GATE EE 2021 Official Paper (Previous Year Paper)

Given data from the chart,

Fail in the first year will give an exam in the second year.

Students will pass on 2^nd attempt.

It means fail in 1^st year students will become pass in 2^nd year.

The number of students who took the exam for the first time in the year 2^nd = 50 + 5 = 55

The number of students who took the exam for the first time in the year 3^rd = 45 + 3 = 48

Concept:

The mean or expectation of a continuous random variable is given by

\(E\left( x \right) = \mathop \smallint \limits_{ - \infty }^\infty xf\left( x \right)dx\)

Where f(x) is the probability density function (PDF).

Application:

Given probability density function of X 

f(x) = 0.01 for 0 ≤ X ≤ 100

The mean or expectation of a continuous random variable is given by

\(E\left( x \right) = \mathop \smallint \limits_{ 0}^{100} x\left( 0.01 \right)dx=50\)

Additional Information

 Properties of a valid probability density function (PDF):

1) \({f_X}\left( x \right) \ge 0,;\forall ;x;\epsilon;R\)

∴ PDF will be bounded between 0 and 1.

2) \(\mathop \smallint \limits_{ - \infty }^\infty {f_X}\left( x \right)dx = {F_X}\left( \infty \right) = 1\)

3) \({F_X}\left( \infty \right) = \begin{array}{*{20}{c}} {{\rm{lim}}}\ {x \to \infty } \end{array}{F_X}\left( x \right)\)

∴ PDF will always be a monotonically increasing function as the probability is always greater than or equal to 0.

The correct answer is 'who'.

Key Points

We use who to refer to the subject of a sentence.

• In the above sentence, 'The people' is the subject of the sentence.
• Thus, we will use the pronoun 'who' as we are referring to the subject 'The people'.

Additional Information

Pronouns are words that take the place of a noun. 

• Relative pronouns are used at the beginning of an adjective clause (a dependent clause that modifies a noun).
• The relative pronouns are "that," "which," "who," "whom," and "whose."
• "Whom" is an object pronoun like "him," "her" and "us."  We use "whom" to ask which person receives an action.
• "Whose" is a possessive pronoun like "his," "her" and "our." We use "whose" to find out which person something belongs to.
• We use "Which" is used for animals in general or things.

Concept:

(aⁿ - bⁿ) is divisible by (a+b), only when 'n' is even.

Application: 

\(\dfrac {{11^{52}-1}} {{11^{13}+1}}=\dfrac {({{11^{13})}^4-1^4}}{{11^{13}+1}}\)

⇒ \(\dfrac {{11^{52}-1}}{{11^{13}+1}}=\dfrac {{[{{{(11^{13})}^2}-1^2}][{{{(11^{13})}^2}+1^2}]}}{{{{11^{13}}}+1}}\)

⇒ \(\dfrac {{11^{52}-1}}{{11^{13}+1}}=\dfrac {{[{{{11^{13}}}-1}][{{{11^{13}}}+1}][{{{(11^{13})}^2}+1^2}]}}{{{{11^{13}}}+1}}\)

⇒ \(\dfrac {{11^{52}-1}}{{11^{13}+1}}=[11^{13}-1][{({11^{13})}^2}+1]\)

∴ (1152 - 1) is exactly divisible by (1113 + 1)

The correct answer is ​To do well in an exam, adequate sleep must be part of the preparation.

Key Points

• Look at the lines: Research has consistently shown that sleep deprivation greatly reduces the ability to recall the material learnt. Hence, cutting down on sleep to study longer hours can be counterproductive.
• Upon the perusal of the above lines, it can be concluded that adequate sleep is needed to do well in the exam as it increases the ability to recall the material learnt.
• Hence, option 1 is the most appropriate answer choice.

Additional Information

• Overlook: to fail to notice or consider something
• Deprivation:  a situation in which you do not have things or conditions that are usually considered necessary for a pleasant life
• Counterproductive: having an effect that is the opposite of what you intend or desire

The arrangement can be written as,

Arrangement 1: Cars S and V also should be parked next to each

Arrangement 2: P and Q cannot be parked next to each other.

Arrangement 3: Q and S must be parked next to each other.

Arrangement 4: R is parked to the immediate right of V.

Arrangement 5: T is parked to the left of U.

From Arrangement 1 and Arrangement 4 the arrangement will be,

From Arrangement 3,

From Arrangement 2 and Arrangement 5 the final arrangement will be,

Conclusion:

V is the only car parked in between S and R: Right

Car P is parked at the extreme end: Right

Q and R are not parked together: Right

There are two cars parked in between Q and V: Wrong

The correct answer is Water.

Key Points

• An Oasis is a small fertile or green (wet) area that springs up out of the desert. Desert is covered with sand.
• An Island is a small land (dry) area that springs up out of the ocean. Ocean is filled with water.

Thus, Oasis is to sand as island is to water. This maintains a similar logical relation.

Length of the side of larger square L₁ = 10 cm

The first small square is cutting midway of the larger square, so  \(L_2={10\over \sqrt 2}; cm\)

The second small square is cutting midway of the first small square, so \(L_3={{10\over (\sqrt 2)^2}}; cm\)

Similarly, 5 square are made, so the length of the smallest square \(L_6={{10\over (\sqrt 2)^5}}; cm\)

Area of the smallest square = (L₆)² = \([{{10\over (\sqrt 2)^5}}]^2; (cm)^2 = 3.125;{cm}^2\)

The area of the smallest square (shaded) = 3.125 cm²

Area of largest square = 100 cm²

The first small square is dividing in half, so its area = \({100\over 2};{cm}^2\)

Similarly, 5 squares are made, area of smallest square = \({100\over 2^5};{cm}^2 = 3.125;{cm}^2\)

Concept:

The formula for the interior angle (θ_i) of a 'n' side regular polygon:

 \(θ _i={{(n-2)\over n}\times180}\)

Calculation:

Given:

n = 10

\(\theta _i={{(10-2)\over 10}\times180} = 144\)

∴ The interior angle = 144°

The pattern which will appear when the transparent sheet is folded along the dotted line is,

Let, X, be the number of tosses,

The probability that a coin toss "head" is p, where 0 < p < 1

The coin is tossed repeatedly until the first "head" appears.

\(E\left[ X \right] = \mathop \sum \limits_{i = 1}^n {X_i}p\left( {{X_i}} \right) = p + 2p\left( {1 - p} \right) + 3p{\left( {1 - p} \right)^2} + \ldots \)

\(E\left[ X \right] = p\left[ {1 + 2\left( {1 - p} \right) + 3{{\left( {1 - p} \right)}^2} + \ldots } \right]\)

Let,

(1 - p) = a

\(E\left[ X \right] = p\left[ {1 + 2a + 3{a^2} + \ldots } \right]\)

Using Binomial expression,

\(\left( {1 + 2a + 3{a^2} + \ldots } \right) = {\left( {1 - a} \right)^{ - 2}}\)

\(E\left[ X \right] = p{\left( {1 - \left( {1 - p} \right)} \right)^{ - 2}}\)

\(E\left[ X \right] = p{p^{ - 2}}\)

\(E\left[ X \right] = \frac{1}{p}\)

Given waveform,

Finding RMS value of clipping (V_cr) waveform over the period from 0 to π.

\({V_{cr}} = \sqrt {\frac{1}{\pi }\left[ {\mathop \smallint \limits_0^{\frac{\pi }{4}} V_m^2{{\sin }^2}\left( t \right)dt + \mathop \smallint \limits_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} {{\left( {\frac{{{V_m}}}{{\sqrt 2 }}} \right)}^2}dt + \mathop \smallint \limits_{\frac{{3\pi }}{4}}^\pi V_m^2{{\sin }^2}\left( t \right)dt} \right]} \)

\({V_{cr}} = \sqrt {\frac{1}{\pi }\left[ {2 \times \mathop \smallint \limits_0^{\frac{\pi }{4}} V_m^2{{\sin }^2}\left( t \right)dt + \mathop \smallint \limits_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} {{\left( {\frac{{{V_m}}}{{\sqrt 2 }}} \right)}^2}dt} \right]} \)

\({V_{cr}} = \sqrt {\left[ {\frac{{2V_m^2}}{2}\mathop \smallint \limits_0^{\frac{\pi }{4}} (1 - \cos \left( {2t} \right))dt + \frac{{2V_m^2}}{2}\mathop \smallint \limits_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} dt} \right]} \)

\({V_{cr}} = \sqrt {\frac{1}{\pi }\left[ {V_m^2\left{ {\frac{\pi }{4} - \frac{1}{2}\sin \left( {\frac{\pi }{2}} \right);} \right} + \left{ {\frac{{V_m^2}}{2}\left( {\frac{{3\pi }}{4} - \frac{\pi }{4}} \right)} \right}} \right]} \)

\({V_{cr}} = \frac{{{V_m}}}{{\sqrt \pi }};\sqrt {\frac{\pi }{4} - \frac{1}{2} + \frac{\pi }{4};;} \)

\({V_{cr}} = 0.584{V_m}\)

RMS value of full-wave rectified waveform (V_fr) = 70.7% of the maximum value.

\({V_{fr}} = 0.707{V_m}\)

The ratio of the RMS value of the full-wave rectified waveform to the RMS value of the clipped waveform:

\(Ratio = \frac{{{V_{fr}}}}{{{V_{cr}}}} = \frac{{0.707{V_m}}}{{0.584{V_m}}} = 1.21\)

KCL: The algebraic sum of currents flowing into a node is equal to the algebraic sum of currents flowing out of that same node.

Apply KCL at node x, assuming all currents leaving the node

\(\frac{{V_x-6}}{{1}}+\frac{{V_x}}{{4}}+\frac{{V_x-V_y}}{{2}}=0\;\4V_x-24+V_x+2V_x-2Vy=0\;\7V_x-2V_y=24;;;....(1)\)

Given V_y = 0 V

Then V_x = 24/7 V

Apply KCL at node y, assuming all currents leaving the node

\(\frac{{V_y-V_x}}{{2}}-2+\frac{{V_y-β V_x}}{{3}}=0\;\3V_y-3V_x-12+2V_y-2β V_x=0\;\5V_y-(3+2β )V_x=12;;.....(2)\)

Substitute Vx and Vy in equation(2), we get

\(-(3+2β )\frac{{24}}{{7}}=12\)

β = - 3.25

Concept:

Two cables are connected in series

• Their conductor resistance will be added in series to get the resultant conductor resistance.
• Their insulation resistance will be added parallelly to get the resultant insulation resistance.

Calculation:

Given, 

Conductor resistance of cable-1 (R₁) = 0.7 Ω

Conductor resistance of cable-2 (R₂) = 0.5 Ω

Insulation resistance of cable-1 (R_i1) = 600 MΩ

Insulation resistance of cable-2 (R_i2) = 900 MΩ

Two cables are joined in series

Resultant conductor resistance of cable (R_eq) = R₁ + R₂ = 0.7 + 0.5 = 1.2 Ω

Resultant insulation resistance of cable (R_ieq) = Ri1 || Ri2 = (600 × 900) / (600 + 900) = 360 MΩ

The ratio of insulation resistance to conductor resistance = Rieq / Req = 300 M = 300 × 10⁶.

Additional Information  

Insulation resistance is inversely proportional to the length of the cable

\(R = \frac{\rho }{{2\pi l}}\log \frac{{{r_2}}}{{{r_1}}}\)

\(\Rightarrow R \propto \frac{1}{l}\)

where, r2 = Radius of cable

r1 = Radius of the core conductor 

l = Length of cable

Given:

Output (Y) is high (1) for the time period of 24 ms.

                              Q₁₅        Q₁₄      Q₁₃  ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅   Q₀

Y = 1 begins at:    0100     0000    0000 ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅  0000

Y = 1 end at:         1111     1111     1111  ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅  1111

Number of states in which Y is low (0) = 2¹⁴

Total number of states = 2¹⁶

Number of states in which Y is high (1) = 2¹⁶ - 2¹⁴

Time when Y is high (1) in 216 - 214 = 24 × 10^-3 s

Total time (T) = \(24;\times 10^{-3} \over 2^{16};-;2^{14} \)

Frequency = \({1\over {T}} = {2^{16};-2^{14} \over {} 24;\times;10^{-3}} = 2.048;MHz \)

Concept:

The force acting on a charged particle moving under the presence of a magnetic field is given by:

F̅mag = q (v̅ × B̅)

v̅ = Velocity vector of the particle

B̅ = Magnetic field vector

q = Charge of the particle

Application:

Given, q = 1 C

v̅ = 10 \(\widehat x\) m/s

\(\overrightarrow B = \left( {10y\widehat x + 10x\widehat y + 10\widehat z} \right)T\)

F̅mag = q (v̅ × B̅)

F̅mag = \(1(10;\widehat x \times \left( {10y\widehat x + 10x\widehat y + 10\widehat z} \right));N\)

F̅_mag = \(100x\widehat z ;+100(-\widehat y) ; N\)

F̅mag (x = 0+) = - 100 y̅ N

Magnitude of F̅_mag = \(|\sqrt{(-100)^2}| = 100;N \)

Additional Information

The Lorentz force equation describes the magnitude of the force that a moving electric charge would feel as a result of being in the presence of a magnetic field B̅ and an electric field E̅.

The Force due to the presence of an electric field is given by:

F̅elec. = q × E̅ 

q = Charge of the particle

E̅ = Electric field vector

Also, the force acting on a charged particle moving under the presence of a magnetic field is given by:

F̅mag. = q (v̅ × B̅)

v̅ = Velocity vector of the particle

B̅ = Magnetic field vector

When the charge is moving in presence of both an electric and magnetic field, the force acting on it will be the sum of the two forces, i.e.

Fnet = F̅elec. + F̅mag. 

Fnet = q × E̅ + q (v̅ × B̅)

Fnet = q (E̅ + v̅ × B̅)

Concept:

Maximum Power Transfer Theorem:

The maximum power transfer theorem states that the maximum power transfer from the AC network to the load when the load impedance is equal to the complex conjugate of Thevenin’s equivalent impedance.

ZL = RL + jXL = Rth – jXth = Zth*

If the load is purely resistive, the load resistance is equal to the magnitude of Thevenin impedance.

\({R_L} = \left| {{Z_{th}}} \right| = \sqrt {R_{th}^2 + X_{th}^2} \)

Application:

Given, 

ω = 1000 rad/sec

XL = jωL = j × 1000 × 4 × 10-3 = j4 Ω 

XC = - j / ωC = - j / (1000 × 0.5 × 10-3)= - j2 Ω 

R = 2 Ω 

To find thevenins impedance, short circuit the voltage source and find equivalent impedance from the load terminal.

Z_th = R || (X_L + X_C) = 2 || j2 =\({2;\times; j2\over 2;+;j2} = {j2\over1;+;j1} \)

For maximum power transfer \({R_L} = \left| {{Z_{th}}} \right| = \sqrt {R_{th}^2 + X_{th}^2} \)

R_L = | Z_th | = \({{\sqrt {(2^2)}\over \sqrt {(1^2+1^2)}}} = {2\over \sqrt 2 } = \sqrt 2; Ω \)

∴ The value of R_L = 1.414 Ω

​Inductance can be measured by Maxwell bridge

Important Points

Given circuit,

The transfer function for series RLC circuit by taking output voltage acroos the capacitor is given by,

\(T\left( s \right) = \frac{{{V_O}\left( s \right)}}{{{V_i}\left( s \right)}} = \frac{1}{{LC{s^2} + RCs + 1}}\)

Put, s = jω

\( \frac{{{V_O}\left( s \right)}}{{{V_i}\left( s \right)}} = \frac{1}{{ - {\omega ^2}LC + j\omega RC + 1}}\)

\( \frac{{{V_O}\left( s \right)}}{{{V_i}\left( s \right)}} = \frac{1}{{ - \left( {{{2000}^2} \times 1m \times 250\mu } \right) + j\left( {250\mu } \right)R + 1}}\)

\( \frac{{{V_O}\left( s \right)}}{{{V_i}\left( s \right)}} = \frac{1}{{0.5R}}\)

From the plot,

20 log|M_r| = 26 dB

|M_r| = 19.95

19.95 × 0.5R = 1

R = 0.1 Ω

Alternate Method

Resonant Peak formula,

\({M_r} = \frac{1}{{2\zeta \sqrt {1 - {\zeta ^2}} }}\)

M_r = 19.95

ζ = 0.025

For RLC series circuit,

\(\zeta = \frac{R}{2}\sqrt {\frac{C}{L}} \)

\(0.025 = \frac{R}{2}\sqrt {\frac{{250\mu }}{{1m}}} \)

R = 0.1 Ω

Given,

\({G_p}(s) = \frac{{2.2}}{{(1 + 0.1s)(1 + 0.4s)(1 + 1.2s)}}\) ...(1)

\({G_c}(s) = K\left( {\frac{{1 + {T_1}s}}{{1 + {T_2}s}}} \right)\)  ....(2)

Maximum steady state error due to external input = 0.1

For step external disturbance input D(s) = 1/s  .....(3)

Assume R(s) = 0,

⇒ C(s) = - E(s)

Then the output expression is given as

C(s) = [E(s)Gc(s) + D(s)]Gp(s)

• E(s) = E(s)Gc(s)Gp(s) + D(s)Gp(s)

Error function \(E(s)=\frac{{-D(s)G_p(s)}}{{1+G_c(s)G_p(s)}}\)   ....(4)

Steady state error is given by

\(e_{ss}=\mathop {\lim }\limits_{s \to 0} s.E(s)\)

From equations (1), (2), (3)&(4)

\(|e_{ss}|=\mathop {\lim }\limits_{s \to 0} {\frac {{s.\frac {{1}}{{s}}G_p(s)}}{{1+G_c(s)G_p(s)}}}\)

\(e_{ss}≥\frac{{2.2}}{{1+2.2K}}\)

\(\frac{{2.2}}{{1+2.2K}}\le0.1\)

K ≥ 9.54

Minimum value of K is 9.54

Concept:

Consider a function y = f(x) on a defined interval of x.

The function attains extreme values (the value can be maximum or minimum or both).

For maxima:

• Local maxima: A point is the local maxima of a function if there is some other point where the maximum value is greater than the local maxima but that point doesn’t exist nearby local maxima.
• Global maxima: It is the point where there is no other point has in the domain for which function has more value than global maxima.

For minima:

• Local minima: A point is the local minima of a function if there is some other point where the minimum value is less than the local minima but that point doesn’t exist nearby local minima.
• Global minima: It is the point where there is no other point has in the domain for which function has less value than global minima.

Stationary Points: Points where the derivative of the function is zero i.e., f’(x) = 0. The points can be:

• Inflection point
• Local maxima
• Local minima

Second derivative test: Let the function has a stationary point x = a

• If \({\left( {\frac{{{d^2}f}}{{d{x^2}}}} \right)_{x = a}} < 0\)  then x = a, is a point of maxima.
• If \({\left( {\frac{{{d^2}f}}{{d{x^2}}}} \right)_{x = a}} > 0\)  then x = a, is a point of minima.

Application:

Given f(x) is a real -valued function such that f'(x0) = 0 for some x0 ∈ (0, 1)

Also given f"(x) > 0 for all x ∈ (0, 1)

So, Then f(x) has exactly one local minimum in (0, 1), called the point of minima.

Given:

Single-phase full-bridge inverter

Supply voltage (V_S) = 325 V

Modulation index (m) = 0.8

The modulation index (m) is defined as the ratio of the peak of the fundamental component (V_01P) of Vab to the applied DC value.

\(m = {{V_{01P} \over V_S}} \)

\(V_{01P} = {{4V_S \over \pi}}sin;d\)

Where d = Half of the conduction angle (α)

\(m={{{{4V_S}\over \pi}sin;d}\over V_S} = {{4}\over \pi}sin;d\)

\(0.8 = {{4\over \pi}sin;d} \)

d = 38.93° 

θ = 90° - d = 90° - 38.93° = 51.07°

Number of buses in the system = N

Number of PQ buses = N_L

Number of slack buses = 1

Number of PV buses = N - 1 - N_L

Newton Raphson method for load flow study in polar form

\(\left[ {\begin{array}{*{20}{c}} {Δ P}\ {Δ Q} \end{array}} \right] = J\left[ {\begin{array}{*{20}{c}} {Δ δ }\ {Δ V} \end{array}} \right]\)

Where \(J = \left[ {\begin{array}{*{20}{c}} H&S\ M&R \end{array}} \right]\)

The submatrix M relates between [ΔQ] and [Δδ]

Number of elements in ΔQ vector = Number of known Q 

Number of elements in ΔQ vector = Number of PQ buses = N_L

Number of elements in Δδ vector = Number of unknown δ = N - 1

Size of matrix M = NL × (N - 1)

Size of other sub-matrix:

H = (N - 1) × (N - 1)

S = (N - 1) ×  N_L

R = NL × NL

Concept:

Condition for orthogonal:

m1 ⋅ m2 = - 1

Where, m1 = Slope of curve-1 

m2 = Slope of curve-2

Calculation:

Given:

Curve-1: 

x2 + y2 = 1

Differentiating both side,

\(2x;+;2y{dy\over{dx}} =0 \)

\({dy\over dx}=m_1={-x\over y}\)

Curve-2: 

(x - 1)2 + (y - 1)2 = r2

Differentiating both side,

\(2(x-1)+2(y-1){dy\over dx} = 0\)

\({dy\over dx}=m_2 ={{1-x}\over {y-1}}\)

Both curves are intersecting each other at the point (u,v), so x and y are replaced by u and v.

m1 ⋅ m2 = - 1

\(m_1 \cdot m_2 = {[{{({-u\over v})}{({1-u})\over {({v-1})}}}]}=-1\)

• u + u2 = - v2 + v

u2 + v2 = u + v

∵ x2 + y2 = 1 ⇒ u2 + v2 = 1

∴ u + v = 1

Given:

Forward path gain = G

Feedback path gain = H

Input signal = R(s)

Output signal = C(s)

Error signal = E(s)

C(s) = G × E(s)

Error signal = Input signal - Feedback signal

E(s) = R(s) - H × C(s)

E(s) = R(s) - H × G × E(s)

E(s) + H × G × E(s) = R(s) 

E(s) [1 + GH] = R(s)

\({E(s)\over R(s)} = \dfrac{1}{ (1+GH)}\)

Concept:

Economic distribution of the loads between the units of the power plant:

To determine the economic distribution of a load amongst the different units of a plant, the variable operating costs of each unit must be expressed in terms of its power output.

The fuel cost is the main cost and it must be expressed in Rs./ hour

The fuel cost of a unit can be expressed as

Fi = a Pi2 + b Pi + c Rs / hour

The incremental cost is expressed as

\(IC_i = \frac{dF_i}{dP_i} = 2aP_i + b\) in Rs / MWh or Rs / kWh.

For the economic operation of the system, the incremental cost of each generator must be equal.

Calculation:

Total delivering power is

P1 + P2 = 260 MW .....(1)

The incremental cost of Generator-1 is 

\(IC_1=\frac{dF_1}{dP_1}=40+0.2P_1\)

The incremental cost of Generator-2 is 

\(IC_2=\frac{dF_2}{dP_2}=32+0.4P_2\)

For the optimum cost, the incremental cost of each generator must be equal.

IC1 = IC2 

⇒ 40 + 0.2 P1 = 32 + 0.4 P2

⇒ 0.2 P₁ - 0.4 P₂ = - 8 .....(2)

By solving equations (1) and (2), we get

P1 = 160 MW

P2 = 100 MW

Concept:

Boost Converter:

It is a switch mode DC to DC converter in which the output voltage is greater than the input voltage. It is also called a step-up converter.

A typical Boost converter is shown below.

\(D = \frac{{{T_{ON}}}}{T}\)

\(\frac{{{V_o}}}{{{V_s}}} = \frac{1}{{\left( {1 - D} \right)}}\)

Where V₀ is the output voltage and V_s is the input voltage.

The average source current (Is) given by

\({I_s} = \frac{{{I_o}}}{{\left( {1 - D} \right)}} \)

Calculation:

Given,

D = 0.6

V_s = 15 volt

From the above concept,

\({V_0} = \frac{{{V_s}}}{{\left( {1 - D} \right)}} = \frac{{15}}{{\left( {1 - 0.6} \right)}} = 37.5;volt\)

Current through the load (I_o),

\({I_o} = \frac{{{V_o}}}{R}\)

R = 10 Ω

\({I_o} = \frac{{37.5}}{{10}} = 3.75;A\)

The average source current (I_s) given by,

\({I_s} = \frac{{{I_o}}}{{\left( {1 - D} \right)}} = \frac{{3.75}}{{\left( {1 - 0.6} \right)}} = 9.375;A\)

V_s = I_s × R_s

\({R_s} = \frac{{{V_s}}}{{{I_s}}} = \frac{{15}}{{9.375}} = 1.6;{\rm{\Omega }}\)

Important Points

Given:

High output voltage level = 5 V

Lower threshold voltage = 1.6 V

Upper threshold voltage = 2.4 V

Resistance (R) = 10 kΩ 

Capacitance (C) = 47 nF

t₁ = Capacitor charging time

t₂ = Capacitor discharging time

Capacitor charging diagram is given as:

In case of charging voltage across capacitor V_c(t) is given as:

V_c(t) = V_cf + (V_ci - V_cf) e^-t/RC

Where V_cf = Final value of voltage across capacitor (= 5 V)

V_ci = Initial value of voltage across capacitor (= 1.6 V)

In case of discharging voltage across capacitor V_c(t) is given as:

V_c(t) = V_ci e^-t/RC

RC = 10 × 10³ × 47 × 10^-9 = 470 × 10^-6

At the end of charging time (t₁), voltage across capacitor is = 2.4 V

V_c(t) = 2.4 = 5 + (1.6 - 5) e^{-t₁/(470 ×} 10^-6)

t₁ = 126.1 × 10^-6 s

At the end of discharging time (t₂), voltage across capacitor is = 1.6 V

V_c(t) = 1.6 = 2.4 e^{-t₂/(470 ×} 10^-6)

t₂ = 190.57 × 10^-6 s

T = t₁ + t₂ = 126.1 × 10^-6 + 190.57 × 10^-6 = 316.67 × 10^-6 s

Frequency = 1/T = 1/(316.67 × 10^-6) = 3157.87 Hz

Hunting:

Oscillations of the rotor about its new equilibrium position, due to sudden application or removal of load is called swinging or hunting in the synchronous machine.

Causes for Hunting:

• Sudden change in load.
• Sudden change in the supply system or in the field system.
• Load containing harmonic torque.

​

Methods for eliminating hunting:

1. By designing the machine with a suitable synchronizing power coefficient.
2. By using the flywheel.
3. By using damper winding.

​

Damper winding or bars:

• Damper winding is made with low resistance copper, aluminum, or brass.
• They are inserted in the slots made under the pole shoes.
• With respect to damper winding, the rotor behaves like a squirrel cage rotor of an induction motor.

​

Functions of damper winding:

1. In alternator to eliminate hunting and to suppress the negative sequence field.
2. In synchronous motor to eliminate hunting and for starting purpose.

Skewing:

• ​In the squirrel cage induction motor, the rotor slots in lamination or rotor core are not made parallel to the rotor shaft. A slight angle is maintained between the rotor slots and the rotor shaft due to some operational advantages. This is called rotor skewing.
• As the rotor slot of the induction motor skewed through some angle so that the bars lie under alternate harmonic poles of the same polarity.​

​The functions of skewed rotor slots in induction motor:

1. The skewed rotor slot increases the length of the copper bar thereby increases the resistance of the rotor bars hence starting torque of the machine can be improved and also the starting current is drawn by the machine can be reduced.
2. It makes the air gap flux distribution uniform thereby reduces harmonic torque produced by the machine.
3. As harmonic torque is reduced, the cogging or magnetic locking phenomenon due to harmonic torque can also be reduced. Here the magnetic locking tendency occurs when rotor teeth remain directly under stator teeth thus they might be magnetically locked.
4. And we can also eliminate a particular harmonic by selecting a proper skew angle.
5. The humming noise can be reduced.

Compensating windings:

• The cross-magnetizing effect of armature reaction may cause trouble in dc machines subjected to large fluctuations in load.
• In order to neutralize the cross magnetizing effect of the armature reaction, a compensating winding is used. The compensating windings consist of a series of coils embedded in slots in the pole faces. These coils are connected in series with the armature.
• The series-connected compensating windings produce a magnetic field, which varies directly with armature current. Because the compensating windings are wound to produce a field that opposes the magnetic field of the armature, they tend to cancel the cross magnetizing effect of the armature magnetic field.

Given circuit,

Assume that, Phase sequence is RYB,

Hence,

V_RY = V∠0°,

V_YB = V∠-120°,

VBR = V∠120°

Now,

\({I_R} = ;\frac{{{V_{RB}}}}{{ - j10}} = \frac{{ - {V_{BR}}}}{{ - j10}} = \frac{{V\angle 120^\circ }}{{j10}}\)

Similarly,

\({I_Y} = ;\frac{{{V_{YB}}}}{{j10}} = \frac{{V\angle - 120^\circ }}{{j10}}\)

Apply KCL at node (N),

I_B = - (I_R + I_Y)

Or, \(\frac{{{I_B}}}{{{I_R}}} = - \left( {1 + \frac{{{I_Y}}}{{{I_R}}}} \right)\)

\( = - \left( {1 + \frac{{\frac{{V\angle - 120^\circ }}{{j10}}}}{{\frac{{V\angle 120^\circ }}{{j10}}}}} \right)\)

\( = - \left( {1 + \frac{{\left( { - 0.5 - j0.866} \right)}}{{\left( { - 0.5 + j0.866} \right)}}} \right)\)

Or, \(\frac{{{I_B}}}{{{I_R}}} = - \left( {1 + \frac{{\left( { - 1 - j\sqrt 3 } \right)}}{{\left( { - 1 + j\sqrt 3 } \right)}}} \right)\)

Or, \(\frac{{{I_B}}}{{{I_R}}} = - \left( {\frac{{\left( { - 1 + j\sqrt 3 } \right) + \left( { - 1 - j\sqrt 3 } \right)}}{{\left( { - 1 + j\sqrt 3 } \right)}}} \right)\)

Or, \(\frac{{{I_B}}}{{{I_R}}} = \left( {\frac{2}{{\left( { - 1 + j\sqrt 3 } \right)}}} \right)\)

Or, \(\frac{{{I_B}}}{{{I_R}}} = \left( {\frac{2}{{\left( { - 1 + j\sqrt 3 } \right)}}} \right) \times \frac{{\left( { - 1 - j\sqrt 3 } \right)}}{{\left( { - 1 - j\sqrt 3 } \right)}} = \frac{{2\left( { - 1 - j\sqrt 3 } \right)}}{4} = \frac{{\left( { - 1 - j\sqrt 3 } \right)}}{2}\)

Hence, \(\left| {\frac{{{I_B}}}{{{I_R}}}} \right| = \frac{{\sqrt {{1^2} + {{\left( {\sqrt 3 } \right)}^2}} }}{2} = 1\)

Concept:

The z-transform of a unit impulse function or Kronecker delta δ [n] ↔ 1

The time-shifting affects the z-transform as:

x[n - n0] = z -n0 X(z)

Application:

Given:

h1[n] = δ[n - 1] + δ[n + 1] 

h2[n] = δ[n] + δ[n - 1] 

If h₁[n]and h₂[n] are cascaded connected then h[n] = h₁[n] * h₂[n]

Where '*' denotes convolution.

h[n] = h1[n] * h2[n]

Taking z-transform both side

H[z] = H₁[z] ⋅ H₂[z]

H[z] = (z^-1 + z) ⋅ (1 + z^-1) = (z^-1 + z-2 + z + 1 )

Taking inverse z-transform both side

h[n] = δ[n-1] + δ[n-2] + δ[n+1] + δ[n]

∴ Impulse response of the cascaded system is δ[n - 2] + δ[n - 1] + δ[n] + δ[n + 1]

Given:

Pole (P) = 8

Frequency (f) = 50 Hz

Rotor resistance (r₂) = 0.08 Ω 

Rotor speed at maximum torque = 650 RPM

Let's consider the slip at maximum torque as s_m,

Slip at maximum torque is given by

\(s_m = \dfrac {r_2}{x_2}\)

Consider additional rotor resistance to be added = r₂'

Rotor reactance = x₂

Synchronous speed N_s = 120f/P = 120 × 50 / 8 = 750 RPM

As the speed of the motor at maximum toque condition is 650 RPM

Slip at maximum torque is given by

\(s_m = \dfrac {N_s - N_r}{N_s} =\dfrac {750-650}{750}= 0.1333\)

At maximum torque

\(s_m =\dfrac {r_2}{x_2} ⇒ 0.1333 = \dfrac {0.08}{x_2}\)

x₂ = 0.6 Ω 

To get the maximum torque at starting the slip at maximum torque should be equal to slip at starting (s=1).

⇒ \(1 =\dfrac {{r_2+r_2'}}{ x_2}\)

x₂ = r₂ + r₂'

0.6 = 0.08 + r₂'

r₂' = 0.52 Ω 

∴ The value of additional rotor resistance to be added (r2') at the time of starting is 0.52 Ω.

Concept:

The current through the inductor in transient is given by:

\(i\left( t \right) = i(\infty ) + \left( {i\left( 0 \right) - i\left( ∞ \right)} \right){e^{ - \frac{{tR}}{L}}}\) ....(1)

i(∞) = Steady-state/final value of the inductor

i(0) = Initial stored current

Also, an inductor does not allow a sudden change in voltage, i.e.

i(0-) = i(0+)

τ = Time Constant

The time constant in the RL series circuit is given by:

\(τ=\dfrac{L}{R}\)

Calculation:

For t < 0:

Initially, for t < 0, the switch is closed for a long time.

The 4 Ω resistor and 30 V source are bypassed and the inductor is short-circuited.

Current flowing in the circuit will be:

\(i_L(0^-)=\dfrac{10}{1}=10~A\)

For t > 0:

After an infinite time when the switch is closed, the inductor is short-circuited and the current will be:

\(i_L(\infty)=\dfrac{10+30}{4+1}=8~A\)

τ = L/R

\(\rm \tau = \frac{{\frac{1}{2}}}{5} = \frac{1}{{10}}\)

From equation (1)

i_L(t) = 8 + [10 - 8] e^-10t

i_L(t) = 8 + 2e^-10t

Concept:

Intermediate Value Theorem:

It states that if 'f' is a continuous function whose domain contains the interval [a, b], then it takes any given value between f(a) and f(b) at some point within the interval.

Considerd an interval of I = [a, b], and a continuous function f(x) then,

If 'u' is the number between f(a) and, f(b), then

\(\min \left[ {f\left( a \right),;f\left( b \right)} \right] < u < \max \left[ {f\left( a \right),f\left( b \right)} \right]\)

\(c \in \left( {a,b} \right);and,;f\left( c \right) = u\)

Calculation:

Given polynomial,

p(x) = x4 - 4x3 + 2

Using intermediate value theorem,

p(0) = 0 - 0 + 2 = 2

p(1) = 1 - 4 + 2 = -1

Since,

p(0) > 0 ....

(1) p(1) < 0 .... (2) From equation (1) and (2), there exist a value 'x' in between 0 and 1 where,

p(x) = 0

Hence, in the open interval (0, 1), the polynomial p(x) has one real root.

Concept:

EMF equation of the DC machine is given as

\(E = \dfrac{{{ZP\phi N}}}{60A } \)   ......(1)

Where,

N is the speed in rpm

ϕ is the flux per pole

P is the number of poles

Z is the number of conductors

Calculation:

Machine acting as a generator: 

Given,

Output power P = VI_L = 100 kW

Terminal voltage (V) = 200 V

⇒ 200 × I_L = 100000

Load current (I_L) = 500 A

Shunt current (I_sh) = V / R_sh = 200/50 = 4 A

Brush drop voltage V_b = 2 V

Armature current (I_a1) = I_L + I_sh = 504 A

Speed (N₁) = 300 RPM

Armature resistance (R_a) = 0.025 Ω

Generated emf, E_g = V + I_a1R_a + V_b

Eg = 200 + 504 × 0.025 + 2 

Eg = 214.6 V

Machine acting as a motor: When the belt broke it behaves as a motor

Input power P = VIL = 10 kW

Terminal voltage (V) = 200 V

⇒ 200 × I_L = 10000

Load current (I_L) = 50 A

Armature current (I_a2) = I_L - I_sh = 46 A

Speed of motor = N₂

Back emf, E_b = V - I_a2R_a - V_b

Eb = 200 - 46 × 0.025 - 2 

Eb = 196.85 V

From equation(1), E ∝ N

\(\dfrac{{E_g}}{{E_b}}=\dfrac{{N_1}}{{N_2}}\)

\(\dfrac{{214.6}}{{196.85}}=\dfrac{{300}}{{N_2}}\)

N₂ = 275.19 RPM

Concept:

A series RC circuit is drawn as:

The voltage across the capacitor using the voltage division rule is:

\({V_c}\left( ω \right) = V_s\left( {\frac{ \frac{{{1}}}{{jω C}}}{{R + \frac{1}{{jω C}}}}} \right)\)

\({V_c}\left( ω \right) = \frac{{{V_s}}}{{jω C}}\left( {\frac{{jω C}}{{1 + Rjω C}}} \right)\)

\({V_c}\left( ω \right) = \frac{{{V_s}}}{{1 + jω RC}}\) ........

(1) Where, ω = 2πf

R is the source resistance.

V_s is the source voltage.

Calculation:

Given, f = 1 kHz, R = 50 Ω 

Output peak to peak voltage under open circuit condition is the source peak to peak voltage

Peak to peak V_s = peak to peak V_o = 10 V

Peak V_s = 5 V

When the capacitor is connected across the open circuit terminals, the voltage across the capacitor is given

Peak to peak V_c = 8 V

Peak V_c = 4 V

∴ From equation(1) 

\({1 + jω RC} = \frac{{{V_s}}}{{V_c}}\)

Now considering magnitude on both sides,

\(\sqrt{1 + (ω RC)^2} = \frac{{{V_s}}}{{V_c}}\ ;\ω RC=\sqrt{\left(\frac{{{V_s}}}{{V_c}}\right)^2-1}\;\C=\frac{{1}}{{2\pi f R}}\sqrt{\left(\frac{{{V_s}}}{{V_c}}\right)^2-1}\;\C=\frac{{1}}{{2\pi \times 1000\times 50}}\sqrt{\left(\frac{{{5}}}{{4}}\right)^2-1}\)

C = 2.38 μF

Concept:

Thevenin’s Theorem:

Thevenin’s theorem states that a linear-bilateral two-terminal circuit can be replaced by an equivalent circuit consisting of a voltage source Vth in series with a resistor Rth.

Where V_th is the open-circuit voltage at the terminals and R_th is the input or equivalent resistance at the terminals when the independent sources are turned off.

Different case of Thevenin's theorem:

Case 1: Circuit with only dependent source,

In this case, R_th is finding by the voltage test method

And, V_th = 0

Case 2: Circuit with both dependent source and independent source,

In this case, Rth is finding by the voltage test method

And, Vth is finding by the same method above discuss.

Calculation:

Given circuit,

Step 1: Finding Vth across terminal ab,

V_th = V_10Ω + 3i₁

5 A = i₁ + 0

∴ i₁ = 5 A

V_th = 10(5) + 3(5) = 65 volt

Step 2: Finding R_th across terminal ab,

To finding R_th we have to open-circuited the current source by its internal resistance and applied test voltage across the terminal ab,

\({R_{th}} = \frac{{{V_T}}}{{{I_T}}}\)

I_T = i₁

V_T = 13i₁ + 2i₁ = 15I_T

\({R_{th}} = \frac{{{V_T}}}{{{I_T}}} = \frac{{15}}{1} = 15\Omega \)

Given:

Let's consider,  V₀ = Load voltage

I₀ = Load current

V_dc = Supply voltage

Duty cycle = D

Duty cycle (D) = 0.75

\({V_0} = {V_{dc}}\left[ {\frac{D}{{1 - D}}} \right] = {20[{0.75\over {1-0.75}}]} = 60; V \)

\(I_0 = {V_0\over R} = {60 \over 10} = 6;A\)

\((I_L)_{avg} = {I_0\over {1-D}} = {6\over {1-0.75}} = 24;A\)

Additional Information

Identification of DC-DC Converter:

• Buck converter: Inductor is connected in series with the combination of the capacitor and resistor.
• Boost converter: Inductor is connected in series with the supply voltage source.
• Buck-Boost converter: Inductor is connected in parallel to the supply voltage source.

Under steady-state operating conditions:

• The average voltage across the inductor is zero. VL = 0
• The average current across the capacitor is zero. IC = 0

Concept:

Bus Admittance Matrix (Y_Bus):

Consider a small power system network consisting of two generating stations, three transmission lines, one load, and a static capacitor connected to load bus 3.

Assumed that the network is symmetrical and operating under the balance conditions.

The node voltage equation of the system can be written as,

\({I_1} = \left( {{y_{12}} + {y_{31}}} \right){V_1} - {y_{12}}{V_2} - {y_{31}}{V_3}\)

\({I_2} = - {y_{12}}{V_1} + \left( {{y_{12}} + {y_{23}}} \right){V_2} - {y_{23}}{V_3}\)

\( - {I_3} = - {y_{31}}{V_1} - {y_{23}}{V_2} + \left( {{y_{31}} + {y_{23}} + {y_{30}}} \right){V_3}\)

Where,

\({y_{12}} = \frac{1}{{{z_{12}}}},;{y_{23}} = \frac{1}{{{z_{23}}}},;{y_{31}} = \frac{1}{{{z_{31}}}}\)

The above equation can be written in form of a matrix,

\(\left[ {\begin{array}{*{20}{c}} {{I_1}}\ {{I_2}}\ { - {I_3}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {{y_{12}} + {y_{31}}}&{ - {y_{12}}}&{ - {y_{31}}}\ { - {y_{12}}}&{{y_{12}} + {y_{23}}}&{ - {y_{23}}}\ { - {y_{31}}}&{ - {y_{23}}}&{{y_{31}} + {y_{23}} + {y_{30}}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{V_1}}\ {{V_2}}\ {{V_3}} \end{array}} \right]\)

\(\left[ {\begin{array}{*{20}{c}} {{I_1}}\ {{I_2}}\ { - {I_3}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {{Y_{11}}}&{{Y_{12}}}&{{Y_{13}}}\ {{Y_{21}}}&{{Y_{22}}}&{{Y_{23}}}\ {{Y_{31}}}&{{Y_{32}}}&{{Y_{33}}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{V_1}}\ {{V_2}}\ {{V_3}} \end{array}} \right]\)

Where,

Y₁₁ = y₁₂ + y₃₁, Y₂₂ = y₂₁ + y₂₃, Y₃₃ = y₃₀ + y₃₁ + y₃₂

Y₁₂ = Y₂₁ = - y₁₂, Y₂₃ = Y₃₂ = -y₂₃, Y₃₁ = Y₁₃ = -y₃₁

The element Y₁₁, Y₂₂, Y₃₃ is termed as self admittances.

The elements Y₁₂, Y₁₃, Y₂₁, Y₂₃, Y₃₁, Y₃₂ termed as mutual admittance.

Calculation:

Given circuit can be drawn as,

Applying KCL at node V,

I₁ + I₂ + I₃ + I₄ = 0

\(\frac{{{E_1} - V}}{{j1}} + \frac{{{E_2} - V}}{{j1}} + \frac{{{E_3} - V}}{{j1}} + \frac{{ - V}}{{j1}} = 0\)

4V = E₁ + E₂ + E₃

\({I_1} = \frac{{{E_1} - V}}{{j1}} = \frac{{{E_1} - \left( {\frac{{{E_1} + {E_2} + {E_3}}}{4}} \right)}}{{j1}}\)

\({I_1} = j\frac{{ - 3}}{4}{E_1} + j\frac{1}{4}{E_2} + j\frac{1}{4}{E_3}\)

\({I_2} = \frac{{{E_2} - V}}{{j1}} = \frac{{{E_2} - \left( {\frac{{{E_1} + {E_2} + {E_3}}}{4}} \right)}}{{j1}}\)

\({I_2} = j\frac{1}{4}{E_1} + j\frac{{ - 3}}{4}{E_2} + j\frac{1}{4}{E_3}\)

\({I_3} = \frac{{{E_3} - V}}{{j1}} = \frac{{{E_3} - \left( {\frac{{{E_1} + {E_2} + {E_3}}}{4}} \right)}}{{j1}}\)

\({I_1} = j\frac{1}{4}{E_1} + j\frac{1}{4}{E_2} + j\frac{{ - 3}}{4}{E_3}\)

The equation of current can be written in form of the matrix,

\(\left[ {\begin{array}{*{20}{c}} {{I_1}}\ {{I_2}}\ {{I_3}} \end{array}} \right] = j\left[ {\begin{array}{*{20}{c}} {\frac{{ - 3}}{4}}&{\frac{1}{4}}&{\frac{1}{4}}\ {\frac{1}{4}}&{\frac{{ - 3}}{4}}&{\frac{1}{4}}\ {\frac{1}{4}}&{\frac{1}{4}}&{\frac{{ - 3}}{4}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{E_1}}\ {{E_2}}\ {{E_3}} \end{array}} \right]\)

\({Y_{bus}} = j\left[ {\begin{array}{*{20}{c}} {\frac{{ - 3}}{4}}&{\frac{1}{4}}&{\frac{1}{4}}\ {\frac{1}{4}}&{\frac{{ - 3}}{4}}&{\frac{1}{4}}\ {\frac{1}{4}}&{\frac{1}{4}}&{\frac{{ - 3}}{4}} \end{array}} \right]\)

\(Y_{bus}=\left[ \begin{array}{20 {c*}} {-\frac 3 4 j}&{\frac 1 4 j}&{\frac 1 4 j}\\ {\frac 1 4 j}&{-\frac 3 4 j}&{\frac 1 4j}\\ {\frac 1 4 j}&{\frac 1 4 j}&{-\frac 3 4 j} \end{array} \right]\)

Concept:

Time Response of the second-order System:

\(T\left( s \right) = \frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{{G\left( s \right)}}{{1 + G\left( s \right)H\left( s \right)}}\)

\(G\left( s \right) = \frac{{\omega _n^2}}{{s\left( {s + 2\zeta {\omega _n}} \right)}},;H\left( s \right) = 1\)

\(T\left( s \right) = \frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{{\frac{{\omega _n^2}}{{s\left( {s + 2\zeta {\omega _n}} \right)}}}}{{1 + \frac{{\omega _n^2}}{{s\left( {s + 2\zeta {\omega _n}} \right)}}}}\)

\(T\left( s \right) = \frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{{\omega _n^2}}{{[{s^2} + 2\zeta {\omega _n}s + \omega _n^2]}}\)

In the above transfer function, the power of 's' is two in the denominator therefore it is called a second-order system.

\({\omega _d} = {\omega _n}\sqrt {1 - {\zeta ^2}}\)

Where ω_d is damped natural frequency in rad/sec.

Calculation:

Given,

\(G_p(s) = \frac {14.4}{s(1 + 0.1s)}\)

Gc(s) = 1

H(s) = 1

G(s) = Gc(s)GP(s)

\(G(s) = \frac {14.4}{s(1 + 0.1s)}\)

The transfer function of the system will be,

\(T\left( s \right) = \frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{{G\left( s \right)}}{{1 + G\left( s \right)H\left( s \right)}}\)

\(T\left( s \right) = \frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{{\frac{{14.4}}{{s\left( {1 + 0.1s} \right)}}}}{{1 + \frac{{14.4}}{{s\left( {1 + 0.1s} \right)}}}}\)

\(T\left( s \right) = \frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{{14.4}}{{0.1{s^2} + s + 14.4}}\)

\(T\left( s \right) = \frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{{14.4}}{{0.1[{s^2} + 10s + 144]}}\)

\(T\left( s \right) = \frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{{144}}{{[{s^2} + 10s + 144]}}\) ....

(1) We know that,

\(T\left( s \right) = \frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{{\omega _n^2}}{{[{s^2} + 2\zeta {\omega _n}s + \omega _n^2]}}\) .... (2) From equation (1) and (2),

\(\omega _n^2 = 144\)

\({\omega _n} = 12\)

\(2\zeta {\omega _n} = 10\)

\(\zeta = \frac{{10}}{{24}} = \frac{5}{{12}}\)

\({\omega _d} = {\omega _n}\sqrt {1 - {\zeta ^2}} = 12\sqrt {1 - {{\left( {\frac{{10}}{{24}}} \right)}^2}} = 10.91;rad/sec\)

Concept:

Linearity: Necessary and sufficient condition to prove the linearity of the system is that the linear system follows the laws of superposition i.e. the response of the system is the sum of the responses obtained from each input considered separately.

y{ax1[t] + bx2[t]} = a y{x1[t]} + b y{x2[t]}

Conditions to check whether the system is linear or not.

• The output should be zero for zero input.
• There should not be any non-linear operator present in the system.

Time-Invariance: If the input to a time-invariant system is shifted in time, its output remains the same signal, but is shifted equally in time.

If the output for an input x(t) is y(t), then a time shift of t0 in the input gives the t0 shift in the output.

x(t) → y(t), then x(t – t0) → y(t – t0)

Application:

Given y(t) = max (0, x(t))

\(y(t)= \left{ {\begin{array}{*{20}{c}} {0,;;;;x(t) < 0}\ {x(t),x(t) > 0} \end{array}} \right.\)

Linearity checking:

For input x₁(t) = -1 , Output y₁(t) = 0

For input x₂(t) = 1 , Output y₂(t) = 1

y(x1(t) + x2(t)) = y(-1 + 1) = y(0) = 0

y1(t) + y2(t) = 0 + 1 = 1

⇒ y(x1(t) + x2(t)) ≠ y1(t) + y2(t)

Hence the system is non linear.

Time-invariance checking:

Let the output is delayed by t_o

\(y(t-t_o)= \left{ {\begin{array}{*{20}{c}} {0,;;;;;;;;;;;x(t-t_o) < 0}\ {x(t-t_o),x(t-t_o) > 0} \end{array}} \right.\)

Let the output of the system be g(t) for a delayed input x(t - t_o)

\(y_1(t)= \left{ {\begin{array}{*{20}{c}} {0,;;;;g(t) < 0}\ {g(t),g(t) > 0} \end{array}} \right.\;\= \left{ {\begin{array}{*{20}{c}} {0,;;;;;;;;;;;x(t-t_o) < 0}\ {x(t-t_o),x(t-t_o) > 0} \end{array}} \right.\;\=y(t-t_o)\)

Hence, the given system is time-invariant

Concept:

The capacitance of a parallel plate capacitor is

\(C = \frac{{ϵ A}}{d}\)

Where ϵ = ϵ_oϵ_r is the dielectric constant. 

Permittivity of free space ϵ_o = 8.854 × 10-14 C2/N-cm2

ϵ_r is the relative permittivity of the dielectric medium

ϵr = 1 for the air medium.

A is the area of cross-section of the plates 

d is the distance between the plates

The electric field intensity between the plates is 

E = V / d

V is the voltage across the plates.

The charge stored in the capacitor is 

Q = CV = \(\frac{{ϵ_o ϵ_r AV}}{d}=ϵ_o ϵ_r AE\)  ....(1)

Calculation:

Given that the plates are initially charged to a potential difference of V volts

Charge stored Q in both cases is the same.

From equation(1), the electric field intensity is inversely proportional to the relative permittivity of the medium

E ∝ 1 / ϵ_r

\(\frac{{E_2}}{{E_1}}=\frac{{\epsilon_{r1}}}{{\epsilon_{r2}}}\)

\(\frac{{E_2}}{{E_1}}=\frac{{5}}{{1}} = 5\)

Concept:

• It is given that the current flowing in the circuit is zero.
• This can happen only when the impedance of the circuit is infinite.
• The given circuit is closed which means to get infinite impedance the denominator part of the impedance must be zero.
• That means the denominator of the impedance of the parallel sub-circuit is zero.

Application:

Given: ω = 5k rad/sec

Z = j5 || (j5 - j/ωC)

\(Z = {j5× ({j5-{j\over \omega C}})\over j5+j5-{j\over \omega C}}\)

By equating the denominator to zero, we get

\(j10-{j\over \omega C} = 0\)

\(C={1\over 10\omega}={1\over 10× 5000}; F\)

C = 20 × 10^-6 F 

∴ The value of C = 20 μF.

Concept:

The working of the Zener diode is explained in the below figures.

Case 1: V_i and R_o are fixed

\(I_s=\frac{{(V_i-V_z)}}{{R_s}}\)   → fixed

\(I_o=\frac{{V_o}}{{R_o}}=\frac{{V_z}}{{R_o}}\)   → fixed

I_z = I_s - I_o  → fixed

Case 2: V_i is variable, R_o is fixed

\(I_{smin}=\frac{{(V_{imin}-V_z)}}{{R_s}},;;;;I_{smax}=\frac{{(V_{imax}-V_z)}}{{R_s}}\)

​\(I_o=\frac{{V_o}}{{R_o}}=\frac{{V_z}}{{R_o}}\)  → fixed​

Izmin = Ismin - I_o

Izmax = Ismax - I_o

Case 3: Vi is fixed, Ro is variable

\(I_s=\frac{{(V_i-V_z)}}{{R_s}}\)  → fixed

​\(I_{omax}=\frac{{V_z}}{{R_{omin}}},;;;I_{omin}=\frac{{V_z}}{{R_{omax}}}\)

Izmin = I_s - Iomax

Izmax = I_s - Iomin

Case 4: Vi and Ro are variable

Izmin = Ismin - Iomax

Izmax = Ismax - Iomin

Calculation:

Given,

Zener voltage V_z = 5 V

For maximum power in Zener.

Power = Vz Izmax

2.5 = 5 × Izmax

Izmax = 0.5 A

Given diode is ideal so, I_zmin = 0 A

Output resistance R_o = V_z / I_o

Since V_z is constant, to get the minimum value of Ro, I_o must be maximum.

From case 4 equations,

Ismin = I_omax + I_zmin 

The current drawn from the source will be minimum when the input voltage is minimum.

Minimum input voltage V_min = 6 V

\(I_{smin}=\frac{6-5}{6}=\frac{{1}}{{6}}~A=I_{omax}\)

So the minimum output resistance can be calculated as

\(R_{omin}=\frac{{5}}{{\frac{{1}}{{6}}}}=30;\Omega\)

Concept:

Electric Potential: It is the amount of work needed to move a unit charge from a reference point to a specific point against the electric field.

If the two points are specified then the total work done in moving the charge from one point to another with respect to the reference point charge is given as,

\(W=-\frac{1}{4\pi {{\epsilon }_{0}}}{{q}_{1}}{{q}_{2}}\left( \frac{1}{{{r}_{1}}}-\frac{1}{{{r}_{2}}} \right)=k{{q}_{1}}{{q}_{2}}\left( \frac{1}{{{r}_{2}}}-\frac{1}{{{r}_{1}}} \right)\)

Where, k = 9 × 109 N-m2 C-2

q1 and q2 = charges

r1 and r2 are distances

Electric Potential due to point charge: Consider a point charge is placed at the distance of ‘r’ from the charge ‘Q’ then the potential due to charge is,

\({{V}_{p}}=\frac{kQ}{r}\)

Calculation:

Given, q₁ = 1 μC, q₂ = 10 μC

Let r₁ be the distance the two-point charges one placed placed at (0, 0, 0) and other at (0, 10 , 0)

r₁ = 10 m

Let r₂ be the final distance between the two-point charges one placed at origin (0, 0, 0) and other at (5, 0, 0)

r₂ = 5 m

Total work done in moving the pont charge from (0, 0, 0) to (5, 0, 0) is

\(W=k{{q}_{1}}{{q}_{2}}\left( \frac{1}{{{r}_{2}}}-\frac{1}{{{r}_{1}}} \right)\)

\(W=9\times 10^9\times 10^{-11}\left( \frac{1}{5}-\frac{1}{10} \right)\)

W = 9 mJ

Concept:

Matrix:

• If A is any square matrix of order n, we can form the matrix [A – λI], where I is the nth order unit matrix.
• The determinant of this matrix equated to zero i.e. |A – λI| = 0 is called the characteristic equation of A.
• The roots of the characteristic equation are called Eigenvalues or latent roots or characteristic roots of matrix A.

Properties of Eigenvalues:

• The sum of Eigenvalues of a matrix A is equal to the trace of that matrix A
• The product of Eigenvalues of a matrix A is equal to the determinant of that matrix A

Calculation:

Given:

p2 + q2 = 1

Let \(A = \left[ {\begin{array}{*{20}{c}} p&{ q}\ { q}&-p \end{array}} \right]\)

|A – λI| = 0

\( \Rightarrow \left| {\begin{array}{*{20}{c}} {p - λ }&{ q}\ { q}&{-p - λ } \end{array}} \right| = 0\)

⇒ (p - λ)(- p - λ) - q² = 0

⇒ - p² - pλ + pλ + λ2 - q² = 0

⇒ λ2 - 1 = 0

⇒ λ = ± 1

Let the Eigenvalues be a, b

Determinant of matrix = -p² - q² = - 1 

Product of Eigenvalues = a × b = -1 

Checking from the given option

⇒ a = 1, b = - 1.

Concept:

Mutual Inductance:

When two coils are placed close to each other, a change in current in the first coil produces a change in magnetic flux, which cuts not only the coil itself but also the second coil as well. The change in the flux induces a voltage in the second coil, this voltage is called induced voltage and the two coils are said to have a mutual inductance.

Consider a pair of coupled inductors with self-inductance L₁ and L₂, magnetically coupled through coupling coefficient k.

Input and output voltage expressions are given as

V₁ = jωL₁I₁ + jωMI₂   ...(1)

V₂ = jωL₂I₂ + jωMI₁   ...(2)

Where,

ω = 2πf

\(M = K\sqrt {{L_1}{L_2}} \)

M = Mutual inductance

L1 = Inductance of coil one

L2 = Inductance of coil two

Calculation:

In the given circuit secondary is open-circuited, so I₂ = 0 A

Given secondary voltage V₂ = 7.3 V_p-p

The output voltage expression from equation(2) is given as

V₂ = jωMI1  ....(3)

The given voltage across the 22 Ω resistor is 5 Vp-p

So primary current is calculated as I₁ = 5 / 22 A_p-p

From equation(3)

|V₂| = 2π f M I₁

7.3 = 2 × π ×  100 × 10³ × M × (5/22)

M = 51.12 μH

Given,

x(t) = 0 for |t| > 1

 x(t) = 1 - |t| for |t| ≤ 1

The waveform of function can be drawn as,

Hence, given function is triangular function,

\(x\left( t \right) = ATri\left( t \right)\)

From standered Fourier transform,

If \(x\left( t \right) = ATri\left( {\frac{t}{T}} \right)\)

Fourier transform of x(t)  is given by

\(X\left( \omega \right) = ATs{a^2}\left( {\frac{{\omega T}}{2}} \right)\)

A = T = 1

\(X\left( \omega \right) = s{a^2}\left( {\frac{\omega }{2}} \right)\)

From the waveform of function,

\(X\left( \omega \right){\left. \right|_{peak}} = X\left( 0 \right) = s{a^2}\left( {\frac{0}{2}} \right) = 1\)

Therefore, the peak value of the sampling function occurs at (ω = 0).

The maximum magnitude (peak) of X(ω) will be 1.

Concept:

Residue Theorem: 

If f(z) is analytic in a closed curve C except at a finite number of singular points within C, then

∫_cf(z) dz = 2πj × [sum of residues at the singular points within C]

Formula to find residue:

1. If f(z) has a simple pole at z = a, then

\(Res;f\left( a \right) = \mathop {\lim }\limits_{z \to a} \left[ {\left( {z - a} \right)f\left( z \right)} \right]\)

2. If f(z) has a pole of order n at z = a, then

\(Res;f\left( a \right) = \frac{1}{{\left( {n - 1} \right)!}}{\left{ {\frac{{{d^{n - 1}}}}{{d{z^{n - 1}}}}\left[ {{{\left( {z - a} \right)}^n}f\left( z \right)} \right]} \right}_{z = a}}\)

Application:

Given (-1 - j), (3 - j), (3 + j) and (-1 + j) are the vertices of a rectangle C in the complex plane

f(z) from the given data is,

\(f(z)=\dfrac{1}{z^2 (z-4)}\)

Poleas of f(z) is

z = 0 of order n = 2, lies in side the closed curve.

z = 4 of order n = 1, lies outside the closed curve.

∴ \(\displaystyle\oint_C \dfrac{dz}{z^2 (z-4)}=2π j;Res;f\left( 0 \right) \) 

⇒ \(Res;f\left( 0 \right) = \frac{1}{{\left( {2 - 1} \right)!}}{\left{ {\frac{{{d^{2 - 1}}}}{{d{z^{2 - 1}}}}\left[ {{{\left( {z } \right)}^2}\dfrac{1}{z^2 (z-4)}} \right]} \right}_{z = 0}}\)

\(Res;f(0)=\dfrac{d}{dz}\left[\dfrac{1}{z-4}\right]_{z=0}=-\dfrac{1}{16}\)

\(\displaystyle\oint_C \dfrac{dz}{z^2 (z-4)}=2π j\left(\dfrac{-1}{16}\right)=\dfrac{-jπ }{8}\)

Additional Information

Cauchy’s Theorem:

If f(z) is an analytic function and f’(z) is continuous at each point within and on a closed curve C, then

\(\mathop \oint \limits_C f\left( z \right)dz = 0\)

Cauchy’s Integral Formula:

If f(z) is an analytic function within a closed curve and if a is any point within C, then

\(f\left( a \right) = \frac{1}{{2\pi i}}\mathop \oint \limits_C \frac{{f\left( z \right)}}{{z - a}}dz\)

\({f^n}\left( a \right) = \frac{{n!}}{{2\pi i}}\mathop \oint \limits_C \frac{{f\left( z \right)}}{{{{\left( {z - a} \right)}^{n + 1}}}}dz\)

Calculation:

Given  state model,

ẋ = - x + u

u = - Kx

ẋ = - x - Kx

ẋ = - (K + 1)x = A x

A = - (K + 1)

Charactestic equation of a state model is given by

|sI - A| = 0

⇒ sI = s

s - [- (K + 1)] = 0

s + K + 1 = 0

s = - 1 - K

Given that a pole is located at -2

⇒ s = -2

-2 = - 1 - K

K = 1

A 100 Hz square wave, switching between 0 V and 5 V is

Apply KVL in the input loop,

V_i = V_c + V_o

V_o = V_i - V_c

From 0 < t < T/2:

capacitor charges to a maximum value of 5 V

V_o = 5 - V_c

From T/2 < t < T:

Input voltage in this period is zero

V_i = 0 V

V_o = - V_c

The peak to peak output voltage across the resistor is given as

V_o(p-p) = (5 - Vc min) - (- Vc max)

6.2 = 5 + Vc max - Vc min

Vc max - Vc min = 1.2 V  .....(1)

From 0 < t < T/2:

V_c(0⁺) = V_c(0^-) = V_c(0) = V_{c min}

V_c(∞) = V_{c max} = 5 V

The voltage transient across the capacitor is 

V_c(t) = V_c(∞) + [V_c(0⁺) - V_c(∞)]e^-t/τ

Vc(t) = 5 + [Vc min - 5]e-T/2τ = V_{c max}

Vc max = 5[1- e-T/2τ] + Vc min e-T/2τ  .....(2)

From T/2 < t < T:

Circuit becomes

In this half cycle, the capacitor starts discharging in an exponential manner and at t = T, it becomes V_{c min}

V_c = V_{c min} = V_{c max} e^-T/2τ     ....(3)

From equations (1), (2), and (3)

5[1- e-T/2τ] + Vc min e-T/2τ - Vc max e-T/2τ = 1.2 

5[1- e-T/2τ] + (Vc min - Vc max) e-T/2τ = 1.2 

5[1- e-T/2τ] + (- 1.2) e-T/2τ = 1.2 

5 - 6.2 e-T/2τ = 1.2 

e-T/2τ = 3.8/6.2 = 0.6819

-T/2τ = 0.4895

Where T = 1/f = 1 / 100 sec

τ = RC = 820 × C

⇒ \(\frac{1}{10\times 2\times 820 \times C}=0.4895\)

C = 12.46 μF

Concept: 

Power Transmission Capability of Synchronous Generator:

Considered a synchronous generator connected to the infinite bus through a transmission line of reactance X_l.

Assumed that the resistance and capacitance of the system are zero.

Where,

V = V∠0 = voltage at infinite bus

E = E∠δ = voltage behind direct axis synchronous reactance of the generator

X_d = synchronous or transient reactance of machine

X_l = line reactance

Complex power delivered by the generator to the system is,

\(S = V{I^*} = V{\left[ {\frac{{E\angle {δ _;} - V}}{{j\left( {{X_d} + {X_l}} \right)}}} \right]^*}\)

Let, X = X_d + X_l

\(S = P + jQ = \frac{{EV}}{X}sinδ + j\left( {\frac{{EV}}{X}cosδ - \frac{{{V^2}}}{X}} \right)\)

Active power transferred to the system,

\(P = \frac{{EV}}{X}sinδ \)

The maximum steady-state power transfer occurs when,

δ = 90°

Reactive power transferred to the system,

\(Q = \left( {\frac{{EV}}{X}cosδ - \frac{{{V^2}}}{X}} \right)\)

Calculation:

Given,

X_d = 0.4 p.u

X_l = 0.1 p.u

X_m = 0.35 p.u

X = X_d + X_l + X_m = 0.4 + 0.1 + 0.35 = 0.85 p.u

E = 1∠δ1 p.u

V = 0.85 ∠δ2 p.u

P = 0.866 p.u

\(P = \frac{{EV}}{X}{\rm{sin}}\left( {{δ _1} - {δ _2}} \right)\)

\(0.866 = \frac{{1 \times 0.85}}{{0.85}}{\rm{sin}}\left( {{δ _1} - {δ _2}} \right)\)

\(\sin \left( {{δ _1} - {δ _2}} \right) = 0.866\)

(δ₁ - δ₂) = 60°

Concept:

Nilpotent Matrix: Any square matrix of order n is said to be nilpotent matrix if there exist least positive integer m such that Am = O, where O is the null matrix of order n.

The determinant of the sum of the nilpotent matrix with the identity matrix of the same order is always unity.

Example:

Consider a nilpotent matrix of order 2

\(A = \left[ {\begin{array}{*{20}{c}} 2&{-1}\ { 4}&{ - 2} \end{array}} \right]\)

 \({A^2} = \left[ {\begin{array}{*{20}{c}} 2&{-1}\ { 4}&{ - 2} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 2&{-1}\ { 4}&{ - 2} \end{array}} \right]= \left[ {\begin{array}{*{20}{c}} 0&0\ 0&0 \end{array}} \right] \) 

So here A + I = \(\left[ {\begin{array}{*{20}{c}} 3&{-1}\ {4}&{ -1} \end{array}} \right]\)

⇒  |A + I| = 1

Calculation:

Given A is a 10 × 10 matrix and A5 is a null matrix,

So, A is a nilpotent matrix of order 10.

Also given  I is the 10 × 10 identity matrix.

Then the determinant of A + I = 1 (unity)

Additional Information

Singular Matrix: Any square matrix of order n is said to be singular if |A| = 0.

Involuntary Matrix: Any square matrix of order n is said to be an involuntary matrix if A2 = I, where I is the identity matrix of order n.

Idempotent Matrix: Any square matrix of order n is said to be an idempotent matrix if A2 = A.

Concept:

Power Stages in an Induction Motor:

Stator iron loss (eddy current loss and hysteresis losses) considered as constant loss and it depends on the supply frequency and magnetic flux density in the iron core.

The iron loss of the rotor is negligible because the frequency of rotor currents under normal running conditions is always small.

\( {{P}_{2}}:{{P}_{cu}}:{{P}_{m}}=1:s:\left( 1-s \right)\)

Where,

s is the slip of motor and P_cu is rotor copper loss.

Calculation:

Given,

Stator input (P₁) = 40 kW

Stator loss = 1 kW

Rotor input (P₂) = 40 kW - 1 kW = 39 kW

f = 50 Hz

P = 6

\({N_s} = \frac{{120f}}{P} = \frac{{120 × 50}}{6} = 1000;RPM\)

N_r = 975 RPM

\(s = \frac{{{N_s} - {N_r}}}{{{N_s}}} = \frac{{1000 - 975}}{{1000}} = 0.025\)

∴ P_m = (1 - s) × P₂ = (1 - 0.025) × 39 = 38.025 kW

 Total friction and windage losses = 2.025 kW

From power flow diagram,

Rotor output (P_o) = (P_m) - (windage friction loss)

P_o = 38.025 kW - 2.025 kW = 36 kW

\(\eta = \frac{{{P_o}}}{{{P_1}}} = \frac{{36}}{{40}} = 0.9\)

Hence, the efficiency of the motor is 90%.

Concept:

Causal Signals:

• Causal Signals are signals that are zero for all negative time.
• Causality in a system determines whether a system relies on future information of a signal x [n+1].
• Talking about “causality” in signals, it means whether they are zero to the left of t = 0 or zero to the right of t = 0.
• A causal signal is zero for t < 0.
• A system is said to be causal if its output depends upon present and past inputs, and does not depend upon future input.

Calculation:

Given a casual signal,

X (Z) = z2 (z - a)-2

\(X(Z) = \frac{{{z^2}}}{{{{\left( {a - z} \right)}^2}}}\)

From standard Z-Transform,

\(n{\left( a \right)^n}u\left( n \right) \leftrightarrow \frac{{az}}{{{{\left( {z - a} \right)}^2}}}\)

\(n{\left( a \right)^{n - 1}}u\left( n \right) \leftrightarrow \frac{z}{{{{\left( {z - a} \right)}^2}}}\)

From the time-shifting property,

\(x\left( {n - {n_0}} \right) \leftrightarrow {z^{ - {n_0}}}X\left( z \right)\)

\(\left( {n + 1} \right){\left( a \right)^{n + 1 - 1}}u\left( {n + 1} \right) \leftrightarrow z\left[ {\frac{z}{{{{\left( {z - a} \right)}^2}}}} \right]\)

\(\left( {n + 1} \right){\left( a \right)^n}u\left( {n + 1} \right) \leftrightarrow \frac{{{z^2}}}{{{{\left( {z - a} \right)}^2}}}\)

Since,

\(X\left( z \right) = \frac{{{z^2}}}{{{{\left( {z - a} \right)}^2}}}\)

Therefore,

\(x\left( n \right) = \left( {n + 1} \right){\left( a \right)^n}u\left( {n + 1} \right)\)

The given signal is causal, therefore,

\(x\left( n \right) = \left( {n + 1} \right){\left( a \right)^n}u\left( n \right)\)

Important Points

Z-Transform basic functions:

Concept: 

Power Transmission Capability of Synchronous Generator:

Considered a synchronous generator connected to the infinite bus through a transmission line of reactance Xl.

Assumed that the resistance and capacitance of the system are zero.

Where,

V = V∠0 = voltage at infinite bus

E = E∠δ = voltage behind direct axis synchronous reactance of the generator

Xd = synchronous or transient reactance of machine

Xl = line reactance

Complex power delivered by the generator to the system is,

\(S = V{I^*} = V{\left[ {\frac{{E\angle {δ _;} - V}}{{j\left( {{X_d} + {X_l}} \right)}}} \right]^*}\)

Let, X = Xd + Xl

Active power transferred to the system,

\(P = \frac{{EV}}{X}sinδ \)

The maximum steady-state power transfer occurs when,

δ = 90°

Reactive power transferred to the system,

\(Q = \left( {\frac{{EV}}{X}cosδ - \frac{{{V^2}}}{X}} \right)\)

Calculation:

Given,

X_s1 = X_s2 = j1.5 p.u.

X_m = j0.5 p.u.

Equivalent reactance (X) will be written as,

\(X = \frac{{{X_{s1}}{X_{s2}} - X_m^2}}{{{X_{s1}} + {X_{s2}} - 2{X_m}}}\)

\(\frac{{{L_1}{L_2} - {M^2}}}{{{L_1} + {L_2} - 2M}}\)

From the above concept, 

The maximum steady-state power transfer occurs when δ = 90°

\({P_m} = \frac{{\left| E \right|\left| V \right|}}{X} = \frac{{\left| E \right|\left| V \right|}}{{j1}} = \left| E \right|\left| V \right|\)

Important Points

The equivalent inductance of parallel aiding connection is

Leq = \(\frac{{{L_1}{L_2} - {M^2}}}{{{L_1} + {L_2} - 2M}}\)

The equivalent inductance of parallel opposing connection is

Leq = \(\frac{{{L_1}{L_2} - {M^2}}}{{{L_1} + {L_2} + 2M}}\)

Concept:

The maximum voltage across a reverse bias diode is known as Peak Inverse Voltage.

PIV for different rectifiers is shown below:

• Half Wave rectifier: Vm
• Full Wave centre tap rectifier: 2Vm
• Full Wave Bridge rectifier: Vm​

Calculation:

The given circuit is a full-wave bridge rectifier.

Given input Vi is a sinusoidal AC voltage having an RMS value of

V_rms = 230 V ± 20%

The worst-case peak-inverse voltage seen across any diode is

Vm = √2 × maximum value of V_rms

V_m = √2 × (230 + 0.2 × 230)

V_m = 390.32 V

Important Points

Concept:

Excitation Table / Transition Table:

D Flip-Flop:

The single input is called the 'DATA' input. If this data input is held HIGH, the flip flop would be 'SET' and when it is LOW the flip flop would change and become 'RESET'.

Application:

Given:

000 → 001 → 011 → 010 → 110 → 111 → 101 → 100 → 000 ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅

Table for the given condition will be

D₁ will have high logic at (1,3,2,6) ⇒ D₁ = ∑m (1,3,2,6)

∴ D₁ = Q̅₂ Q₀ + Q₁ Q̅₀

Concept:

Separation of Iron loss in Transformer:

In transformer iron loss (P_i) has two components namely Hysteresis loss (P_h) and Eddy current loss (P_e).

P_i = P_h + P_e

\({P_i} = {k_h}fB_m^n + {k_e}{f^2}B_m^2\)  ….

(1) Where,

K_h and k_e are loss coefficient constant and their value depends on the type of material

B_m is the maximum value of flux density

f is supply frequency

\({P_h} \propto f\)

\({P_h} = af\)

\({P_e} \propto {f^2}\)

\({P_e} = b{f^2}\)

Where a and b are constants.

\({P_i} = af + b{f^2}\)

\(\frac{{{P_i}}}{f} = a + bf\)

Note: For the separation of these two losses, the no-load test is performed on the transformer.

Calculation:

Given,

V₁ = 440 V

f1 = 50 Hz

V2 = 220 V

f₂ = 25 Hz

\(\frac{{{V_1}}}{{{f_1}}} = \frac{{{V_2}}}{{{f_2}}} = 8.8\)

Hence, (V / f) ratio is constant,

P₁ = 2500 W

P₂ = 850 W

From the above concept,

P_i = af + bf²

2500 = 50a + 2500b ....

(1) 850 = 25a + 625b .... (2) From equation (1) and (2),

a = 18 and b = 0.64

Iron loss at 50 Hz,

Hysteresis loss (P_h) = af = 18 × 50 = 900 W

Eddy current loss (P_e) = bf² = 0.64 × 2500 = 1600 W

Concept:

Even Signal:

A signal f(t) is called even signal if, 𝑓(𝑡) = 𝑓(−𝑡)

It is given by,

\({f_e}\left( t \right) = 2\mathop \smallint \limits_0^{\frac{T}{2}} {f_e}\left( t \right)dt \)

Differentiation and Integration Properties of Fourier transform:

If,

\(x\left( t \right)\mathop \leftrightarrow \limits^{F.T} X\left( \omega \right)\)

Then differentiation property states that,

\(\frac{{dx\left( t \right)}}{{dt}}\mathop \leftrightarrow \limits^{F.T} j\omega X\left( \omega \right)\)

\(\frac{{{d^n}x\left( t \right)}}{{d{t^n}}}\mathop \leftrightarrow \limits^{F.T} {\left( {j\omega } \right)^n}X\left( \omega \right)\)

Then integration property states that,

\(\smallint x\left( t \right)dt\mathop \leftrightarrow \limits^{F.T} \frac{1}{{j\omega }}X\left( \omega \right)\)

\(\smallint \smallint \smallint \ldots \smallint x\left( t \right)dt\mathop \leftrightarrow \limits^{F.T} \frac{1}{{{{\left( {j\omega } \right)}^n}}}X\left( \omega \right)\)

Calculation:

Given function is even,

\(\dfrac{dF(ω)}{dω} = -ω F(ω)\) ....

(1) From differentiation property,

\(tf\left( t \right) = \frac{{jdF\left( \omega \right)}}{{d\omega }}\)

Applying IFT to the above equation,

\( - jtf\left( t \right) = \frac{{jdf\left( t \right)}}{{df}}\)

\(\frac{{df\left( t \right)}}{{dt}} = - tf\left( t \right)\) .... (2) From equation (1) and (2) it is clear that the f (t) is the change of Gaussian function, it can be written as,

\(f\left( t \right) = \frac{1}{{\sqrt {2\pi } }}{e^{\frac{{ - {t^2}}}{2}}}\) .... (3) Note:

\(\left[ {{e^{ - a{t^2}}} \leftrightarrow \sqrt {\frac{\pi }{a}} {e^{\frac{{ - {\omega ^2}}}{{4a}}}}} \right]\)

\(\left[ {{e^{ - \frac{{{t^2}}}{2}}} \leftrightarrow \sqrt {2\pi .} {e^{\frac{{ - {\omega ^2}}}{2}}}} \right]\)

From equation (3),

\(\frac{1}{\sqrt{{2\pi }}}{e^{\frac{{ - {t^2}}}{2}}} \leftrightarrow {e^{;\frac{{ - {\omega ^2}}}{2}}}\)

\(f\left( t \right) = \frac{1}{\sqrt{{2\pi }}}{e^{\frac{{ - {t^2}}}{2}}}\)

\(f\left( 0 \right) = \frac{1}{\sqrt{{2\pi }}}\)

\(f\left( 0 \right) = 0.3989\)

\(\therefore {\bf{f}}\left( 0 \right) < 1\)

Concept:

Mutual Inductance:

When two coils are placed close to each other, a change in current in the first coil produces a change in magnetic flux, which cuts not only the coil itself but also the second coil as well. The change in the flux induces a voltage in the second coil, this voltage is called induced voltage and the two coils are said to have a mutual inductance.

Consider a pair of coupled inductors with self-inductance L1 and L2, magnetically coupled through coupling coefficient k.

Input and output voltage expressions are given as

V1 = jωL1I1 + jωMI2   ...(1)

V2 = jωL2I2 + jωMI1    ...(2)

Where,

ω = 2πf

\(M = K\sqrt {{L_1}{L_2}} \)

M = Mutual inductance

L1 = Inductance of coil one

L2 = Inductance of coil two

Calculation:

The given circuit in the Laplace domain is 

Apply KVL in the input loop,

V₁(s) = (4 + 6s)I₁(s) + sI₂(s)  ...(1)

Apply KVL in the output loop

I₂(s)[5 +4s] + sI₁(s) = 0

\(I_2(s)=\dfrac{{-sI_1(s)}}{{4s + 5}}\)

Substitute I₂(s) in equation(1)

\(V_1(s)=(4+6s)I_1(s)-\dfrac{{s^2I_1(s)}}{{4s + 5}}\)

\(\dfrac{{V_1(s)}}{{I_1(s)}}=\dfrac{{(6s+4)(4s+5)-s^2}}{{4s+5}}\)

\(Z_{in}(s)=\dfrac{{24s^2+16s+30s+20-s^2}}{{4s+5}}\)

\(Z_{in}(s)=\dfrac{{23s^2+46s+20}}{{4s+5}}\)

Given:

β = 100

V_C = 5 V

\(I_C = {V_C \over R_C}={5\over3.3;k}=1.515;mA \)

\(I_B = {I_C\over β}=0.01515;mA\)

I_E = (1 + β) I_B = 1.53 mA

• 12 + 1.2 × 10³ × 1.53 × 10^-3 + 0.7 + V_B = 0

V_B = 9.464 V

\(I_X =\dfrac {{12-V_B}}{4.7;k}=\dfrac{{12-9.464}}{4.7;k}=0.54;mA\)

I_Y = I_X + I_B = (0.54 + 0.01515) mA = 0.5547 mA

\(R_2=\dfrac{V_B}{I_Y}=\dfrac{9.464}{0.5547\times10^{-3}}=17.06;kΩ\)

∴ The value of R₂ = 17.06 kΩ.

Concept:

The general form of a cubic equation is ax³ + bx² + cx + d = 0.

Where a, b, c, and d are constants and a ≠ 0.

Let the roots be p, q, and r

• The sum of the roots (p + q + r) = - b/a
• The product of the roots (pqr) = - d/a
• The sum of the product of any two roots (pq + qr + rp) = c/a

Calculation:

Given p(z) = z3 + (1 + j) z2 + (2 + j) z + 3

Sum of the roots (p + q + r) = - (1 + j)

Product of the roots (pqr) = - 3

Sum of the roots is complex, so all the roots cannot be real.

Concept:

The magnetic field forms a closed loop, i.e. the amount of field leaving a point equals the amount entering. i.e. Magnetic monopoles do not exist.

Since the divergence of a field gives the net outflow of a field and is calculated as \(∇ .\vec F\)

So, ∇ ⋅ B = 0

B = Magnetic flux density

Since B is related to H (magnetic field intensity) as:

B = μH

so, ∇ ⋅ H = 0

Application:

Option: 1

B = 10x x̂  - 30z ŷ + 20y ẑ  

\(∇ ⋅ B = \frac{{\partial {B_x}}}{{\partial x}} + \frac{{\partial {B_y}}}{{\partial y}} + \frac{{\partial {B_z}}}{{\partial z}} = \frac{{\partial 10x}}{{\partial x}} + \frac{{\partial ( - 30z)}}{{\partial y}} + \frac{{\partial 20y}}{{\partial z}} = 10\)

∇ ⋅ B ≠ 0, so function does not represent magnetic field.

Option: 2

B = 10y x̂ + 20x ŷ - 10z ẑ 

\(\nabla ⋅ B = \frac{{\partial {B_x}}}{{\partial x}} + \frac{{\partial {B_y}}}{{\partial y}} + \frac{{\partial {B_z}}}{{\partial z}} = \frac{{\partial 10y}}{{\partial x}} + \frac{{\partial 20x}}{{\partial y}} + \frac{{\partial ( - 10z)}}{{\partial z}} = - 10\)

∇ ⋅ B ≠ 0, so function does not represent magnetic field.

Option: 3

B = 10x x̂ + 20y ŷ - 30z ẑ 

\(\nabla ⋅ B = \frac{{\partial {B_x}}}{{\partial x}} + \frac{{\partial {B_y}}}{{\partial y}} + \frac{{\partial {B_z}}}{{\partial z}} = \frac{{\partial 10x}}{{\partial x}} + \frac{{\partial 20y}}{{\partial y}} + \frac{{\partial ( - 30z)}}{{\partial z}} = 10 + 20 - 30 = 0\)

∇ ⋅ B = 0, so function represent magnetic field.

Option: 4

B = 10z x̂ + 20y ŷ - 30x ẑ 

\(\nabla ⋅ B = \frac{{\partial {B_x}}}{{\partial x}} + \frac{{\partial {B_y}}}{{\partial y}} + \frac{{\partial {B_z}}}{{\partial z}} = \frac{{\partial 10z}}{{\partial x}} + \frac{{\partial 20y}}{{\partial y}} + \frac{{\partial ( - 30x)}}{{\partial z}} = 20\)

∇ ⋅ B ≠ 0, so function does not represent magnetic field.

Concept:

Fortescue’s Theorem:

A unbalance set of ‘n’ phasors may be resolved into (n - 1) balance n-phase system of different phase sequence and one zero phase sequence system.

A zero-phase sequence system is one in which all phasors are of equal magnitude and angle.

Considered three phasors are represented by a, b, c in such a way that their phase sequence is (a b c).

The positive phase sequence will be (a b c) and the negative phase sequence will be (a c b).

Assumed that subscript 0, 1, 2 refer to zero sequences, positive sequence, negative sequence respectively.

Current I_a, I_b, I_c represented an unbalance set of current phasor as shown,

Each of the original unbalance phasor is the sum of its component and it can be written as,

I_a = I_a0 + I_a1 + I_a2

I_b = I_b0 + I_b1 + I_b2

I_c = I_c0 + I_c1 + I_c2

For a balance position phase sequence (a b c) we can write the following relation,

I_a0 = I_b0 = I_c0

I_b1 = α² I_a1

I_c1 = α I_a1

I_b2 = α I_a2

I_c2 =α² I_a2

From the above equation I_a, I_b, I_c can be written in terms of phase sequence component,

I_a = I_a0 + I_a1 + I_a2

I_b = I_b0 +  α² I_a1 + α I_a2

I_c = I_a0 + α I_a1 + α² I_a2

The above equation can be written in form of Matrix as shown,

\(\left[ {\begin{array}{*{20}{c}} {{I_a}}\ {{I_b}}\ {{I_c}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&1&1\ 1&{{\alpha ^2}}&\alpha \ 1&\alpha &{{\alpha ^2}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{I_{a0}}}\ {{I_{a1}}}\ {{I_{a2}}} \end{array}} \right]\)

Calculation:

Given,

IB0 = 0.1 ∠0° p.u

IA0 = IB0 = IC0 = 0.1 ∠0° p.u

IA = 1.1 ∠0° p.u.

IC = (1 ∠120° + 0.1) p.u.

From the above concept,

I_A = I_A0 + I_A1 + I_A2

I_A - I_A0 = I_A1 + I_A2 = 1.1 ∠0° + 0.1 ∠0° p.u = 1 ∠0° p.u

I_A1 + I_A2 = 1 ∠0° p.u ....

(1) I_C = I_C0 + I_C1 + I_C2 = I_A0 + α I_A1 +α² I_A2

1 ∠120° + 0.1 = 0.1 ∠0° + α I_A1 +α² I_A2

α I_A1 +α² I_A2 = 1 ∠120° .... (2) Adding equation (1) and (2),

(I_A1 + I_A2) + (α I_A1 +α² I_A2) = 1 ∠0° + 1 ∠120°

I_A1(α + 1) + I_A2(α² + 1)  = 1 ∠0° + 1 ∠120° (Since, 1 + α + α2 = 0 ⇒ 1 + α = -α²)

IA1(-α²) + IA2(-α)  = 1 ∠0° + 1 ∠120°

α2 I_A1 + α I_A2 = 1 ∠-120° .... (3) From above concept,

I_B = I_A0 + α² I_A1 + α I_A2

I_B = 0.1 ∠0° + 1 ∠-120°