GATE EE 2015 Official Paper: Shift 2 (Previous Year Paper)

Given \(f\left( x \right) = 1-\left| x \right|\) in \(- 1 \le x \le 1\)

Since, \(\left| x \right| > 0,;for;all;x\)

So, \(\left| x \right|\)can take minimum value of 0 and maximum value of 1. Also on increasing value of \(\left| x \right|\), value of \(f\left( x \right)\) decreases. At \(\left| x \right| = 0\), we obtain

\(f\left( x \right) = 1 - \left| x \right|\)

\(= 1 - 0 = 1\) (function has maximum value)

Again, at \(\left| x \right| = 1\), we obtain

\(f\left( x \right) = 1 - \left| x \right|\)

\(= 1 - 1 = 0\) (function has minimum value)

Thus, at \(\left| x \right| = 0\) value of \(f\left( x \right)\) is maximum and at \(\left| x \right| = 1\) value of \(f\left( x \right)\) is minimum.

The word ‘personnel’ means 'people who are employed in an organization'.

Out of the given options, this meaning is accurately represented only in the sentence from option 3. The other options are instead referring to the word 'personal'.

Hence, option 3 is the answer.

Here are the meanings of the given words:

• Fabric - cloth produced by weaving or knitting fibres
• Textile - a type of cloth or woven fabric
• Fibre - a thread or filament from which textile is formed
• Apparel - clothing in general

We can see that the word that most accurately defines the given set of words is ‘apparel’.

Hence, option 4 is correct.

The given problem requires the total weight of 10 poles, each of same weight.

We check the given statements for evaluation of the total weight.

Statement I:

Given that one fourth of weight of a pole is 5 kg.

Let weight of a pole \(= x;kg\)

\(\begin{array}{l} \frac{x}{4} = 5\ x = 20;kg \end{array}\)

Total weight of 10 poles \(= 10x = 200;kg\)

Statement II:

The total weight of 10 poles is 160 kg more than the total weight of two poles.

Let weight of a pole \(= x;kg\)

So, total weight of 10 poles \(= 10x\)

\(\begin{array}{l} 10x = 2x + 160\ 8x = 160\ x = 20;kg \end{array}\)

Hence, total weight of 10 poles \(= 10x = 20 \times 10 = 200;kg\)

Thus, the given problem can be solved by using any of the two statements.

'Completely’ and ‘just’ are considered adverbs of emphasis and should be placed in front of the verbs they describe. Even if a verb is in its negative form, for eg. "don't know", the adverb should be placed before the entire form.

The only option that satisfies this criteria in both cases is option C.

Hence, option C is the correct answer.

Here, we have to determine the value of \(fg\left( {h\left( {2,5,7,3} \right),4,6,8} \right)\).

Given,

\(\begin{array}{l} h\left( {p,q,r,s} \right) = remainder;of\frac{{p \times q}}{{r \times s}}if;\left( {p \times q} \right) > \left( {r \times s} \right)\ ;\left( {or} \right)remainder;of\frac{{r \times s}}{{p \times q}};if;\left( {r \times s} \right) > \left( {p \times q} \right) \end{array}\)

So, we obtain \(h\left( {2,5,7,3} \right) = remainder;\left( {\frac{{7 \times 3}}{{2 \times 5}}} \right) = 1\)

Therefore, we get

\(fg\left( {h\left( {2,5,7,3} \right),4,6,8} \right) = fg\left( {1,4,6,8} \right)\)

Again, we have

\(\begin{array}{l} f\left( {p,q,r,s} \right) = max\left( {p,q,r,s} \right)\ g\left( {p,q,r,s} \right) = min\left( {p,q,r,s} \right) \end{array}\)

So, we obtain

\(\begin{array}{l} fg\left( {1,4,6,8} \right) = f;\left( {1,4,6,8} \right) \times g\left( {1,4,6,8} \right)\ = max\left( {1,4,6,8} \right) \times min\left( {1,4,6,8} \right)\ = 8 \times 1\ = 8 \end{array}\)

Hence, the value of desired function is

\(fg\left( {h\left( {2,5,7,3} \right),4,6,8} \right) = fg\left( {1,4,6,8} \right) = 8\)

From the given data, we draw the schematic as

Let coordinate of N = (0,0)

Coordinate of M = (0,4)

Coordinate of O = \(\left( { - \frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right)\)

Coordinate of P = \(\left( { - \frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }} - 2} \right)\)

Hence by distance formula, we obtain

\(\begin{array}{l} = \sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2}} \ = \sqrt {{{\left( {0 - \frac{1}{{\sqrt 2 }}} \right)}^2} + {{\left( {4 + 2 - \frac{1}{{\sqrt 2 }}} \right)}^2}} \ = 5.34;km \end{array}\)

Out of all the given sentences, only the sentence in option 1 is entirely correct with respect to grammar and usage.

Here are the errors in the other sentences:

• Option 2: The verb 'were' should actually be 'was' since the subject 'information' is singular.
• Option 3: The word 'real' should instead be 'really' - an adverb to describe the adjective 'useful'.
• Option 4: 'Would not had been' is an ungrammatical tense construction.

Hence, the correct answer is option 1.

We redraw this triangle as

In triangle, the internal bisector of an angle bisect the opposite side in the ratio of the other two sides. So, we have

\(\begin{array}{l} \frac{{QS}}{{SR}} = \frac{{PQ}}{{PR}}\ \frac{{QS}}{{SR}} = \frac{r}{q}\ \frac{{QS + SR}}{{SR}} = \frac{{r + q}}{q}\ \frac{{SR}}{{QR}} = \frac{q}{{r + q}}\ SR = \frac{{qp}}{{r + q}} \end{array}\)

By using cosine formula in triangle \({\rm{\Delta }}PSR\),

\(\begin{array}{l} \cos 60^\circ = \frac{{{{\left( {PR} \right)}^2} + {{\left( {PS} \right)}^2} - {{\left( {SR} \right)}^2}}}{{\left( {PR} \right)\left( {PS} \right)}}\ \frac{1}{2} = \frac{{{{\left( q \right)}^2} + {{\left( {PS} \right)}^2} - {{\left( {\frac{{qp}}{{r + q}}} \right)}^2}}}{{\left( q \right)\left( {PS} \right)}}\ PS = \frac{{qr}}{{r + q}} \end{array}\)

Here \(P,R,S,T,U\) are in A.P. So, the difference between two consecutive numbers will be equal, i.e.

\(d = R - P = S - R = T - S = U - T\)

Where \(d\) is common difference for the given A.P. Now, we check the given sequences

Sequence I:

\(2P,;2R,;2S;,;2T;,;2U\)

The differences between two consecutive numbers are obtained as

\(\begin{array}{l} D = 2R - 2P = 2\left( {R - P} \right) = 2d\ D = 2S - 2R = 2\left( {S - R} \right) = 2d\ D = 2T - 2S = 2\left( {T - S} \right) = 2d \end{array}\)

Since, the differences are same in each case, so it is in A.P.

Sequence II:

\(P - 3,;R - 3,;S - 3,;T - 3,;V - 3\)

The differences between two consecutive numbers are obtained as

\(\begin{array}{l} D = \left( {R - 3} \right) - \left( {P - 3} \right)\ = R - P - 3 + 3 = R - P = d\ D = \left( {S - 3} \right) - \left( {R - 3} \right)\ = S - R - 3 + 3 = S - R = d \end{array}\)

Again, the differences are same in each case, so it is in A.P.

Sequence III:

\({P^2},;{R^2},;{S^2},;{T^2},;{V^2}\)

The differences between two consecutive numbers are obtained as

\(\begin{array}{l} D = {R^2} - {P^2} = \left( {R - P} \right)\left( {R + P} \right)\ = d\left( {R + P} \right)\ D = {S^2} - {R^2} = \left( {S - R} \right)\left( {S + R} \right)\ = d\left( {S + R} \right) \end{array}\)

In this case, the differences are not same, so it is not in A.P.

Thus, sequences I and II will be in A.P.

Concept:

Let f(z) = u + iv be the analytic function,

Cauchy-Riemann equations are 

vy = ux

vx = - uy

Calculation:

Given \(f\left( z \right) = g\left( z \right) + h\left( z \right)\)

Let \(g\left( z \right) = gu\left( z \right) + j;gv\left( z \right)\)

\(h\left( z \right) = hu\left( z \right) + j;hv\left( z \right)\)

If g and h are differentiable, then it will satisfy C-R equations. So, we have

\(\begin{array}{*{20}{c}} {g{u_x} = g{v_y},}&{g{u_y} = - g{v_x}}\ {h{u_x} = h{v_y},}&{huy = - h{v_x}} \end{array}\)

\(f\left( z \right) = \left( {gu\left( z \right) + hu\left( z \right)} \right) + j;(\left( {gv\left( z \right) + hv\left( z \right)} \right)\)

By observing the above equations, we get

\(\begin{array}{*{20}{c}} {\left( {g{u_x} + h{u_x}} \right) = \left( {g{v_y} + h{v_y}} \right)}\ {\left( {g{v_y} + h{v_y}} \right) = - \left( {g{v_x} + h{v_x}} \right)} \end{array}\)

Hence, form above two equations f(z) is differentiable (because it satisfies C- R equation).

Concept:

Consider the system of m linear equations

a11 x1 + a12 x2 + … + a1n xn = b1

a21 x1 + a22 x2 + … + a2n xn = b2

…

am1 x1 + am2 x2 + … + amn xn = bm

The above equations containing the n unknowns x1, x2, …, xn. To determine whether the above system of equations is consistent or not, we need to find the rank of the following matrices.

\(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}& \ldots &{{a_{1n}}}\ {{a_{21}}}&{{a_{22}}}& \ldots &{{a_{2n}}}\ \ldots & \ldots & \ldots & \ldots \ {{a_{m1}}}&{{a_{m2}}}& \ldots &{{a_{mn}}} \end{array}} \right]\) and \(\left[ {A{\rm{|}}B} \right] = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}& \ldots &{{a_{1n}}}&{{b_1}}\ {{a_{21}}}&{{a_{22}}}& \ldots &{{a_{2n}}}&{{b_2}}\ \ldots & \ldots & \ldots & \ldots & \ldots \ {{a_{m1}}}&{{a_{m2}}}& \ldots &{{a_{mn}}}&{{b_m}} \end{array}} \right]\)

A is the coefficient matrix and [A|B] is called an augmented matrix of the given system of equations.

We can find the consistency of the given system of equations as follows:

(i) If the rank of matrix A is equal to rank of an augmented matrix and it is equal to the number of unknowns, then the system is consistent and there is a unique solution.

The rank of A = Rank of augmented matrix = n

(ii) If the rank of matrix A is equal to rank of an augmented matrix and it is less than the number of unknowns, then the system is consistent and there are an infinite number of solutions.

The rank of A = Rank of augmented matrix < n

(iii) If the rank of matrix A is not equal to rank of the augmented matrix, then the system is inconsistent, and it has no solution.

The rank of A ≠ Rank of an augmented matrix

Explanation:

If determinant of coefficient matrix is non-zero then there is a unique solution. If determinant is non – zero. So all the Eigen values are non-zero. If Eigen values are non – zero then question are linearly independent.

we have a set of three linear equations with three variables.

i.e., [A]_3×3 [X]_3×1 = [B]_3×1

Statement P:

In this case,

P(A) = P(A : B) = Number of variables

⇒ P(A) = P(A : B) = 3

∴ three independent rows exist in matrix A irrespective of matrix B.

Also, |A| ≠ 0     ---(i)

Hence, P = Q = S

Statement R:

Let λ_1, λ₂ & λ₃ are eigen values of matrix A

using the properties,

λ1 λ2 λ₃ = |A|   ----(ii)

from equation (i) & (ii),

λ1 λ2 λ3 ≠ 0

⇒ λ1 ≠ 0, λ2 ≠ 0 & λ3 ≠ 0

Therefore, from above discussion, it is inferred that,

P = Q = R = S

Stoke’s theorem: The line integral of a vector around closed path L is equal to the integral of curl over the open surface is enclosed by the closed path L.

\(\oint \vec A \cdot d\vec l\)=\(\iint \left( {\vec \nabla \times \vec A} \right).d\vec s\)

Gauss's theorem: The total flux coming out of a closed surface is equal to the change enclosed.

\(\oint D.ds = Q\)

Divergence theorem: The total outward flux of a vector field \(\vec F\) through a closed surface is equal to the volume integral of the divergence of \(\vec F\)

\(\mathop \oint \limits_s \vec F.d\vec s = \mathop \smallint \limits_v \left( {\vec \nabla .\vec F} \right)dv\)

Cauchy's integral theorem: \(\oint f\left( z \right).dz = 0\)

Concept:

\(x\left( t \right)\mathop \leftrightarrow \limits^{LT} X\left( s \right)\)

\(\frac{{dx\left( t \right)}}{{dt}}\mathop \leftrightarrow \limits^{L.T} sX\left( s \right)\)

\(sX\left( s \right)\mathop \leftrightarrow \limits^{I.L.T} \frac{{dx\left( t \right)}}{{dt}}\)

\(L\left{ {\frac{{x\left( t \right)}}{{t}}} \right} = \frac{}{}\mathop \smallint \limits_s^0 \left{ {x\left( t \right)} \right}ds\ \)

Calculation:

Given that Laplace transform \(f\left( t \right) = 2\sqrt {\frac{t}{\pi }}\) is \({s^{ - \frac{3}{2}}}\)

Given as \(g\left( t \right) = \frac{1}{{\sqrt {\pi t} }}\)

\(\begin{array}{l} \Rightarrow g\left( t \right) = \frac{{2\sqrt {\frac{t}{\pi }} }}{{2t}} = \frac{{f\left( t \right)}}{{2t}}\ L\left{ {g\left( t \right)} \right} = L\left{ {\frac{{f\left( t \right)}}{{2t}}} \right} = \frac{1}{2}\mathop \smallint \limits_s^0 \left{ {f\left( t \right)} \right}ds\ = \frac{1}{2}\mathop \smallint \limits_s^\infty {s^{ - \frac{3}{2}}}ds = \frac{1}{2}\left( {\frac{{{s^{\frac{{ - 3}}{2} + 1}}}}{{\frac{{ - 3}}{2} + 1}}} \right)_s^\infty \ = \frac{1}{2}\left( { - 2} \right)\left[ {0 - {s^{ - \frac{1}{2}}}} \right] = {s^{ - \frac{1}{2}}} = \frac{1}{{\sqrt s }} \end{array}\)

• Permanent Magnet Moving Coil (PMMC) is only used for DC measurements.
• Moving Iron (MI) type instruments can be used for both AC & DC measurements. But Moving Iron connected through current Transformer block DC supply. So that Moving Iron connected through current Transformer are only used for AC measurements.
• Rectifier type instruments are used for both AC & DC measurements.
• Induction type instruments are only used for AC measurements.

Concept:

In a two-wattmeter method, for lagging load

The reading of first wattmeter (W1) = VL IL cos (30° + ϕ)

The reading of second wattmeter (W2) = VL IL cos (30° - ϕ)

Total power in the circuit (P) = W1 + W2

Calculation:

Given,

Total power consumed by load W₁ + W₂ = 5 kW

Power factor cos ϕ = 0.707

⇒ ϕ = cos^-1 (0.707) = 45° 

The reading of first wattmeter (W1) = VL IL cos (30° + ϕ)

W1 = VL IL cos (30° + 45°) = 0.2588 V_L I_L

The reading of second wattmeter (W2) = VL IL cos (30° - ϕ)

W2 = VL IL cos (30° - 45°) = 0.966 VL IL

The total power consumed by the load is 

W1 + W2 = 5 kW = (0.2588 + 0.966) VL IL

⇒ VL IL = 4.082 kW

Therfore the reading in each wattmeter is 

W₁ = 0.966 VL IL = 0.966 × (4.082 kW) 

W₁ = 3.94 kW

W2 = 0.2588 VL IL = 0.2588 × (4.082 kW)

W₂ = 1.06 kW

Concept:

Voltage Division Rule:

The voltage applied across the resistor R1 using voltage division rule is given by:

\({V_{R1}} = \frac{{V\left( {{R_1}} \right)}}{{{R_1} + {R_2}}}\)

The voltage applied across the resistor R2 using voltage division rule is given by:

\({V_{R2}} = \frac{{V\left( {{R_2}} \right)}}{{{R_1} + {R_2}}}\)

Calculation:

Given circuit diagram

Given V_bus = 100 kV

By using the voltage division rule 

\({V_{out}} = {V_{bus}}\left[ {\frac{{{X_{{c_1}}}}}{{{X_{{c_1}}} + {X_{{c_2}}}}}} \right] = {V_{bus}}\left[ {\frac{{\frac{1}{{{C_2}}}}}{{\frac{1}{{{C_1}}} + \frac{1}{{{C_2}}}}}} \right] = {V_{bus}}\left[ {\frac{{{c_1}}}{{{c_1} + {c_2}}}} \right]\)

Given, 

C₁ = 1 μF ± 10%, C2 = 9 μF ± 10%, 

The value of capacitances are 

\({C_1} = 1 ± \frac{10}{{100;}}\left( {1} \right) = 1 ± 0.1;{\rm{μ F}}\)

\({C_2} = 9 ± \frac{10}{{100;}}\left( {9} \right) = 9 ± 0.9;{\rm{μ F}}\)

Now the sum of capacitances will be

⇒ C₁ + C₂ = 1 ± 0.1 μF + 9 ± 0.9 μF

⇒ C1 + C2 = 10 ± 1 μF = 10 μF ± 10%

\(\Rightarrow \frac{{{C_1}}}{{{C_1} + {C_2}}} = \frac{{1μF ± 10% }}{{10μF ± 10% }} = 0.1μF ± 20%\)

∴ V_out = 100 × 10³ (0.1 μF + 20%)

V_out = 10 kV ± 20% = (10 + 2) kV or (10 - 2) kV

Vout = 12 kV or 8 kV

Therefore the maximum RMS output voltage Vout (in kV), considering the capacitor tolerance, is 12 kV

Concept:

• Once the capacitor gets charged to a particular voltage, the capacitor behaves like a voltage source of that particular voltage and the diode responsible for the charging of the capacitor will always be off.
• The voltage across the capacitor will always only depend on the peak to the peak value of the input and not on the shape of the input waveform.
• If the given input waveform is anything other than a square wave, then during the analysis consider it as a square wave between the same two levels (+Vm, -Vm)

Explanation:

During the positive half cycle, the circuit under steady-state is as shown below.

The voltage across R, |V_R| = 2 V_m

During the negative half-cycle, the circuit under steady-state is as shown below.

The voltage across R, |V_R| = 2 V_m

The voltage across resistance R = 2 V_m = 2 × 100 = 200 V

Given y= 0 or x-z plane

x and z are tangential components, y is normal component

y < 0, ϵ₁ = 2

E₁ = 3a_x + 4a_y + 2a_z

Normal component is E_n1 = 4a_y 

y > 0, ϵ2 = 5

ϵ₁ E_n1 = ϵ2 En2

⇒ E_n2 = (2/5) (4a_y) 

\({E_2} = 3{a_x} + \frac{2}{5}\left( {4{a_y}} \right) + 2{a_z}\)

E₂ = 3a_x + 1.6 a_y + 2 a_z

Given,

\(\vec H = {10^7}\hat z\frac{A}{m}\)

Magnetic flux density B = μ H

Velocity V = r.ω 

\({V_{emf}} = \oint \left( {\bar V \times \bar B} \right) \cdot dl\)

\(= \mathop \oint \limits_L^; (rω {a_\phi } \times μ ;H{a_z}) \cdot dl\)

\(= \oint rω μ H{a_p} \cdot dl\)

\( = rω μ H\left( {\frac{{2π r}}{2}} \right) \)

V_emf = ω μ H π r²

By substituting all the values in the above equation

= 6.28 × 4π × 10^-7 × 10⁷ × π (1)²

V_emf = 247.92 V

Concept:

An ideal Integrator circuit is shown below:

Note:

• At high frequencies, the capacitor is short circuit so the output is 0.
• At zero frequency (0 Hz) or DC, the capacitor acts like an open circuit due to its reactance (1/ωC) thus blocking any output voltage feedback.
• As a result, very little negative feedback is provided from the output back to the input of the amplifier at low frequencies.
• Therefore with just capacitor C, in the feedback path, at zero frequency the op-amp is connected like a normal open-loop amplifier with very high open-loop gain (open-loop gain of the op-amp is ideally infinite).
• This high gain can cause op-amp to be unstable and possible voltage rail saturation.

To avoid undesirable conditions due to high gain. The circuit connects a high-value resistance in parallel with a continuously charging and discharging capacitor C.  The addition of this feedback resistor, R2 across the capacitor, C gives the circuit the characteristics of an inverting amplifier with finite closed-loop voltage gain given by \(\frac{R_2}{R_1}\)

Therefore, at very low frequencies, the op-amp acts like an inverting amplifier and at very high frequencies the feedback resistor also acts like a short circuit (∵ the impedance of the capacitor is 0) and thus producing output = 0. 

Explanation:

\(\begin{array}{l} {Z_1} = 1k;{Z_2} = \frac{{1 \times {{10}^3} \times \frac{1}{{0.1 \times {{10}^{ - 6}}s}}}}{{{{10}^3} + \frac{1}{{0.1 \times {{10}^{ - 6}}s}}}} = \frac{{{{10}^3}}}{{j0.1 \times {{10}^{ - 3}}\omega + 1}}\ \omega = 2\pi \times 2000;rad/sec\ {Z_2} = \frac{{{{10}^3}}}{{1 + j0.1 \times 2\pi \times 2000 \times {{10}^{ - 3}}}} = \frac{{{{10}^3}}}{{1 + j1.25}}\ {V_0} = \frac{{ - {Z_2}{V_i}}}{{{Z_1}}} = \frac{{{{10}^3} \times 2\sin \left( {2\pi \times 2000t} \right)}}{{\left( {1 + j1.25} \right) \times {{10}^3}}} = - \frac{{2\sin \left( {2\pi \times 2000t} \right)}}{{1.6\angle 51.3}}\ {V_0} = - 1.25\sin \left( {2\pi 2000t - 51.3} \right) \end{array}\)

Amplitude of the output voltage = 1.25V

We have \({V_C}; = ;2V\)

\({I_c}{R_c} = 10 - 2 = 8\)

We have \({V_C}; = ;4V\)

\(\begin{array}{l} {I_c}R_c' = 10 - 4 = 6\ \Rightarrow \frac{{{I_c}R_c'}}{{{I_c}{R_c}}} = \frac{6}{8} \Rightarrow \frac{{R_c'}}{{{R_c}}} = \frac{3}{4} = 0.75 \end{array}\)

Concept:

The SOP representation of the circuit is:

F = Σm (minterms)

Minterm: a minterm of n variables is a product of the variables in which each appears exactly once in true or complemented form.

The POS representation of the circuit:

F = ΠM (max terms)

Maxterm: a maxterm of n variables is a sum of the variables in which each appears exactly once in true or complemented form.

Calculation:

Given,

\(F=ABC+\bar A\bar B C+A\bar BC+\bar ABC+\bar A\bar B\bar C\)

For the function, we form the K-map as:

Hence, the function in the form of minterms is expressed as:

f(A, B, C) = Σm (0,1,3,5,7)

Now, we put 0 in each block of the K-map excluding the blocks corresponding to the terms in the above function.

Grouping the 0’s in K-map, we obtain the max terms as

F = πM(2, 4, 6)

F = πM(2, 4, 6) = \(\left( {A + \bar B + C} \right)\left( {\bar A + B + C} \right)\left( {\bar A + \bar B + C} \right)\)

In the given figure, (a) is low pass filter and (b) high pass filter. Also, we have

\({f_1} < {f_2}\)

For \(f < {f_1}\), we have the op- omp circuit as

Again, for \({f_1} < f < {f_2}\), the op – amp circuit is

for \(f > {f_2}\), the op – omp circuit is

Hence, the output graph is

Thus, the resultant circuit represents a band stop filter.

In Bipolar transistor, saturation mode occurs when

1. Collector – base (CB) is forward biased.
2. Base-emitter (BE) is also forward biased.

Hence, for the saturation condition, bath the junction should be in forward-bias, as shown in the figure below.

Analysis:

In the given line diagram

Line 1 is tripped at time t1 by opening the circuit breakers at the two ends (although there is no fault)

As the line is disconnected the reactance of the system increases

The maximum power transfer increases (as power is inversely proportional to the reactance of the line \(P_{max}=\frac{EV}{X}\))

It is also given that mechanical power input to the generator and the voltage magnitude E is constant.

Therefore the new electrical power is to be the same as the old power, so δ value should be increased (given in the line reactance diagram the steady-state value of δ before the transient is positive)

The rotor angle transient will be

Concept:

For a 3 bus power system

For the above figure, the admittance matrix is as shown below.

\({y_{bus}} = \left[ {\begin{array}{*{20}{c}} {{y_{11}} + {y_{12}} + {y_{13}}}&{ - {y_{12}}}&{ - {y_{13}}}\ { - {y_{12}}}&{{y_{21}} + {y_{22}} + {y_{23}}}&{ - {y_{23}}}\ { - {y_{31}}}&{ - {y_{23}}}&{{y_{31}} + {y_{32}} + {y_{33}}} \end{array}} \right]\)

Diagonal elements of the Bus Admittance matrix are known as self-admittances and the off-diagonal elements are known as mutual admittances.

Calculation:

Given bus admittance matrix of the uncompensated line is

\(Y=\left[ {\begin{array}{*{20}{c}} { - j6}&{j3}&{j4}\ {j3}&{ - j7}&{j5}\ {j4}&{j5}&{ - j8} \end{array}} \right]Pu\)

By comparing the above matrix with standard 3 bus matrix

y₁₃ = -j4

y₃₂ = -j5

⇒ y₃₁ + y₃₂ + y₃₃ = -j8

⇒ y_33­ = j

After compensating, \({y_{33}} = \frac{j}{2}\)

Y_33(new) = -8.5 j.

Given as zero initial condition:

I(0^-) = I(0⁺) = 0

When the switch is closed the inductor will act as an open circuit

∴ I(0⁺) = 0

After some time inductor will start charging and at t → ∞

I(∞) = V/R

Now The current through the inductor is given by:

Putting all the value in the above equation

Differentiate the equation first time then we get

Differentiate the equation a second time then we get

Hence option (4) is correct.

CONCEPT:

• Wheatstone bridge: It is an arrangement of four resistance that can be used to measure one of them in terms of rest. Here arms AB and BC are called ratio arms and arms AC and BD are called conjugate arms.
• The bridge is said to be balanced when deflection in the galvanometer is zero i.e. no current flows through the galvanometer or in other words VB = VD.
• In the balanced condition

Calculation:

Let the given circuit diagram nodes be marked as a, b, c, and d as shown below

Now by rearranging the nodes the circuit will be a follows

Here the ratio of arms is equal. The network is balanced Wheatstone Bridge.

The potential at node b is equal to the potential at node c, therefore no current will pass from node b to node c 

Hence, I = 0

Therefore the current through 2Ω resistor is 0 A

Concept:

Referred value in Transformer:

In order to simplify the calculation, it is theoretically possible to transfer the voltage, current, and impedance of one winding to the other winding and combined them to a single value for each quantity.

Considered a transformer has turns ration 'a' which is given by,

\(a = \frac{{{N_2}}}{{{N_1}}} = \frac{{{V_2}}}{{{V_1}}} = \frac{{{I_1}}}{{{I_2}}}\)

Where,

I1 and I2 are primary and secondary current respectively.

V1 and V2 are primary and secondary voltage respectively.

N1 and N2 are numbers of turn in primary and secondary respectively.

For an ideal transformer:

Input Power = Output Power

\(\frac{{V_1^2}}{{{Z_1}}} = \frac{{V_2^2}}{{{Z_2}}}\)

\(\frac{{V_2^2}}{{V_1^2}} = \frac{{{Z_2}}}{{{Z_1}}}\)

\({\left( {\frac{{{V_2}}}{{{V_1}}}} \right)^2} = \frac{{{Z_2}}}{{{Z_1}}}\)

\({a^2} = \frac{{{Z_2}}}{{{Z_1}}}\)

Equivalent secondary Impedance in terms of primary Impedance:

\({Z_1} = \frac{{{Z_2}}}{{{a^2}}}\)

Equivalent primary Impedance in terms of secondary Impedance:

\({Z_2} = {a^2}{Z_1}\)

Calculation:

Given,

Zin is resistive and equal to 2.5 Ω

Angular frequency ω = 5000 rad/s

Inductive reactance X_L = jωL = j 5000 × 1 × 10^-3 = j 5 Ω 

Capative reactance X_C = 1 / jωC = 1 / j(5000 × 10 × 10^-6) = - j 20 Ω 

The input impedance of the transformer from the secondary side of the transformer will be

Z' = XL + R/a² = j 5 + 2.5/a²

Circuit diagram will be as follows  

The input impedance of the transformer from the primary side of the transformer will be

\(⇒ {Z_{in}} = \frac{1}{{{b^2}}}\left[ {\frac{{2.5}}{{{a^2}}} + j5} \right] - j20\)

\(⇒ {Z_{in}} = \left[ {\frac{{2.5}}{{{a^2}\times {b^2}}} +\frac{{j5}}{{{b^2}}}} \right] - j20\)

It is given that the input impedance is resistive, therefore there won't be any reactance term in the input impedance

As there is no reactance in the input impedance, make reactance equal to zero in the above equation

\(⇒- 20 + \frac{5}{{{b^2}}} = 0 \)

⇒ b² = 0.25

⇒ b = 0.5

Now the given resistive term of the input impedance is equal to 2.5 Ω 

\(⇒ \frac{{2.5}}{{{a^2}{b^2}}} = 2.5 \)

\(⇒ \frac{{2.5}}{{{a^2}\times{0.5^2}}} = 2.5 \)

⇒ a = 2

Therefore the value of a = 2.0, b = 0.5

Concept

Under steady state load torque will be equal to The developed torque by the motor.

Explanation

Given,

no load speed (ω0) = 200 rad/sec

Rated torque (T_r) = 500 Nm

Rated Speed (ωr) = 180 rad/sec

Load torque (TL) = 2.78 ω_r

using equation of straight line

Developed torque can be expressed as,

\(T=\frac{200-ω_r}{20}\times500\)

At steady state,

load torque = developed torque

⇒  T_L = T

⇒ 2.78 ω_r = \(\frac{200-ω_r}{20}\times500\)

⇒ ω_r = 180 rad/sec

Therefore, steady state speed is 180 rad/sec

Concept:

Exact Equivalent Circuit

Here, R1 is the winding resistance of the stator

X1 is the inductance of the stator winding

R0 is the core loss component

XM is the magnetizing reaction of the winding

R2/s is the power of the rotor, which includes output mechanical power and loss of rotor

Important:

If we draw the circuit with refereed to the stator then the circuit will look like-

Here all the other parameters are same except R2’ is the rotor winding resistance with referred to stator winding. R2(1-s) / s is the resistance which shows the power which is mechanical power output or useful power. The power dissipated in that resistor is the useful power output or shaft power.

Explanation:

We redraw the given circuit as

Given \({V_g} = 210;V,;s = 0.05\)

Air gap voltage is the voltage across AB, i.e.

\(\begin{array}{l} {V_{AB}} = {V_g} = 210{V_{\left( {rms} \right)}}\ {I_2} = \frac{{{V_{AB}}}}{z} = \frac{{210}}{{1 + j0.22}}\ = 205.0953;\angle - 12.40 \end{array}\)

Hence, torque \(= \frac{3}{{{\omega _s}}}I_2^2\frac{{{r_2}}}{s}\)

\(= \frac{3}{{314.15}} \times {\left( {205.0953} \right)^2} \times 1 = 401.09;N - m.\)

Wave wound

For wave wound, number of parallel pat (A₁) = 2

∴ \(E_1=\frac{pϕ Nz}{60\ A_1}\)

⇒ \(E_1∝\frac{1}{A_1}\) {p, ϕ, N & z are constants}

Lap wound

for lap wound,

Number of parallel path = Number of poles

i.e., A₂ = P = 4

∴ \(E_2∝\frac{1}{A_2}\)

So,

\(\frac{E_2}{E_1}=\frac{A_2}{A_1}\)

⇒ \(E_2=\frac{E_1}{2}\)

It is given that armature resistance (R_a) = 0

so, E ∝ V

⇒ \(\frac{E_2}{E_1}=\frac{V_1}{V_2}\)

⇒ \(V_2=\frac{E_2}{E_1}V_1\)

⇒ \(V_2=\frac{V_1}{2}\)

so, power (p) = V I_a

⇒ \(I_a=\frac{P}{V}\)

⇒ \(I_a\propto\frac{1}{V}\)

so,

\(\frac{I_{a_2}}{I_{a_1}}=\frac{V_1}{V_2}\ {V_2=\frac{V_1}{2}}\)

\(I_{a_2}=2I_{a_1}\)

The power of the machine will remain same as P_lap = P_wave

The machine with lap wound armature coils will have rated voltage.

\(V_2=\frac{V_1}{2}=\frac{230}{2}=115\) V

∴ P_lap = P_wave = p = 5 kW

\( g\left( t \right); = ;{e^{ - 2t}};\left[ {sin;5t; + ;cos;5t} \right]\)

Taking Laplace transform, we get

\(G\left( s \right) = \frac{5}{{{{\left( {s + 2} \right)}^2} + {5^2}}} + \frac{{s + 2}}{{{{\left( {s + 2} \right)}^2} + {5^2}}} \)

DC gain means:

 \({\left| {G\left( s \right)} \right|_{s=0}};\)

\(G\left( 0 \right) = \frac{5}{{{2^2} + {5^2}}} + \frac{2}{{{2^2} + {5^2}}} = \frac{7}{{29}}\)

From the Nyquist plot, we have

\(\begin{array}{l} {F_{G1}}\left( S \right) = \frac{{{K_1}}}{s}\ {F_{G2}}\left( S \right) = {K_2}s; \end{array}\)

So, the product of \({F_{G1}}\) and \({F_{G2}}\) is obtained as

\(\begin{array}{l} F\left( s \right) = {F_{G1}}\left( s \right).{F_{G2}}\left( s \right)\ = \frac{{{k_1}}}{s}.{k_2}s = {k_1}{k_2}; \end{array}\)

Triangle is bounded by \(x = y,x = 0,;y = 1\) is \(xy\) plane.

Required volume \(= \smallint \mathop \smallint \limits_{0AB} f\left( {x,y} \right)dxdy\)

\(\begin{array}{l} = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = x}^1 {e^x}dxdy = \mathop \smallint \limits_{x = 0}^1 {e^x}.\left( y \right)_x^1dx\ = \mathop \smallint \limits_{x = 0}^1 {e^x}\left( {1 - x} \right)dx = \mathop \smallint \limits_{x = 0}^1 \left( {{e^x} - x{e^x}} \right)dx\ = \left( {{e^x}} \right)_0^1 - \left( {{e^x}\left( {x - 1} \right)} \right)_0^1\ = \left( {{e^1} - 1} \right) - \left[ {0 - \left( { - 1} \right)} \right] = e - 2 = 0.72 \end{array}\)

Given events \({H_R}{H_S},;{T_R}{T_S},{H_R}{T_S},{T_R}{H_S}\)

If coins are independent, corresponding probabilities will be

\(\begin{array}{*{20}{c}} {\frac{1}{2}.\frac{1}{2},\frac{1}{2}.\frac{1}{2}}&{\frac{1}{2}.\frac{1}{2},}&{\frac{1}{2}.\frac{1}{2}} \end{array}\)

\(= \frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4}\) respectively

But given probabilities are 0.28, 0.18, 0.3, 0.24 respectively. We cannot decide whether R is fair or S is fair

⇒ The coin tosses are dependent.

Concept:

Linear Differential Equation: Linear differential equations are those in which the dependent variable, its derivatives occur only in the first degree and they are not multiplied together. It is of the form:

\(\frac{{{{\rm{d}}^{\rm{n}}}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^{\rm{n}}}}} + {{\rm{k}}_1}\frac{{{{\rm{d}}^{{\rm{n}} - 1}}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^{{\rm{n}} - 1}}}} + {{\rm{k}}_2}\frac{{{{\rm{d}}^{{\rm{n}} - 2}}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^{{\rm{n}} - 2}}}} + \ldots + {{\rm{k}}_{\rm{n}}}{\rm{y}} = {0}\)

Where, k1, k2, …kn are the constants.

The solution of the equation is given as:

y = C.F 

where C.F is the complementary function.

The above linear differential equation in the symbolic form is represented as

(Dn + k1 Dn-1 + k2 Dn-2 +…+ kn) y = 0

For different roots of the auxiliary equation, the solution (complementary function) of the differential equation is as shown below.

Calculation:

\(\begin{array}{l} \frac{{di}}{{dt}} - 0.2i = 0\ \left( {D - 0.2} \right)i\left( t \right) = 0\ D; = ;0.2\ i\left( t \right) = k.{e^{0.2t}}, - 10 < t < 10 \end{array}\)

At \(t; = ;4\);   \(10 = K.{e^{0.8}}\)

\(\begin{array}{l} K = 4.493\ \therefore i\left( { - 5} \right) = 4.493 \times {e^{ - 1}}\ i;\left( { - 5} \right); = ;1.65 \end{array}\)

Concept:

The Fourier Transform of a continuous-time signal x(t) is given as:

\(X\left( \omega \right) = \mathop \smallint \limits_{ - \infty}^{\infty} x(t) ~{e^{ - j\omega t}}~dt \)

Analysis:

Given:

x(t) = e^j10t  defined from t = -1 to 1. 

\( X\left( \omega \right) = \mathop \smallint \limits_{ - 1}^1 {e^{j10t}}.{e^{ - j\omega t}}dt = \mathop \smallint \limits_{ - 1}^1 {e^{j\left( {10 - \omega } \right)t}}dt\)

\(X(\omega) = \left. {\frac{{{e^{j\left( {10 - \omega } \right)t}}}}{{j\left( {10 - \omega } \right)}}} \right|_{ - 1}^1 = \frac{{2\sin \left( {\omega - 10} \right)}}{{\left( {\omega - 10} \right)}} \)

Concept:

Relative Static Error: 

The ratio of absolute error to the true value is called relative static error or percentage error.

\(R.S.E = \frac{{\left| {{A_m} - {A_t}} \right|}}{{{A_t}}} × 100\)

Calculation:

Given, 

Load consumes 30 V, 25 A, 0.8 power factor

True power consumed by the load 

P_true = VI cos ϕ

P_true = 30 ×  25 × 0.8 = 600 W

The given connection is an LC connection, therefore the losses will be due to pressure coil

Given, 

Pressure coil resistance = 1000 Ω 

Pressure coil loss = V²/R = (30)² / 1000 = 0.9 W

Measured power  = True Power + Pressure coil losses

P_meas = 600 + 0.9 = 600.9 W

\(R.S.E = \frac{{\left| {{A_m} - {A_t}} \right|}}{{{A_t}}} × 100=\frac{600.9-600}{600}\times 100\)

Percentage error = 0.15%

Concept:

current waveform for a buck converter 

Boundary of continuous & discontinuous conduction mode

For continuous conduction,

I_L(max) = I_L + \(\frac{Δ I_L}{2} \)     ---(1)

IL(min) = IL + \(\frac{Δ I_L}{2} \)     ---(2)

Calculation:

given information,

f = 100 kHz

D = 0.6

V₀ = 36 V

V_s = 60 V

At the boundary of continuous conduction

I_L(min) = 0

From eq(2)

\(⇒ 0 = I_L - \frac{Δ I_L}{2}\)

\(\rm ⇒ I_L = \frac{Δ I_L}{2}\)

The inductor ripple current is given as,

\(\rm Δ I_L = \frac{V_0(V_S - V_0)}{fLV_s}\)

\(\rm ⇒ Δ I_L = \frac{36(60-36)}{100 \times 10^3 \times 5 \times 10^{-3} \times 60}\)

⇒ ΔI_L = 28.8 mA

∵ \( I_L = \frac{\Delta I_L}{2}\)

\(\rm \Rightarrow I_L = \frac{28.8}{2} = 14.4 mA\)

\(\Rightarrow \frac{V_0}{R} = 14.4 mA \left( \because I_L = \frac{V_0}{R} \right)\)

\(\Rightarrow R = \frac{36}{14.4 \times 10^{-3}}\)

∴ R = 2500 Ω

Concept:

Boost converter:

Average voltage across inductor is always zero.

Average output voltage \((V_0) = \frac{V_s}{1 - D}\)

Calculation:

Given,

Mode I: When switch 's' : ON,

using KVL in loop '1'

\(V_L = L \frac{di}{dt}\)

\(\rm ⇒ 15 = L \frac{di}{dt}\)

\(\rm ⇒ \frac{\Delta I}{T_{ON}} = \frac{15}{L}\)

Mode II: when switch 's' : OFF

using KVL in loop '2'

V_in - V₀ = V_L

⇒ V₀ = 45 + V_in = 45 + 15

⇒ V₀ = 60 V

As observed that V₀ > V_in so, it is step up chopper

\(\rm V_0 = \frac{V_{in}}{ 1 - D}\)

\(\Rightarrow \rm 1 - D = \frac{15}{60}\)

∴ D = 0.75

Concept:

A diode is forward biased when its voltage is anywhere on the +_ve voltage side of the origin.

During the positive half cycle of an A.C. sinusoidal waveform, the thyristor is forward biased (anode positive).

Explanation:

Given information,

supply voltage (V_s) = 100 sin(100πt)

firing angle (α) = 30°

Case I: For 0 < ωt < α

                  None of the device working

                  so, output voltage (V₀) = 0

Case II: For α < ωt < π,

                 T₁D₂ : forward biased

                 D₃D₄ : Reversed biased

given circuit can be redrawn as,

using KVL,

V₀ = V_s

Case III: For π < ωt < 2π

              D₃D₄ : forward biased

              T₁D₂: Reversed biased

     using KVL, V₀ = -V_s

Average output voltage is given as,

\(\rm V_0 = \frac{1}{2 \pi} \left[\displaystyle\int_{\pi/6}^\pi V_m \sin (\omega t) d(\omega t) + \displaystyle\int_{\pi}^{2\pi} -V_m \sin (\omega t) d(\omega t) \right]\)

\(\Rightarrow \rm \rm V_0 = \frac{1}{2 \pi} \left[\displaystyle\int_{\pi/6}^\pi 100 \sin (\omega t) d(\omega t) + \displaystyle\int_{\pi}^{2\pi} -100 \sin (\omega t) d(\omega t) \right]\)

\( \rm \Rightarrow \rm \rm V_0 = \frac{1}{2 \pi} \left[ -100 \left[ \cos \omega t \right]_{\pi/6}^\pi + 100 \left[ \cos \omega t \right]_{\pi}^{2\pi} \right]\)

\(\Rightarrow \rm V_0 = \frac{1}{2 \pi} \left[ -100 \left(\cos \pi - \cos \frac{\pi}{6} \right) + 100 ( \cos 2 \pi - \cos \pi) \right]\)

\(\Rightarrow \rm V_0 = \frac{1}{2 \pi} \left[ -100 \left( -1 - \frac{\sqrt 3}{2} \right) + 100(1+1) \right]\)

∴ V₀ = 61.53 V

Concept:

• BIBO (Bounded Input Bounded output stable system is a type of system in which for given bounded input output is bounded.
• On applying unit step input to the system then the response which we get is called unit step response.
• Condition for absolutely integrable.

\(\rm\displaystyle\int_{-\infty}^{\infty}|x(t)| dt < \infty\)

• Bounded signals are those whose value always remains less than a certain value & finite support implies that the duration of the signal is finite.

Explanation:

• Since the system is BIBO stable & the unit step function is bounded, the unit step response will also be bounded.

Note:

\(\rm\displaystyle\int_{-\infty}^{\infty}|u(t)| dt = \rm\displaystyle\int_{-\infty}^{\infty}1 dt = \infty\)

Therefore unit step response is NOT absolutely integrable.

Concept:

 The first difference of any signal t[n] is t[n] - t[n-1]

Explanation: 

As per the question,

\(y\left( n \right)\) is first difference of \(x\left( n \right)\) 

So, 

      \(y\left( n \right){\rm{;}} = {\rm{;}}x\left( n \right){\rm{;}}-{\rm{;}}x\left( {n - 1} \right)\)

⇒ \(Y\left( z \right){\rm{;}} = {\rm{;}}x\left( Z \right)\left( {1 - {z^{ - 1}}} \right){\rm{;}} = {\rm{;}}X\left( z \right){\rm{;}}-{\rm{;}}{z^{ - 1}}{\rm{;}}X\left( z \right)\)

\(\begin{array}{l} Y\left( z \right){\rm{;}} = {\rm{;}}\left[ {2x{\rm{;}} + {\rm{;}}4{\rm{;}}-{\rm{;}}4{z^{ - 1}}{\rm{;}} + {\rm{;}}3{z^{ - 2}}\left] {{\rm{;}}-{\rm{;}}} \right[2{\rm{;}} + {\rm{;}}4{z^{ - 1}}{\rm{;}} - {\rm{;}}4{z^{ - 2}}} \right]{\rm{;}}\ = 2x{\rm{;}} + {\rm{;}}4{\rm{;}}-{\rm{;}}4{z^{ - 1}} + {\rm{;}}3{z^{ - 2}}{\rm{;}}-{\rm{;}}2{\rm{;}}-{\rm{;}}4{z^{ - 1}}{\rm{;}}-{\rm{;}}4{z^{ - 2}}{\rm{;}}-{\rm{;}}3{z^{ - 3}}\ = {\rm{;}}2z{\rm{;}} + {\rm{;}}2{\rm{;}}-{\rm{;}}8{z^{ - 1}}{\rm{;}} + {\rm{;}}7{z^{ - 2}}{\rm{;}}-{\rm{;}}3{z^{ - 3}} \end{array}\)

The force F₁ acting will be \({F_1} = \frac{{Q \times \left( { - Q} \right)}}{{4\pi \epsilon{_0}{{\left( {2d} \right)}^2}}}\left( {\frac{{2d{{\hat a}_x}}}{{2d}}} \right)\)

The force F2 acting will be \({F_2} = \frac{{Q \times \left( { - Q} \right)}}{{4\pi \epsilon{_0}{{\left( {2d} \right)}^2}}}\left( {\frac{{2d{{\hat a}_y}}}{{2d}}} \right)\)

The force F₃ acting will be \({F_3} = \frac{{Q \times Q}}{{4\pi \epsilon{_0}{{\left( {\sqrt 8 ;d} \right)}^2}}}\left( {\frac{{2d\hat a + 2d\hat y}}{{\sqrt 8 d}}} \right)\)

The net force on Q due to the remaining charges will be

F = F₁ + F₂ + F₃

\(F = \frac{{Q \times \left( { - Q} \right)}}{{4\pi \epsilon{_0}{{\left( {2d} \right)}^2}}}\left( {\frac{{2d{{\hat a}_x}}}{{2d}}} \right)+\frac{{Q \times \left( { - Q} \right)}}{{4\pi \epsilon{_0}{{\left( {2d} \right)}^2}}}\left( {\frac{{2d{{\hat a}_y}}}{{2d}}} \right)+\frac{{Q \times Q}}{{4\pi \epsilon{_0}{{\left( {\sqrt 8 ;d} \right)}^2}}}\left( {\frac{{2d\hat a + 2d\hat y}}{{\sqrt 8 d}}} \right)\)

\(F= \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q^2}}}{{{{\left( {2d} \right)}^3}}}\left[ { - 2d{a_x} - 2d{a_y} + \frac{1}{{2\sqrt 2 }}\left( {2d{a_x} + 2d{a_y}} \right)} \right] \)

\(F = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q^2}}}{{{d^2}}};\left[ {\frac{{1 - 2\sqrt 2 }}{{8\sqrt 2 }}{a_x} + \frac{{1 - 2\sqrt 2 }}{{8\sqrt 2 }}{a_y}} \right]\)

By comparing the above equation with \(\frac{{{Q^2}}}{{4\pi {\varepsilon _0}}}\frac{K}{{{d^2}}}\)

\(\Rightarrow K = \frac{{1 - 2\sqrt 2 }}{{8\sqrt 2 }}\hat i + \frac{{1 - 2\sqrt 2 }}{{8\sqrt 2 }}\hat j \)

Given counter is 

The same clock pulse is given to both the flip-flops, therefore it is synchronous counter.

For every four cycles, the output is repeated 

So, dividing 333 by 4, we get 83.25

So, at 332 (83 × 4 = 332) clock pulse the output will be 0 0 (Q₁Q₀) 

At 333^rd clock pulse, Q₁ Q₀ = 0 1

Given maxterm

F (A,B,C,D) = π (1,5,12,15)

The minterms of the given max terms will be

f (A,B,C,D) = Σm (0,2,3,4,6,7,8,9,10,11,13,14) 

Therefore, the correct inputs to pins 0, 1, 2, 3, 4, 5, 6, 7 in order is D’, 1, D’, 1, 1, 1, D, D’

Concept:

An astable multivibrator is also called a Square wave generator or Free running oscillator.

It generates a square wave output.

The output waveform:

It generates a square wave output with time (T ) = \(\ T = 2{R_c}\log \frac{{\left( {1 + \beta } \right)}}{{\left( {1 - \beta } \right)}}\)

Calculation:

Astable multivibrator produces square wave.

\(\begin{array}{l} \beta = \frac{{{R_2}}}{{{R_1} + {R_2}}} = \frac{2}{4} = 0.5\ T = 2{R_c}\log \frac{{\left( {1 + \beta } \right)}}{{\left( {1 - \beta } \right)}} = 2 \times 1 \times {10^3} \times 0.25 \times {10^{ - 6}} \times \log \left( {\frac{{1 + 0.5}}{{1 - 0.5}}} \right)\ T = 0.55;ms \end{array}\)

Square wave of period 0.55 ms.

Concept:

The condition for optimum operation and economic dispatch is

\(\frac{{d{F_1}}}{{d{P_1}}} = \frac{{d{F_2}}}{{d{P_2}}} = \frac{{d{F_3}}}{{d{P_3}}} \ldots \ldots \ldots \ldots = \frac{{d{F_n}}}{{d{P_n}}} = \lambda \)

Where,

\(\frac{{dF}}{{dP}}\) is incremental fuel costs in Rs/MWh for a power plant.

The above equation shows that the criterion for the most economical division of load between plants is that all the unit is must operate at the same incremental fuel cost.

Calculation:

\(\begin{array}{l} \frac{{d{C_1}}}{{d{P_1}}} = 76 = 0.2{P_1} + 50 \Rightarrow {P_1} = 130\ \frac{{d{C_2}}}{{d{P_2}}} = 68.8 = 0.24{P_2} + 40 \Rightarrow {P_2} = 120 \end{array}\)

P₁ + P₂ = 250

For total cost minimization

\(\frac{{d{C_1}}}{{d{P_1}}} = \frac{{d{C_2}}}{{d{P_2}}}\)

⇒ 0.2P₁ + 50 = 0.24P₂ + 40

⇒ P₂ = 136.36

Concept:

GMR:

Geometrical Mean Radius GMR is basically the Geometric Mean of distances between the strands of a conductor. It is the fictitious radius of a conductor having no internal flux linkage but only external flux linkage.

GMR of a conductor having 3- strands each of radius, r placed in the form of a triangle is given below

GMR = [{D11 × D12 × D13} {D21 × D22 × D23} {D31 × D32 × D33}]1/9

Where

r = radius of a conductor.

D12 = Distance between conductor 1 and 2.

D23 = Distance between conductor 2 and 3.

  :

  :

D32 = Distance between conductor 4 and 3.

D11, D22, D33 = r’ = GMR of conductor 1, 2 and 3 respectively

Calculation:

Given 

D₁₁ = e-0.25R = 0.7788 R

D12 = D13 = D21 = D23 = D31 = D32 = 3R

GMR = [0.7788 R × 3R × 3R]^1/3

= 1.9137 R = kR

⇒ k = 1.913

\({i_2} = 0\)

Since entire current flows through fault,    Primary kVA = Secondary kVA

\(\sqrt 3 \times 33000 \times {I_L} = \sqrt 3 \times 11000 \times \left( {300 \times \frac{5}{{500}}} \right)\),     \({I_L} = 1A\)

\(\begin{array}{l} {I_L} = \sqrt 3 {I_{Ph}}\ {I_{ph}} = {i_1} = \frac{1}{{\sqrt 3 }}A \end{array}\)

Explanation:

Given figure can be redrawn as,

With primary star (γ) side,

Phase voltage (v_a) = \(\rm \frac{line voltage (V_{ab})}{\sqrt{3}}=\frac{230}{\sqrt{3}}\)

At secondary delta (Δ) side,

Phase voltage (V_A) = line voltage (V_AB) = 115 V

Transformation ratio (k) = \(\frac{V_A}{V_a}=\frac{115}{\frac{230}{\sqrt{3}}}=0.87\)

Phase current (I_A) on Δ side is,

\(I_A=\frac{I_s\rm(line)}{\sqrt{3}}=\frac{100}{\sqrt{3}}=57.7\)

∴ I_s lags I_A by 30°

So, I_A = 57.7 ∠30°

Now, \(k=\frac{I_p}{I_A}\)

⇒ I_P = k I_A = 0.87 × 57.7∠30°

⇒ = 50.199 ∠30°

Hence, the correct answer is (A).

Concept:

Nodal Analysis:

• Nodal Voltage Analysis uses the “Nodal” equations of Kirchhoff’s first law to find the voltage potentials around the circuit. So by adding together all these nodal voltages the net result will be equal to zero.
• If there are “n” nodes in the circuit there will be “n-1” independent nodal equations and these alone are sufficient to describe and hence solve the circuit.
• At each node point write down Kirchhoff’s first law equation(KCL), that is: “the currents entering a node are exactly equal in value to the currents leaving the node” then express each current in terms of the voltage across the branch using Ohm's law.
• So the nodal analysis is primarily based on the application of KCL and Ohm's law.
• For “n” nodes, one node will be used as the reference node and all the other voltages will be referenced or measured with respect to this common node.

Let a network and find voltage V using nodal analysis

Applying nodal analysis at node V

\(\frac{{V - {V_1}}}{{{R_1}}} + \frac{{V - {V_2}}}{{{R_2}}} + \frac{{V - {V_3}}}{{{R_3}}} = 0\)

Calculation:

The given circuit diagram 

Given, 

V1 = 100 ∠0°V, V2 = 100 ∠-120°V, V3 = 100 ∠+120°V

Apply nodal analysis at the node where the current i is entering

\(⇒ - i = \frac{{\left( {{V_1} - {V_3}} \right)}}{{ - j}} + \frac{{\left( {{V_2} - {V_3}} \right)}}{j}\)

Substitute the value of V₁, V2, and V3 in the above equation

\(⇒ - i = \frac{{100\angle 0^\circ - 100\angle 120^\circ }}{{1\angle - 90^\circ }} + \frac{{100\angle - 120^\circ - 100\angle 120^\circ }}{{1\angle 90^\circ }}\)

\(⇒ - i = \frac{{100\angle 60^\circ}}{{1\angle - 90^\circ }} + \frac{{-100\angle 0^\circ }}{{1\angle 90^\circ }}\)

⇒ i = 173.2 ∠-60°

Concept:

Fourier-series of symmetrical square wave (duty cycle = 50%)

x(-t) = -x(t) so it is odd symmetrical wave.

From fourier series property

a₀ = a_n = 0

only bn coefficient exist for odd symmetrical wave

\({b_n} = \frac{2}{T}\mathop \smallint \nolimits_{\frac{{ - T}}{2}}^{\frac{T}{2}} x\left( t \right)\sin n{\omega _0}t;dt\)

\({b_n} = \frac{{2 \times 2}}{T}\mathop \smallint \nolimits_0^{\frac{T}{2}} x\left( t \right)\sin n{\omega _0}t;dt\)

\({b_n} = \frac{{4V}}{T}\mathop \smallint \nolimits_0^{\frac{T}{2}} \sin n{\omega _0}t;dt\)

\({b_n} = \frac{{4V}}{T}\left[ {\frac{{ - \cos n{\omega _0}t}}{{n{\omega _0}}}} \right]_0^{\frac{T}{2}}\)    ω0T = 2π

\( = \frac{{4V}}{{n{\omega _0}T}}\left[ { - \cos n{\omega _0}t} \right]_0^{\frac{T}{2}}\)

\({b_n} = \frac{{4V}}{{2n\pi }}\left[ {1 - \cos n\pi } \right]\)

\({b_n} = \frac{{2V}}{{n\pi }}\left{ {1 - {{\left( { - 1} \right)}^n}} \right}\)

if n = even,   b_n = 0

n = odd, \(b_n = \frac{{2V}}{{n\pi }}\left( 2 \right) = \frac{{4V}}{{n\pi }}\)

We can represent x(t) = \(\mathop \sum \limits_{n = 1,3,5..}^\infty \frac{{4V}}{{n\pi }}\sin n{\omega _0}t\)

Trigonometric fourier series expansion of x(t)

Calculation:

Given:

In the ques it is given that square wave is symmetrical with duty cycle of 50%

V = 15 V

T = 0.4 m sec

⇒ fourier series expansion = \(\mathop \sum \limits_{n = 1,3,5,..}^\infty \frac{{4; \times ;15}}{{n\pi }}\sin n{\omega _0}t\)

Fundamental frequency \(({\omega _0}) = \frac{{2\pi }}{T} = \frac{{2\pi }}{{0.4; \times ;{{10}^{ - 3}}; \times ;\pi }}\) 

= 5000 rad/sec

We want to find the amplitude of capacitor voltage at 5000 rad/sec or at a fundamental frequency so n = 1

From fourier series expansion \(V\left( t \right) = \frac{{4\pi V}}{\pi } = \frac{{4; \times ;15}}{\pi } = 19.09;V\)

V_max = 19.09 V

\({X_C} = \frac{1}{{{\omega C}}} = \frac{1}{{500; \times ;4; \times ;{{10}^{ - 6}}}} = 50;{\rm{\Omega }}\) 

X_L = ωL = 5000 × 10^-2 = 50 Ω

Since X_C = X_L → Resonance conduction, so reactive component will cancel each other

Current:

\(I = \frac{V}{R} = \frac{{19.09}}{5} = 3.818;Amp\) 

V_c = I X_C = (3.818) (-j50)

= -j190.9 = 190.9 ∠ -90° V

Amplitude V_C = +190.9 V

Here, we consider the mutual inductance for two identical coils connected in series as shown below.

Coefficient of coupling is given by

\(\begin{array}{l} K \le \frac{M}{{\sqrt {{L_1}{L_2}} }}\ M \ge K\sqrt {{L_1}{L_2}} \end{array}\)

Here, we have

\({L_1} = {L_2}\)

So, we get

\(M \ge KL\)

Therefore, the equivalent inductance is given by

\({L_{eq}} = {L_1} + {L_2} - 2M\) (when polarity is different)

\(\begin{array}{l} = 2L-2M\ = 2L-2KL \end{array}\)

For minimum value of \({L_{eq}}\), we use maximum value of k, i.e. \(K = 1\)

Hence, we get

\({L_{eq}} = 2L-2L = 0\)

Given \({L_{eq}} = \alpha L\)

So, the minimum value of \(\alpha\) is zero.

Concept: 

The MMF balance equation for three winding transformer 

N1 I1 = N2 I2 + N3 I3

Explanation:

The MMF balance equation is,

\(\begin{array}{l} {I_1}{N_1} = {I_2}{N_2} + {I_3}{N_3}\ {I_1}.100 = 2\angle 30 \times 50 + 2\angle 150 \times 50\ {I_1} = 1\angle 30 + 1\angle 150 = 1\angle 90^\circ \end{array}\)

Concept:

Equivalent circuit of separately excited D.C motor,

V = E_b + I_aR_a

Explanation:

Let,

Torque T₁ at N₁ = 1000 rpm

Torque T₂ = at N₂ = 500 rpm

∴ T ∝ N² {Given}

⇒ \(\frac{T_2}{T_1}=\left(\frac{N_2}{N_1}\right)^2\)

⇒ \(\frac{T_2}{T_1}=\left(\frac{500}{1000}\right)^2\)

⇒ \(\frac{T_2}{T_1}=\frac{1}{4}\)

Load torque can be expressed as,

T = k ϕ. I_a

T ∝ ϕ I_a

T ∝ I_a {∴ ϕ = constant}

⇒ \(\frac{T_2}{T_1}=\frac{Ia_2}{Ia_1}\)

⇒ \(\frac{Ia_2}{Ia_1}=\frac{1}{4}\)

⇒ \(Ia_2=\frac{10}{4}=2.5 \) A

Therefore, 

\(E_{b_1}=V_1-I{a_1}R_a=100-10=90\) V

∴ \(E_b=\frac{NP\phi Z}{60\ A}\)

⇒ \(\frac{E_{b_1}}{E_{b_2}}=\frac{N_1}{N_2}\)

⇒ \(E_{b_2}=E_{b_1}\frac{N_2}{N_1}\)

⇒ \(E_{b_2}=\frac{90×500}{1000}\)

⇒ \(E_{b_2}=45\) V

Therefore,

\(V_2=E_{b_2}+I_{a_2}R_a\)

⇒ V₂ = 45 + 2.5 × 1

⇒ V₂ = 47.5 V

Hence, The armature voltage is 47.5 V

Concept

Emf equation for synchronous motor,

E_f = V_t - I_aZ_s

\((I_aZ_s)^2=E_f^2+V_t^2-2E_fV_t\cos \delta\)

Explanation

power factor (cos ϕ) = unity       (Current is in phase with voltage)

Emf equation of synchronous motor is given by,

Ef = Vt - I_aZ_s

\(E_f=V_t-I_{a_1}(R_a+jX_s)\)

[∴ R_a = 0 Ω]

⇒ \(E_f=V_t-jI_{a_1}X_s\)

⇒ \(E_f=\frac{11\times10^3}{\sqrt{3}}-j100\times50\)

⇒ \(E_f=(6.35-j5)\times10^3\) V

⇒ E_f = 8.082 ∠-38.21° kV

Excitation is kept constant. Load angle will increase on increasing the load torque in this case, motor will operate in lagging power factor condition

\((I_{a_2}Z_s)^2=E^2_f+V_t^1-2E_fV_t\cos \delta_2\)

\((6\times10^3)^2=(8.08\times10^3)^2+(6.35\times10^3)^2-2\times8.08\times10^3\times6.35\times10^3\times\cos \delta_2\)

⇒ \(\cos \delta_2=\frac{(8.08)^2+(6.35)^2-(6)^2}{2\times8.08\times6.35}=0.678\)

⇒ \(\delta_2=\cos^{-1}(0.678)=47.28\)

Here, in the synchronous motor, the load angle is negative -47.2⁰.

Concept:

Torque equation:

The torque equation of a three-phase induction motor is given by,

\(T = \frac{{180}}{{2\pi {N_s}}}\left( {\frac{{sV^2{R_2}}}{{\left( {R_2^2 + {s^2}X_2^2} \right)}}} \right)\)

Where Ns is the synchronous speed

V = supply voltage

R2 = rotor resistance

X2 = rotor reactance

s is the slip

By the above expression, we can say that the torque of an induction motor depends on rotor resistance and slip.

Condition for maximum torque:

The condition to get the maximum torque at starting is,

\({s_m} = \frac{{{R_m}}}{{{X_m}}} = 1\)

Xm = R­m

Where,

Rm = Motor resistance per phase

Xm = Motor reactance per phase

At the starting of the three-phase slip ring induction motor

Slip (s) = 1 (At the starting Nr = 0)

Therefore, \({s_m} = \frac{{{R_m}}}{{{X_m}}} = 1\)

⇒ Xm = Rm

the starting torque of an induction motor is maximum when rotor resistance equals rotor reactance.

Explanation:

The condition for maximum torque in three phase induction motor,

\(\begin{array}{l} {s_m} = \frac{{{r_2}}}{{x_2}} \ 0.15 = \frac{{{r_2}}}{{x_2}} = \frac{{0.03}}{{{x_2}}} \Rightarrow {x_2} = 0.2{\rm{\Omega }} \end{array}\)

For \({T_{est}} = {T_{emax}}\)

\(\begin{array}{l} \frac{{{T_{est}}}}{{{T_{em}}}} = \frac{2}{{\frac{1}{{{S_{mT}}}} + {S_{mT}}}} = 1 \Rightarrow {s_{mT}} = 1\ 1 = \frac{{r_2'}}{{{x_2}}} \Rightarrow r_2' = {x_2} = 0.2{\rm{\Omega }} \end{array}\)

Extra resistance =0.2 – 0.03 = 0.17 Ω /p

Assume, ABC phase sequence in clockwise direction

Phasor diagram of above Dd_o connection is shown below

Here voltage \(V_{A_2A_1}&V_{a_2a_1}\) are in same phase.

Above connection is Dy11 '' its phasor diagram shown below is,

V_ab leads V_AB by +30° or V₂ leads V₁ by +30°.

Hence, phase difference between V₁ & V₂ is 30°.

Here integration time step is h, so

\(\begin{array}{l} {x_{k + 1}} = {x_k} + h\ {y_{k + 1}} = {y_k} - h{x_k} \end{array}\)

Again, we may write

\(\begin{array}{l} {y_k} = \frac{{{x_{k + 1}} - {x_k}}}{h}\ \Rightarrow {y_{k + 1}} = \frac{{{x_{k + 1}} - {x_k}}}{h} - h{x_k}\ = \frac{{h + {x_k} - {x_k}}}{h} - h{x_k}\ = ; - h{x_{k + 1}} \end{array}\)

Now, we form the Routh array for the above equation

\(\begin{array}{*{20}{c}} {{s^1}}\ {{s^0}} \end{array}\left| {\begin{array}{*{20}{c}} { - h}&0\ 1&0 \end{array}} \right.\)

Hence, according to Routh – Hurwitz criterion system is stable for

\(\begin{array}{l} - ;h > 0\ \Rightarrow h < 0 \end{array}\)

Thus, system is not stable for \(h > 0\)

Concept:

Transfer function:

The transfer function is defined as the ratio of the Laplace transform of the output variable to the input variable with all initial conditions zero.

TF = \(\left. {\frac{{L\left[ {output} \right]}}{{L\left[ {input} \right]}}} \right|\) Initial conditions = 0

TF = C(s) / R(s)

Calculation:

Given transfer function 

\(G\left( s \right) = \frac{{1 - 2s}}{{1 + s}}\)

Input of the system r(t) = u(t)

Apply laplace transform

R(s) = 1/s

Response of the system of unit step is

C(s) = R(s) × G(s)

⇒ \(Y\left( s \right) = \frac{{\left( {1 - 2s} \right)}}{{\left( {1 + s} \right)}}.\frac{1}{s}\)

⇒ \(Y\left( s \right) = \frac{A}{{\left( s \right)}} + \frac{B}{{\left( {s + 1} \right)}}\)

A = 1, B = -3

Apply inverse Laplace transform 

⇒ y(t) = u(t) - 3 e^-t u(t)

⇒ y(t) = (1 - 3e^-t) u(t)

By plotting the above we get as follows

Concept:

Break-in/away points: These exist when there are multiple roots on the root locus diagram.

At the breakpoints gain K is either maximum and/or minimum.

So, the roots of \(\frac{{dK}}{{ds}}\) are the breakpoints.

Calculation:

Given open-loop system,

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

⇒ 1 + G(s) = 0

⇒ \(1+\frac{K}{{s\left( {s + 1} \right)\left( {s + 2} \right)}}=0\)

⇒ k = - (s³ + 3s² + 2s)

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

\(\frac{{dK}}{{ds}} = 3s^2+6s+2=0\)

The values of s are - 0.42, - 1.577

When we substitute both the values in transfer function only - 0.42 satisfies 

Therefore, the breakaway point of the root loci on the real axis occurs at -0.42

\(\frac{{d{x_1}}}{{dt}} = 2{x_1} + {x_2} + 1,;;;\frac{{d{x_2}}}{{dt}} = - 2{x_1} + 1\)

\(y; = ;3{x_1}\)

Considering the standard equation

\(\begin{array}{l} {x_i}; = ;Ax; + ;BU,;;y; = ;Cx; + ;DU\ \left[ {\begin{array}{*{20}{c}} {{{\dot x}_1}}\ {{{\dot x}_2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{r}} 2&1\ { - 2}&0 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}}\ {{x_2}} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 1\ 1 \end{array}} \right]\left[ 1 \right]\ y = \left[ {\begin{array}{*{20}{c}} 3&0 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}}\ {{x_2}} \end{array}} \right] \end{array}\)

Transform function \(C{\left( {SI;-;A} \right)^{ - 1}};B\)

\(G\left( s \right) = \left[ {\begin{array}{*{20}{c}} 3&0 \end{array}} \right]{\left[ {\left[ {\begin{array}{*{20}{c}} s&0\ 0&s \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 2&1\ { - 2}&0 \end{array}} \right]} \right]^{ - 1}}\left[ {\begin{array}{*{20}{c}} 1\ 1 \end{array}} \right]\)

\(\Rightarrow \left[ {\begin{array}{*{20}{c}} 3&0 \end{array}} \right]{\left[ {\begin{array}{*{20}{c}} {s - 2}&{ - 1}\ 2&s \end{array}} \right]^{ - 1}}\)

\( \Rightarrow \frac{1}{{{s^2} - s + 2}}\left[ {\begin{array}{*{20}{c}} 3&0 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {s + 1}\ {s - 4} \end{array}} \right]\)

\(\Rightarrow \frac{{3\left( {s + 1} \right)}}{{{s^2}-2s + 2}}\)