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

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 %

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. 

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 \(0A\)

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.

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).

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 V_BE = 0.7 V

EXPLANATION:

Apply K.V.L in the input section

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

The correct answer is 20μA 

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.

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.

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< V_in < V

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

\(V_o=-V_s\)

(ii) V < Vin < V_m

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

\(V_o=0\)

So, the output waveform is:

So, the correct answer is option 2.

Concept

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

\(V_L=V_S\)

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

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

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

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

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

where, V_S = Source voltage

T_ON = ON period

L = Inductor

Calculation

Given, VS = 220 V

TON = 100 μs

L = 220 μH

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

I_o = 100 A

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)

\(Z=\overline{ABC}\)

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

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

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

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

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

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

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

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

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.

Concept:

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

D₃ is ON and remaining OFF.

When all diodes are opposite to the previous case.

D₁ is ON and remaining OFF

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

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

V₀ = 1.6 V

For V₂ = 2 V

V₀ = 2.6 V

For V₁ = 3 V

Both will be ON

V₀ = 3.6 V

For V₁ = 4 V diode D₁ will be OFF.

V₀ = 3.6 V

For V₁ = 5 V diode D₁ will be OFF.

V₀ = 3.6 V

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

Option B is correct.

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.

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 V_in is the peak voltage of the AC source, and V_knee is the knee voltage of a diode. The factor of 4 in front of the V_knee 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.

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: \(V'_{H} =\frac{ I R_H B₀}{ (√2 d)}\)

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 \(\frac{\delta e} {e}\approx 0.05\)

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 \(0A\)