Electronics and Experimental Methods MCQ Quiz - Objective Question with Answer for Electronics and Experimental Methods - Download Free PDF

Last updated on Apr 22, 2025

Latest Electronics and Experimental Methods MCQ Objective Questions

Electronics and Experimental Methods Question 1:

The average speed of an object can be determined from multiple measurements taken by different individuals.

To assess the accuracy and precision of the measurements, different error analyses are performed, such as:

  • Mean absolute error (the difference between the measured value and the true value)
  • Relative error (the ratio of the absolute error to the true value)
  • Percentage error (the relative error expressed as a percentage)

Given Data:

  • Measured speeds by six students: 15.5 m/s, 12.7 m/s, 17.8 m/s, 12.4 m/s, 22.2 m/s, and 19.6 m/s.

Now, determine the correct match from the following options:

List-I List-II
(P) Mean absolute error (1) ± 0.1796
(Q) Mean value (2) 3.0
(R) Relative error (3) 16.7
(S) Percentage error (4) ± 17.96 

  1. P → 1, Q → 2, R → 3, S → 4.
  2. P → 2, Q → 3, R → 1, S → 4.
  3. P → 1, Q → 3, R → 2, S → 4.
  4. P → 2, Q → 4, R → 3, S → 1.

Answer (Detailed Solution Below)

Option 2 : P → 2, Q → 3, R → 1, S → 4.

Electronics and Experimental Methods Question 1 Detailed Solution

Concept:

Mean Absolute Error, Mean Value, Relative Error, and Percentage Error:

  • The Mean Value is the average of the measurements.
  • The Mean Absolute Error is the average of the absolute differences between each measurement and the mean value.
  • The Relative Error is the mean absolute error divided by the mean value.
  • The Percentage Error is the relative error multiplied by 100.

Calculation:

Given measurements: 15.5 m/s, 12.7 m/s, 17.8 m/s, 12.4 m/s, 22.2 m/s, 19.6 m/s

1. Mean Value:

The mean value is calculated as:

Mean Value = (15.5 + 12.7 + 17.8 + 12.4 + 22.2 + 19.6) / 6 = 100.2 / 6 = 16.7 m/s

2. Mean Absolute Error:

The mean absolute error is calculated as:

Mean Absolute Error = (|15.5 - 16.7| + |12.7 - 16.7| + |17.8 - 16.7| + |12.4 - 16.7| + |22.2 - 16.7| + |19.6 - 16.7|) / 6

Mean Absolute Error = (1.2 + 4.0 + 1.1 + 4.3 + 5.5 + 2.9) / 6 = 18.0 / 6 = 3.0 m/s

3. Relative Error:

The relative error is calculated as:

Relative Error = 3.0 / 16.7 = 0.1796

4. Percentage Error:

The percentage error is calculated as:

Percentage Error = 0.1796 × 100 = 17.96 %

Summary of Results:

  • Mean Value: 16.7 m/s
  • Mean Absolute Error: 3.0 m/s
  • Relative Error: 0.1796
  • Percentage Error: 17.96 %

Electronics and Experimental Methods Question 2:

Consider the LCR circuit shown below, where the resistance is \( R = 0.05 \, \Omega \), the inductance is \( L = 1 \, \text{H} \), and the capacitance is \( C = 0.04 \, \text{F} \).

F4 Vinanti Teaching 28.11.22 D5

If the input voltage \( v_{\text{in}} \) is a square wave with an angular frequency of \( 1 \, \text{rad/s}, \) determine the frequency (rad/s)  of  waveform that best approximates the output voltage \( v_{\text{out}}. \)

Answer (Detailed Solution Below) 5

Electronics and Experimental Methods Question 2 Detailed Solution

Concept:

An LC circuit is a type of electric circuit that is made up of an inductor which is expressed by the letter L and a capacitor represented by the letter C.

Calculation:

Given, vin = 1rad/s

L = 1H, C = 0.04F

Resonant angular frequency (ωr )

ωr = \(1 \over \sqrt{LC} \)

\(1 \over \sqrt{0.04} \) = 5 rad/s

Thus, for an input frequency of 1 rad/s, the LC circuit will oscillate in a sinusoidal fashion it can only oscillate harmonically, at 5 rad/s. 

Electronics and Experimental Methods Question 3:

In the circuit shown below, the diode \( D \) exhibits a forward voltage drop of \( 0.7 \, \text{V} \) when conducting, and no current flows through it in reverse bias.

F1 Teaching Arbaz 23-10-23 D11

Given that the input voltage \( V_{\text{in}} \) is a sinusoidal signal with a frequency of 50 Hz and an RMS value of 1 V, determine the approximate maximum current flowing through the diode.

Answer (Detailed Solution Below) 0

Electronics and Experimental Methods Question 3 Detailed Solution

Explanation:

Case 1 When diode is in forward biasing

Given, \(V_{rms}=1V\)

Now, peak voltage \(V_p=\sqrt{2} V_{rms}=\sqrt{2}\times 1=1.414\)

  • Current in \(20\Omega=\frac {voltage difference} {resistance}=\frac{1.414-0.7} {20}\approx0.03A\)

This is low current

  • Current in \(10\Omega=\frac {voltage difference} {resistance}=\frac{0.7-0} {10}=0.07A\)

This is high current as comparison to \(20\Omega\) resistor in forward biasing 

So, Forward biasing is not possible in this circuit.

Case 2 When the circuit is in Reverse biasing

This circuit is possible in reverse biasing.

In Reverse biasing the current in the diode should be zero.

So, the correct answer is \(0A\)

Electronics and Experimental Methods Question 4:

A circuit with an operational amplifier is shown in the diagram below.

qImage670bcf824cd3b7bc6ce9806318-4-2025 IMG-767 -4

Determine the output voltage waveform \( V_{\text{out}} \) that corresponds most closely to the given circuit.

  1. qImage670bcf834cd3b7bc6ce9806718-4-2025 IMG-767 -5
  2. qImage670bcf844cd3b7bc6ce9806818-4-2025 IMG-767 -6
  3. qImage670bcf834cd3b7bc6ce9806418-4-2025 IMG-767 -7
  4. qImage670bcf844cd3b7bc6ce9806a18-4-2025 IMG-767 -8

Answer (Detailed Solution Below)

Option 3 : qImage670bcf834cd3b7bc6ce9806418-4-2025 IMG-767 -7

Electronics and Experimental Methods Question 4 Detailed Solution

Explanation:

The circuit is standard op amp multivibrator circuit. Let's assume that the capacitor is initially uncharged and output of the op amp is saturated at the positive saturation voltage.

qImage6711f3b1d553ae06af41dddf18-4-2025 IMG-767 -9

The capacitor, C starts to charge up from the output voltage, Vout through resistor, R. As soon as the capacitors charging voltage at the op amps inverting ( ) terminal is equal to or greater than the voltage at the non inverting terminal (the op amps output change state and be driven to the opposing negative saturation voltage. Once the op amps inverting terminal reaches the new negative reference voltage, Vref at the non inverting terminal, the op amp once again change state and the output is driven to the opposing supply voltage, +V(sat).

The correct option is (3).

Electronics and Experimental Methods Question 5:

Question:

In the collector feedback circuit shown in the figure below, the base-emitter voltage \( V_{BE} = 0.7 \, \text{V} \), and the transistor's current gain \( \beta = \frac{I_C}{I_B} = 100 \).

F2 Technical Mrunal 24.02.2023 D14

Calculate the value of the base current \( I_B \). in μA. 

Answer (Detailed Solution Below) 20

Electronics and Experimental Methods Question 5 Detailed Solution

CONCEPT:

Kirchhoff's Voltage Law: 

The algebraic sum of all voltage differences around any closed loop is zero.

In a transistor, the voltage between the base and emitter VBE = 0.7 V

EXPLANATION:

Apply K.V.L in the input section

F2 Technical Mrunal 24.02.2023 D15

\( \begin{aligned} &-20.7 \mathrm{~V}+(I_C \times 5 K\Omega)+I_B \times 500 K\Omega+0.7 \mathrm{~V}=0 \\ & -20.7 \mathrm{~V}+(\beta I_B \times 5 ×10^3)+(I_B \times 500 ×10^3)+0.7 \mathrm{~V}=0 \\ & I_B=\frac{20}{100 \times 5 ×10^3+500×10^3}=20 μ \mathrm{A} \end{aligned}\)

The correct answer is 20μA 

Top Electronics and Experimental Methods MCQ Objective Questions

The output of the following is:

F1 Engineering Mrunal 13.03.2023 D7

Assume ideal diode.

  1. F1 Engineering Mrunal 13.03.2023 D8
  2. F1 Engineering Mrunal 13.03.2023 D9
  3. F1 Engineering Mrunal 13.03.2023 D10
  4. F1 Engineering Mrunal 13.03.2023 D11

Answer (Detailed Solution Below)

Option 2 : F1 Engineering Mrunal 13.03.2023 D9

Electronics and Experimental Methods Question 6 Detailed Solution

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Concept

A diode conducts when the positive terminal of the battery is connected to the anode and the negative terminal of the battery is connected to the cathode. Under this condition, the diode is said to be in a forward-biased condition.

In forward-biased conditions, the diode is replaced by short-circuit.

qImage640971a5d09d3adcafed8066

A diode does not conduct when the positive terminal of the battery is connected to the cathode and the negative terminal of the battery is connected to the anode. Under this condition, the diode is said to be in a reverse-biased condition.

In reverse-biased conditions, the diode is replaced by an open circuit.

qImage640971a6d09d3adcafed8073

Note: When the battery of the same polarity is present on both sides of the diode, then the battery with greater voltage magnitude will decide the biasing of the diode.

Calculation

Case 1: During +ve half cycle:

Diode is forward-biased, hence the diode is replaced by short-circuit.

\(V_o=-V_s\)

Case 2: During -ve half cycle:

(i) 0< Vin < V

Diode is forward-biased, hence the diode is replaced by short-circuit.

\(V_o=-V_s\)

(ii) V < Vin < Vm

Diode is reverse-biased, hence the diode is replaced by an open-circuit.

\(V_o=0\)

So, the output waveform is:

F1 Engineering Mrunal 13.03.2023 D9

So, the correct answer is option 2.

The source voltage for the circuit shown is 220 V and the load inductance is 220 μH. If the switch is closed for a time t1 = 100 μs, then the load current will be:

F1 Engineering Mrunal 13.03.2023 D47

  1. 10 A
  2. 100 A
  3. 0.1 A
  4. 0.01 A

Answer (Detailed Solution Below)

Option 2 : 100 A

Electronics and Experimental Methods Question 7 Detailed Solution

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Concept

When S1 is closed, the main diode is ON and Dm is OFF.

F1 Engineering Mrunal 13.03.2023 D48

\(V_L=V_S\)

\(L{di_L\over dt}=V_S\)

\(\int_{0}^{I_L}di_L={1\over L}\int_{0}^{T_{ON}}V_S\space dt\)

\(I_L={V_S\times T_{ON}\over L}\)

In the above circuit, the inductor current is the output current.

\(I_o={V_S\times T_{ON}\over L}\)

where, VS = Source voltage

TON = ON period

L = Inductor

Calculation

Given, VS = 220 V

TON = 100 μs

L = 220 μH

\(I_o={220\times 100\times 10^{-6}\over 220\times 10^{-6}}\)

Io = 100 A

From the output Z of the logic circuit shown, determine Z̅.
F2 Vinanti Engineering 03.05.23 D1 V2

  1. QR̅
  2. Q̅R
  3. R̅Q̅ + P
  4. P̅ + QR

Answer (Detailed Solution Below)

Option 1 : QR̅

Electronics and Experimental Methods Question 8 Detailed Solution

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Concept

1.) De-morgan's law: \(\overline{A+B}=\overline{A}\space\overline{B}\)

2.) \(\overline{A}{A}=0\)

3.) \(\overline{A}+{A}=1\)

Explanation

Consider the input of the last NAND gates to be A, Q, and C.

where, A is the output of the first NAND gate

C is the output of the second NAND gate

Then, the output Z is given by:

\(Z=\overline{AQC}\)..........(i)

Output A of the first NAND gate is given by:

\(A=\overline{P\overline{Q}}\)

Output C of the second NAND gate is given by:

\(C=\overline{QR}\)

Putting the value of A, and C in equation (i)

\(Z=\overline{ABC}\)

\(Z=\overline{{\overline{(P\overline{Q})}}(Q)\overline{QR}}\)

\(Z=P\overline{Q}+\overline{Q}+QR\)

\(Z=\overline{Q}(P+1)+QR\)

\(Z=\overline{Q}+QR\)

\(Z=(\overline{Q}+Q)(\overline{Q}+R)\)

\(Z=\overline{Q}+R\)

\(Z=\overline{Q\overline{R}}\)

\(\overline Z={Q\overline{R}}\)

Wires that carry equal and opposite currents are twisted because-

  1. It is easy to coil 
  2. The magnetic field at distant points are reduced by currents in opposite directions and distance from centres is nullified 
  3. While laying, the two wires may fall apart
  4. It reduces the magnetic field away from it

Answer (Detailed Solution Below)

Option 2 : The magnetic field at distant points are reduced by currents in opposite directions and distance from centres is nullified 

Electronics and Experimental Methods Question 9 Detailed Solution

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The correct option is 2

Concept:

  • Twisting wires that carry equal and opposite currents serve to minimize the interference caused by external magnetic fields. Twisting the wires causes the distance from a point in the surroundings to each wire to change continuously. At some points, the point is closer to one wire, and at others, it's closer to the other wire.
  • Since the wires carry equal and opposite currents, the magnetic field produced by each wire at any one point in the surroundings is in opposite directions. This means that the effects of the magnetic fields produced by the two wires tend to cancel each other out because of the twist, leading to a significant reduction in magnetic field disturbances overall.
  • Furthermore, this phenomenon is also beneficial in reducing electromagnetic emission from the wires, hence reducing electromagnetically induced noise in surrounding electronic components or systems.
  • So, statement 4) "It reduces the magnetic field away from it" is also somewhat true, but statement 2) better explains why the phenomenon occurs, specifically addressing the role of the counterbalanced currents and the physical twisting of the wires.

Consider the AND logic circuit in which V2 = 3 V and V1 lie between 0 to 5 V. The output voltage is V0. The cut-in voltage of diode D1 and D2 is 0.6 V. The output voltage V0 versus V1 corresponding to the below network is

F1 S.B 1.9.20 Pallavi D8

  1. F1 S.B 1.9.20 Pallavi D9
  2. F1 S.B 1.9.20 Pallavi D10
  3. F1 S.B 1.9.20 Pallavi D11
  4. F1 S.B 1.9.20 Pallavi D12

Answer (Detailed Solution Below)

Option 2 : F1 S.B 1.9.20 Pallavi D10

Electronics and Experimental Methods Question 10 Detailed Solution

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Concept:

Whenever diodes are tied together, the diode which experiences the more potential difference will ON first.

F1 S.B 1.9.20 Pallavi D13

D3 is ON and remaining OFF.

F1 S.B 1.9.20 Pallavi D14 (1)

When all diodes are opposite to the previous case.

F1 S.B 1.9.20 Pallavi D15

D1 is ON and remaining OFF

F1 S.B 1.9.20 Pallavi D16

If all are given the same voltage then all will be ON

F1 S.B 1.9.20 Pallavi D17

Calculation:

Given voltage levels are V1 = 0 to 5 V and V2 = 3V

Until V1 < 3 V diode D2 is OFF

For V1 = 1 V

F1 S.B 1.9.20 Pallavi D18

V0 = 1.6 V

For V2 = 2 V

F1 S.B 1.9.20 Pallavi D19

V0 = 2.6 V

For V1 = 3 V

Both will be ON

F1 S.B 1.9.20 Pallavi D20

V0 = 3.6 V

For V1 = 4 V diode D1 will be OFF.

F1 S.B 1.9.20 Pallavi D21

V0 = 3.6 V

For V1 = 5 V diode D1 will be OFF.

F1 S.B 1.9.20 Pallavi D22

V0 = 3.6 V

So the output is constant at 3.6 V for V1 ≥ 3 V

Option B is correct.

Which feedback device translate physical motion in to electrical data? 

  1. Transducer
  2. Encoder
  3. Digital system monitoring
  4. None of the above

Answer (Detailed Solution Below)

Option 2 : Encoder

Electronics and Experimental Methods Question 11 Detailed Solution

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Concept

  • Transducers are devices that convert one form of energy (an input signal—usually physical, like motion) to another (an output signal—usually electrical).(usually analog)
  • An encoder is a sensing device that provides feedback.
  • Encoders convert the motion to an digital signals.
  • Digital construction monitoring simply means an automated method of monitoring.

In the circuit shown below, four silicon diodes and four capacitors are connected to a sinusoidal voltage source of amplitude Vin > 0.7 V and frequency 1 kHz. If the knee voltage for each of the diodes is 0.7 V and the resistances of the capacitors are negligible, the DC output voltage Vout after 2 seconds of starting the voltage source is closest to

F1 Teaching Arbaz 23-10-23 D24

  1. 4Vin - 0.7 V
  2. 4Vin - 2.8 V
  3. Vin - 0.7 V
  4. Vin - 2.8 V

Answer (Detailed Solution Below)

Option 2 : 4Vin - 2.8 V

Electronics and Experimental Methods Question 12 Detailed Solution

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Explanation:

The circuit appears to be a full wave rectifier with a 4-stage voltage multiplier, often known as a ladder network or Cockcroft–Walton generator. This kind of circuit is used to generate a high DC voltage from an AC power source. It is made by a combination of diodes and capacitors, allowing for increasing the voltage value.

  • Given that the knee voltage for the silicon diodes is 0.7 volts, this will have to be subtracted for every diode as we move through the voltage multiplier.
  • The output voltage is approximately equal to \(4V_{in} - 4V_{knee}=4V_{in}-4(0.7V)=4V_{in}-2.8V\), where Vin is the peak voltage of the AC source, and Vknee is the knee voltage of a diode. The factor of 4 in front of the Vknee originates from the fact that there are 4 diodes in the current path from the source to the output, and each diode subtracts its knee voltage from the overall voltage.

The Hall coefficient RH of a sample can be determined from the measured Hall voltage \(V_H=\frac{1}{d} R_H B I+R I \text {, }\)where d is the thickness of the sample, B is the applied magnetic field. I is the current passing through the sample and R is an unwanted offset resistance. A lock-in detection technique is used by keeping I constant with the applied magnetic field being modulated as B = B0sinΩt, where B0 is the amplitude of the magnetic field and Ω is frequency of the reference signal. The measured VH is

  1. \(B_0\left(\frac{R_H I}{d}\right)\)
  2. \(\frac{B_0}{\sqrt{2}}\left(\frac{R_H I}{d}\right)\)
  3. \(\frac{I}{\sqrt{2}}\left(\frac{R_H B_0}{d}+R\right)\)
  4. \(I\left(\frac{R_H B_0}{d}+R\right)\)

Answer (Detailed Solution Below)

Option 2 : \(\frac{B_0}{\sqrt{2}}\left(\frac{R_H I}{d}\right)\)

Electronics and Experimental Methods Question 13 Detailed Solution

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Explanation:

  • A lock-in amplifier primarily recognizes the component of the signal that is in phase with the reference signal (B = B₀sinΩt in this case) and measures its amplitude and phase, not just the amplitude.
  • When \(sinΩt\) is at its peak amplitude, the Hall voltage \(V_H\) is also at its peak: \(V_{H_{max}}= \frac{I R_H B₀}{ d}+ R I\) .
  • However, the DC offset term RI will get removed by the lock-in amplifier. What remains is the AC part of the signal, which follows the sinusoidal modulation.
  • The observed amplitude of this sinusoidal signal, having accounted for our removal of the DC offset, is the root-mean-square (RMS) value.
  • The RMS value of a sinusoidal function A sin(x) is \(\frac{A}{√2}\).
  • Recognising that, and applying it to the signal we're talking about, we get: \(V'_{H} =\frac{ I R_H B₀}{ (√2 d)}\)

A DC motor is used to lift a mass M to a height h from the ground. The electric energy delivered to the motor is VIt, where V is the applied voltage, I is the current and t the time for which the motor runs. The efficiency e of the motor is the ratio between the work done by the motor and the energy delivered to it. If M = 2.00 ± 0.02 kg, h = 1.00 ± 0.01 m, V = 10.0 ± 0.1 V, I = 2.00 ± 0.02 A and t = 300 ± 15s, then the fractional error |δe/e| in the efficiency of the motor is closest to
 

  1. 0.05
  2. 0.09
  3. 0.12
  4. 0.15

Answer (Detailed Solution Below)

Option 1 : 0.05

Electronics and Experimental Methods Question 14 Detailed Solution

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Concept:

Every measurement has an air of uncertainty in it, so error is measured by standard deviation \(\sigma_i\) of the measurement.

\(\sigma^2_x=\left(\dfrac{\delta{x}}{\delta{a}}\right)^2\sigma^2_a+\left(\dfrac{\delta{x}}{\delta{b}}\right)^2\sigma^2_b+\left(\dfrac{\delta{x}}{\delta{c}}\right)^2\sigma^2_c\)

 

Explanation:

Given, efficiency of the motor is the ratio between the work done by the motor and the energy delivered to it.

So, \(e=\frac {W} {E}=\frac {mgh} {VIt}\)

Given, \(M=2.00\pm0.02kg, h=1.00\pm0.01m, V=10.0\pm0.1 V, I=2.00\pm 0.02A, t=300\pm15s\)

Using propagation of errors, we get,

  • \(\frac {\delta e} {e} =\sqrt {\left(\dfrac{\delta{m}} {m}\right)^2+\left(\dfrac{\delta{h}} {{h}}\right)^2+\left(\dfrac{\delta{V}}{{V}}\right)^2+\left(\dfrac{\delta{I}} {{I}}\right)^2+\left(\dfrac{\delta{t}} {{t}}\right)^2}\)    

 

  • \(\frac {\delta e} {e} =\sqrt {\left(\dfrac{{0.02}} {2}\right)^2+\left(\dfrac{{0.01}} {{1}}\right)^2+\left(\dfrac{{0.01}}{{1}}\right)^2+\left(\dfrac{{0.02}} {{1}}\right)^2+\left(\dfrac{{15}} {{300}}\right)^2}\)
  • \(\frac {\delta e} {e}=\sqrt{(0.01)^2\times 4+(0.05)^2}\)
  • \(\frac{\delta e} {e}=\sqrt29 \times 10^{-2}\)
  • \(\frac{\delta e} {e}\approx 0.05\)

So, the correct answer is \(\frac{\delta e} {e}\approx 0.05\)

 

In the circuit below, there is a voltage drop of 0.7 V across the diode D in forward bias, while no current flows through it in reverse bias.

F1 Teaching Arbaz 23-10-23 D11

If Vin is a sinusoidal signal of frequency 50 Hz with an RMS value of 1 V, the maximum current that flows through the diode is closest to

  1. 1 A
  2. 0.14 A
  3. 0 A
  4. 0.07 A

Answer (Detailed Solution Below)

Option 3 : 0 A

Electronics and Experimental Methods Question 15 Detailed Solution

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Explanation:

Case-1-When diode is in forward biasing

Given, \(V_{rms}=1V\)

Now, peak voltage \(V_p=\sqrt{2} V_{rms}=\sqrt{2}\times 1=1.414\)

  • Current in \(20\Omega=\frac {voltage difference} {resistance}=\frac{1.414-0.7} {20}\approx0.03A\)

This is low current

  • Current in \(10\Omega=\frac {voltage difference} {resistance}=\frac{0.7-0} {10}=0.07A\)

This is high current as comparison to \(20\Omega\) resistor in forward biasing 

So, Forward biasing is not possible in this circuit.

Case-2-When the circuit is in Reverse biasing-

This circuit is possible in reverse biasing.

In Reverse biasing the current in the diode should be zero.

So, the correct answer is \(0A\)

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