Electronics and Experimental Methods MCQ Quiz in தமிழ் - Objective Question with Answer for Electronics and Experimental Methods - இலவச PDF ஐப் பதிவிறக்கவும்

Concept:

Resistance in series: When two or more  resistances are connected in series, the total or equivalent resistance (\(R_{eq}\)) is the sum of the individual resistances, and the uncertainties (standard deviations) are propagated similarly.

\(R_{eq} = R_1 + R_2 +R_3+...\)

Where, R₁, R₂, R₃, ... are resistances in series.

Propagation of Uncertainty: The resultant uncertainty (standard deviation) of two or more resistances in series is given by the square root of the sum of the squares of the individual standard deviations.

\( \sigma_R = \sqrt{Δ R_1^2+Δ R_2^2+Δ R_3^2+...} \)

Where, Δ R₁ is the uncertainty in R₁.

 Δ R2 is the uncertainty in R₂ and so on...

Explanation:

Given,  R₁ = 100 and  Δ R₁=5Ω,  R₂ = 150 and  Δ R2=15Ω:

\(R_{eq}=R_1+R_2=100+150=250 \Omega \)

Resultant standard deviations\( \sigma_R = \sqrt{(5 \Omega)^2 + (15 \Omega)^2} = \sqrt{250\Omega } \approx 15.8 \Omega \)

Hence, 

\(R_{eq}=250 \pm 15.8\Omega\)

The correct answer is Option 3.

Concept:

Reactance of Inductive Coil: The reactance (\(X_L\)) of an ideal inductor is given by:

\( X_L = ω L\)

where, ω = 2πf which is the angular frequency, and (L) is the inductance.

Effect of Stray Capacitance: Stray capacitance in an inductive coil causes the coil to behave as a resonant circuit at certain frequencies, effectively forming a parallel LC circuit.

Effective Reactance (\(X_{\text{eff}}\)):

1.   Below the resonant frequencies (ω<ω_c ), the inductive reactance ( \(X_L\) ) dominates.
2. At the resonant frequency which is (ω=ωc ), the inductive reactance and capacitive reactance cancel each other out, leading to a minimum in the effective reactance.
3. Above the resonant frequency (ω>ωc ), the coil's behaviour is dominated by the capacitive reactance ( \(X_C\) ), and the effective reactance starts to decrease.

Hence, the effective reactance of an inductive coil changes due to the presence of stray capacitance as the frequency of the applied signal changes.

Explanation:

Because of the stray capacitance, the effective reactance of an inductive coil: Initially increases with frequency because of the inductive reactance ( \(X_L\) ). Decreases at high frequencies because the capacitive reactance ( \(X_C\) ) becomes significant and ultimately dominates.

So, the effective reactance of an inductive coil decreases because of stray capacitances as the frequency increases.

The correct answer is Option 2.

Concept:

1. The least count (LC) of the instrument is 0.1 mm.
2. The number of divisions on the Vernier scale (VSD) is 20.
3. The number of divisions on the main scale (MSD) that coincide with the Vernier scale is 19.

The least count (LC) of a Vernier caliper is given by the formula:

" id="MathJax-Element-28-Frame" role="presentation" style="position: relative;" tabindex="0"> \(LC=Value of 1 Main Scale Division (MSD)−Value of 1 Vernier Scale Division (VSD)\)

$\text{Calculation:}$

Here,

20 VSD = 19 MSD

1VSD = \(\frac{19}{20}\) MSD

L.C. = 1 MSD – 1 VSD

0.1 mm = 1 MSD - \(\frac{19}{20}\) MSD

0.1 = \(\frac{1}{20}\) MSD

1 MSD = 2mm

∴ The correct option is 3

Explanation:

This is the tank circuit of the Hartley oscillator. 

• There are two inductors in the series connection. The equivalent inductance is : \(L=12\mu H+8\mu H=20\mu H\)
• The capacitance is given as : \(C=0.05\mu f\)
• The formula for the frequency is given as : 

\(f=\frac{1}{2\pi \sqrt{LC}}\)

\(f=\frac{1}{2\pi \sqrt{(20\mu H\times0.05\mu f)}}\)

\(f=\frac{1}{2\pi\mu}Hz\)

\(f=\frac{10^6}{2\times3.14 }=159235Hz=159KHz\)

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 \(0A\)

Concept:

• The voltage being applied causes an electric field in the region between the plates of the deflector. This electric field will cause a force on the electrons and hence will cause them to deviate from their straight path.
• Since the applied voltage is sinusoidally varying time, the electric field and hence the force on the electrons also alternates in direction with time. So, at different points in time, the electrons will be deflected upward and downward.
• When the voltage is zero, the force on the electrons is also zero and hence they travel in a straight line. As the magnitude of the voltage increases from zero, more and more electrons are deflected from the straight path. Hence, the intensity of the beam coming through the aperture (which is at the center of the plates) decreases.
• When the magnitude of the voltage is maximum, the deflection is maximum and hence the intensity is minimum. Then as the magnitude of the voltage decreases back to zero, fewer and fewer electrons are deflected and the intensity increases.

Explanation:

The voltage is given as : \(V_i=V_msin(\omega t)\)

Now, the input voltage will be zero at time :

\(V_msin(\omega t)=0\)

\(\omega t=n\pi(n=0,1,2,3,...)\)

\(n=\frac{\omega t}{\pi}(n=0,1,2,3,...)\)

So, peaks will be obtained at : \(\frac{\omega t}{\pi}=0,1,2,3,...\)

Graph in the option(a) is the correct answer.

Concept:

• AND : When both the inputs are high, it will show high. If one is high and one is low, it will show low.
• OR : It will show high even when one input is low and another is high.
• NOT : Converts high to low and vice-versa.

Explanation:

• Shut off signal should work when the level goes down the threshold value. Suppose the level goes down then the L will be at 0 and the "Y" should be at high.
• So, there are two cases where we want "Y" to be at high. One is when the level goes down and another case where the level goes down and the switch "S" is also at low. So, the cases are :
• 1) L= 0 and S = 1 , 2) L = 0 and S = 0
• Now, "Z" clearly is NOT here as it is used to reverse the signals in the image(as it is applied at the individual path and it is the only circuit that takes one input, the rest needs two inputs)
• Now, "Z" will reverse the signals so, the signals will be: 1) L= 1 and S = 0, 2) L = 1 and S = 1 at "X" point. 
• Now, in any case, the "y" should be at high so, it suggests that "Y" should be "OR".
• For "Y" to be OR, "X" should be "AND".