its resistance.

36. An instrumentation system does not include

A. Transducer
B. Instrumentation amplifier
C. Automatic process controller
D. Tester

Answer: D

Except for the tester, the remaining blocks form the input, intermediate, and output stages of the instrumentation system.

37. Transmission lines are used for

A. Output signal
B. Input signal
C. Signal transfer
D. All of the mentioned

Answer: C

Transmission lines are the connecting line between the blocks and permit signal transfer from unit to unit.

37. The length of the transmission lines are

A. Longer than 10 meters
B. Shorter than 10 meters
C. Equals to 10 meters
D. None of the mentioned

Answer: D

The length of the transmission lines depends primarily on the physical quantities being monitored and on system requirements.

38. Why output of the transducer is not directly connected to the indicator or display?

A. Low-level output is produced
B. High-level output is produced
C. No output is produced
D. Input is fed directly

Answer: A

Many transducers do not produce output with sufficient strength to permit their users directly. Therefore, the low-level output signal of the transducer needs to be amplified.

39. What are the features of an instrumentation amplifier?

A. Low noise
B. High gain accuracy
C. Low thermal and time drift
D. All of the mentioned

Answer: D

The features of an instrumentation amplifier are

Low noise
High gain accuracy
Low thermal and time drift

Instrumentation amplifiers are intended for precise low-level signal amplification because of the features mentioned.

40. What is the disadvantage of using LH0036 instrumentation op-amp?

A. Extremely stable
B. Relatively expensive
C. Accurate
D. All of the mentioned

Answer:b

LH0036 is a very precise special-purpose circuit in which most electrical parameters are minimized and performance is optimized. So, it is relatively expensive. [/bg_collapse]

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