GATE EE 2016 Official Paper: Shift 2 (Previous Year Paper)

The context of the sentence requires a phrasal verb that will mean to ‘endure’. The phrasal verb from the options with that meaning is ‘bear with’. ‘Bore with’ is in the past tense and cannot be used because the blank is preceded by ‘to’. Thus the base form of the verb needs to be used. ‘Bare’ means to disclose or uncover and thus does not fit the context.

The correct spelling is ‘manageable’

In the group of given numbers, all are prime numbers but 33 is not.

There are two facts. R2D2 is a robot who can repair aeroplanes and no other robot can do this work. The other functions of R2D2 is not known and thus inference 1 does not follow.

We do not know whether R2D2 can repair anything else other than aeroplanes. Thus option 3 does not follow.

There can be other robots and thus option 4 is eliminated.

Only option 2 is the logical inference that follows.

\(\begin{array}{l} \left| {9y - 6} \right| = 3\ 9y - 6 = \pm 3 \end{array}\)

Case 1:

\(9y - 6 = 3 \Rightarrow y - 1\)

Case 2:

\(9y - 6 = - 3 \Rightarrow y = \frac{1}{3}\)

Substitute \(y = 1\)

\({y^2} - \frac{{4y}}{3} = - \frac{1}{3}\)

Substitute \(y = \frac{1}{3}\)

\({y^2} - \frac{{4y}}{3} = - \frac{1}{3}\)

Installed capacity \(\ge 200;tonnes \Rightarrow large;plant\)

Installed capacity \(< 200;tonnes \Rightarrow small;plat\)

Fro given multiple pie chart, the large plants are 1,4,8 & 9

Total production of large plants \(= 160 + 190 + 230 + 190 = 770;tonnes\)

Total production of small plants \(= 150 + 160 + 120 + 100 + 120 = 650;tonnes\)

The difference between total production of large plants and small plants in tonnes

\(= 770 - 650 = 120\)

Let us examine the options one by one:

1. Various engineering disciplines have not been talked about in the paragraph and thus this inference does not follow. We cannot make a general statement with respect to various disciplines when only mechanical engineering has been talked about.
2. The poll taken was for students and the gender of the students participating in the poll has not been mentioned. Thus this inference does not hold.
3. This inference hold because the research study found that the women with a high degree in mechanical engineering were successful in their profession. Thus it can be a suitable profession for them.
4. We cannot infer anything about the number of women pursuing mechanical engineering because no information has been provided regarding the same.

Thus option 3 is the answer.

Let us examine the options one by one.

1. No inference can be drawn about the degrees of the students because no such information has been provided in the passage.
2. Since the institutes are being set up to cater to the technological development needs, it can be inferred they have a national objective. Thus this options follows
3. Though SITs re being established in collaboration with IITs it is not always true that they cannot be set up independently. Thus this option does not hold.
4. The passage states that the SITs are being set up keeping in mind the technological developments. Thus this inference definitely follows.

Hence options 2 and 4 are logical inferences.

n = No. of attempts = 10

x = free throws successfully = 6

\(p = 60% = 0.6\)

The probability that he will successfully make exactly 6 free throws in 10 attempts

\(\begin{array}{l} = {n_{{c_x}}}{p^x}{q^{n - x}}\ = {10_{{c_6}}}{\left( {0.6} \right)^6} \times {\left( {0.4} \right)^4}\ = 210 \times 0.046656 \times 0.0256\ = 0.2508 \end{array}\)

Calculation:

• The last digit of 211870 will be dependent on the last digit of 211. Since digit 1(The last digit of 211) is having cyclicity 1 when it is multiplied by itself one-time results in the same as 1.

So the last digit of  211870  will always be 1.

• Similarly, the last digit of 146127 will be dependent on the last digit of 146. Since digit 6(The last digit of 146) is having cyclicity 1, when it is multiplied by itself one-time results in the same as 6.

So the last digit of  146127  will always be 6.

• The last digit of 3424 will be dependent on the cyclicity of digit 3. Since digit 3 is having cyclicity 4 means the last digit of 3¹ and 3⁵ will be the same as 3.

Here the power of  3424 is multiple of 4 so the last digit of  3424 will be the same as the last digit of 3⁴ which will be 1.

\(\Rightarrow \begin{array}{l} {211^{870}} + {146^{127}} \times {3^{424}} = {1} + {6} \times 1\ \end{array}\)

Units digit = 7

Concept:

3 variable K-maps:

• For a 3-variable Boolean function, there is a possibility of 8 output minterms.
• The general representation of all the minterms using 3-variables is shown below.

Calculation:

Given k-map is as shown

It can be grouped as follows

Therefore in SOP form (sum of products) output is Y = A + C̅

Concept:

Capacitive reactance(X_c) 

\(X_c = \frac{1}{ω C}\)

Where C = capacitance in farad, ω = angular frequency in rad/sec

For finding the type of filter find its low frequency and high-frequency response :

At Low-Frequency open circuit, the Capacitors.

At a High-frequency short circuit, the capacitors

For ideal op-amp according to the virtual short concept 

V- = V+

Calculation:

At low frequencies i.e. at \({\rm{ω }} = {\rm{;}}0{\rm{;}} \Rightarrow {{\rm{X}}_{\rm{c}}} = \frac{1}{{{\rm{ω c}}}} = \infty\)

∴ The capacitor acts like an open circuit

Due to the virtual short concept V- = V+ = 0 (Virtual Ground)

Apply KCL at node 'a' and we get,

\(\therefore {{\rm{V}}_{{\rm{out}}}} = \frac{{ - {{\rm{R}}_2}}}{{{{\rm{R}}_1}}}{{\rm{V}}_{{\rm{in}}}}\)

At high frequencies i.e. at \({\rm{ω }} = \infty \Rightarrow {{\rm{X}}_{\rm{c}}} = \frac{1}{{{\rm{ω c}}}} = 0\)

I.e. capacitor acts like a short circuit

Input directly connected to the output so

\(\therefore {\rm{;}}{{\rm{V}}_{{\rm{out}}}}{\rm{;}} = {\rm{;}}0\)    (Due to virtual short concept)

So the given circuit passes only low frequencies hence it acts as a low pass filter.

Here \({{\rm{Z}}_{\rm{L}}}{\rm{;}} = {\rm{;R}} - {\rm{j}}{{\rm{X}}_{\rm{C}}}{\rm{;}} = {\rm{;}}\left( {80{\rm{;}}-{\rm{;j}}40} \right){\rm{;k\Omega ;}}\)

Given \({{\rm{N}}_1}{\rm{;}}:{\rm{;}}{{\rm{N}}_2}{\rm{;}} = {\rm{;}}1{\rm{;}}:{\rm{;}}100\)

We know \(\frac{{{{\rm{Z}}_{{\rm{in}}}}}}{{{{\rm{Z}}_{\rm{L}}}}} = {\left( {\frac{{{{\rm{N}}_1}}}{{{{\rm{N}}_2}}}} \right)^2}\)

\(\therefore {{\rm{Z}}_{{\rm{in}}}} = \left( {80 - {\rm{j}}40} \right) \times {10^3} \times \frac{1}{{{{10}^4}}} = \left( {8 - {\rm{j}}4} \right){\rm{\Omega }}\) 

Now the circuit becomes,

\(\therefore {\rm{I}} = \frac{{\rm{V}}}{{\rm{Z}}} = \frac{{100}}{{\left| {8 - {\rm{j}}4 + {\rm{j}}10} \right|}} = \frac{{100}}{{\left| {8 + {\rm{j}}6} \right|}} = \frac{{100}}{{10}} = 10\)

\(\therefore {\rm{;}}{{\rm{I}}_{{\rm{RMS}}}}{\rm{;}} = {\rm{;}}10{\rm{;A}}\)

Concept:

Laplace transform is an important tool for converting differential equations in the time domain to algebraic equations in the frequency domain.

This analysis involves three important processes:

1) The transformation from the time domain to the frequency domain.

2) Manipulate the algebraic equations to form a solution.

3) The inverse transformation from the frequency to the time domain.

Mathematically, the Laplace transform of a function f(t) is defined as:

\(L[f(t)]=F(s)=\mathop \smallint \limits_0^\infty f\left( t \right){e^{ - st}}dt\)

s = σ + jω is the complex frequency 

Applications of the Laplace Transform:

1. Solve differential equations (both ordinary and partial)

2. Application in RLC circuit analysis.

Calculation:

Given \(\frac{{\rm{d}}}{{{\rm{dt}}}}{\rm{y}}\left( {\rm{t}} \right) + \frac{1}{6}{\rm{y}}\left( {\rm{t}} \right) = 3{\rm{x}}\left( {\rm{t}} \right)\)

Apply Laplace transform we get

\({\rm{sY}}\left( {\rm{s}} \right) + \frac{1}{6}{\rm{Y}}\left( {\rm{s}} \right) = 3{\rm{X}}\left( {\rm{s}} \right)\) But \({\rm{X}}\left( {\rm{s}} \right) = \frac{3}{{\left( {{\rm{s}} + \frac{1}{3}} \right)}}\) 

\(\begin{array}{l} \therefore {\rm{Y}}\left( {\rm{s}} \right) = \frac{{3{\rm{X}}\left( {\rm{s}} \right)}}{{\left( {{\rm{s}} + \frac{1}{6}} \right)}} = \frac{9}{{\left( {{\rm{s}} + \frac{1}{6}} \right)\left( {{\rm{s}} + \frac{1}{3}} \right)}}\ {\rm{Y}}\left( {\rm{s}} \right) = \frac{9}{{\left( {{\rm{s}} + \frac{1}{6}} \right)\left( {\frac{{ - 1}}{6} + \frac{1}{3}} \right)}} + \frac{9}{{\left( {{\rm{s}} + \frac{1}{3}} \right)\left( {\frac{{ - 1}}{3} + \frac{1}{6}} \right)}}\ {\rm{Y}}\left( {\rm{s}} \right) = \frac{{54}}{{{\rm{s}} + \frac{1}{6}}} - \frac{{54}}{{\left( {{\rm{s}} + \frac{1}{3}} \right)}} \end{array}\)

Apply Inverse Laplace transform we get

\({\rm{y}}\left( {\rm{t}} \right) = 54\left[ {{{\rm{e}}^{ - \frac{1}{6}{\rm{t}}}} - {{\rm{e}}^{ - \frac{1}{3}{\rm{t}}}}} \right]{\rm{u}}\left( {\rm{t}} \right)\)

Given:

f_m = 5 kHz, 

ω_c = 2πfc = 2000π 

f_c = 1 kHz

Maximum frequency in x(t) is given as:

f_c + f_m = 6 kHz

Concept:

Function of a Complex Variable: ω = f (z) = u (x, y) + i v (x, y) Where x, y ϵ R  and  z = x + iy

u (x, y), v (x, y) are real valued functions. 

Analytic function: A function f (z) is said to be Analytic in a region R of z-plane if it is differentiable at every point of R. 

Necessary Condition for function f (z) to be Analytic: 

1. \(\frac{\partial u}{\partial x},\ \frac{\partial u}{\partial y}, \ \frac{\partial v}{\partial x}, \ \frac{\partial v}{\partial y}\)  exist 2. f (z) to satisfy the Cauchy-Riemann (C-R) equation.

 \(\frac{\partial u}{\partial x}\ =\ \ \frac{\partial v}{\partial y}, \ \frac{\partial u}{\partial y}\ =\ - \ \frac{\partial v}{\partial x}\)

Calculation: 

\(\begin{array}{l} {\rm{z;}} = {\rm{;x;}} + {\rm{;iy;}} \Rightarrow {\rm{;}}{{\rm{z}}^{\rm{*}}}{\rm{;}} = {\rm{;x;}}-{\rm{;iy}}\ \therefore {\rm{;f}}\left( {\rm{z}} \right){\rm{;}} = {\rm{;}}\left( {{\rm{x;}} + {\rm{;iy}}} \right){\rm{;}} + {\rm{;}}\left( {{\rm{x;}} - {\rm{;iy}}} \right){\rm{;}} = {\rm{;}}2{\rm{x}} \end{array}\)

\(\therefore ;f\left( z \right)\) is finite everywhere and it is continuous and differentiable.

But it doesn't satisfy the CR equations.

So we can say that \(f\left( z \right)\) is continuous but not analytic.

Concept:

CAYLEY-HAMILTON THEOREM:

Statement: Every square matrix satisfies its own characteristic equation.

The Cayley–Hamilton theorem states that substituting the matrix A for x in polynomial, p(x) = det(xI_n – A), results in the zero matrices, such as:

p(A) = 0

It states that a ‘n x n’ matrix A is demolished by its characteristic polynomial det(tI – A), which is monic polynomial of degree n.

Uses of Cayley-Hamilton theorem:

(1) To calculate the positive integral powers of A

(2) To calculate the inverse of a square matrix A

Calculation:

\({P^3} = P\)

From Cauley Hamilton theorem \({\lambda ^3} = \lambda\) 

\(\begin{array}{l} \Rightarrow \lambda ;\left( {{\lambda ^2}-1} \right) = 0\ \lambda = 0,; + 1,; - 1 \end{array}\)

Concept:

 Laplace transform of the first two derivatives.

L{y^''} = s²Y(s) - sy(0) - y'(0)

L{y'} = sY(s) - y(0)

Calculation:

Given \({\rm{y''}}\left({\rm{t}} \right){\rm{;}} + {\rm{;}}2{\rm{y'}}\left({\rm{t}} \right){\rm{;}} + {\rm{;y}}\left( {\rm{t}} \right){\rm{;}} = {\rm{;}}0\)

Apply Laplace transform we get

\(\left[ {{{\rm{s}}^2}{\rm{Y}}\left( {\rm{s}} \right){\rm{;}}-{\rm{;sy}}\left( 0 \right){\rm{;}}-{\rm{;y'}}\left( 0 \right)\left] {{\rm{;}} + {\rm{;}}2} \right[{\rm{sy}}\left( {\rm{s}} \right){\rm{;}}-{\rm{;y}}\left( 0 \right)} \right]{\rm{;}} + {\rm{;Y}}\left( {\rm{s}} \right){\rm{;}} = {\rm{;}}0\)

\(\begin{array}{l} \left( {{{\rm{s}}^2}{\rm{;}} + {\rm{;}}2{\rm{s;}} + {\rm{;}}1} \right){\rm{;Y}}\left( {\rm{s}} \right){\rm{;}}-{\rm{;}}1{\rm{;}} = {\rm{;}}0\ {\rm{Y}}\left( {\rm{s}} \right) = \frac{1}{{{{\left( {{\rm{s}} + 1} \right)}^2}}} \Rightarrow {\rm{y}}\left( {\rm{t}} \right) = {\rm{t}} \cdot {{\rm{e}}^{ - {\rm{t}}}}{\rm{u}}\left( {\rm{t}} \right) \end{array}\)

Concept:

When two points (x₁, y₁. z₁) and (x₁, y₁. z₂) are mentioned find the relation in terms of the third variable in terms of x,y, and z:

\(\dfrac{{{{x}} - x_1}}{{x_2 - x_1}} = \dfrac{{{{y}} - y_1}}{{y_2 - y_1}} = \dfrac{{{{z}} - z_1}}{{z_2 - z_1}} = {{t}}\)

Put the value of z,y, and z and use the end-points of one variable.

Calculation:

Given:

\({\rm{I}} = \smallint \left( {2{\rm{x}}{{\rm{y}}^2}{\rm{dx}} + 2{{\rm{x}}^2}{\rm{ydy}} + {\rm{dz}}} \right)\), A (0, 0, 0) and B(1, 1, 1).

Equation of line i.e. path

\( \frac{{{\rm{x}} - 0}}{{1 - 0}} = \frac{{{\rm{y}} - 0}}{{1 - 0}} = \frac{{{\rm{z}} - 0}}{{1 - 0}} = {\rm{t}} \)

\( \therefore {\rm{;x;}} = {\rm{;y;}} = {\rm{;z;}} = {\rm{;t;and;t;}}:{\rm{;}}0{\rm{;}} \to {\rm{;}}1\)

\( \therefore {\rm{I}} = \mathop \smallint \limits_0^1 \left( {2{{\rm{t}}^3}{\rm{dt}} + 2{{\rm{t}}^3}{\rm{dt}} + {\rm{dt}}} \right) \)

\(= 4 \cdot \left[ {\frac{{{{\rm{t}}^4}}}{4}} \right]_0^1 + \left[ {\rm{t}} \right]_0^1 = 1 + 1 = 2\)

\( \therefore \smallint \left( {2{\rm{x}}{{\rm{y}}^2}{\rm{dx}} + 2{{\rm{x}}^2}{\rm{ydy}} + {\rm{dz}}} \right) = 2 \)

Concept

\(\begin{array}{l} {\rm{X}}\left( {\rm{t}} \right) = {a_0} + \mathop \sum \limits_{n = 1}^\infty {a_n}cosn{\omega _0}t + {b_n}sinn{\omega _0}t\ Where;{a_0} = dc;value;of;signal = \mathop \smallint \limits_0^{{T_0}} x\left( t \right)dt\ {a_n} = \frac{2}{{{T_0}}}\mathop \smallint \limits_0^{{T_0}} ;x\left( t \right).cos;n{\omega _0}t;\ {b_n} = \frac{2}{{{T_0}}}\mathop \smallint \limits_0^{{T_0}} ;x\left( t \right).sin;n{\omega _0}t = - {b_{ - n}} \to odd;function;w.r.t;n;;; \end{array}\)

Solution:

Given that \(f\left( { - x} \right) = - f\left( x \right)\) 

So, cosine term and DC term will be zero. It has only sine terms.

\(\therefore {\rm{f}}\left( {\rm{x}} \right) = \mathop \sum \limits_{{\rm{k}} = 1}^\infty {{\rm{b}}_{\rm{k}}} \cdot \sin \left( {{{\rm{k}}_{\rm{x}}}} \right)\)

It is a series RL circuit and

\({{\rm{V}}_{\rm{S}}} = \sqrt {{\rm{V}}_{\rm{R}}^2 + {\rm{V}}_{\rm{C}}^2}\)

\(\therefore 100 = \sqrt {{{\left( {60} \right)}^2} + {{\left( {80} \right)}^2}}\)

Instantaneous voltage, v(t) = 100 sin (ωt)

Instantaneous current, i(t) = 10 sin (ω t – 60°) + 2 sin (3 ωt) + 5 sin (5 ωt)

\(Average;Power = \frac{1}{T}\mathop \smallint \nolimits_0^T v\left( t \right).i\left( t \right)dt\)

\(= \frac{1}{T}\left[ {\mathop \smallint \nolimits_0^T \left{ {\left( {100\sin \left( {\omega t} \right)} \right).\left( {10\sin \left( {\omega t} \right) - 60^\circ } \right) + \left( {2\sin \left( {3\omega t} \right) + 5\sin \left( {5\omega t} \right)} \right)} \right}dt} \right]\)

\(= \frac{1}{T}\left[ {\mathop \smallint \nolimits_0^T \left( {100\sin \left( {\omega t} \right)} \right).\left[ {10\sin \left( {\omega t - 60^\circ } \right)} \right]dt + 0 + 0} \right]\)

(∵ In case of average power multiplication of different frequency component is zero)

\(= \frac{1}{{2T}}\left[ {\mathop \smallint \nolimits_0^T 1000\left( {\cos \left( {\omega t - \omega t} \right) + 60^\circ } \right) - \cos \left( {\omega t + \omega t - 60^\circ } \right)dt} \right]\)

[∵ 2 sin A. sin B = cos (A - B) – cos (A + B)]

\(= \frac{1}{{2T}}\left[ {\mathop \smallint \nolimits_0^T 1000\left[ {\cos 60^\circ - \cos \left( {2\omega t - 60^\circ } \right)} \right]dt} \right]\)

\(= \frac{1}{{2T}}\left[ {\mathop \smallint \nolimits_0^T 1000\left( {\frac{1}{2} - 0} \right)dt} \right]\)

[∵ integration of cosine function over a complete cycle is zero]

∴ Average power = 250 watt

Common mistake:

Generally students use formulae P_avg = V_rms.I_rms which is not correct when the phase difference is given. The correct formulae for calculating active power is P_avg = V_rms.I_rms cos ϕ

Whereas apparent power, S = V_rms.I_rms

Given the network is:

In the given network,

R₂, R₄, and R₅ are directional overcurrent relays. R₁, R₃ and R₆ are non-directional overcurrent relays.

For the fault on line 2 i.e. L₂, R3 and R4 must be operated. If R₄ is not operated then R₁ and R₆ will operate.

Therefore, back up for R₄ are R1 and R6

The relay R₃ is directly connected to relay R₄, so that relay R₃ can’t provide remote backup.

Number of buses \(\left( n \right); = 100\)

Generator busses \(\left( m \right) = 10\)

Order of Jacobian matrix \(= \left( {2n - 1 - m} \right) \times \left( {2n-1-m} \right)\)

\(\begin{array}{l} = \left( {200-1-10} \right) \times \left( {200-1-10} \right)\ = 189 \times 189 \end{array}\)

Surge impedance \(= \sqrt {\frac{L}{C}}\)

\(= \sqrt {\frac{{1.6 \times {{10}^{ - 3}}}}{{10 \times {{10}^{ - 9}}}}} = 400{\rm{\Omega }}\)

Surge impedance loading \(= \frac{{{V^2}}}{{{Z_c}}} = {\left( {\frac{{400 \times {{10}^3}}}{{400}}} \right)^2}\)

\(= 400;MW\)

Given Load \(= 300;MW\)

Load is lesser than surge impedance loading.

So in order to maintain rated voltage at receiving end shunt inductance is required.

Concept:

Capacitors in Series: When capacitors are connected in series, the equivalent capacitance (C_eq) is less than the smallest individual capacitance in the series combination.

The voltage across each capacitor can be different, but the charge (Q) on each capacitor is the same.

Equivalent Capacitance: \(\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \cdots + \frac{1}{C_n} \ \)

Voltage Distribution: The total voltage (V_total) across the series combination is the sum of the voltages across each capacitor:

\( V_{total} = V_1 + V_2 + V_3 + \cdots + V_n \ \)

Here, V_i is the voltage across capacitor (C_i).

Since the same charge (Q) is on each capacitor:

\(Q = C_1 V_1 = C_2 V_2 = C_3 V_3 = \cdots = C_n V_n \ \)

Capacitors in Parallel: When capacitors are connected in parallel, the total (equivalent) capacitance (C_{eq}) is the sum of the individual capacitances. The voltage (V) across each capacitor is the same, but the charge on each capacitor can be different.

Equivalent Capacitance: \(C_{eq} = C_1 + C_2 + C_3 + \cdots + C_n \ \)

Voltage Distribution: Each capacitor has the same voltage applied to it: \( V_{total} = V_1 = V_2 = V_3 = \cdots = V_n \ \)

Each capacitor stores a charge (Q_i) given by:  \(Q_i = C_i V \) 

The total stored charge (Q_total) is the sum of the individual charges:  

\(Q_{total} = Q_1 + Q_2 + Q_3 + \cdots + Q_n \ \)

Equivalent capacitance: \(\frac{1}{C_{eq}} = \sum \frac{1}{C_i}\ \)

Explanation:

From the given figure, it is clear that two capacitors are connected in parallel. So, the voltage across them is same.

d is also same for both capacitors.

Hence, the electric field is also the same by the following formula.

\({\rm{E}} = \frac{{{{\rm{ }}}{{\rm{ }}_{\rm{}}}{\rm{V}}}}{{\rm{d}}}\)

So, electrical filed in the region B is 4 kV/cm

Fault current is inversely proportional to reactance during fault.

\(\Rightarrow I_f\propto\frac{1}{X}\)

\(\Rightarrow \frac{{intial;current}}{{final;current}} = \frac{{Synchronous;reactance}}{{Sub - transient;reactance}} = \frac{1}{{0.2}}\) \(= 5\)

Concept:

For an LTI system, the following results hold true:

• The same form of the sinusoidal signal is produced at the output.
• The only change will be in magnitude and phase.

The following block diagram shows the output

h(t) is the impulse response of the LTI system.

f r(t) = A sin (ωot), and the open loop transfer function be G(s) = G(jωo) = |G(jωo)| ∠G(jωo)

The output signal c(t) can be expressed as:

c(t) = A |G(jωo)|⋅ sin (ωot + ∠G(jωo))

Analysis:

Given input \(x\left( t \right) = cost\) i.e. \(\omega = 1\)

\(\begin{array}{l} \therefore {\rm{H}}\left( {\rm{s}} \right) = \frac{1}{{{\rm{s}} + 1}}\ {\rm{H}}\left( {{\rm{j\omega }}} \right) = \frac{1}{{{\rm{j\omega }} + 1}} = \frac{1}{{1 + {\rm{j}}}} = \frac{1}{{\sqrt 2 }}\angle - 45^\circ = {\rm{M}}\angle \phi \ \therefore {\rm{;output;}} = {\rm{;}}1.{\rm{M}}.{\rm{;cos}}\left( {{\rm{t;}} + {\rm{;}}\phi } \right){\rm{;}} \end{array}\)

\(\therefore {\rm{M}} = {\rm{A}} = \frac{1}{{\sqrt 2 }}\) and \(\phi = - 45^\circ\) 

\(\therefore ;A = 0.707\)

Concept:

In a three-phase diode bridge rectifier, each diode conducts for 120° out of 360°.

The average value of the current through each diode \({{I}_{DA}}=\frac{{{I}_{0}}}{3}\)

The RMS value of the current through each diode \({{I}_{DR}}=\frac{{{I}_{0}}}{\sqrt{3}}\)

Calculation:

In a three phase diode bridge rectifier,

RMS value of diode current \(= \frac{{{I_0}}}{{\sqrt 3 }};\)

Given that \({I_0} = 100\)

\(\Rightarrow \frac{{{I_0}}}{{\sqrt 3 }} = \frac{{100}}{{\sqrt 3 }} = 57.7;A\)

Important Points:

In a three-phase full converter,

The average value of the source current (ISA) = 0

The rms value of the source current \({{I}_{SR}}={{I}_{0}}\sqrt{\frac{2}{3}}\)

The average value of the current through each diode \({{I}_{DA}}=\frac{{{I}_{0}}}{3}\)

The rms value of the current through each diode \({{I}_{DR}}=\frac{{{I}_{0}}}{\sqrt{3}}\)

Concept:

Buck - Boost Converter: 

The buck–boost converter is a type of DC to DC converter that has an output voltage magnitude that is either greater than or less than the input voltage magnitude. It is equivalent to a Fly-back converter using a single inductor instead of a fly-back Transformer. The opposite voltage is the opposite polarity of that of the input.

The steady-state output voltage Vout of buck-boost converter is

\({V_{out}}; = -;{V_{in}}\left( {\frac{D}{{1; - ;D}}} \right)\)

Here D = duty cycle

• When Chopper is ON, Current flows from Sending Voltage, Vs to Chopper, CH, Inductor, L and back to Vs
• Energy is stored in the inductor during TON
• When the chopper is OFF, the inductor stored energy now discharges through the path, load, diode D, and L.

Explanation:

When switch is OFF, diode D is ON as shown in below figure,

Using KVL,

peak voltage across switch = 24 + 36 = 60 V

Given circuit is a lag compensator and transfer function is given as

\(\frac{{{{\rm{V}}_0}\left( {\rm{s}} \right)}}{{{{\rm{V}}_{{\rm{in}}}}\left( {\rm{s}} \right)}} = \frac{{\left( {1 + \frac{1}{{\rm{s}}}} \right)}}{{10 + \frac{1}{{\rm{s}}}}} = \frac{{{\rm{s}} + 1}}{{10{\rm{s}} + 1}} = \frac{{\left( {1 + {\rm{sT}}} \right)}}{{\left( {1 + {\rm{\beta sT}}} \right)}}\)

On comparison we get, \(T = 1\)

\(\beta T = 10; \Rightarrow ;\beta = 10\)

The frequency at which maximum lead occurs is \({{\rm{\omega }}_{\rm{m}}} = \frac{1}{{{\rm{T}}\sqrt {\rm{\beta }} }}\)

\(\therefore {{\rm{\omega }}_{\rm{m}}} = \frac{1}{{1\sqrt {10} }} = 0.316\frac{{{\rm{rad}}}}{{{\rm{sec}}}}\)

Single-phase capacitor run induction motor

• The direction of rotation of a single-phase capacitor run induction motor is reversed by changing the direction of the rotating magnetic field produced by the main and starter winding or auxiliary winding.
• This can be accomplished by reversing the polarity of the starter or auxiliary winding.
• Basically, it is done by interchanging the connections on either end of the starter or auxiliary winding.
• Sometimes this reversal is only done by the auxiliary winding & sometimes reversal is accomplished by winding, switch, and capacitor. While the order of the switch and the capacitor does not affect the reversal, as long as they are in series connection.

Important Points

Advantages of capacitor run induction motor:

• A Centrifugal switch is not required.
• It has high efficiency
• It has a high power factor because of a permanently connected capacitor.
• It has a higher pull-out torque

Limitations:

• Electrolytic capacitors cannot be used for continuous running. Therefore, paper-spaced oil-filled capacitors are used.
• It has relatively low starting torque compared to the capacitor start induction motor.

Concept:

Kirchoff’s second law:

• This law is also known as loop rule or voltage law (KVL) and according to it “the algebraic sum of the changes in potential in the complete traversal of a mesh (closed-loop) is zero”, i.e. Σ V = 0.
• This law represents “conservation of energy” as if the sum of potential changes around a closed loop is not zero, unlimited energy could be gained by repeatedly carrying a charge around a loop.

For any load, current enter through positive polarity and exit from negative polarity and for source it will be reversed.

Calculation:

Apply KVL in the loop as shown below,

The current flow through 2 Ω is 5 A as shown in fig and it will give a drop of 10 V.

From the figure we get,

• V_out + 10 + 10 = 0

Vout = 20 V

Concept:

Nodal Analysis:

Nodal analysis is a method of analyzing networks with the help of KCL equations.

For a network of N nodes, the number of simultaneous equations to be solved to get the unknowns

= Number of KCL equations

= N - 1

Mesh Analysis:

Mesh analysis is a method of analyzing networks with the help of KVL equations.

For a network having N nodes and B branches, the number of simultaneous equations to be solved to get the unknowns

= Number of KVL equations

= number of independent loop equations

= B - N + 1

Where, B = no of the branch, N = No of node

Calculation:

Number nodes n = 5

Branches b = 7

Number of KCL equations = 4

Number of KVL equations = 3

As shown in the figure 

The direction of electric field is always outward  to corresponding surface

Electric field at point D is zero, because it is at the center of electrode.

Electric field at point C have subtractive effect due to second electrode ( i.e. E = E₂ - E' )

Electric field at point A have addetive effect due to second electrode ( i.e. E = E1 + E' ) 

So that at point A, the electric field is maximum.

Concept:

De Morgan’s law states that:

\(\overline {\left( {{A_1}.{A_2} \ldots {A_n}} \right)} = \left( {\overline {{A_1}} + \overline {{A_2}} + \ldots + \overline {{A_n}} } \right)\)

Also,

\(\overline {\left( {{A_1} + {A_2}+ \ldots + {A_n}} \right)} = \left( {\overline {{A_1}} .\overline {{A_2}} ...\overline {{A_n}} } \right)\)

Analysis:

Given

\({\rm{F}} = \overline {\left( {{\rm{a}} + {\rm{̅ b}} + {\rm{c}} + {\rm{̅ d}}} \right) + \left( {{\rm{b}} + {\rm{̅ c}}} \right)}\)

Applying the De-Morgans theorem in the above function F

\(\begin{array}{l} = \overline {\left( {{\rm{a}} + {\rm{̅ b}} + {\rm{c}} + {\rm{̅ d}}} \right)} \cdot \overline {\left( {{\rm{b}} + {\rm{̅ c}}} \right)} \ \end{array}\)

= a̅.b.c̅.d.b̅. c 

As, b.b̅  = c.c̅ = 0

∴ F = a̅.b.c̅.d.b̅. c  = 0

Hence option (4) is correct

Concept: 

Virtual ground -

The concept of the virtual ground is stated as if anyone of the i/p terminals is grounded physically the other i/p terminal will also be at ground potential even though, it is not grounded physically.

• One key feature of an Op-Amp is the differential input, and when put together in a circuit, this can form a virtual ground.
• The virtual ground concept is helpful for the analysis of Op Amps. This concept makes Op-Amp circuit analysis much easier.

Calculation:

Voltage at Non-inverting terminal, \({{\rm{v}}_ + } = \frac{{{{\rm{v}}_1}\left( 4 \right)}}{{4 + 1}} + \frac{{{{\rm{v}}_2}\left( 1 \right)}}{{1 + 4}}\)

\({{\rm{v}}_ + } = \frac{{\left( {4{{\rm{v}}_1} + {{\rm{v}}_2}} \right)}}{5}\)

Voltage at inverting \({{\rm{v}}_ - } = {{\rm{v}}_ + } = \left( {\frac{{4{{\rm{v}}_1} + {{\rm{v}}_2}}}{5}} \right)\)

Apply nodal at inverting terminal we get

\(\begin{array}{l} \left( {\frac{{4{{\rm{v}}_1} + {{\rm{v}}_2}}}{5}} \right) - {{\rm{v}}_3} + \frac{{\left[ {\left( {\frac{{4{{\rm{v}}_1} + {{\rm{v}}_2}}}{5}} \right) - {{\rm{v}}_{{\rm{out}}}}} \right]}}{9} = 0\ \left( {\frac{{4{{\rm{v}}_1} + {{\rm{v}}_2}}}{5}} \right)\left[ {1 + \frac{1}{9}} \right] = {{\rm{v}}_3} + \frac{{{{\rm{v}}_{{\rm{out}}}}}}{9}\ 8{V_1} + 2{V_2}-9{V_3} = {V_{out}}\ \therefore ;{V_{out}} = 8{V_1} + 2{V_2}-9{V_3} \end{array}\)

Concept:

• Lower limit of y(t) is sum of lower limit of x(t) & h(t).
• Upper limit of y(t) is sum of upper limit of x(t) & h(t).

Explanation:

Fourier transform of x1(t) = x1(ω)

Maximum frequency components of

x1(ω) = B, rad/sec

Fourier transform of x2(t) = x2(ω)

maximum frequency component of x2(ω) = B2 rad/sec

∴ x(t) = x1(t). x2(t)

Using multiplication property in Fourier transform,

\(F{x(t)}=x(ω)=\frac{1}{2\pi}{x_1(ω)*x_2(ω)}\)

So, maximum frequency component of x(ω) = B1 + B2 rad/sec

Impulse response  \(h(t)=e^{-2|t|}\)

Taking Fourier transform of h(t) is,

\(H(ω)=\frac{4}{4+ω^2}\)

Maximum frequency component of H(ω) = ∞

∴ y(t) = x(t) * h(t)

Fourier Transform, Y(ω) = x(ω) H(ω)

\(=\frac{1}{2\pi}{x_1(ω) *x_2(ω)}.H(ω)\)

Highest frequency component of

Y(ω) = min {x(ω), H(ω)}

Y(ω) = min {B1 + B2, ∞}

Y(ω) = B1 + B2

Minimum sampling rate = Nyquist rate

= 2 × maximum frequency component of Y(ω)

= 2 (B1 + B2) rad/sec

Concept:

Duality Property

If \(\rm x(t)\overset{F.T.}{\leftrightarrow}x(\omega)\)

then \(\rm x(t)\overset{F.T.}{\leftrightarrow}2\pi x(-\omega)\)

Fourier time of signal x(t) is defined as,

\(x(\omega)=\displaystyle\int^{\infty}_{-\infty}x(t).e^{-j\omega t}dt\)

putting t = 0;

\(x(0)=\displaystyle\int^{\infty}_{-\infty}x(t) dt\)

Explanation:

We know that \(\mathop \smallint \limits_{ - \infty }^\infty {\rm{x}}\left( {\rm{t}} \right) \cdot {\rm{dt}} = {\rm{X}}\left( 0 \right)\)

Also we know

Where \(X\left( 0 \right) = 1\)

\(\therefore 2\mathop \smallint \limits_{ - \infty }^\infty \left( {\frac{{{\rm{sin}}2{\rm{\pi t}}}}{{{\rm{\pi t}}}}} \right) \cdot {\rm{dt}} = 2{\rm{X}}\left( 0 \right) = 2\left( 1 \right) = 2\)

​

Concept:

 Laplace transform of the first two derivatives.

L{y''} = s2Y(s) - sy(0) - y'(0)

L{y'} = sY(s) - y(0)

Calculation:

Given \(\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}} - 4 \cdot \frac{{{\rm{dy}}}}{{{\rm{dx}}}} + 4{\rm{y}} = 0\)

Apply Laplace transform we get

\(\begin{array}{l} \left[ {{s^2}Y\left( s \right)-sy\left( 0 \right)-;y\left( 0 \right)\left] { - 4} \right[sY\left( s \right)-y\left( 0 \right)} \right] + 4Y\left( s \right) = 0\ \left( {{s^2} - 4s + 4} \right)Y\left( s \right) = 1\ {\rm{Y}}\left( {\rm{s}} \right) = \frac{1}{{{{\left( {{\rm{s}} - 2} \right)}^2}}}\ y\left( t \right) = t \cdot {e^{2t}}\ y\left( 1 \right) = {e^2} = 7.389 \end{array}\)

Explanation:

Line integral:

\(\smallint {\vec{F}} \cdot {\rm{d\vec r}} = \smallint 5{\rm{xzdx}} + \smallint\left( {3{{\rm{x}}^2} + 2{\rm{y}}} \right){\rm{dy}} + \smallint{{\rm{x}}^2}{\rm{z}} {\rm{dz}}\)

\( x = t \Rightarrow dx = dt\)

\(y = {t^2} \Rightarrow dy = 2t;dt\)

\( z = t \Rightarrow dz = dt\)

\(\smallint {\rm{\vec F}} \cdot {\rm{d\vec r}} = \mathop \smallint \limits_0^1 5{{\rm{t}}^2}{\rm{dt}} + \left( {3{{\rm{t}}^2} + 2{{\rm{t}}^2}} \right)2{\rm{tdt}} + {{\rm{t}}^2} \cdot {\rm{t}} \cdot {\rm{dt}} = \frac{5}{3} + \frac{{10}}{4} + \frac{1}{4} = \frac{5}{3} + \frac{{11}}{4}\)

\(\therefore \smallint {\rm{\vec F}} \cdot {\rm{d\vec r}} = \frac{{53}}{{12}}{\rm{;}} = {\rm{;}}4.42\)

Given \({\rm{P}} = \left[ {\begin{array}{*{20}{c}} 3&1\ 1&3 \end{array}} \right]\)

\(\left[ {\begin{array}{*{20}{c}} {\rm{a}}\ {\rm{b}} \end{array}} \right] = {\rm{P}}\left[ {\begin{array}{*{20}{c}} {\rm{x}}\ {\rm{y}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 3&1\ 1&3 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\rm{x}}\ {\rm{y}} \end{array}} \right]\)

∴ a = 3x + y and b = x + 3y

Given \({a^2} + {b^2} = 1\)

\(\begin{array}{l} \therefore ;{\left( {3x + y} \right)^2} + {\left( {x + 3y} \right)^2} = 1\ 10{x^2} + 10{y^2} + 12xy-1 = 0 \end{array}\)

\({x^2} + {y^2} + 1.2xy-0.1 = 0\)

\( \Rightarrow \frac{{{{\left( {x + y} \right)}^2}}}{{1/8}} + \frac{{{{\left( {x - y} \right)}^2}}}{{1/2}} = 1\)

It represents an ellipse a < b

Major axis is x + y = 0 and minor axis is x – y = 0

∴ So the given equation is the tilted version of ellipse with angle \(45^\circ\) with minor axis along \(\left( {\begin{array}{*{20}{c}} 1\ 1 \end{array}} \right)\)

Concept:

A probability density function is a function of continuous random variable, whose integral across an interval gives the probability that the value of the variable lies within the same interval.

For a given probability density function f(x) to be valid, the integration of it over the complete range must equal 1.

Mathematically, it must satisfy:

\(\mathop \smallint \limits_{ - \infty }^\infty f\left( x \right) = 1\)

Calculation:

Given \({{\rm{f}}_{\rm{X}}}\left( {\rm{x}} \right) = \frac{3}{2}{{\rm{e}}^{ - 3{\rm{x}}}}{\rm{u}}\left( {\rm{x}} \right) + {\rm{a}} \cdot {{\rm{e}}^{4{\rm{x}}}}{\rm{u}}\left( { - {\rm{x}}} \right)\) 

We know \(\mathop \smallint \limits_{ - \infty }^\infty {{\rm{f}}_{\rm{x}}}\left( {\rm{x}} \right) \cdot {\rm{dx}} = 1\) 

\(\begin{array}{l} \mathop \smallint \limits_{ - \infty }^0 {\rm{a}} \cdot {{\rm{e}}^{4{\rm{x}}}} + \mathop \smallint \limits_0^\infty \frac{3}{2} \cdot {{\rm{e}}^{ - 3{\rm{x}}}} = 1\ {\rm{a}} \cdot \left[ {\frac{{{{\rm{e}}^{4{\rm{x}}}}}}{4}} \right]_{ - \infty }^0 + \frac{3}{2} \cdot \left( {\frac{{{{\rm{e}}^{ - 3{\rm{x}}}}}}{{ - 3}}} \right)_0^\infty = 1\ \frac{{\rm{a}}}{4}\left[ {1 - 0} \right] - \frac{1}{2}\left[ {0 - 1} \right] = 1\ \frac{{\rm{a}}}{4} = \frac{1}{2} \Rightarrow {\rm{a}} = 2\ {\rm{P}}\left( {{\rm{X}} \le 0} \right) = \mathop \smallint \limits_{ - \infty }^0 {{\rm{f}}_{\rm{X}}}\left( {\rm{x}} \right) \cdot {\rm{dx}}\ = \mathop \smallint \limits_{ - \infty }^0 2 \cdot {{\rm{e}}^{4{\rm{x}}}} \cdot {\rm{dx}} = 2 \cdot \left[ {\frac{{{{\rm{e}}^{4{\rm{x}}}}}}{4}} \right]_\infty ^0 = \frac{1}{2}\left[ {1 - 0} \right] = \frac{1}{2} \end{array}\)

Concept:

Driving point impedance or the input impedance is the impedance seen from the input side.

Example:

A simple RC driving point impedance circuit is shown below.

Now driving point impedance is calculated as

\(z\left( s \right) = R + \frac{1}{{sC}}\)

Calculation:

Given circuit diagram

Let V_A be a node

Apply Nodal Analysis we get,

\(⇒- {{\rm{I}}_{\rm{s}}} + \frac{{{{\rm{V}}_{\rm{A}}}}}{3} - 4{{\rm{V}}_1} + \frac{{{{\rm{V}}_{\rm{A}}}}}{6} = 0 \)

\(⇒ \frac{{3{{\rm{V}}_{\rm{A}}}}}{6} = 4{{\rm{V}}_1} + {{\rm{I}}_{\rm{s}}}\) → 1

But \({{\rm{I}}_{\rm{s}}} = \frac{{\left( {{{\rm{V}}_{\rm{s}}} - {{\rm{V}}_{\rm{A}}}} \right)}}{2}\) and \(-{V_S} + {V_1} + {V_A} = 0\)

⇒ VA = VS - V₁ → 2

⇒ 2I_S = VA + V₁ - VA = V₁ = 2I_S → 3

From equations 1, 2, and 3 we get

⇒ VA = 16I_S + 2I_S

⇒ V_S - 2I_S = 18I_S

⇒ VS = 20IS

 \(\frac{V_s}{I_s} = \frac{20}{1}=20 Ω\) 

Therefore the driving point impedance of the given circuit is 20 Ω

Concept:

Z parameters:

V₁ = Z₁₁ I₁ + Z₁₂ I₂

V₂ = Z₂₁ I₁ + Z₂₂ I₂

Calculation:

From the z-matrix we get,

V₁ = 40I₁ + 60I₂

V1 = 80I1 + 100I₂ 

V₂ = -I₂R_L = -20I₂ 

∴ (20 - 10I₁) = 40I₁ + 60I₂

20 = 50(-120/80)I₂ + 60I₂

20 = I₂[60 - 600/8]

∴ I₂ = -1.33 Amp

Average power delivered to \({{\rm{R}}_{\rm{L}}} = {\rm{I}}_2^2 \cdot {{\rm{R}}_{\rm{L}}} \)

 = (1.33)² × 20 = 35.56 watt 

Admittance (Y) matrix of a 2 port network can be written as:

\(\left[ Y \right]= {\left[ Z \right]^{ - 1}}\)

[Z] = Impedance matrix of the two-port network represented as:

\(\left[ {\rm{Z}} \right]{\rm{}} = {\rm{}}\left[ {\begin{array}{*{20}{c}} {{Z_{11}}}&{{Z_{12}}}\ {{Z_{21}}}&{{Z_{22}}} \end{array}} \right]\)

Also,

\({\left[ Z \right]^{ - 1}}= \frac{1}{{\left( {{Z_{11}}{Z_{22}} - {Z_{12}}{Z_{21}}} \right)}}\left[ {\begin{array}{*{20}{c}} {{Z_{22}}}&{-{Z_{21}}}\ {{-Z_{12}}}&{{Z_{11}}} \end{array}} \right]\)

In the given circuit diagram,

The inductive reactance X_L = jωL = j2πfL

\(\begin{array}{l} {X_L} = jω L\ = j\left( {2π \times 50} \right) \times 10 \times {10^{ - 3}}\ = jπ ;\Omega \end{array}\)

When the star connection of capacitors are converted into the delta 

Star to delta conversion in capacitors:

\({C_1} = \frac{{\frac{1}{{{C_A}}}}}{{\frac{1}{{{C_A}}}.\frac{1}{{{C_B}}} + \frac{1}{{{C_B}}}.\frac{1}{{{C_C}}} + \frac{1}{{{C_C}}}.\frac{1}{{{C_A}}}}}\)

\({C_2} = \frac{{\frac{1}{{{C_B}}}}}{{\frac{1}{{{C_A}}}.\frac{1}{{{C_B}}} + \frac{1}{{{C_B}}}.\frac{1}{{{C_C}}} + \frac{1}{{{C_C}}}.\frac{1}{{{C_A}}}}}\)

\({C_3} = \frac{{\frac{1}{{{C_C}}}}}{{\frac{1}{{{C_A}}}.\frac{1}{{{C_B}}} + \frac{1}{{{C_B}}}.\frac{1}{{{C_C}}} + \frac{1}{{{C_C}}}.\frac{1}{{{C_A}}}}}\)

As all the capacitors are of equal value, the equivalent value of capacitor when converted from star to delta will be C' = C/3

Capacitive reactance of each phase will be XC = 1/jωC' = 3/jωC

The equivalent circuit diagram will be,

For all line currents are zero

\(\begin{array}{l} \Rightarrow \left| {{X_L}} \right| = \left| {{X_C}} \right|\ \Rightarrow \frac{3}{{ω c}} = π \ \Rightarrow C = \frac{3}{{2π \left( {50} \right)\left( π \right)}} = 0.00304 = 3;mF \end{array}\)

Concept:

The transient equation for the voltage across a capacitor is given by,

\({V_c}\left( t \right) = V\left( ∞ \right) + \left[ {V\left( 0 \right) - V\left( ∞ \right)} \right]{e^{ - \frac{t}{τ }}}\)

V(∞) is the final voltage

V(0) is the initial voltage

τ is the time constant

The time constant of RC series circuit τ = R × C 

Calculation:

Given circuit diagram:

Time constant  = RC = 5 × 5 × 10 ^-6 = 25 × 10^-6 secs

Equation for the voltage across capacitor is given by:

\( V\left( t \right) = {V_0};{e^{ - \frac{t}{{RC}}}}\) (as there is no active source in the network the capacitor dicharges and voltage acorss capacitor will be zero at time t = ∞)

\( V\left( t \right)= 4 \times {e^{ - \frac{t}{{25 \times {{10}^{ - 6}}}} }}\)

Charge in the capacitor at time t is q(t) = C v(t)

q(t) = 20 × 10-6 × e-t/(25×10^-6)

At 25 μs,

 \( q_{1}\left( t \right)= 20 \times {e^{ - \frac{25 \times {{10}^{ - 6}}}{{25 \times {{10}^{ - 6}}}} }} =7.35 μ C\)

At 100 μs,

\( q_{2}\left( t \right)= 20 \times {e^{ - \frac{100 \times {{10}^{ - 6}}}{{25 \times {{10}^{ - 6}}}} }} =0.36 μ C\)

The charge lost Q = q₁ - q₂ = (7.35 - 0.36) μC

The charge lost = 7 μc

\({X_{1cg}} = j0.2 + \left( {j0.6} \right)\left( {j0.6} \right) = j0.5\)

Maximum power transfer \(\left( {{P_{m1}} = \frac{{1.0 \times 1.0}}{{0.5}}} \right) = 2;p.u.\)

During fault at the middle of line – 2.

Convert delta to star,

Convert star to delta

\(\begin{array}{l} {P_{{m_2}}} = \frac{{1.0 \times 1.0}}{{1.143}}\ \Rightarrow \frac{{{P_{{m_1}}}}}{{{P_{{m_2}}}}} = \frac{2}{{\frac{1}{{1.143}}}} = 2 \times 1.143 = 2.286 \end{array}\)

Given \({\rm{G}}\left( {\rm{s}} \right) = \frac{{{\rm{K}}\left( {{\rm{s}} + 1} \right)}}{{{\rm{s}}\left( {1 + {\rm{sT}}} \right)\left( {1 + 2{\rm{s}}} \right)}}\)

Characteristic equation 1 + G(s) H(s) = 0

As it is a unity feedback control system H(s) = 1

\(1+G\left( s \right)H\left( s \right) = 1+\frac{{K\left( {s + 1} \right)}}{{s\left( {1 + Ts} \right)\left( {1 + 2s} \right)}}=0\)

⇒ s + 2s² + s²T + 2s³T + Ks + K = 0

⇒ (2T)s³ + (2 + T)s² + (1 + K)s + K = 0

From R-H stability criteria 

\(\begin{array}{*{20}{c}} {{s^3}}\ {{s^2}}\ {{s^1}}\ {{s^0}} \end{array}\left| {\begin{array}{*{20}{c}} 2T&2+T\ 1+K&{K}\ \frac{{(2+T) \left( {1 + K} \right) - K(2T)}}{1+K }&0\ K&{} \end{array}} \right.\)

For closed-loop system to be stable 

2T > 0, 1 + K > 0, \(\frac{{(2+T) \left( {1 + K} \right) - K(2T)}}{1+K }>0\), K > 0

\(0 < K < \frac{{T + 2}}{{T - 2}};\)

Alternate Method

Consider \({\rm{a}}{{\rm{s}}^3}{\rm{;}} + {\rm{;b}}{{\rm{s}}^2}{\rm{;}} + {\rm{;cs;}} + {\rm{;d;}} = {\rm{;}}0\) from the R – H criteria the system is stable when

\(\begin{array}{l} {\rm{bc;}} > {\rm{;ad;;and;a}},{\rm{;b}},{\rm{;c}},{\rm{;d;}} > {\rm{;}}0\ \therefore {\rm{;}}\left( {2{\rm{;}} + {\rm{;T}}} \right){\rm{;}}\left( {1{\rm{;}} + {\rm{;K}}} \right){\rm{;}} > {\rm{;}}2{\rm{TK}}\ 2{\rm{;}} + {\rm{;}}2{\rm{K;}} + {\rm{;T;}} + {\rm{;TK;}}-{\rm{;}}2{\rm{TK;}} > {\rm{;}}0\ \left( {2{\rm{;}} + {\rm{;T}}} \right){\rm{;}} + {\rm{;K}}\left( {2{\rm{;}} - {\rm{;T}}} \right){\rm{;}} > {\rm{;}}0\ {\rm{K}}\left( {2{\rm{;}} - {\rm{;T}}} \right){\rm{;}} > {\rm{;}} - \left( {2{\rm{;}} + {\rm{;T}}} \right)\ {\rm{K}} > \frac{{ - \left( {2 + {\rm{T}}} \right)}}{{\left( {2 - {\rm{T}}} \right)}} ⇒ {\rm{K}} < \frac{{\left( {{\rm{T}} + 2} \right)}}{{\left( {{\rm{T}} - 2} \right)}} \end{array}\)

∴ Condition for stability: \(0 < {\rm{K}} < \left( {\frac{{{\rm{T}} + 2}}{{{\rm{T}} - 2}}} \right)\)

For a long transmission line receiving end voltage is higher than sending end voltage, this indicated Ferranti effect

V_s = A V_R (at the lightly loaded condition)

Where 

V_s is the sending end voltage

V_R is the receiving end voltage

Also from the ABCD parameters, the approximate value of A is

\(A=1+\frac{YZ}{2}=1-\frac{\omega^{2}l^{2}}{2V^{2}}\)

\( A = \frac{{{V_S}}}{{{V_R}}} = \frac{{400 \times {{10}^3}}}{{420 \times {{10}^3}}} = 0.95\)

\(A=0.95=1 - \frac{{{\omega ^2}{l^2}}}{{2{V^2}}}\)

Where v= Velocity of traveling Wave

           C= Speed of light.

\(⇒ 1 - \frac{{{{\left( {2\pi \times 50} \right)}^2}{l^2}}}{{2 \times {{\left( {C} \right)}^2}}} = 0.95 \)

⇒ l = 294.8 km

Therefore the length of the transmission line will be 294.8km

Given,

Apparent power S = 500 kVA

Power factor cos ϕ = 0.8 

sin ϕ = √(1 - cos² ϕ) = 0.6

Reactive power of motor Q = S sin ϕ = 500 × 0.6

Q = 300 kVAr

Given real power of motor P = 100 kW 

Total power  = √(P² + Q²)

Total power \(= \sqrt{{{100}^2} + {{300}^2}} = 316.22 kVA\) 

⇒ Power factor = \(cos \phi = \frac{P}{S}\)

\(cos \phi= \frac{{100}}{{\sqrt {{{100}^2} + {{300}^2}} }} = 0.316\) leading

Magnetic stored energy,

\({f_e} = - \frac{\partial }{{\partial x}}w\left( {\lambda x} \right)\)

\(\lambda = \frac{{\sqrt i }}{g}\)

\( \Rightarrow \sqrt i = \lambda g\)

\( \Rightarrow i = {\lambda ^2}{g^2}\)

\(w\left( {\lambda ,x} \right) = \mathop \smallint \limits_0^\lambda id\lambda \)

\( = \mathop \smallint \limits_0^\lambda {\lambda ^2}{g^2}d\lambda \)

\( = {g^2}\frac{{{\lambda ^3}}}{3}\)

\({f_e} = - \frac{\partial }{{\partial x}}w\left( {\lambda x} \right)\)

\( = - \frac{\partial }{{\partial x}}\left( {{g^2}\frac{{{\lambda ^3}}}{3}} \right)\)

\( = - \frac{{{\lambda ^3}}}{3}\frac{\partial }{{\partial x}}\left( {{g^2}} \right)\)

x = g = 0.1 m

\( = - \frac{{{\lambda ^3}}}{3}2g\)

\(\left| {{f_e}} \right| = \frac{2}{3}{\lambda ^3}g\)

\( = \frac{2}{3}{\left( {\frac{{\sqrt i }}{g}} \right)^3}g\)

\( = \frac{2}{3}\frac{{i\sqrt i }}{{{g^2}}}\)

\( = \frac{2}{3} \times \frac{{2\sqrt 2 }}{{{{\left( {0.1} \right)}^2}}}=188.56~N\)

At starting slip, \(S = 1\)

\(\begin{array}{l} Current;drawn;\alpha ;voltage\ \Rightarrow \frac{{{I_{s{t_1}}}}}{{{I_{s{t_2}}}}} = \frac{{{V_1}}}{{{V_2}}}\ \Rightarrow \frac{{120}}{{{I_{s{t_2}}}}} = \frac{{415}}{{110}}\ \Rightarrow {I_{s{t_2}}} = \frac{{110 \times 120}}{{415}} = 31.807A \end{array}\)

Current rating in single-phase transformer \(= \frac{{2 \times {{10}^3}}}{{100}} = 20A\) 

\(KVA;rating = 300 \times 20 = 6;KVA\)

Alternate Method

Consider the transformation ratio for Autotransformer is \(K = \frac{{{N_1}}}{{{N_2}}} \)

The kVA rating of autotransformer = \(\frac{1}{{1 - \frac{1}{K}}}\) × two winding transformers; for step-up transformer

Note: Maximum kVA rating of Auto-transformer occurs in Additive polarity.

Calculation:

Voltage rating of Autotransformer in Additive polarity = 200/300 V

Transformation ratio \(\left( K \right) = \frac{{200}}{{300}} = \frac{2}{3}\)

The new kVA rating is given as,

\(= \frac{1}{{1 - \frac{2}{3}}} \times 2=6\ kVA\)

Concept:

Single-phase full converter with RLE load:

During α to π, both vs and is are positive, power, therefore, flows from ac source to load. During the interval π to (π + α), vs is negative but is positive, the load,

therefore, returns some of its energy to the supply system. But the net power flow is from ac source to dc load because (π – α) > α.

The average value of the output voltage is given by:

\({V_0} = \frac{{2{V_m}}}{\pi }\cos \alpha \)

• If α > 90°, V0 is negative. If the load circuit emf E is reversed, this source E will feed power back to ac supply. This operation of full converter is known as the inverter operation of the converter.
• The full converter with a firing angle delay greater than 90° is called a line-commutated inverter. Such an operation is used in the regenerative braking mode of a dc motor in which case then E is counter emf of the dc motor.
• During 0 to α, ac source voltage vs is positive, but ac source current is negative, power, therefore, flows from dc source to ac source.
• From α to π, both vs and is are positive, power, therefore, flows from ac source to dc source. But the net power flow is from dc source to ac source because of (π – α) < α.

Explanation:

The average value of the output voltage is given by:

\({V_0} = \frac{{2{V_m}}}{\pi }\cos \alpha \)

Applying KVL at output,

V_o = - E + RI 

\(\begin{array}{l} \frac{{2{V_m}}}{\pi } \times cos\alpha = E + RI\ \Rightarrow \frac{{2 \times 200 \times \pi }}{\pi }\cos 120 = - 800 + \left( {20} \right)\left( {{I_0}} \right)\ \Rightarrow - 200 = - 800 + 20{I_0}\ \Rightarrow {I_0} = 30;A \end{array}\)

Hence, the real power feedback to source is,

Power \(= 200 \times 30 = 6;kW\)

Concept:

Calculation:

A three phase voltage source inverter supplying equivalent delta load.

Equivalent star load resistance is,

\(\begin{array}{l} {R_{ph}} = \frac{{30}}{5} = 10;{\rm{\Omega }}\ {V_{ph}} = \frac{{\sqrt 2 }}{3}{V_{dc}} = \frac{{\sqrt 2 }}{3} \times 600 = 200\sqrt 2 V = 282.84;V\ P = 3\frac{{{V_p}{h^2}}}{R} = 3 \times \frac{{{{\left( {200\sqrt 2 } \right)}^2}}}{{10}} = 24kW \end{array}\)

Concept:

In a boost converter, the output voltage is given by,

\(\frac{V_0}{V_s}=\frac{1}{1−D}\)

Calculation:

Input dc voltage (V_s) = 360 V

Output voltage (V₀) = 400 V

⇒ \(\frac{400}{360}=\frac{1}{1-D}\)

⇒ 400 - 400 D = 360

⇒ 400 D = 40

D = 0.1

Output power = V_sI_s = 400 × I₀ = 4 kW

⇒ I0 = 10 A

From the power balance equation, input power is equal to output power.

V_sI_s = V_oI_o

or, 360 Is = 4000

Is = 11.1  A

Neglecting ripple in Is

IRMS = \(I_s \times \sqrt{D}=11.1\times \sqrt{0.1}\approx3.5\ A\)

DC power, P_dc = 5 kW

AC power, P_ac = V_s1 I_s1 cos θ

= 220 × I_s1 × 1 = 220 I_s1

DC power is equal to AC power.

220 I_s1 = 5000

⇒ I_s1 = 22.72 A

X_S = ωL = 2πfL = 2π × 50 × 5 × 10^-3

From the given phasor diagram,

\(\tan \delta = \frac{{{I_s}{X_s}}}{{{V_s}}}\)

\( = \frac{{22.72 \times 2\pi \times 50 \times 5 \times {{10}^{ - 3}}}}{{220}} = 0.162\)

⇒ δ = 9.2°

Given \({\rm{\dot x}} = {\rm{Ax}}\)

\(\begin{array}{l} {\rm{sX}}\left( {\rm{s}} \right){\rm{;}}-{\rm{;x}}\left( 0 \right){\rm{;}} = {\rm{;A;X}}\left( {\rm{s}} \right)\ {\rm{X}}\left( {\rm{s}} \right){\rm{;}}\left[ {{\rm{sI;}} - {\rm{;A}}} \right]{\rm{;\alpha ;}} \Rightarrow {\rm{;X}}\left( {\rm{s}} \right){\rm{;}} = {\rm{;}}{\left( {{\rm{sI;}} - {\rm{;A}}} \right)^{ - 1}}{\rm{;}} \cdot {\rm{;\alpha ;}}\ {\rm{x}}\left( {\rm{t}} \right){\rm{;}} = {\rm{;}}{{\rm{e}}^{{\rm{At}}}}{\rm{;\alpha ;}} \end{array}\)

The Eigen value corresponding to \(\alpha\) is \({\lambda _1}\)

\(\therefore {\rm{x}}\left( {\rm{t}} \right) = {{\rm{e}}^{{{\rm{\lambda }}_1}{\rm{t}}}} \cdot {\rm{\alpha }}\)

Standard 2^nd order system is \({\rm{T}}\left( {\rm{s}} \right) = \frac{{{\rm{K\omega }}_{\rm{n}}^2}}{{{{\rm{s}}^2} + 2{\rm{\xi }}{{\rm{\omega }}_{\rm{n}}}{\rm{s}} + {\rm{\omega }}_{\rm{n}}^2}}\) 

Given \({\rm{\xi ;}} = {\rm{;}}0.5{\rm{;and;}}{{\rm{\omega }}_{\rm{n}}}{\rm{;}} = {\rm{;}}10\)

Steady state value \({\left. {{\rm{T}}\left( {\rm{s}} \right)} \right|_{{\rm{s}} = 0}} = {\rm{K}} = 1.02\)

\(\therefore {\rm{T}}\left( {\rm{s}} \right) = \frac{{\left( {1.02} \right)\left( {100} \right)}}{{{{\rm{s}}^2} + 10{\rm{s}} + 100}} = \frac{{102}}{{{{\rm{s}}^2} + 10{\rm{s}} + 100}}\)

∇ - side phase voltage \(= 110;kV\)

Y – side phase voltage \(= \frac{{11}}{{\sqrt 3 }}kV\)

\(\frac{{\nabla - side;phase;current}}{{Y - side;phase;current}} = \frac{{11}}{{\sqrt 3 \left( {110} \right)}} = \frac{1}{{10\sqrt 3 }}\)

Given \({\rm{G}}\left( {\rm{s}} \right) = \frac{{{\rm{Ks}}}}{{\left( {{\rm{s}} - 1} \right)\left( {{\rm{s}} - 4} \right)}}\)

Characteristic equation is 1 + G(s) H(s) = 0

\({1+\rm{G}}\left( {\rm{s}} \right)H(s) = 1+ \frac{{{\rm{Ks}}}}{{\left( {{\rm{s}} - 1} \right)\left( {{\rm{s}} - 4} \right)}}\)

⇒ (s - 1)(s - 4) + Ks = 0

∴ \( {\rm{K}} = \frac{{ - \left( {{\rm{s}} - 1} \right)\left( {{\rm{s}} - 4} \right)}}{{\rm{s}}}\)

For finding the breakaway point \( \frac{{{\rm{dK}}}}{{{\rm{ds}}}} = 0\)

\(⇒ \frac{{{\rm{dK}}}}{{{\rm{ds}}}}=- \left[ {\frac{{{\rm{s}}\left( {2{\rm{s}} - 5} \right) - \left( {{{\rm{s}}^2} - 5{\rm{s}} + 4} \right)}}{{{{\rm{s}}^2}}}} \right] = 0\)

⇒ s² - 4 = 0

⇒ s = ± 2

By plotting on s plane 

Centroid = \(\frac{{\left( {1 + 4} \right) - 0}}{1} = 5\)

∴ The valid break-away point is between 1 and 4.

∴ Valid breakaway point is \(s = 2\)

∴ Gain at breakaway point \(= \frac{{{\rm{product;of;distance;from;poles}}}}{{{\rm{product;of;distance;from;zeros}}}}\)

\(= \frac{{\left( 1 \right)\left( 2 \right)}}{2} = 1\)

For \(L - G\) fault,

\({I_f} = \frac{3}{{{x_1} + {x_2} + {x_0}}}\)

Positive require a network is

\(\Rightarrow {X_1} = j0.2;p.u.\)

Negative recurrence network is

\(\Rightarrow {X_2} = j0.15;p.u\)

Zero sequence network is

\(\begin{array}{l} \Rightarrow {X_0} = j0.15;p.u.\ {I_f} = \frac{3}{{j0.2 + j0.15 + j0.15}} = \frac{3}{{j0.5}} = 6p.u \end{array}\)

Concept:

Induction type energy meter:

• The basic principle of induction type energy meter is electromagnetic induction.
• When alternating current flow through two suitably located coils produces a rotating magnetic field which is cut by the metallic disc suspended near to the coils thus emf induce in the disc which is circulated eddy current in it.
• With the interaction of the rotating magnetic field and eddy currents, torque is developed and causes the disc to rotate.

Meter constant(K):

\(K = \frac{{No.;of;revolutions}}{{Energy;in;kWh}}\)

Explanation:

Given,

K = 1200, t = 30 seconds = 30/3600 hr

Number of revolutions = 20

Energy of meter in kWh (E) = kW × t

K = 20 / (kW × t)

1200 = 20 / (kW × (1 / 120))

Load in kW = 2 kW

Concept:

EMF induced in a moving conductor. The emf induced in a straight conductor of length l moving with velocity v perpendicular to a magnetic field B is

E = Blv

Where B, l, and v are mutually perpendicular.

The emf is in volts

B is in webers/m2

l is in meters

v is in m/sec.

If the velocity vector v makes an angle θ with the direction of the magnetic field, then

E = Blv sin θ

EMF induced in the coil would be zero when its plane is at right angles to the plane of the field, even though it will have maximum flux linked with it. i.e. θ = 0°

However, the coil will have a maximum EMF induced in it when its plane lies parallel to the plane of the field even though it will have minimum flux linked with it. i.e. θ = 90°

Calculation:

In the given diagram magnetic field is perpendicular to the direction of rotation

θ = 90° 

E = Blv sin θ

Given, 

E = 1V, B = 1T, l = 1m

⇒ 1 = 1 × 1 × v × sin 90° 

⇒ v = 1 m/s

\(V = 1 × \frac{{60}}{{2\pi }} = 9.549;rpm \)