Sinusoidal Steady State Analysis MCQ Quiz - Objective Question with Answer for Sinusoidal Steady State Analysis - Download Free PDF

Explanation:

Half-Wave Rectifier or Series Clipper Using a Single Diode

Definition: A half-wave rectifier is an electronic circuit that allows only one half of the input alternating current (AC) signal to pass through, effectively clipping the other half. This is achieved using a single diode, which conducts during the positive half-cycle of the AC signal and blocks during the negative half-cycle. A series clipper, similarly, is a circuit designed to clip (restrict) a portion of the input signal based on the diode's characteristics and configuration.

Problem Statement: Determine the peak value of the output signal of the half-wave rectifier or series clipper realized using a single diode having a forward voltage of 0.6 V, if the input applied to the circuit is a periodic sinusoidal signal of 10 V peak-to-peak voltage.

Solution:

To solve this problem, let us analyze the circuit step by step:

1. Input Signal Characteristics:

• The input is a sinusoidal signal with a peak-to-peak voltage of 10 V.
• The peak-to-peak voltage indicates the total voltage swing from the positive peak to the negative peak.
• For a sinusoidal signal, the peak voltage (V_peak) is half of the peak-to-peak voltage:

V_peak = (Peak-to-Peak Voltage) ÷ 2 = 10 ÷ 2 = 5 V

This means the input signal swings from +5 V (positive peak) to -5 V (negative peak).

2. Diode Characteristics:

• The diode used in the circuit has a forward voltage drop of 0.6 V. This is the minimum voltage required for the diode to conduct.
• When the input voltage exceeds 0.6 V (during the positive half-cycle), the diode will conduct, allowing the signal to pass through.
• During the negative half-cycle, the diode will block the signal, and the output voltage will be zero.

3. Output Signal Analysis:

• During the positive half-cycle of the input signal:
• The diode conducts when the input voltage exceeds 0.6 V.
• The output voltage is equal to the input voltage minus the forward voltage drop of the diode.
• At the peak of the input signal (+5 V), the output voltage is:

V_output = V_input - V_forward = 5 V - 0.6 V = 4.4 V

• During the negative half-cycle of the input signal:
• The diode is reverse-biased and does not conduct.
• The output voltage is zero.

4. Peak Value of the Output Signal:

The peak value of the output signal is the maximum voltage achieved during the positive half-cycle, which is:

V_{peak, output} = 4.4 V

Correct Answer: Option 4

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: 5 V

This option assumes that the output voltage is equal to the peak voltage of the input signal. However, it fails to account for the diode's forward voltage drop (0.6 V). The correct peak output voltage is 4.4 V, not 5 V.

Option 2: 9.4 V

This option is incorrect because it exceeds the peak voltage of the input signal. The input signal has a peak voltage of 5 V, so the output signal cannot have a peak voltage of 9.4 V.

Option 3: 10 V

This option is incorrect because it corresponds to the peak-to-peak voltage of the input signal, not the peak output voltage. The diode's forward voltage drop further reduces the output peak voltage to 4.4 V.

Option 4: 4.4 V

This is the correct option as it accurately accounts for the input signal's peak voltage and the diode's forward voltage drop.

Conclusion:

A half-wave rectifier or series clipper using a single diode clips the negative half-cycle of the input signal and reduces the positive half-cycle by the diode's forward voltage drop. For an input sinusoidal signal with a peak-to-peak voltage of 10 V and a diode with a forward voltage of 0.6 V, the peak value of the output signal is 4.4 V.

Laplace Transform and Its Applicability to Signals

Definition: The Laplace transform is a widely used integral transform in mathematics and engineering that converts a time-domain function into a complex frequency-domain representation. It is particularly useful in analyzing and solving linear time-invariant systems such as electrical circuits, mechanical systems, and control systems.

Mathematical Representation:

The Laplace transform of a time-domain function f(t) is defined as:

L{f(t)} = F(s) = ∫₀^∞ f(t) e^-st dt

Where:

• t: Time variable (in the time domain).
• s: Complex frequency variable (s = σ + jω).
• e^-st: Exponential decay factor.

Uses and Advantages:

• The Laplace transform simplifies the analysis of differential equations by converting them into algebraic equations.
• It provides a systematic way to handle initial conditions in dynamic systems.
• It is extensively used in control systems, signal processing, and communication engineering.

Correct Option Analysis:

The correct option is:

Option 2: Continuous time domain.

The Laplace transform is primarily applicable to signals in the continuous time domain. This is because the integral definition of the Laplace transform requires the signal to be defined over a continuous range of time, typically from 0 to infinity. In engineering and physics, most applications of the Laplace transform deal with continuous-time systems, such as analog electrical circuits, mechanical vibrations, and control systems.

Reasoning:

• In continuous time systems, signals are functions of a continuous variable (time t), and the Laplace transform effectively captures their frequency-domain characteristics.
• It is especially useful for analyzing linear systems where the system's behavior can be expressed using differential equations.
• The Laplace transform simplifies the analysis by converting these differential equations into algebraic equations in the s-domain (complex frequency domain).

Application:

• Analysis of electrical circuits with capacitors and inductors.
• Modeling and analysis of mechanical systems, such as damped harmonic oscillators.
• Control system design and stability analysis.
• Signal processing tasks such as filtering and system identification.

Important Information

To further understand the analysis, let’s evaluate the other options:

Option 1: Integral time.

This option is incorrect. The term "integral time" is not a standard term in signal processing or control systems. If it refers to discrete-time signals, then it is unrelated to the Laplace transform, as the Laplace transform is specifically defined for continuous-time signals. For discrete-time signals, the Z-transform is used instead of the Laplace transform.

Option 3: Non-linear.

This option is incorrect. The Laplace transform is primarily applicable to linear systems and signals. Non-linear systems cannot be analyzed directly using the Laplace transform because their behavior does not satisfy the principle of superposition (additivity and homogeneity). For non-linear systems, other mathematical tools such as perturbation methods, numerical simulation, or specific transforms might be needed.

Option 4: Digital.

This option is incorrect. Digital signals are typically discrete-time signals, and the Laplace transform is not applicable to them. Instead, the Z-transform is used for analyzing and processing discrete-time signals in the digital domain. The Z-transform is analogous to the Laplace transform but is specifically designed for signals defined at discrete time intervals.

Option 5: (Blank).

This option is invalid as it does not provide any specific information for analysis. It is not relevant to the question.

Conclusion:

The Laplace transform is an essential mathematical tool for analyzing continuous-time systems and signals. Its ability to convert time-domain differential equations into frequency-domain algebraic equations makes it invaluable in engineering and scientific applications. While the Laplace transform is highly effective for continuous-time systems, it is not applicable to discrete-time or digital signals, nor is it suitable for analyzing non-linear systems. Instead, other transforms like the Z-transform or specific mathematical tools are used in such cases. Understanding the scope and limitations of the Laplace transform is crucial for its correct application in engineering and science.

Concept

The inverse Laplace Transform of 

$\frac{k}{s+a}=k{e}^{-at}$

where, k = Constant

Calculation

Given, $\mathrm{I}(\mathrm{s})=\frac{1.5}{\mathrm{s}+4}$

On comparison, k = 1.5 and a = 4

$\frac{1.5}{s+4}=1.5e^{-4t}$

Concept

The average and RMS value of the full and half-wave rectifier are given by:

Full wave rectifier:

$V_{o(avg)}=\frac{2V_m}{\pi }$

$V_{o(rms)}=\frac{V_m}{\sqrt{2}}$

Half-wave rectifier:

$V_{o(avg)}=\frac{V_m}{\pi }$

$V_{o(rms)}=\frac{V_m}{2}$

Calculation

Given, V_m = 20V

The average output voltage is given by:

$V_{o(avg)}=\frac{2\times 20}{\pi }$

$V_{o(avg)}=\frac{40}{\pi }V$

Concept

The current through a capacitor is given by:

$I_C=C\frac{d{V}_C}{dt}$

where, $\frac{d{V}_C}{dt}=$ Rate of change of capacitor voltage

The voltage across a capacitor is given by:

$V_c(t)=\frac{1}{C}\int I_c(t)dt$

Calculation,

Given,

cos ϕ₁ = 0.8

∴ sin ϕ₁ = 0.6

cos ϕ₂ = 0.6

∴ sin ϕ₂ = 0.8

Asuume,

|I₁| = |I₂| = |I|

In complex form the phasor quantity is written as,

i = I(cos ϕ ± jsin ϕ)

Both current can be written as,

i₁ = |I|(0.8 + j0.6)

i₂ = |I|(0.6 + j0.8)

Resultant current (i) will be phasor sum of i₁ and i₂,

i = |I|(0.8 + j0.6) + |I|(0.6 + j0.8) = |I|(1.4 + j1.4)

Phase angle can caculated as,

$\tan \varphi =\frac{imaginarrycomponent}{realcomponent}=\frac{1.4}{1.4}=1$

ϕ = tan^-1 (1) = 45° 

Hence,

Power factor = cos 45° = 0.707

Alternate Method

When two circuits having the same magnitudes of impedances are joined in parallel. The power factor of the combination is 

$\cos \varphi =\frac{1}{\sqrt{1+\left({\cos }^2{\varphi }_1+{\cos }^2{\varphi }_2\right)}}$

The power factor of the first circuit = 0.8 (cos ϕ₁)

Power factor of second circuit = 0.6 (cos ϕ₂)

The power factor of parallel combination,

$\cos \varphi =\frac{1}{\sqrt{1+\left({\cos }^2{\varphi }_1+{\cos }^2{\varphi }_2\right)}}$

Kirchhoff’s Current Law (KCL): It states that the algebraic sum of currents entering a node or closed boundary is zero.

Mathematically, KCL implies that

$\sum _{n=1}^N{i}_n=0$

Where N is the number of branches connected to the node

And in is the nth current entering or leaving the node.

By this law current entering a node may be regarded as positive and current leaving a node may be regarded as negative or vice-versa.

Hence, the sum of currents entering a node is equal to the sum of the currents leaving the node.

Calculation:

Given that, three currents i1, i2, and i3 meet at a node,

From the above concept,

i1,+ i2, + i3 = 0

or,  i3 = - (i1 + i2)

Since the current is given in phasor form, hence the addition of current i₁ and i₂ can be done by using the parallelogram method,

We have,

i₁ = 10 sin (400t + 60°) A .... (1)

i₂ = 10sin (400t - 60°) A .... (2)

The phasor diagram can be drawn as,

By using the parallelogram method,

 |ir| = $\sqrt{i_1^2+i_2^2+2i_1{i}_{2}cos\theta }$ .... (3)

Where, it is the resultant current

We have, θ = (60° + 60°) = 120°

From equation (1), (2) & (3),

|i_r| = $\sqrt{{10}^2+{10}^2+200cos{120}^{\circ }}$

|ir| = $\sqrt{{10}^2+{10}^2+200\times -0.5}=10$

Since i₁ & i₂ has the same phase angle as well as the magnitude,

∴ i_r = 10sin 400t A

Since the sum of all current is zero, hence,

i₃ = - i_r = - 10sin 400t A

Concept:

The impedance(Z) of a circuit can be written as:

Z = R ± jX

R = Resistance

X = Reactance

The admittance(Y) can be written as:

Y = G ± jB   ---(1)

G = Conductance

B = Susceptance

$Y=\frac{1}{Z}$

Calculation:

Z = 3 + j4

$Y=\frac{1}{Z}=\frac{1}{3+j4}$

$=\frac{\left(3-j4\right)}{\left(3+j4\right)\left(3-j4\right)}$

$Y=\frac{3}{25}-\frac{j4}{25}$

Comparing this with Equation (1), we get:

$G=\frac{3}{25}$

Mistake Point:

The conductance will not be 1/R. For a given impedance, the conductance will be:

$G=\frac{R}{|Z{|}^2}$

R = Resistance 

|Z|² = Square of the magnitude of impedance

Concept:

The average value of a periodic waveform is given as

$F_{avg}=\frac{1}{T}\underset{0}{\overset{T}{\int }}f\left(t\right)dt$

It is also calculated by calculating the area of the given waveform.

$F_{avg}=\frac{A}{T}$

Where A is the area

RMS value of a periodic waveform is given as

$F_{rms}=\sqrt{\frac{1}{T}\underset{0}{\overset{T}{\int }}{\left[f\left(t\right)\right]}^{2}dt}$

Where T is the time-period of the given waveform.

Calculation:

Time period of the given waveform (T) = π

The area of the given waveform is,

$A=\left(\frac{1}{2}\times \alpha \times F_m\right)+F_m\left(\pi -2\alpha \right)+\left(\frac{1}{2}\times \alpha \times F_m\right)$

= F_m (π – α)

Concept:

Equation of current is I = Im sin ωt

I_m = Maximum current

ω = Angular frequency = 2πf

f = Frequency

Calculation:

Given, I = 100 A

f = 50 Hz

100 = 200 sin 2 × π × 50 × t

100 × π × t = 30° = π/6

t = 1/600 s   

t = 1.666 ms

Concept:

RMS (Root mean square) value:

• RMS value is based on the heating effect of wave-forms.
• The value at which the heat dissipated in AC circuit is the same as the heat dissipated in DC circuit is called RMS value provided, both the AC and DC circuits have equal value of resistance and are operated at the same time.

• RMS value 'or' the effective value of an alternating quantity is calculated as:

  $V_{rms}=\sqrt{\frac{1}{T}\underset{0}{\overset{T}{\int }}{v}^2\left(t\right)dt}$

  T = Time period

Note:

• Average or mean value of alternating current is that value of steady current which sends the same amount of charge through the circuit in a certain interval of time as is sent by alternating current through the same circuit in the same interval of time

• The ratio of the maximum value (peak value) to RMS value is known as the peak factor or crest factor.
• $Peakfactor=\frac{maximumvalue}{rmsvalue}$
• The ratio of RMS value to the average value is known as the form factor.
• $Formfactor=\frac{rmsvalue}{averagevalue}$

Calculation:

For a rectangular wave, the RMS value is equal to the average value. RMS value is also equal to peak value.

For given waveform RMS value is 100 V.

IMPORTANT EVALUATIONS:

Concept:

The Laplace transform resistance, inductor, and capacitance are given by:

1. Resistance: R
2. Inductor: sL
3. Capacitor: $\frac{1}{sC}$

​Calculation:

The circuit diagram in the Laplace domain is given below:

Applying voltage division rule across capacitor:

$V_{out}(s)=V_{in}\times \frac{\frac{1}{s}}{\frac{1}{s}+2s}$

Concept:

KCL in DC circuits:

According to Kirchhoff’s current law (KCL), the algebraic sum of the electric currents meeting at a common point is zero. I.e. the sum of currents entering a node is equal to the sum of currents leaving the node. It is based on the conservation of charge.

KCL in AC circuits:

Kirchhoff’s current law as applied to the ac circuit is defined as the phasor sum of currents entering the node is equal to the phasor sum of currents leaving the node.

Calculation:

By applying KCL at the node,

I₁ = I₂ + I₃

I₁ = 1∠10° + 1∠70°

= (cos 10 + j sin 10) + (cos 70 + j sin 70)

= (cos 10 + cos 70) + j(sin 10 + sin 70)

= 1.3268 + j1.1133 A

⇒ I₁ = 1.732∠ 40° A

The correct answer is option 4):​$\sqrt{\frac{5}{18}}A$

​Concept:

The magnitude of the total current through the circuit

I = $\sqrt{I_R^2+(I_L-I_C{)}^2}$

where

I_R is the resistive component of current

I_L  is the Inductive component of current

I_C  is the Capacitive component of current

IR = $\frac{V}{R}$

V is the voltage

R is the resistance

I_C = $\frac{V}{X_c}$

X_c is the capacitive reactance

I_L = $\frac{V}{X_L}$

X_L  is the inductive reactance

Calculation:

The impedance of a network is given as

I = $\sqrt{I_R^2+(I_L-I_C{)}^2}$

I_R = $\frac{20}{120}$

= $\frac{1}{6}$

I_c = $\frac{20}{40}$

= $\frac{1}{2}$

I = $\sqrt{(}\frac{1}{36}+\frac{1}{4})$

= $\sqrt{\frac{4+36}{36\times 4}}$

=$\sqrt{\frac{10}{36}}$

= $\sqrt{\frac{5}{18}}A$

Given that,

V = 200 Sin (2000t + 50°) V

i = 4 cos (2000t + 13.2°) A

We know that,

sin (90 + ϕ) = cos ϕ 

⇒ V = 200 sin (2000t + 50°) V  --------- (i)

⇒ i = 4 sin (2000t + 13.2°+ 90°) A

⇒ i = 4 sin (2000t + 103.2°) A    ------- (ii)

From, (i) and (ii) we can observe that,

Current 'i' is leading voltage V by 53.2° 

Hence, the element is practical capacitor