GATE EE 2018 Official Paper (Previous Year Paper)

The given sentence has two blanks. The first blank needs a word that means 'path' or 'way' and the second blank needs an adjective that can describe the sand.

Option 1 is correct as course means path or way and coarse means rough which logically and grammatically fit the first and the second blank, respectively.

Explanation:

Correct answer: Option 4 (convoluted)

The given sentence states that there is a common misunderstanding among existing authors that the quality of sentence structure closely resembles the thought process or even affects it.

'Convoluted' means something that is extremely complex and difficult to follow. It is generally used in the context of an argument or a story. 

The other options (especially 'simple' and 'clear') do not provide a correct meaning if placed in the blank.

Hence, convoluted is the correct answer.

NOTE: 'Detailed' can imply something that is comprehensive. It does not mean complicated. Hence, it cannot be the correct answer.

The three roots of the equation f(x) = 0 are x = {-2, 0, 3}

f(x) = (x – (-2)) (x – 0) (x – 3)

⇒ f(x) = x (x + 2) (x – 3)

f(x – 3) = (x – 3) (x – 3 + 2) (x – 3 – 3) = (x – 3) (x – 1) (x – 6)

Given that, f (x - 3) = 0

⇒ (x – 3) (x – 1) (x – 6) = 0

⇒ x = 1, 3, 6

Given function:

\(f\left( k \right) = \frac{{{{\left( {k + 2} \right)}^2}}}{{k - 3}}\)

For k = 4:

\(f\left( 4 \right) = \frac{{{{\left( {4 + 2} \right)}^2}}}{{4 - 3}} = 36\)

For k = 8:

\(f\left( 8 \right) = \frac{{{{\left( {8 + 2} \right)}^2}}}{{8 - 3}} = 20\)

For k = 28:

\(f\left( {28} \right) = \frac{{{{\left( {28 + 2} \right)}^2}}}{{28 - 3}} = 36\)

F (a, b) = (a - b)²

F (1, 3) = (1 – 3)² = 4

G (a, b) = |a - b|

G (1, 3) = |1 – 3| = 2

G (F (1, 3), G (1, 3)) = G (4, 2)

G (4, 2) = |4 – 2| = 2

An e-mail password must contain three characters.

One character will be one numeral from 0 to 9.

Number of choices \(= {10_{{C_1}}} = 10\)

Second character will be upper case letter from the English alphabet.

Number of choices \(= {26_{{C_1}}} = 26\)

Third character will be lower case letter from the English alphabet.

Number of choices \(= {26_{{C_1}}} = 26\)

We can arrange these three characters in 3! ways i.e. 6 ways.

Total number of distinct passwords = 10 × 26 × 26 × 6 = 40,560

A M C F → E Q G J (A + 4 = E, M + 4 = Q, C + 4 = G, F + 4 = J)

N K U F → R O Y J (N + 4 = R, K + 4 = O, U + 4 = Y, F + 4 = J)

D H L P → H L P T (D + 4 = H, H + 4 = L, L + 4 = P, P + 4 = T)

Total number of children in a class = 12

Given that class has two more boys than girls.

So, number of boys = 7

Number of girls = 5

We need to pick a group of three children randomly so that the group contains more girls than boys.

Case 1: 3 girls and 0 boys

Probability to select 3 girls, \({P_1} = \frac{{{5_{{C_3}}}}}{{{{12}_{{C_3}}}}} = \frac{{10}}{{220}}\)

Case 2: 2 girls and 1 boy

Probability to select 2 girls and 1 boy, \({P_2} = \frac{{{5_{{C_2}}}{7_{{C_1}}}}}{{{{12}_{{C_3}}}}} = \frac{{10 \times 7}}{{220}} = \frac{{70}}{{220}}\)

The probability that the group accompanying the teacher contains more girls than boys is,

\(P = {P_1} + {P_2} = \frac{{10}}{{220}} + \frac{{70}}{{220}} = \frac{{80}}{{220}} = \frac{4}{{11}}\)

Let the total number of marbles = 100

Number of blue marbles = 40

Number of black marbles = 25

Number of red marbles = 20

Number of yellow marbles = 15

And the cost of each marble = 1 unit

The total cost of marbles as per current design = 100 units

The cost of each marble after the cost of each marble is increased by 25% = 1.25 unit

Now, the designer decided to reduce equal numbers of marbles of each colour to keep the total cost unchanged.

Let the number of marbles of each colour to reduce = n

Now, the number of blue marbles = 40 – n

Number of black marbles = 25 - n

Number of red marbles = 20 - n

Number of yellow marbles = 15 - n

The total cost remains same as previous.

⇒ [(40 – n) + (25 – n) + (20 – n) + (15 – n)] [1.25] = 100

⇒ 100 – 4n = 80

⇒ n = 5

Now, the number total marbles = 100 – 4(5) = 80

The number of blue marbles = 40 – 5 = 35

The percentage of blue marbles in the new design \(= \frac{{35}}{{80}} \times 100 = 43.75;%\)

P is unable to row when someone else is in the boat. So, P can row during return trip only.

Q is unable to row with anyone else except R. So, Q can row during return and can row forward trip with R.

Trip 1 – Forward trip:

Q and R are involved in this trip and Q rowed the trip.

Trip 1 – Return trip:

R rowed the trip

Trip 2 – Forward trip:

P and R are involved in this trip and R rowed the trip.

Trip 2 – Return trip:

P rowed the trip

Trip 1 – Forward trip:

S rowed the trip

So, R rowed twice.

Concept:

Voltage regulation is the change in secondary terminal voltage from no load to full load at a specific power factor of load and the change is expressed in percentage.

E2 = no-load secondary voltage

V2 = full load secondary voltage

Voltage regulation for the transformer is given by the ratio of change in secondary terminal voltage from no load to full load to no load secondary voltage.

Voltage regulation \(= \frac{{{E_2} - {V_2}}}{{{E_2}}}\)

It can also be expressed as,

Regulation \(= \frac{{{I_2}{R_{02}}\cos {\phi _2} \pm {I_2}{X_{02}}\sin {\phi _2}}}{{{E_2}}}\)

• sign is used for lagging loads and

• ve sign is used for leading loads

Per unit Voltage regulation = %R cos ϕ ± %X sin ϕ

Explanation:

Given that,

Voltage drop across series impedance (% Z) = 5%

Voltage drop due to resistance (% R) = 3%

Power factor = 0.8 lagging

We know that, \(% Z = \sqrt {{{\left( {% R} \right)}^2} + {{\left( {% X} \right)}^2}} \)

\(\Rightarrow \left( {% X} \right) = \sqrt {{{\left( {% Z} \right)}^2} - {{\left( {% R} \right)}^2}} = \sqrt {{5^2} - {3^2}} = 4% \)

Voltage regulation = %R cos ϕ ± %X sin ϕ

+ve is for lagging loads

-ve is for leading loads

Voltage regulation = (3) (0.8) + (4) (0.6) = 4.8%

Concept:

In a salient pole synchronous motor, the power flow is given by,

\(P = \frac{{EV}}{{{X_d}}}\sin \delta + \frac{{{V^2}}}{2}\left[ {\frac{1}{{{X_q}}} - \frac{1}{{{X_d}}}} \right]\sin 2\delta \)

The torque equation is given by

\(T = \frac{{60}}{{2\pi {N_S}}}\left[ {\frac{{EV}}{{{X_d}}}\sin \delta + \frac{{{V^2}}}{2}\left[ {\frac{1}{{{X_q}}} - \frac{1}{{{X_d}}}} \right]\sin 2\delta } \right]\)

Here,

\(\frac{{EV}}{{{X_d}}}\sin \delta\) is electromagnetic torque

\({\rm{and}}\frac{{{V^2}}}{2}\left[ {\frac{1}{{{X_q}}} - \frac{1}{{{X_d}}}} \right]\sin 2\delta = Reluctance;torque.\)

Explanation:

As we know that Vt = terminal voltage depends on the load.

Xd and Xq depend on the power factor angle which also depends on the load.

So, we cannot change the value of the three-parameter.

The value of sin 2δ always varies in the range +1 to -1.

For the maximum value of TR sin 2δ should be maximum

⇒ sin 2δ = + 1

⇒ 2δ = sin-1 (1) = 90°

⇒ δ = 45°

Torque angle δ is 45° for maximum reluctance torque to be produced in a synchronous motor.

Concept:

Freewheeling diode:

• A freewheeling diode placed across the inductive load will provide a path for the release of energy stored in the inductor while the load voltage drops to zero.
• The freewheeling diode prevents the load voltage from becoming negative. Whenever load voltage tends to go negative, FD comes into play.
• As a result, the load current is transferred from the main thyristor to FD, allowing the thyristor to regain its forward blocking capability.
• The advantages are input power factor is improved, load current waveform is improved and better load performance.

Explanation:

The given circuit represents a single phase fully controlled rectifier.

Anti-parallel diode works as freewheeling diode. As the freewheeling diode is present in converter, the output voltage can’t be negative whether the converter is working under continuous mode or discontinuous mode. The output current is always unidirectional as it is A.C. to D.C. converter.

Hence, V₀ ≥ 0 and i₀ ≥ 0 

Note-

For continuous conduction mode : 

V0 ≥ 0 and i0 > 0

Thyristor allows the conduction of current from anode to cathode. Hence, the current cant flow in the given direction.

Triac can allow current in both directions. Hence, the current can flow in a given direction.

The MOSFET is a bidirectional device. Hence, the current can flow in a given direction.

GTO is a unidirectional device. Hence, the current can't flow in the given direction.

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)

\(\phi = {\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)\)

Calculation:

Given that,

\({W_2} = \frac{{{W_1}}}{2}\)

We know that,

\(\phi = {\tan ^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)\)

\( = {\tan ^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - \frac{{{W_1}}}{2}} \right)}}{{\left( {{W_1} + \frac{{{W_1}}}{2}} \right)}}} \right)\)

\(= {\tan ^{ - 1}}\left( {\frac{{\sqrt 3 \left( {2{W_1} - {W_1}} \right)}}{{2{W_1} + {W_1}}}} \right)\)

\(= {\tan ^{ - 1}}\left( {\frac{{\sqrt 3 {W_1}}}{{3{W_1}}}} \right) = {\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} \right) = 30^\circ \)

Power factor, cos ϕ = cos 30° = 0.866

Given that,

Sending end voltage = V₁

Receiving end voltage = V₂

Series impedance = Z

reactance = X

When only X is present:

\({P_{max1}} = \left| {\frac{{{V_1}{V_2}}}{X}} \right|\)

When Z is present:

\({P_{max2}} = \frac{{{V_1}{V_2}}}{{\left| Z \right|}} - \frac{{V_2^2}}{{\left| Z \right|}}\cos \phi \)

P_max2 is less than P_max1

Hence, the steady-steady stability limit for the transmission line will be less than \(\left| {\frac{{{V_1}{V_2}}}{X}} \right|\).

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:

Given that,

Number of nodes (N) = 8

Number of independent loops (L) = 5

Let the number of branches are B.

We know that,

L = B – N + 1

⇒ 5 = B – 8 + 1 ⇒ B = 12

Given that,

V₁ (t) = 100 cos ωt

V₂(t) = 100 cos(ωt + π/18) = 100 cos(ωt +10^∘)

V₃(t) = 100 cos(ωt + π/36) = 100 cos(ωt + 5°)

Phasor diagram of all the three voltages as shown in  below fig,

From the above diagram,

V₂ (t) is leading V₁(t) by 10° and

V₃ (t) is leading V₁(t) by 5°

V₂ (t) is leading V₃ (t) by 5°.

We know that active power always flows from a leading angle to a lagging angle.

Hence, current flows from V₂ (t) to V₁ (t) and V₃ (t).

Hence, P₂ > 0, P₁ < 0 P₃ < 0.

The standard second order system is given by \(\frac{{\omega _n^2}}{{{s^2} + 2\xi {\omega _n}s + \omega _n^2}}\)

Where ξ is damping ratio.

If ξ = 1, then system is critically damped.

If ξ < 1, then system is under damped.

If ξ > 1, then system is order damped.

\(P:\frac{{15}}{{{s^2} + 5s + 15}}\)

By comparing with standard second order transfer function,

ω_n² = 15 ⇒ ω_n = √15

\(2\xi {{\rm{\omega }}_n} = 5 \Rightarrow \xi = \frac{5}{{2\sqrt {15} }} < 1\)

So, it is underdamped system.

\(Q:\frac{{25}}{{{s^2} + 10s + 25}}\)

ω_n² = 25 ⇒ ω_n = 5

2 ξ ω_n = 10 ⇒ ξ = 1

So, it si critically damped system.

\(R:\frac{{35}}{{{s^2} + 18s + 35}}\)

ω_n² = 35 ⇒ ω_n = √35

\(2\xi {{\rm{\omega }}_n} = 18 \Rightarrow \xi = \frac{9}{{\sqrt {35} }} > 1\)

So, it is overdamped system.

Given that a positive charge of 1 nc is placed at (0, 0, 0.2).

According to method image charge concept, there will be an image charge below the conducting ground plane.

Electric field is given by

\(\bar E = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{Q\bar R}}{{{{\left| {\bar R} \right|}^3}}}\)

Electric field at point P due to point charge + 1 nc,

\(\bar E = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{1 \times {{10}^{ - 9}}\left[ {\left( {0 - 0} \right){{\bar a}_x} + \left( {0 - 0} \right){{\bar a}_y} + \left( {0.1 - 0.2} \right){{\bar a}_z}} \right]}}{{{{\left( {0.1} \right)}^3}}}\)

\(= \frac{1}{{4\pi \times 8.85 \times {{10}^{ - 12}}}} \times \frac{{1 \times {{10}^{ - 9}}}}{{{{\left( {0.1} \right)}^3}}}\left( { - 0.1} \right){\bar a_z}\)

= -899.18 V/m

Electric field at point P, due to point charge -1 nc,

\({\bar E_2} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{ - 1 \times {{10}^{ - 9}}\left[ {\left( {0 - 0} \right){{\bar a}_x} + \left( {0 - 0} \right){{\bar a}_y} + \left( {0.1 + 0.2} \right){{\bar a}_z}} \right]}}{{{{\left( {0.3} \right)}^3}}}\)

\( = \frac{1}{{4\pi \times 0.85 \times {{10}^{ - 12}}}} \times \frac{{\left( { - 1 \times {{10}^{ - 9}}} \right)}}{{{{\left( {0.3} \right)}^3}}}\left( {0.3} \right){\bar a_z} = - 99.90;V/m\)

E̅ = E̅₁ + E̅₂ = -899.18 V/m – 99.90 V/m

= -999.08 V/m

Given that,

A function f(x) is said to be differentiable at x =a if,

Left derivative = Right derivative = Well defined

i.e.,

 \(\rm\lim _{x \rightarrow a^{-}} f'(x)=\lim _{x \rightarrow a^{+}} f'(x)\)

Analysis:

f(x) = x², x ≥ 0

= -x², x ≤ 0

f^'(x) = 2x, x ≥ 0

= -2x, x < 0

f^'(x) = 2|x|

f^’(x) is continuous but not differentiable at x = 0.

Hence f(x) is differentiable but its first derivative is not differentiable at x = 0.

Option(4) is the correct answer.

Given that,

ϕ = xy² + yz² + zx²

directional vector (p) = I + 2j + 2K

Directional derivative \( = \nabla \phi .\hat P = \nabla \phi .\frac{{\vec P}}{{\left| {\vec P} \right|}}\)

\(\nabla \phi = \frac{{\partial \phi }}{{\partial x}}\hat i + \frac{{\partial \phi }}{{\partial y}}\hat i + \frac{{\partial \phi }}{{\partial z}} \hat k\)

= (y² + 2xz) î + (2xy + z²) ĵ + (2yz + x²) k̂

∇ϕ at the point (2, -1, 1) is

∇ϕ = ((-1)² + 2(2)(1)) î + (2(2)(-1) + (1)²) ĵ + (2(-1)(1) + (2)²)k̂

= 5î - 3ĵ + 2k̂

Directional derivative = \(\left( {5i - 3j + 2k} \right).\frac{{i + 2j + 2k}}{{\sqrt {{1^2} + {2^2} + {2^2}} }}\)

\(= \frac{{5 - 6 + 4}}{3} = 1\)

Hence Option(1) is the Correct Answer.

Given that,

\(f\left( Z \right) = \mathop \oint \limits_C^; \frac{{Z + 1}}{{{Z^2} - 4}}\)

C : |Z – (-2)| = 1

⇒ C : |Z + 2| = 1

poles of f(Z) are z² – 4 = 0

⇒ Z = ±2

Z = 2 is lies outside the curve C.

f(Z) = 2πi [residue at Z = -2]

\(= 2\pi i;\begin{array}{*{20}{c}} {it}\ {Z \to - 2} \end{array}\left( {Z + 2} \right)\frac{{\left( {Z + 1} \right)}}{{\left( {Z + 2} \right)\left( {Z - 2} \right)}}\)

\(= 2\pi i \times \frac{{\left( { - 2 + 1} \right)}}{{\left( { - 2 - 2} \right)}} = 2\pi i;\left( {\frac{{ - 1}}{{ - 4}}} \right) = \frac{{\pi i}}{2}\)

\(Y = \overline {\overline {AB} .\overline {CD} } = \overline {\overline {AB} } + \overline {\overline {CD} } = AB + CD\)

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:

By voltage division, \({V_b} = {V_o}\left( {\frac{{{R_2}}}{{{R_1} + {R_2}}}} \right)\)

By virtual grand concept, V_a = V_b = V_in

By applying KCL at V_a,

\(\frac{{{V_a} - {V_o}}}{Z} - {i_{in}} = 0\)

\(\Rightarrow {i_{in}} = \frac{{{V_a} - {V_o}}}{Z}\)

\(\Rightarrow {i_{in}} = \frac{{{V_{in}} - {V_{in}}\left( {\frac{{{R_1} + {R_2}}}{{{R_2}}}} \right)}}{Z}\)

\(\Rightarrow {i_{in}}\left( Z \right) = {V_{in}}\left( {1 - \frac{{{R_1}}}{{{R_2}}} - 1} \right)\)

\(\Rightarrow \frac{{{V_{in}}}}{{{i_{in}}}} = - \frac{{{R_2}Z}}{{{R_1}}}\)

A signal is said to be Eigen function if the output is scalar multiple of input signal and that scalar is referred as Eigen value.

Let input signal is x(t) and it is an Eigen function then

output = K x(t) where K is Eigen value.

Given that,

trace (A) = 4

trace (A²) = 5

Matrix A is non-singular 2 × 2 square matrix.

Let λ₁ and λ₂ are the Eigen values of matrix A.

from the properties of Eigen values,

(i) sum of Eigen values = trace of the matrix.

(ii) product of Eigen values = Determinant of the matrix

(iii) If λ is an Eigen value of matrix A, then λ² will be an Eigen value of matrix A².

λ₁ + λ₂ = 4,

λ₁² + λ₂² = 5

(λ₁ + λ₂)² = λ₁² + λ²₂ + 2 λ₁ λ₂

⇒ (4)² = 5 + 2 λ₁ λ₂

⇒ λ₁ λ₂ = 5.5

Hence, determinant of matrix A = 5.5

f(x) = x – [x]

for 0.25 < x < 1, f(x) = x

for 1 < x < 1.25, f(x) = x – 1

\(\mathop \smallint \limits_{0.25}^{1.25} f\left( x \right)dx = \mathop \smallint \limits_{0.25}^1 x;dx + \mathop \smallint \limits_1^{1.25} \left( {x - 1} \right)dx\)

\(= {\left[ {\frac{{{x^2}}}{2}} \right]_{0.25}}^1 + {\left[ {\frac{{{x^2}}}{2} - x} \right]_1}^{1.25}\)

\(= \frac{1}{2}\left[ {1 - \frac{1}{{16}}} \right] + \left[ {\left( {\frac{{25}}{{32}} - \frac{5}{4}} \right) - \left( {\frac{1}{2} - 1} \right)} \right]\)

\(= \frac{1}{2}\left[ {\frac{{15}}{{16}}} \right] + \left[ {\frac{{25}}{{32}} - \frac{5}{4} + \frac{1}{2}} \right]\)

\(= \frac{{15 + 25 - 40 + 16}}{{32}} = 0.5\)

\({h_{11}} = \frac{{{V_1}}}{{{I_1}}}\) when V₂ = 0

When V₂ = 0,

Apply nodal analysis at V₀,

\(\frac{{{V_0}}}{1} + \frac{{{V_0}}}{1} - {I_1} + 2{I_1} = 0\)

\(\Rightarrow 2{V_0} + {I_1} = 0 \Rightarrow {V_0} = \frac{{ - {I_1}}}{2}\)

Apply KVL in input loop,

• V₁ + I₁ + V₀ = 0

\(\Rightarrow - {V_1} + {I_1} - \frac{{{I_1}}}{2} = 0\)

\(\Rightarrow {V_1} = \frac{{{I_1}}}{2} \Rightarrow \frac{{{V_1}}}{{{I_1}}} = {h_{11}} = 0.5\)

Given that, positive sequence impedance (Z₁) = (1 + j10) Ω

Zero sequence impedance (Z₀) = (4 + j31) Ω

We know that, Z₁ = Z_s – Z_m

and Z₀ = Z_s + 2Z_m

Z₁ = 1 + j10 = Z_s – Z_m → (1)

and Z₀ = 4 + j31 = Z_s + 2Z_m → (2)

from equations (1) and (2)

⇒ Z_m = 1 + j7

Imaginary part of Z_m = 7 Ω.

Given that,

Z₁ = j 0.1 pu

Z₂ = j 0.05 pu

Z₀ = j 0.01 pu

Z_n = j 0.01 pu

For LG fault, fault current is given by

\({I_f} = \frac{{3{E_{R1}}}}{{{Z_1} + {Z_2} + {Z_0} + 3{Z_n}}} = \frac{{3 \times 1.0}}{{j0.1 + j0.05 + j0.01 + j0.03}}\)

\(= \frac{3}{{j0.19}} = - j;15.789;pu\)

I_f in Amperes = (I_f in pu) × I_base

\(= \left( {{I_f};in;pu} \right) \times \frac{{VA}}{{\sqrt 3 V}}\)

\(= 15.789 \times \frac{{125}}{{\sqrt 3 \times 15.5}}\)

= 73.51 kA.

Size of Y_BUS matrix = 1000 × 1000

Number of non-zero elements = 8000.

Number of diagonal elements = number of buses = 1000.

Number of non-zero off diagonal elements = 8000 – 1000 = 7000

Minimum number of branches \(= \frac{{7000}}{2} = 3500\)

Concept:

Fourier series representations of supply current of single-phase semi converter is

\({i_s}\left( t \right) = \mathop \sum \limits_{n = 1,;3, \ldots }^\infty \frac{{4{I_0}}}{{n\pi }}\cos \frac{{n \propto }}{2}\sin \left( {n\omega t - \frac{{n\alpha }}{2}} \right)\)

Explanation:

Fundamental component is,

\({I_{S1}} = \frac{{4{I_0}}}{\pi }\cos \frac{\alpha }{2}\)

RMS value of fundamental component is,

\({I_{S1}} = \frac{{4{I_0}}}{\pi }\cos \frac{\alpha }{2} \times \frac{1}{{\sqrt 2 }}\)

\(= \frac{{2\sqrt 2 {I_0}}}{\pi }\cos \frac{\alpha }{2}\)

From the given wave form,

firing angle (α) = 30°

\( \Rightarrow {I_{S1}} = \frac{{2\sqrt 2 \times 20}}{\pi }\cos \frac{{30}}{2} = 17.39;A\)

Explanation:

At zero speed (N = 0)

induced voltage (E) = 0

∴ T = k ϕ Ia1

⇒ T = kIa1 {∴ ϕ = Constant}

E = Vt - Ia1Ra

Ia1 = (100 - 0)/0.05 = 2000 A

Therefore,

K = (T/Ia1) = 500/2000 = 0.25 

At no -load condition, load torque (TL) = 0

E2Ia2 = T ω 

⇒ I_a2 = 0

So, induced voltage

E₂ = k ϕ ω₂ = k ω₂ = 150

and k = 0.25

ω = E₂/k = 150/0.25 = 600 rad/sec

Therefore, no load speed of the motor is 600 rad/sec.

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

Steady-state error for the Unit step input is,

\({E_{ss}} = \frac{A}{{1 + {k_p}}}\)

\({k_p} = \mathop {\lim }\limits_{s \to 0} G(s)~H(s)\)

\( {k_p} = \mathop {\lim }\limits_{s \to 0} \frac{1}{(s+1)(s+2)}\)

= \(\frac{1}{2}\)

\({E_{ss}} = \frac{1}{{1 + \frac{1}{2}}} = \frac{2}{3} = 0.67\)

Given a transformer with toroidal care of permeability μ.

radius of the core = r

Mean radius of the core = R

cross sectional area of the core (A) = π r²

mean length of core (L) = 2πR

\({V_p} = {N_p}\frac{{d\phi }}{{dt}} = {N_p}\frac{d}{{dt}}\left( {\frac{{{N_p}{i_p}}}{\mathbb{R}}} \right)\)

\( = \frac{{N_p^2}}{\mathbb{R}}\frac{d}{{dt}}\left( {I\sin \omega t} \right)\)

\(= \frac{{N_p^2}}{\mathbb{R}} \times \omega I\cos \omega t\)

\(\Rightarrow V\cos \omega t = \frac{{N_p^2}}{\mathbb{R}}\omega I\cos \omega t\)

we know that, \(\mathbb{R} = \frac{l}{{\mu A}}\)

\( \Rightarrow V = \frac{{N_p^2\omega I}}{{\left( {\frac{L}{{\mu A}}} \right)}}\)

\( \Rightarrow V = \frac{{N_p^2;\mu A;\omega I}}{L} = \frac{{N_p^2\omega I\mu \pi {r^2}}}{{2\pi R}}\)

\(\Rightarrow R = \frac{{N_p^2WI\mu {r^2}}}{{2V}}\)

Given that,

meter current (I_m) = 1A

Internal resistance (R_m) = 50 mΩ

Inductance (L_m) = 0.1 mH

required meter current (I) = 10 A

\(\frac{{{I_m}}}{{{I_{sh}}}} = \frac{{{Z_{sh}}}}{{{Z_m}}} = \sqrt {\frac{{R_{sh}^2 + X_{sh}^2}}{{R_m^2 + X_m^2}}} \)

\(\Rightarrow \frac{{{I_m}}}{{{I_{sh}}}} = \sqrt {\frac{{R_{sh}^2 + {{\left( {\omega {L_{sh}}} \right)}^2}}}{{R_m^2 + {{\left( {\omega {L_m}} \right)}^2}}}} \)

\(\Rightarrow \frac{{{I_m}}}{{{I_{sh}}}} = \frac{{\sqrt[{{R_{sh}}}]{{1 + {{\left( {\frac{{\omega {L_{sh}}}}{{{R_{sh}}}}} \right)}^2}}}}}{{{R_m}\sqrt {1 + {{\left( {\frac{{\omega {L_m}}}{{{R_m}}}} \right)}^2}} }}\)

\(\Rightarrow \frac{{{I_m}}}{{{I_{sh}}}} = \frac{{{R_{sh}}}}{{{R_m}}}.\sqrt {\frac{{1 + {{\left( {\omega {\tau _{sh}}} \right)}^2}}}{{1 + {{\left( {\omega {\tau _m}} \right)}^2}}}} \)

In order to operate the meter at all frequencies,

\({\tau _{sh}} = {\tau _m}\)

\(\Rightarrow \frac{{{I_m}}}{{{I_{sh}}}} = \frac{{{R_{sh}}}}{{{R_m}}}\)

\( \Rightarrow {R_{sh}} = \frac{{50 \times {{10}^{ - 3}} \times 1}}{9} = 5.55;m{\rm{\Omega }}\)

Time constant of shunt coil,

\({\tau _{sh}} = {\tau _m} = \frac{{{L_m}}}{{{R_m}}} = \frac{{0.1}}{{50}} = 2;msec.\)

In line to line fault,

fault current with fault impedance z_f is

\({I_{{f_1}}} = \frac{{\sqrt 3 {E_a}}}{{{z_1} + {z_2} + {z_f}}}\)

fault current with zero fault impedance is,

\({I_f} = \frac{{\sqrt 3 {E_a}}}{{{z_1} + {z_2}}}\)

Given that, \({I_{{f_1}}} = k{I_f}\)

\(\Rightarrow \frac{{\sqrt 3 {E_a}}}{{{z_1} + {z_2} + {z_f}}} = k{I_{{f_1}}} = k\frac{{\sqrt 3 {E_a}}}{{{z_1} + {z_2}}}\)

⇒ z₁ + z₂ = k (z₁ + z₂ + z_f)

⇒ z₁ + z₂ - k z₁ - k z₂ = k z_f

\(\Rightarrow {z_f} = \frac{{\left( {{z_1} + {z_2}} \right)\left( {1 - k} \right)}}{k}\)

At Generator-2, load demand is 20 pu.

G₂ is supplying 15 pu and remaining 5 pu is supplied by G₁.

\({P_s} = \left| {\frac{{{V_s}{V_R}}}{{{X_L}}}} \right|\sin \delta\)

\(\Rightarrow 5 = \frac{{1 \times 1}}{{0.1}}sin\delta\)

⇒ δ = 30°

\({Q_s} = \frac{{V_s^2}}{{{X_L}}} - \frac{{{V_s}{V_R}}}{{{X_L}}}cos\delta \)

\(\Rightarrow {Q_s} = \frac{{{1^2}}}{{0.1}} - \frac{{1 \times 1}}{{0.1}}cos30^\circ = 1.339;pu\)

\({Q_R} = \left| {\frac{{{V_s}{V_R}}}{{{X_L}}}} \right|cos\delta - \frac{{V_R^2}}{{{X_L}}}\)

\(= \frac{{1 \times 1}}{{0.1}}cos30^\circ - \frac{{{1^2}}}{{0.1}} = - 1.339pu\)

Q_loss = Q_s - Q_R

= 2.679 pu

At Generator 1,

Q_G1 = Q_load + Q_S

= 5 + 1.339 = 6.339 pu

QG2 = Qload + QR

= 10 + 1.339 = 11.339 pu

P = 1.4 sinδ + 0.15 sin 2δ

P = 0.8

⇒ f(δ) = 1.4 sinδ + 0.15 sin 2δ – 0.8

f^'(δ) = 1.4 cosδ + 0.3 cos2δ

Given that, δ₀ = 30°

According to Newton-Raphson method,

\({\delta _1} = {\delta _0} - \frac{{f\left( {{\delta _0}} \right)}}{{f'\left( {{\delta _0}} \right)}}\)

f(δ₀) = 1.4 sin30° + 0.15 sin 60° - 0.8 = 0.0299

f^’(δ₀) = 1.4 cos30° + 0.3 cos 60° = 1.3624

\({\delta _1} = {\delta _0} - \frac{{f\left( {{\delta _0}} \right)}}{{f'\left( {{\delta _0}} \right)}}\)

\(= 30^\circ - \frac{{0.0299}}{{1.3624}}\left( {\frac{{180}}{\pi }} \right) = 28.74^\circ \)

Concept:

We replace the capacitor with an initial voltage to its s-domain equivalent as shown:

Similarly, an inductor with an initial current of i(0^-) is represented as:

Analysis:

Since for the given circuit, the initial voltage and current across the capacitor is zero, the equivalent circuit will have v_c(0^-) = 0 and i(0^-) = 0, and the circuit will be redrawn as:

\(I\left( s \right) = \frac{{\frac{5}{s}}}{{s + \frac{1}{5}}}\)

\(I\left( s \right) = \frac{5}{{{s^2} + 1}}\)

Taking the inverse Laplace transform, we get:

i(t) = 5 sint

The capacitor voltage will now be:

\({V_C}\left( t \right) = \frac{1}{C}\mathop \smallint \limits_0^t 5\sin tdt\)

\(= \frac{1}{1}\mathop \smallint \limits_0^t 5\sin tdt\)

V_C(t) = 5(1 – cos t)

The current and voltage are now plotted as shown:

Given circuit diagram:

For an infinite ladder, we will replace the common repeating term with the input impedance 

Impedance j5 Ω and -j1 Ω are in series. Therefore the impedance will be j4 Ω 

Now the equivalent impedance Z_eq will be

\({Z_{eq}} = j9 + \frac{{{z_{eq}}\left( {j4} \right)}}{{{z_{eq}} + j4}}\)

\(\Rightarrow \left( {{z_{eq}} - j9} \right)\left( {{z_{eq}} + j4} \right) = {z_{eq}}\left( {j4} \right)\)

\(\Rightarrow z_{eq}^2 + 36 - j5;{z_{eq}} = {z_{eq}}\left( {j4} \right)\)

\(\Rightarrow z_{eq}^2 - j9{z_{eq}} + 36 = 0\)

\(\Rightarrow \left( {{z_{eq}} - j12} \right)\left( {{z_{eq}} + j3} \right) = 0\)

\(\Rightarrow {z_{eq}} = j12;{\rm{Ω ;}}\left( {or} \right) - j3\)

Z_eq can’t be negative as the given circuit is inductive in nature.

⇒ Z_eq = j12 Ω

\(\frac{{d{x_1}\left( t \right)}}{{dt}} = \mathop {\mathop x\nolimits_1 }\limits^. \left( t \right) = {x_2}\left( t \right) - {x_1}\left( t \right)\)

\(\frac{{d{x_2}\left( t \right)}}{{dt}} = \mathop {\mathop x\nolimits_2 }\limits^. \left( t \right) = {x_1}\left( t \right) - {x_2}\left( t \right)\)

\(\left[ {\begin{array}{*{20}{c}} {\mathop {\mathop x\nolimits_1 }\limits^. \left( t \right)}\ {\mathop {\mathop x\nolimits_2 }\limits^. \left( t \right)} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 1}&1\ 1&{ - 1} \end{array}} \right];\left[ {\begin{array}{*{20}{c}} {{x_1}\left( t \right)}\ {{x_2}\left( t \right)} \end{array}} \right]\)

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

\(\phi \left( t \right) = {e^{at}} = {L^{ - 1}}\left[ {{{\left( {sI - A} \right)}^{ - 1}}} \right]\)

\(sI - A = \left[ {\begin{array}{*{20}{c}} s&0\ 0&s \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} { - 1}&1\ 1&{ - 1} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {s + 1}&{ - 1}\ { - 1}&{s + 1} \end{array}} \right]\)

\({\left[ {sI - A} \right]^{ - 1}} = \frac{1}{{{{\left( {s + 1} \right)}^2} - 1}}\left[ {\begin{array}{*{20}{c}} {s + 1}&1\ 1&{s + 1} \end{array}} \right]\)

\(= \frac{1}{{s\left( {s + 2} \right)}}\left[ {\begin{array}{*{20}{c}} {s + 1}&1\ 1&{s + 1} \end{array}} \right]\)

\(= \left[ {\begin{array}{*{20}{c}} {\frac{{s + 1}}{{s\left( {s + 2} \right)}}}&{\frac{1}{{s\left( {s + 2} \right)}}}\ {\frac{1}{{s\left( {s + 2} \right)}}}&{\frac{{s + 1}}{{s\left( {s + 2} \right)}}} \end{array}} \right]\)

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

\(\phi \left( t \right) = {L^{ - 1}}\left[ {{{\left( {sI - A} \right)}^{ - 1}}} \right]\)

\(= \frac{1}{2}\left[ {\begin{array}{*{20}{c}} {1 + {e^{ - 2t}}}&{1 - {e^{ - 2t}}}\ {1 - {e^{ - 2t}}}&{1 + {e^{ - 2t}}} \end{array}} \right]\)

\(x\left( t \right) = \phi \left( t \right);x\left( 0 \right) = \frac{1}{2}\left[ {\begin{array}{*{20}{c}} {1 + {e^{ - 2t}}}&{1 - {e^{ - 2t}}}\ {1 - {e^{ - 2t}}}&{1 + {e^{ - 2t}}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}\left( 0 \right)}\ {{x_2}\left( 0 \right)} \end{array}} \right]\)

\(x\left( t \right) = \frac{1}{2};\left[ {\begin{array}{*{20}{c}} {{x_1}\left( 0 \right) + {x_2}\left( 0 \right) + {e^{ - 2t}}\left( {{x_1}\left( 0 \right) - {x_2}\left( 0 \right)} \right)}\ {{x_1}\left( 0 \right) + {x_2}\left( 0 \right) + {e^{ - 2t}}\left( {{x_2}\left( 0 \right) - {x_1}\left( 0 \right)} \right)} \end{array}} \right]\)

\({x_{1f}} = \mathop {{\rm{lt}}}\limits_{t \to \infty } {x_1}\left( t \right) = \frac{1}{2}\left( {{x_1}\left( 0 \right) + {x_2}\left( 0 \right)} \right)\)

\({x_{2f}} = \mathop {{\rm{lt}}}\limits_{t \to \infty } {x_2}\left( t \right) = \frac{1}{2}\left( {{x_1}\left( 0 \right) + {x_2}\left( 0 \right)} \right)\)

Given that, x₁(0) < x₂ (0) < ∞

⇒ x_1f = x_2f < ∞

Concept:

According to RH stability criteria there are two necessary conditions:

First is there should not be any sign changes in the first column of the tabulation and Second Is there should not be any missing coefficients in the equation 1 + G(s) H(s).

The RH criteria can be solved as following:

\({s^3} - - - {\rm{;;;}}{{\rm{a}}_0} \searrow {\rm{;;;;;}} \swarrow {{\rm{a}}_2}{\rm{;;;;;;}}{{\rm{a}}_4}\)

\({s^2} - - - {\rm{;;;}}{{\rm{a}}_1}{\rm{;}} \nearrow {\rm{;;;}} \nwarrow {{\rm{a}}_3}{\rm{;;;;;;;}}0\)

\({s^1} - - - {\rm{;;;}}\frac{{{{\rm{a}}_1}{{\rm{a}}_2} - {{\rm{a}}_0}{{\rm{a}}_3}}}{{{{\rm{a}}_1}}}{\rm{;}} = {\rm{A;;;;;;;;;;}}0{\rm{;}}\)

\({s^0} - - - {\rm{;;;}}\frac{{{\rm{A}}{{\rm{a}}_3}}}{{\rm{A}}} = {{\rm{a}}_3}\)

Calculation:

From R-H stability criteria

S⁷ + S⁶ + 7 S ⁵ + 14S⁴ + 31S³ + 73S² + 25S + 200

\(\left. {\begin{array}{*{20}{c}} {{S^7}}\ {{S^6}}\ {\begin{array}{*{20}{c}} {{S^5}}\ {{S^4}}\ {\begin{array}{*{20}{c}} {{S^3}}\ {{S^2}}\ {\begin{array}{*{20}{c}} {{S^1}}\ {{S^0}} \end{array}} \end{array}} \end{array}} \end{array}} \right|\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} 1\ 1\ { - 7} \end{array}}&{\begin{array}{*{20}{c}} 7\ {14}\ { - 42} \end{array}}&{\begin{array}{*{20}{c}} {31}\ {73}\ { - 175} \end{array}}&{\begin{array}{*{20}{c}} {25}\ {200}\ 0 \end{array}}\ 8&{48}&{200}&{}\ {\begin{array}{*{20}{c}} {32}\ {24}\ {\begin{array}{*{20}{c}} { - 170.66}\ {200} \end{array}} \end{array}}&{\begin{array}{*{20}{c}} {96}\ {200}\ {\begin{array}{*{20}{c}} 0\ 0 \end{array}} \end{array}}&{\begin{array}{*{20}{c}} 0\ 0\ {\begin{array}{*{20}{c}} 0\ 0 \end{array}} \end{array}}&{\begin{array}{*{20}{c}} {}\ {}\ {\begin{array}{*{20}{c}} {}\ {} \end{array}} \end{array}} \end{array}\)

Number of sign changes = 4

Number of left-hand poles = 7 – 4 = 3

\(f\left( z \right) = \frac{{{z^2}}}{{{{\left( {{z^2} - 3z + 2} \right)}^2}}}\)

\( = \frac{{{z^2}}}{{{{\left( {z - 1} \right)}^2}{{\left( {z - 2} \right)}^2}}}\)

Poles of f(z) are z = 1, 2

Both Z = 1 and 2 are inside the curve |z| = 4

\(\mathop \smallint \limits_C f\left( z \right)dz = 2\pi i\) [sum of residues at z = 1 and z = 2]

If f has a pole of order n at z = a, then Residue of f(z) at z = a is

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

Residue at z = 1,

\(\mathop {{\rm{It}}}\limits_{z \to 1} \frac{d}{{dz}}\left[ {\frac{{{z^2}.;{{\left( {z - 1} \right)}^2}}}{{{{\left( {z - 1} \right)}^2}{{\left( {z - 2} \right)}^2}}}} \right]\)

\( = \frac{{{{\left( {1 - 2} \right)}^2}\left( 2 \right) - {{\left( 1 \right)}^2}\left( {2\left( { - 1} \right)} \right)}}{{{{\left( {1 - 2} \right)}^4}}} = 4\)

Residue at z = 2,

\(= \mathop {{\rm{lt}}}\limits_{z \to 2} \frac{d}{{dz}}\left[ {\frac{{{z^2}}}{{{{\left( {z - 1} \right)}^2}}}} \right]\)

\(= ;\mathop {{\rm{lt}}}\limits_{z \to 2} \left[ {\frac{{{{\left( {z - 1} \right)}^2}\left( {2z} \right) - {z^2}\left( {2\left( {z - 1} \right)} \right)}}{{{{\left( {z - 1} \right)}^4}}}} \right]\)

\(= \frac{{{{\left( {2 - 1} \right)}^2} - 4\left( {2\left( 1 \right)} \right)}}{{{{\left( {2 - 1} \right)}^4}}} = - 4;\)

\(\mathop \smallint \limits_C f\left( z \right)dz = 2\pi i\left( {4 - 4} \right) = 0\)

From the given circuit diagram,

\({D_A} = \overline {{Q_B}{Q_c}} ,;{D_B} = {Q_A},;{D_C} = {Q_B}\)

The output Q_A, Q_B, Q_C is repeating for every 5 clock pulses.

Hence frequency will be divided by 5.

F = m₀ + m₂ + m₃ + m₅ + m₇

m₁ is a don’t care. 

\(F = \bar A + C\)

Concept:

Energy of the discrete time signal x(n) is,

                         \({E} = \mathop \sum \limits_{n = - \infty }^\infty {\left| {{x}\left( n \right)} \right|^2}\)

Energy of the continuous time signal x(t) is,

                         \(E = \mathop \smallint \limits_{ - \infty }^\infty {\left( {x\left( t \right)} \right)^2}dt\)

Explanation:

\({x_1}\left( t \right) = \left{ {\begin{array}{*{20}{c}} {\left| t \right|}&{ - 1 \le t \le 1}\ 0&{otherwise} \end{array}} \right.\)

After sampling the above signal for T = 0.25 sec, the discrete signal will be

Energy of \({x_1}\left( n \right) = {E_1} = \mathop \sum \limits_{n = - \infty }^\infty {\left| {{x_1}\left( n \right)} \right|^2}\)

= 0² + 2 [0.25² + 0.5² + 0.75² + 1²]

= 3.75

\({x_2}\left( t \right) = \left{ {\begin{array}{*{20}{c}} {1 - \left| t \right|}&{ - 1 \le t \le 1}\ 0&{otherwise} \end{array}} \right.\)

After sampling the above signal for T = 0.25 sec, the discrete signal will be

Energy of \({x_2}\left( n \right) = {E_2} = \mathop \sum \limits_{n = - \infty }^\infty {\left| {{x_2}\left( n \right)} \right|^2}\)

= 1² + 2 [0.75² + 0.5² + 0.25²]

= 2.75

⇒ E₁ > E₂

Concept:

The Energy of the signal is given as,

\(E = \mathop \smallint \limits_{ - \infty }^\infty {\left( {x\left( t \right)} \right)^2}dt\)

Explanation:

x(t) = [(t-1) u (t-1)] – [(t-2) u (t-2)] – [(t-3) u (t-3)] + [(t -4) u (t-4)]

We know that, 

                       r(t) = t u (t)

x(t) = r(t - 1) - r(t - 2) - r(t - 3) + r(t - 4)

Energy of the signal, x(t) is

\(E = \mathop \smallint \limits_{ - \infty }^\infty {\left( {x\left( t \right)} \right)^2}dt\)

\(= \mathop \smallint \limits_1^2 {\left( {t - 1} \right)^2}dt + \mathop \smallint \limits_2^3 {1^2}dt + \mathop \smallint \limits_3^4 {\left( { - t + 4} \right)^2}dt\)

\(= \mathop \smallint \limits_1^2 \left( {{t^2} - 2t + 1} \right)dt + \mathop \smallint \limits_2^3 1;dt + \mathop \smallint \limits_3^4 \left( {{t^2} + 16 - 8t} \right)dt\)

\(= \left[ {\frac{{{t^3}}}{3} - \frac{{2{t^2}}}{2} + t} \right]_1^2 + \left[ t \right]_2^3 + \left[ {\frac{{{t^3}}}{3} + 16t - \frac{{8{t^2}}}{2}} \right]_3^4\)

\(= \left[ {\frac{1}{3} - 1 + 1} \right] + \left[ {3 - 2} \right] + \left[ {\frac{{37}}{3} + 16 - 28} \right]\)

\( = \frac{1}{3} + 1 + \frac{{37}}{3} - 12\)

\(= \frac{{1 + 3 + 37 - 36}}{3} = \frac{5}{3}\)

Concept:

\(e^{-at}\rm \overset{F.T.}{\leftrightarrow}\frac{1}{a+j\omega}\)

Frequency differentiation property

\(tx(t)\rm \overset{F.T.}{\leftrightarrow}j\dfrac{d}{d\omega}x(\omega)\)

Explanation:

\(X\left( \omega \right) = \frac{1}{{{{\left( {10 + j\omega } \right)}^2}}}\)

By applying inverse Fourier transform,

⇒ x(t) = t e^-10t u(t)

\(x\left( t \right) = {e^{ - at}}u\left( t \right) \leftrightarrow \frac{1}{{a + j\omega }}\)

By using frequency differentiation property,

\(t;{x_1}\left( t \right) = t{e^{ - at}}u\left( t \right) \leftrightarrow j\frac{d}{{d\omega }} \times \left( \omega \right) = \frac{1}{{{{\left( {a + j\omega } \right)}^2}}}\)

\(\Rightarrow t{e^{ - 10t}}u\left( t \right) \leftrightarrow \frac{1}{{{{\left( {10 + j\omega } \right)}^2}}}\)

x(t) = te^-10t u(t)           

Taking logarithm (to base e ) on both sides,                              

\(\left| {lnx\left( t \right)} \right| = \left| {ln\left[ {t{e^{ - 10t}}u\left( t \right)} \right]} \right|\)

\(at;t = 1, = \left| {ln\left[ {{e^{ - 10}}} \right]} \right| = \left| { - 10} \right| = 10\)

This is an active network with an independent source.

∴ The Thevenin voltage is obtained as:

\({R_{th}} = \frac{{{V_{oc}}}}{{{I_{sc}}}}\)

V_OC = Open circuit voltage across the terminal

I_SC = Short circuit current across the terminal

Apply kVL in loop 1

10.7 – 10 I_B – 0.7 – I_E = 0

10 – 10 I_B – (β +1) I_B = 0

\({I_B} = \frac{{10}}{{10 + 101}} = \frac{{10}}{{111}}\)

\({I_E} = \frac{{101 \times 10}}{{111}} = \frac{{1010}}{{111}}\)

\({V_{oc}} = {I_E} \times 1k\Omega = \frac{{1010}}{{111}}V;\)

10.7 – 10 I_B – 0.7 = 0

\({I_B} = \frac{{10}}{{10}} = 1mA\)

I_E = (1+β) I_B = 101 mA

I_E = I_SC = 101 mA

\({R_{th}} = \frac{{1010}}{{111 \times 101}} = \frac{{10}}{{111}}k{\rm{\Omega }} = 90.09;{\rm{\Omega }}\)

\(\mathop \oint \limits_C^; \left( {{y^2} + 2yx} \right)dx + \left( {2xy + {x^2}} \right)dy\)

According to Green’s theorem

\(\mathop \oint \limits_C^; Mdx + Ndy = \mathop \int!!!\int \limits_S^; \left( {\frac{{\partial N}}{{\partial x}} - \frac{{\partial M}}{{\partial y}}} \right)dxdy\)

Here, M = y² + 2yx

N = 2xy + x²

\(\frac{{\partial M}}{{\partial y}} = 2y + 2x\)

\(\frac{{\partial N}}{{\partial x}} = 2y + 2x\)

\(\mathop \oint \limits_C^; \left( {{y^2} + 2yx} \right)dx\left( {2xy + {y^2}} \right)dy \)

\(= \mathop \int!!!\int \limits_C^; \left( {2y + 2x - 2y - 2x} \right)dxdy = 0\)

f(x) = 3x³ – 7x² + 5x + 6

f’(x) = 9x² – 14x + 5 = 0

\(\Rightarrow x = 1,\frac{5}{9}\)

f”(x) = 18x-14

At x = 1, f”(x) = 18 – 14 = 4 > 0

\(At;x = \frac{5}{9},f''\left( x \right) = 18\left( {\frac{5}{9}} \right) - 14 = - 4 < 0;\)

\(f\left( x \right)at;x = \frac{5}{9}\) is,

\(f\left( x \right) = 3{\left( {\frac{5}{9}} \right)^3} - 7{\left( {\frac{5}{9}} \right)^2} + 5\left( {\frac{5}{9}} \right) + 6 = 7.13\)

f(x) at x = 0 is

f(x) = 3(0)³ – 7(0)² + 5(0) + 6 = 6

f(x) at x = 2 is,

f(x) = 3(2)³ – 7(2)² + 5(2) + 6 = 6

= 24 – 28 + 10 + 6 = 12

The maximum value of f(x) over the interval [0,2] is 12

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

B = A³ – A² - 4A + 5I

|A - λI| = 0

\(\Rightarrow \left| {\begin{array}{*{20}{c}} {1 - \lambda }&0&{ - 1}\ { - 1}&{2 - \lambda }&0\ 0&0&{ - 2 - \lambda } \end{array}} \right| = 0\)

⇒ (1- λ) (2 - λ) (-2 - λ) – 1 (0) = 0

⇒ λ = 1, λ = 2, λ = -2

If λ is an Eigen value of A, then λⁿ will be an Eigen value of Aⁿ.

If λ is an Eigen value of A, then k λ will be an Eigen value of kA where k is a scalar.

Therefore,

Eigen values of A are 1, 2, -2

Eigen values of A² are 1,4, 4

Eigen values of A³ are 1, 8, -8

Eigen values of 4A are 4, 8, -8

Eigen values of 5I are 5, 5, 5

B = A3 – A2 - 4A + 5I

λ_1B = λ1³– λ12– 4λ1 + 5 = 1 - 1 - 4 + 5 = 1

λ2B = λ23– λ22– 4λ2 + 5 = 8 - 4 - 8 + 5 = 1

λ3B = λ33– λ32– 4λ3 + 5 = -8 - 4 + 8 + 5 = 1

Eigen values of B are 1, 1, 1

Determinant of matrix = product of Eigen values = 1

We know that,

Capacitance if a parallel plate capacitor is,

\(C = \frac{{{\epsilon_0}A}}{d}\)

\({C_1} = \frac{{{\epsilon_0}{A_1}}}{{{d_1}}} = 60;pF\)

When a dielectric slab whose thickness is half the distance between placed on one of the plates covering, the capacitance becomes = 86 pF

C₂ = 86 pF

\({C_2} = \frac{{{\epsilon_0}{A_1}}}{{\frac{{{d_1}}}{2}}},\frac{{{\epsilon_0}{\epsilon_r}{A_1}}}{{\frac{{{d_1}}}{2}}}\)

\(= \frac{{2{\epsilon_0}{A_1}}}{{{d_1}}},\frac{{2{\epsilon_0}{A_1}{\epsilon_r}}}{{{d_1}}}\)

= 120, 120 ϵ_r

\({C_2} = \frac{{120 \times 120 \times {\epsilon_r}}}{{120 + 120{\epsilon_r}}}\)

\(\Rightarrow 87 = \frac{{120 \times 120{\epsilon_r}}}{{120\left( {1 + {\epsilon_r}} \right)}}\)

⇒ 86 + 86 ϵ_r = 120 ϵ_r

\(\Rightarrow {\epsilon_r} = \frac{{86}}{{34}} = 2.52\)

\(G\left( s \right)H\left( s \right) = \frac{k}{{{{\left( {s + 1} \right)}^2}\left( {s + 2} \right)}}\)

From the given figure

steady state value of y(t) = 0.8

By using final value theorem,

steady state y(t) value of \(= \mathop {{\rm{lt}}}\limits_{s \to 0} s;y\left( s \right)\)

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

\(\frac{{Y\left( s \right)}}{{R\left( s \right)}} = \frac{k}{{\frac{{{{\left( {s + 1} \right)}^2}\left( {s + 2} \right)}}{{1 + \frac{k}{{{{\left( {s + 1} \right)}^2}\left( {s + 2} \right)}}}}}}\)

\(\Rightarrow \frac{{Y\left( s \right)}}{{R\left( s \right)}} = \frac{k}{{{{\left( {s + 1} \right)}^2}\left( {s + 2} \right) + k}}\)

As the input is unit step function,

R(s) = 1/s

\(\Rightarrow Y\left( s \right) = \frac{k}{{s\left( {{{\left( {s + 1} \right)}^2}\left( {s + 2} \right) + k} \right)}}\)

\(\mathop {{\rm{lt}}}\limits_{s \to 0} s;Y\left( s \right) = 0.8\)

\(\Rightarrow \frac{k}{{\left( 1 \right)\left( 2 \right) + k}} = 0.8\)

⇒ k = 1.6 + 0.8k ⇒ k = 8

V_an = 100 ∠0° V

V_bn = 100 ∠-120° V

V_cn = 100 ∠-240° V

I_n = 0

Apply KCL at neutral position,

I_a + I_b + I_c – I_n = 0

\(\Rightarrow \frac{{100\angle 0^\circ }}{R} + \frac{{100\angle - 120^\circ }}{{j10}} + \frac{{100\angle - 240^\circ }}{{ - j10}} = 0\)

\(\Rightarrow \frac{{100\angle 0^\circ }}{R} + 10\angle - 210 + 10\angle - 150 = 0\)

\(\Rightarrow \frac{{100}}{R} + \left( { - 8.66 + j5} \right) + \left( { - 8.66 - j5} \right) = 0\)

\(\Rightarrow R = \frac{{100}}{{2\left( {8.66} \right)}} = 5.77;{\rm{\Omega }}\)

Concept:

The power triangle is as shown below.

P = Active power (or) Real power (or) Average power in W = Vrms Irms cos ϕ

Q = Reactive power in VAR = Vrms Irms sin ϕ

S = Apparent power in VA = Vrms Irms

If V(t) = A₁ + B₁ cos ω₁t + C₁ cos ω₂ t ....

I(t) = A2 + B2 cos (ω1t + ϕ₁)  + C2 cos (ω2t + ϕ₂)  ....

Average power P = A₁A₂ + B₁B₂ cos ϕ1 + C₁C₂ cos ϕ₂ ...

Calculation:

Given,

V(t) = 5 – 10 cos (ωt + 60°) V

i(t) = 5 + X cos (ωt) A

ω = 100 π radian/sec

P_avg = V_rms I_rms cosϕ = 0

\(\Rightarrow \left( 5 \right)\left( 5 \right) - \frac{{10}}{{\sqrt 2 }} \times \frac{X}{{\sqrt 2 }} \times \cos 60^\circ = 0\)

\(\Rightarrow 25 = \frac{{10X}}{2} \times \frac{1}{2}\)

⇒ X = 10 A

Concept:

The average output voltage of a single-phase semi-converter is given by

\({V_0} = \frac{{{V_m}}}{2\pi }\left( {3 + \cos \alpha } \right)\)

Vm is the maximum value of supply voltage

α is the firing angle

Calculation:

Average output voltage,

\({V_o} = \frac{{{V_m}}}{{2\pi }}\left( {3 + \cos \alpha } \right)\)

Power transfer to the 80 V battery,

P = E I_o = 1600 W

\(\Rightarrow {I_O} = \frac{{1600}}{{80}} = 20A\)

V_m = 80 π

\({V_o} = \frac{{{V_m}}}{{2\pi }}\left( {3 + cos\alpha } \right)\)

\(= \frac{{80\pi }}{{2\pi }}\left( {3 + cos\alpha } \right)\)

The relation between Vo and E is given by,   

V_o = E + I_oR

⇒ 120 + 40 cos α = 80 + 20 (2)

⇒ cos α = 0 ⇒ α = 90°

Method 1:

Input DC voltage (V_s) = 100 V

Load resistance (R) = 1 Ω

V_o = 50 V

\(\Rightarrow {I_o} = \frac{{{V_O}}}{R} = 50;A\)

Input power = output power

⇒ V_sI_s = V_oI_o

⇒ (100) (I_s) = (50) (50)

⇒ I_s = 25 A

Average values of \({i_{s1}} = \frac{{{I_s}}}{2} = 12.5;A\)

Method 2:

In the given circuit two buck converters are connected in parallel.

Duty cycle ratio D = Ton / T = 0.5 / 1 = 0.5

Case-I: when S1 → ON, S2 → open

Output voltage be V01

Case-II: S2 → ON, S1 → open

Output voltage be V02

In both cases, the output voltage is the same given as

V01 = V02 = V0 = DVdc = 0.5 × 100 = 50 V

\( {I_0} = \frac{{{V_0}}}{R} = \frac{{50}}{1} = 50A\)

In a buck converter average value of the inductor current is the same as that of the load current, but here two buck converters are connected in parallel so that the current is divided equally given as

\( {I_{L1}} = {I_{L2}}= \frac{{50}}{2} = 25A\)

so the average value of the switch current is given as

Is1(avg) = DIL1(avg) = 0.5 × 25 = 12.5 A

Is2(avg) = DIL2(avg) = 0.5 × 25 = 12.5 A

∴ Average value of switch S1 current  = 12.5 A

Concept:

Copper loss:

A loss in a transformer that takes place in the winding resistance of a transformer is known as the copper loss. These are variable losses.

Copper loss = i2R

Iron loss:

Iron losses = Eddy current loss + hysteresis loss

Eddy current loss = Ket2f2B2

Hysteresis loss = KhfBx

These are constant losses.

Efficiency:

The efficiency of the Transformer \((\eta ) = \frac{{XS\cos \phi }}{{XS\cos \phi + {P_i} + {X^2}{P_e}}}\)

Where, X = Fraction of load

S = Apparent power in kVA

Wi = Iron losses

Wcu = Copper losses

Explanation:

Primary resistance = 0.3 Ω/phase

Secondary resistance = 0.02 Ω/phase

Iron losses = 10 kW    

Primary line current \(\left( {{I_1}} \right) = \frac{{900 \times {{10}^3}}}{{\sqrt 3 \times 3 \times {{10}^3}}} = 173.2;A\)

Primary line current \(\left( {{I_{ph1}}} \right) = \frac{{{I_1}}}{{\sqrt 3 }} = 100;A\)

Secondary line current \(\left( {{I_2}} \right) = \frac{{900 \times {{10}^3}}}{{\sqrt 3 \times \sqrt 3 \times {{10}^3}}} = 300;A\)

Secondary phase current = I_ph2 = I₂ = 300 A

Copper losses on HV side \(= 3 \times I_{ph1}^2{R_1}\)

= 3 × (100)² × (0.3) = 9 kW

Copper losses on LV side \(= 3 \times I_{ph2}^2{R_1}\)

= 3 × (300)² × (0.02) = 5.4 kW

Efficiency \(\left( \eta \right) = \frac{{kVA}}{{kVA + losses}}\)

\(= \frac{{900 \times {{10}^3}}}{{900 \times {{10}^3} + 9 \times {{10}^3} + 5.4 \times {{10}^3} + 10 \times {{10}^3}}}\)

\(= \frac{{900}}{{924.4}} \times 100\)

= 97.365%

Voltage (V) = 200 V

\({I_{a1}} = {I_L} = 10A\)

where  I_a = Armature current,

I_L = Load current 

for DC series motor  Ia = IL

Series resistance (R_s) = 1Ω

Speed (N₁) = 1000 rpm

T₂ = 1.44 T₁

E_b1 = V – I_a1R_s = 200 – (10)(1) = 190 V

E_b = Back emf,

R_s= Total resistance,

We know that, in a series motor

\(T ∝ ϕ {I_a} ∝ I_a^2\)

ϕ ∝ I_a  ( ∵ magnetic circuit is linear)

where,

ϕ = Magnetic flux,

T= torque

\(\Rightarrow \frac{{{T_1}}}{{{T_2}}} = \frac{{I_{a1}^2}}{{I_{a2}^2}}\)

⇒ I_a2 = 12 A

E_b2 = V – I_a2 R_s = 200 – 12(1) = 188 V

E_b ∝ Nϕ ∝ NI_a

\( \Rightarrow \frac{{{N_1}}}{{{N_2}}} = \frac{{{E_{b1}}}}{{{E_{b2}}}} \times \frac{{{I_{a2}}}}{{{I_{a1}}}}\)

\( \Rightarrow {N_2} = 1000 \times \left( {\frac{{188}}{{190}}} \right) \times \left( {\frac{{10}}{{12}}} \right)\)

\(\Rightarrow {N_2} = 824.5;rpm\)

Duty Cycle is defined as:

\(Duty;Cycle = \frac{{{T_{ON}}}}{{{T_{ON}} + {T_{OFF}}}}\)

T_ON = D × T

Putting on the respective values:

\({T_{ON}} = 0.4 \times \frac{1}{f} = 0.4 \times \frac{1}{{5 \times {{10}^3}}} = 0.8;msec\)

T_OFF = T – T_ON = 0.2 – 0.08 = 0.12 msec

CASE I : When Switch is ON

\({V_L} = L\frac{{di}}{{dt}} = 50\)

\(\frac{{di}}{{dt}} = \frac{{50}}{L} = \frac{{50}}{{5 \times {{10}^{ - 3}}}}\)

\(\frac{{{\rm{\Delta }}I}}{{DT}} = \frac{{50}}{{5 \times {{10}^{ - 3}}}}\)

\({\rm{\Delta }}I = \frac{{50}}{{5 \times {{10}^{ - 3}}}} \times DT\)

\(= \frac{{50}}{{5 \times {{10}^{ - 3}}}} \times 0.4 \times \frac{1}{{5 \times {{10}^3}}}\)

ΔI = 0.8 A

CASE II: Checking whether we have continuous or discontinuous conduction by finding T_{OFF new}

During T_OFF → Switch is open and the diode is ON.

\({V_L} = L\frac{{di}}{{dt}} = - 100\)

\(\frac{{{\rm{\Delta }}I}}{{{T_{OFF}}}} = \left( { - \frac{{100}}{L}} \right)\)

\({T_{OFF}} = - \frac{{L \times {\rm{\Delta }}{I_L}}}{{100}} = \frac{{ - 5 \times {{10}^{ - 3}} \times 0.8}}{{100}}\)

T_OFF = 4 × 10^-5 = 0.04 msec

As T_(OFF)new < T_OFF, conduction will be discontinuous.

\({I_L};or;{I_{s\left( {avg} \right)}} = \frac{1}{2}\left[ {\frac{{{\rm{\Delta }}I \times {T_{ON}}; + ;{\rm{\Delta }}I \times {T_{\left( {OFF} \right)new}}}}{T}} \right]\)

\( = \frac{1}{2} \times 0.8;\left( {{{10}^{ - 3}} \times 0.12 \times f} \right)\)

I_s(avg) = 0.4 × 0.12 × 10³ × 5 × 10³ = 0.24 A

P_avg = V_s.I_s(avg) = 12 W

Output power is calculated as:

\({I_{0\left( {avg} \right)}} = \frac{1}{2} \times \frac{{{\rm{\Delta }}I \times {T_{\left( {OFF} \right)new}}}}{T}\)

\({I_{0\left( {avg} \right)}} = \frac{1}{2} \times 0.8 \times 0.04 \times {10^{ - 3}} \times 5 \times {10^3}\)

\({I_{0\left( {avg} \right)}} = 0.4 \times 0.04 \times 5 = 0.08;A\)

Hence, P₀ = V₀ × I₀ = 150 × 0.08 = 12 W

Concept:

The single-phase induction motor circuit is given by double-revolving field theory which states that a stationary pulsating magnetic field might be resolved into two rotating fields, both having equal magnitude but opposite in direction.

So, the net torque induced is zero at standstill. Here, the forward rotation is called the rotation with slip s and the backward rotation is given with a slip of (2 – s).

The equivalent circuit is:

Here, Zf shows the forward impedance and Zb shows the backward impedance. Also, the sum of forward and backward slip is 2 so in case of a backward slip, it is replaced by (2 – s).

R1 = Resistance of the stator winding

X1 = Inductive reactance of the stator winding

Xm = Magnetising reactance

r'2 = Rotor resistance with referred to stator

X2/2 = Rotor inductive forward and backward reactance with referred to stator

\(\frac{{{{\rm{r'}}_2}}}{{\rm{2s}}}\) = Component of the rotor with resistance r'2 offered to the forward rotating flux wave

\(\frac{{{{\rm{r'}}_2}}}{{2(2 - {\rm{s})}}}\) = Component of the rotor with resistance r'2 offered to the backward rotating flux wave

Explanation:

\({R_1} = R_2' = X_{11}' = X_{12}' = 12{\rm{\Omega }}\)

X_m = 240 Ω 

\({Z_{eq}} = \left[ {12 + j132} \right] + \frac{{\left( {3 + j6} \right)\left( {j120} \right)}}{{3 + j126}}\)

= 138.56 ∠83.9° Ω

No load power factor = cos ϕ

= cos 83.90° = 0.106 lagging

Calculation:

Given circuit diagram

Inductive reactance X_L = jωL = j(100)L Ω

Capacitive Reactance \({X_C} = \frac{1}{{j{\rm{\omega C}}}} = \frac{1}{{j\left( {100} \right)\left( {100 \times {{10}^{ - 6}}} \right)}} \)

\(X_C= - j100{\rm{\Omega }}\)

Resistance R = 100 Ω

The equivalent impedance of the circuit Z is

\(Z = j{\rm{\omega L}} + \frac{{\left( {100} \right)\left( { - {\rm{j}}100} \right)}}{{\left( {100 - j100} \right)}}\)

\(Z = j{\rm{\omega L}} + \left( {\frac{{ - 100{\rm{j}}}}{{1 - {\rm{j}}}}} \right)\left( {\frac{{1 + j}}{{1 + j}}} \right)\)

\(= j{\rm{\omega L}} + \left( {\frac{{ - 100{\rm{j}} + 100}}{2}} \right)\)

Z = j100 L – 50j + 50

i(t) is in phase with V(t)

So, imaginary part of Z must be equal to zero

⇒ 100 L – 50 = 0

⇒ L = 0.5

The magnitude of z = Real part of Z, i.e.

|Z| = 50 Ω