Linear Integrated Circuit MCQ [Free PDF] – Objective Question Answer for Linear Integrated Circuit Quiz

61. RT1, RT2, RT3, and RT4 are unstrained gage resistance. If the resistance change in each gage is 0.3Ω. Choose the correct option?

amplifier 3 q8

1. RT1 and RT3 increase by 0.3Ω
2. RT1 and RT3 increase by 0.3Ω
3. RT2 and RT4 decrease by 0.3Ω
4. All of the above

Answer: A

When external weight is applied to the platform. One pair of elements elongates and the other pair compresses. Therefore, RT1 and RT3 are △R and the increase in RT2 and RT4 is also by △R.


62. Which factor is responsible for the gain of the op-amp to roll off after a certain frequency is reached

A. Capacitive effect
B. Resistive effect
C. Inductor effect
D. None of the mentioned

Answer: A

The gain of the op-amp roll-off is due to the presence of a capacitive component, in the equivalent circuit of the op-amp. The frequency increases as the reactance of the component decrease.


63. What remedy can be followed to maintain the operating frequency in the op-amps?

A. Use LC circuit
B. Use resistor
C. Use capacitor
D. Use transistor

Answer: C

Capacitors are used in the high-frequency model of the op-amp at the output terminal to maintain or control the frequency.


64. How does the physical characteristic of semiconductors account for the increase in the frequency of op-amps?

A. Transistor values
B. Junction capacitance
C. Dopant concentration
D. None of the mentioned

Answer: B

Op-amps are composed of BJTs and FETs which contain junction capacitors. These junction capacitors are very small and take finite values at a higher frequency. So, when the reactances of these capacitors decrease, the frequency increases.


65. How are op-amp with three break frequencies are represented

A. Using two capacitors
B. Using three capacitors
C. Using one capacitor
D. All of the mentioned

Answer: B

Op-amps with more than one break frequency may be represented by using as many capacitors as the break frequency they have.


65. In what way does the internal construction of the op-amp contributes to the capacitive effect in op-amp?

A. Formation of junction capacitor
B. Due to internally connected capacitors
C. Formation of stray capacitor
D. None of the mentioned

Answer: C

In op-amps, a number of transistors, resistors, and capacitors are integrated on the same substrate. Substrates act as an insulator and separate these components. So, the various components are connected by conducting paths. However, when two conducting paths are separated by an insulator, it acts as a capacitor (stray capacitor).


66. The gain of the differential amplifier is -125. Assume the voltage applied to bridge circuit Vdc=+10v and the unstrained resistance of four elements of the strain gage is 200Ω. When a certain weight is placed on the platform, the output voltage is Vo=5v. Determine the change in resistance of each strain gauge for the analog weight scale. (Assuming the output is initially nulleD.

A. 1Ω
B. 0.8Ω
C. 0.3Ω
D. 1.83Ω

Answer: B

The output voltage of the analog weight scale

Vo =  Vdc×(△R/R)×(RF/R1)

= △R=( Vo/ Vdc)×(R1/RF)×R

=( 5v×200Ω)/(125×10v) =0.8Ω.


67. Assume that the increase and decrease in the resistance of the strain gage element are by the same number of △R. Determine the unbalanced voltage equation.

A. V = – Vdc×(△R/R)
B. V = – Vdc×(△R/R+△R)
C. V = Vdc×(△R/R)
D. V = – Vdc×[(△R+ R)/R].

Answer: A

When the unstrained gage resistance is the same, then the output voltage of the strain gage bridge circuit (unbalanced voltage) is given as V= -Vdc×(△R/R).


68. Voltage to current converter is also called as

A. Current series positive feedback amplifier
B. Voltage series negative feedback amplifier
C. Current series negative feedback amplifier
D. Voltage series positive feedback amplifier

Answer: C

Voltage to current converter is also called a current series negative feedback amplifier because the feedback voltage across the internal resistor applied to the inverting terminal depends on the output current and is in series with the input difference voltage.


69. Given voltage to current converter with the floating load. Determine the output current?

Given voltage to current converter with floating load. Determine the output current?

A. 3mA
B. 6mA
C. 4mA
D. 2mA

Answer: D

Output current, Io = Vin /R1 = 10/5kΩ =2mA.


70. Which of the following application uses voltage for the current converter?

A. Low voltage dc and ac voltmeter
B. Diode match finding
C. Light-emitting diode
D. All of the mentioned

Answer: D

In all the applications mentioned above, the input voltage Vin is converted into an output current of Vin/R1, or the input voltage appears across the resistor.


71. The op-amp in low voltage DC voltmeter cannot be nullified due to

A. D’Arsonaval meter movement
B. Offset voltage compensating network
C. Selection of switch
D. Gain of amplifier

Answer: A

The op-amp sometimes cannot be nullified because the output is very sensitive to even slight variation in wiper position of D’Arsonaval meter movement (ammeter with a full-scale deflection of 1mA.


72. What is the maximum input voltage that has to be selected to calibrate a dc voltmeter with a full-scale voltage range of 1-13v.

A. ≤ ±14v
B. ≥ ±13v
C. ≤ ±15v
D. = ±14v

Answer: A

The maximum input voltage has to be ≤ ±14v, to obtain the maximum full-scale input voltage of 13v.


73. Higher input voltage can be measured in low voltage DC voltmeter using

A. Smaller resistance value
B. Higher resistance value
C. Random resistance value
D. All of the mentioned

Answer: B

Higher resistance values are required to measure relatively higher input voltage. For example, if the range of switch is at x10 position in the low voltage dc voltmeter then, the corresponding resistance value would be 10kΩ. So, it requires a 10v input to get a full-scale deflection (if 1v causes a full-scale deflection in the ammeter with a full-scale deflection of 1mA.


74. In the diagram given below, determine the deflection of the ammeter with a full-scale deflection of 1mA when the switch is at X2kΩ. Consider resistance of the offset voltage compensating network to be 10Ω.

deflection of the ammeter with a full scale deflection of 1mA when the


A. Full-scale deflection in the ammeter
B. Half scale deflection in the ammeter
C. Quarter scale deflection in the ammeter
D. No deflection occurs in the ammeter

Answer: B



R1=10+2kΩ ≅2kΩ

Io = Vin/R1= 1v/2kΩ =0.5mA.

This means that 2v causes half-scale deflection of the ammeter.


75. How to modify a low voltage DC voltmeter to a low voltage ac voltmeter

A. Add a full-wave rectifier in the feedback loop
B. Add a half-wave rectifier in the feedback loop
B. Add a square wave rectifier in the feedback loop
B. Add a sine wave rectifier in the feedback loop

Answer: A

A combination of an ammeter and a full-wave rectifier can be employed in the feedback loop to form an ac voltmeter.


76. Determine the full-scale range for the input voltage if the resistance in series with meters is 1kΩ, 2kΩ, 47kΩ, and full-scale meter movement is 1mA in a low voltage AC voltmeter?

A. 1.0 to 7.48 Vrms
B. 1.1 to 7.48 Vrms
C. 1.2 to 7.48 Vrms
D. 1.3 to 7.48 Vrms

Answer: B

The minimum and maximum values of resistors are 1kΩ and 6.8kΩ. So, the range for the input voltages are

Vin(rms)|min = 1.1×R1/Io = (1.1×1kΩ)/1mA =1.1v.

Vin(rms)|max =1.1×R1/ Io =(1.1×6.8kΩ)/1mA = 7.48v.

Thus, the full-scale input voltage ranges from 1.1 to 7.48 Vrms.


77. Determine the current through the diode, when the switch is in positions 1, 2& 3. Assuming op-amps initially nulled.

Determine the current through the diode, when the switch is in position 1, 2& 3. Assuming op-amps initially nulled.

A. Io (LED. =4.01mA; Io (Zener) =4.01mA; Io (rectifier) =8.33mA
B. Io (LED. =25mA; Io (Zener) =4.01mA; Io (rectifier) =4.01mA
C. Io (LED. =16.67mA; Io (Zener) =16.66mA; Io (rectifier) =4.01mA
D. Io (LED. =8.33mA; Io (Zener) =8.33mA; Io (rectifier) =8.33mA

Answer: D

All the diodes are connected one after another in the feedback path. Therefore, the current through the diode remains the same.

Io =Vin/R1 =1.5/180 =8.33mA.


78. A diode match finder circuit has an input voltage of 2.6v and an output voltage is 5.78v. Calculate the voltage drop across diode 1N4735

A. 2.22v
B. 8.38v
C. 3.18v
D. 15.02v

Answer: C

The output voltage Vo =Vin+ VD.

∴ the voltage drop across 1N4735, VD =Vo – Vin = 5.78-2.6 =3.18v.


79. Find the voltage drop across the zener diode in the zener diode tester from the given specifications: IZk=1mA, VZ =6.2v, input voltage= 1.2v, output voltage =3.2v and resistance in series with meter =150Ω.

A. 6.2mA
B. 8mA
C. 21.33mA
D. Cannot be determined

Answer: A

Current through the zener

Io=Vin/R1 =1.2v/150Ω =8mA.

Since Io > IZk the voltage across the zero will be approximately equal to 6.2v. As the current is larger than the knee current (IZk) of the Zener, it blocks VZ volts.


80. Which among the following is preferred to display devices in digital applications?

A. Matched Zener diode
B. Matched LEDs
C. Matched rectifier diode
D. All of the mentioned

Answer: B

Matched LEDs with equal brightness at a specific value of current are useful as indicators and display devices in digital applications.

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