GATE EE 2010 Official Paper (Previous Year Paper)

Given:

Total number of people in the room = 25

Number of people who play Hockey = 15

Number of people who play Football = 17

Number of people who play both = 10

Calculation:

Number of people who play only Hockey = 15 - 10 = 5

Number of people who play only Football = 17 - 10 = 7

⇒ 

⇒ The total number of person who play either of the two = 5 + 10 + 7 = 22

⇒  The number of persons playing neither hockey nor football = 25 - 22 = 3

∴ The required result will be 3.

The number of persons playing neither hockey nor football = ξ - n(A ⋃ B) - n(A) - n(B)

The number of persons playing neither hockey nor football = 25 - 10 - 7 - 5

The number of persons playing neither hockey nor football = 3

Unemployed → doesn't have a job to work 

Worker → Worker always works, and mostly has a job.

Workers are employed and Unemployed.

It implies a somewhat opposite relationship.

(1) Fallow → uncultivated land to improve its fertility.

Land → Land is usually cultivated.

Lands are cultivated and uncultivated.

It also implies a somewhat opposite relationship.

(2) Unaware and Sleeper (someone who sleeps is always unaware)

(3) Wit → A message whose ingenuity or verbal skill or incongruity has the power to evoke laughter.

Jester → A clown, joker

(4) Renovated → Make brighter or prettier or better

House → A place to lives humans

Hence, option (1) is the correct answer.

The correct answer is 'conserve'.

Key Points

• Let's explore the options:

 

• Thus, from the above-given explanation, it is clear that 'conserve' will be used in the blank.
• Therefore, the correct answer is option 4.

The correct answer is Indirect.

Key Points

• The word 'Circuitous' means not being forthright or direct in language or action.
• The synonyms of the word are "circular, indirect, roundabout".
• The antonyms of the word are "direct, straight, straightforward".
• From the Synonym of the given word 'Circuitous', we can say that the word 'Indirect' is its opposite meaning.
• The word 'Indirect' means not having a direct connection with something.
• Hence, we can say that 'Indirect' is the closest in meaning to the given word.
• Let's see the meaning of other given options-

Important Points

• The word 'Circuitous' is derived from the Medieval Latin "circuitosus", from Latin "circuitous".
• Its first known use was in the year 1664.
• Examples of 'Circuitous' in a sentence:
• "Such devices have an extra lead that uses a circuitous route to reach the otherwise hard-to-reach left ventricle."
• "So the agonised, circuitous discussions continue and continue, leading nowhere."

The correct answer is 'betrayed'.

Key Points

• Let's explore the options:

 

• Thus, from the above-given explanation, it is clear that 'betrayed' will be used in the blank.
• Therefore, the correct answer is option 3.

1) H + G > I + S

2) |G - S| = 1

Meaning G and S will be next to each other in order.

So the option (1) is eliminated.

Because S and G are not next to each other, G is not the oldest, and S is not the youngest.

Assume ages are 6, 5, 4, and 2.

From option (2):

S = 6, G = 5, H = 4, I = 2

G + H > I + S

Option (2) satisfy all the conditions.

Now;

Options (3) and (4) are both impossible since in both those cases you would have I + S > H + G.

Hence, option (2) is the correct answer.

GIVEN:

Time taken by 5 skilled workers to build the wall = 20 days

Time taken by 8  semi-skilled workers to build the wall = 25 days

Time taken by 10 unskilled workers to build the wall = 30 days

Calculation:

Time taken by 2 skilled workers to build the wall = 20 × 5/2 days = 50 days

Time taken by 6 semi-skilled workers to build the wall = 25 × 8/6 days = 200/6 days

Time taken by 5 unskilled workers to build the wall = 30 × 10/5 days = 60 days 

Time taken by the three working together = ${{1} \over {1\over50}+{6\over200}+{1\over60}}$ = 1/(1/15)

Time taken by the three working together = 15 days

The correct answer is 'People in military establishments like to use chemical agents in war.'

Key Points

• Let's explore the options:
• In statement 1, let's take a look at the keyword 'civil strife' which means A war between factions or regions of the same country, which isn't correct in the context of passage, therefore option 1 is incorrect.
• In statement 2, it is mentioned that the chemical agents are useful in modern warfare but the passage doesn't have any facts that can justify the usefulness of the chemical agents, and hence incorrect.
• In statement 3, it is mentioned that chemical agents are undesirable, but in the passage it is mentioned that 'Chemical agents that do their work silently appear to be suited to such warfare; and regretfully, there exist people in military establishments who think that chemical agents are useful tools for their cause.' which conveys that they are in fact desirable, thus the statement 3 is incorrect.
• In statement 4, it is mentioned that 'People in military establishments like to use chemical agents in war.' and in the passage, it is also mentioned that 'Chemical agents that do their work silently appear to be suited to such warfare; and regretfully, there exist people in military establishments who think that chemical agents are useful tools for their cause', therefore statement 4 is correct.

Given digits: 2, 2, 3, 3, 3, 4, 4, 4, 4

Case 1:

The first digit is 4.

4 _ _ _

Rest of the places are filled by 2, 2, 3, 3, 3, 4, 4, 4

\( \Rightarrow \begin{array}{*{20}{c}} 4&_&_&_\ \downarrow & \downarrow & \downarrow & \downarrow \ 1&3&3&3 \end{array}\)

The no. of cases = 1 × 3 × 3 × 3 – 1 = 26

Case 2:

The first digit is 3:

Rest of the places are filled by 2, 2, 3, 3, 4, 4, 4, 4

The exception cases are = 222, 333

Therefore:\(\begin{array}{*{20}{c}} 3&_&_&_\ \downarrow & \downarrow & \downarrow & \downarrow \ 1&3&3&3 \end{array}\)

No. of cases =  1 × 3 × 3 × 3 – 2

= 25

Case 3:

First digit is 2

No number can be possible which is greater than 3000

Total no. of cases = 26 + 25 = 51

By observation of summation, it can be seen that numbers are not in decimal.

The value of the answer is actually the remainder obtained by actually finding the sum of original numbers.

Let us proceed step by step.

First, take the unit digit position,

1. 7 + 6 = 13 ⇒ Rem (13/8) = 5

2. 3 + 7 + 1 (Carry from 13) = 11 ⇒ Rem (11/8) = 3

3. 1 + 2 + 1 (Carry from 11) = 4

Now, write in reverse order;

(137)8 + (276)8 = (435)8

Similarly;

1. 1 + 2 = 3

2. 3 + 7 = 10 ⇒ Rem (10/8) = 2

3. 7 + 6 + 1 (Carry from 10) = 14 ⇒ Rem (14/8) = 6

Leaves carry 1.

Write it reverse order;

(731)8 + (672)8 = (1623)8

Hence, "1623" is the correct answer.

Concept:

$\smallint f\left( x \right)g\left( x \right){\bf{dx}} = {\bf{f}}\left( {\bf{x}} \right)\smallint g\left( x \right);dx - \smallint \left[ {\left( {\frac{d}{{dx}}f\left( x \right)} \right)\smallint g\left( x \right)dx} \right]dx$

Calculation:

Given:

$\smallint x{e^{x;}}dx = x\smallint {e^x};dx - \smallint \left[ {\left( {\frac{d}{{dx}}x} \right)\smallint {e^x};dx} \right]dx$

$= {\rm{x}}{{\rm{e}}^{\rm{x}}} - \smallint 1{\rm{;}}{{\rm{e}}^{\rm{x}}}{\rm{;dx}}$

$= {\rm{x}}{{\rm{e}}^{\rm{x}}} - {{\rm{e}}^{\rm{x}}}{\rm{;}}$

Applying limits,

\(\mathop \smallint \limits_0^1 {\rm{x}}{{\rm{e}}^{\rm{x}}}{\rm{;dx}} = \left[ {{\rm{x}}{{\rm{e}}^{\rm{x}}} - {{\rm{e}}^{\rm{x}}}} \right]_0^1\)

= [1 × e¹ – e¹] – [0 × e⁰ – e⁰]

= [e¹ – e¹] – [0 - 1]

= 1

Concept:

For execution of CALL instruction 

Step 1:

Current contents pf PC is pushed on tp stack and SP is updated.

Step 2:

The 16 bit target address given in instruction is moved into PC.

Concept:

V(t) = V_m sin(ωt + ϕ )   ...(1)

Where 

V_m = Peak value

ω = Angular frequency

ϕ = Phase angle

As we know

ω = 2 π f  rad/sec

A time period (T) is the time taken for one complete cycle of the waveform. It is the reciprocal of frequency (f). 

Time period

 $T = \frac{1}{f}$

Explanation:

Given

x(t) = 8 sin $\left( {0.8;π t + \frac{π }{4}} \right)$

Compare the x(t) with equation 1.

ω = 0.8 π 

2πf = 0.8π 

f = 0.4

Time period

 $T = \frac{1}{f}$

T = 1 / 0.4

T = 2.5 sec

Input-output relationship,

\(y\left( t \right) = \mathop \smallint \limits_{ - \infty }^{5t} x\left( \tau \right)d\tau ,\ t > 0\)

Causality:

y(t) depends on x(5t) , t > 0

System is non-causal

For example, t = 3

y(3) depends on x(15) (future value of input)

Linearity:

o/p is integration of  i/p which is a linear function, so system is linear

For t < 0, the switch was closed for a long time so equivalent circuit is

Voltage across capacitor at t = 0

${V_C}\left( o \right) = \frac{5}{{4 \times 1}} = 4V$

Now switch is opened, so equivalent circuit is

For capacitor at t = 0⁺

V_C(0⁺) = V_C(0) = 4V

Current in 4 Ω resistor at t = 0⁺,

${i_1} = \frac{{{V_C}\left( {{0^ + }} \right)}}{4} = 1A$

So, current in capacitor at t = 0⁺,

i_C(0⁺) = i₁ = 1A

The given wave form f(t) is half wave symmetric or it has rotation symmetric.

i.e f(t) = -f (t-T/2) or the wave form over any half period is a replica of the adjacent wave form but with opposite values.

∴ In the given wave form, only odd harmonics are present.

Even harmonics are apsent.

∴ Amplitude of the second harmonic (even harmonic) component is zero.

Concept:

Thevenin's Theorem:

Any two terminal bilateral linear DC circuits can be replaced by an equivalent circuit consisting of a voltage source and a series resistor.

To find V_oc: Calculate the open-circuit voltage across load terminals. This open-circuit voltage is called Thevenin’s voltage (V_th).

To find I_sc: Short the load terminals and then calculate the current flowing through it.

 This current is called Norton current (or) short circuit current (i_sc).

To find R_th: Since there are Independent sources in the circuit, we can’t find Rth directly. We will calculate Rth using Voc and Isc and it is given by

\({{\rm{R}}{{\rm{th}}}} = \frac{{{{\rm{V}}{{\rm{oc}}}}}}{{{{\rm{i}}_{{\rm{sc}}}}}}\)  

Application:

Given: Load line equation = V + i = 100

To obtain open-circuit voltage (Vth) put i = 0 in load line equation 

⇒ Vth = 100 V

To obtain short-circuit current (isc) put V = 0 in load line equation

⇒ isc = 100 A

So, ${R_{th}} = \frac{{{V_{th}}}}{{{i_{sc}}}} = \frac{{100}}{{100}} = 1{\rm{\Omega }}$

Equivalent circuit is

Current (i) = 100/2 = 50 A

 

Applying loop-law in the given circuit.

• V + i × R = 0
• V + I × 1 = 0

⇒ V = i

Given Load line equation is V + i = 100

Putting V = i 

then i + i = 100 

⇒ i = 50 A

Concept:

Wattmeter reads the power and it is given by

P = VPC ICC cos ϕ

VPC is the voltage across pressure coil

ICC is current flows through the current coil

ϕ is the phase angle between VPC and ICC

Application:

The circuit representation of the given question is as shown below.

• The potential coil is connected across Z2.
• It reads the voltage across Z2 only.
• So, Wattmeter reads only power consumed by Z2.

Concept:

• For range extension of voltage measurement in moving coil instrument, a resistance is connected in series with coil resistance.
• Because for a constant value of current, resistance connected in series connection has a higher voltage drop compared to parallel or shunt connection.
• For range extension of current measurement in moving coil instrument, a resistance is connected in parallel or shunt with coil resistance.
• Because for a constant value of voltage, resistance connected in parallel connection has a higher value of current flow compared to series connection.

 

Formula:

Rse = Rm(M – 1)

${R_m} = \frac{{{V_m}}}{{{I_m}}}$

${R_{sh}} = \frac{{{R_m}}}{{M - 1}}$

M= Multiplying factor = (Required full scale deflection) / (Initial full scale deflection)

Where,

Rsh = Series resistance

Rm = Meter resistance

Vm = Potential difference across meter

Im = Meter current

Calculation:

Given -

Rm = 0.2 Ω

$M = \frac{{25}}{{5}} = 5$ 

$R_{sh} = \frac{{0.2}}{{5 - 1}} = 0.05 ;\Omega$

Rsh = 0.05 Ω

$\begin{array}{l} \frac{{dA}}{A} = 10\% ,\ \beta = \frac{9}{{100}},\ A = 100\ \frac{{d{A_f}}}{{{A_f}}} = \frac{1}{{1 + \beta A}}\frac{{dA}}{A}\ \left( {{A_f} = \frac{A}{{1 + \beta A}} = 10} \right)\ \therefore \frac{{d{A_f}}}{{{A_f}}}\ = \frac{1}{{1 + \beta A}}\frac{{dA}}{A} \end{array}$

= 0.1 × 10% = 1%

System is given as

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

Step input R(s) = 1/s

o/p , Y(s) = H(s) R(s)

$\begin{array}{l} = \frac{2}{{\left( {s + 1} \right)}}\frac{{1}}{s}\ = \frac{2}{s} - \frac{2}{{s + 1}} \end{array}$

Taking inverse laplace transform,

y(t) = (2-2e^-t) u(t)

Final value of y(t),

\(y\left( {ss} \right)\left( t \right) = \begin{array}{*{20}{c}} {lt}\ {t \to \infty } \end{array}\ y\left( t \right) = 2\)

Let time taken for step response to reach 98% of its final value is t_s

So,

2-2e^-ts = 2×0.98

0.02 = e^-ts

t_s = ln 50 = 3.91 sec.

Concept:

$p{u_{new}} = p{u_{old}}{\left( {\frac{{k{V_{old}}}}{{k{V_{new}}}}} \right)^2}\left( {\frac{{MV{A_{new}}}}{{MV{A_{old}}}}} \right)$

Where kVold = old base value of voltage,

kVnew = New base value of voltage,

MVAold = Old base value of machine rating,

MVAnew = New base value of machine rating,

Calculation:

For Generator 1:

G1:25 kV, 100 MVA, X = 9%

Reactance of generator (Xold) = 0.09 p.u.

Old base value of voltage (kVold) = 25 kV

Old base value of MVA (MVAold) = 100 MVA

New base value of voltage (kVnew) = 25 kV

New base value of MVA (MVAnew) = 200 MVA

${X_{G1_{new}}} = 0.09 \times {\left( {\frac{{25}}{{25}}} \right)^2} \times \left( {\frac{{200}}{{100}}} \right)$ = .18

For transformer 1:

T1: 20 kV/220 kV, 90 MVA, X = 12%

Reactance of Transformer 1 (Xold) = 0.12 p.u.

Old base value of voltage (kVold) = 25 kV

Old base value of MVA (MVAold) = 90 MVA

New base value of voltage (kVnew) = 25 kV

New base value of MVA (MVAnew) = 200 MVA

${X_{T1_{new}}} = 0.12 \times {\left( {\frac{{25}}{{25}}} \right)^2} \times \left( {\frac{{200}}{{90}}} \right)$ = .27

For Line:

Line1:220 kV, X = 150 ohms

X_line pu = 150 (200/220²) = .62

For Transformer 2:

T2: 220 kV/25 kV, 90 MVA, X = 12%

${X_{T2_{new}}} = 0.12 \times {\left( {\frac{{25}}{{25}}} \right)^2} \times \left( {\frac{{200}}{{90}}} \right)$ = .27

For Generator 2:

G2:25 kV, 100 MVA, X = 9%

${X_{G2_{new}}} = 0.09 \times {\left( {\frac{{25}}{{25}}} \right)^2} \times \left( {\frac{{200}}{{100}}} \right)$ = .18

Given that

Turn's ratio a = N₁ / N₂ = 1 / 2

The load is purely resistive, so the load power factor is unity. And the phase angle is equal to zero.

Load current on the secondary side is I₂ = 1 A

Load current referred to the primary side I₁^' = I₂ / a = 1 / (1 / 2) = 2 A

Core losses and leakage reactance are neglected.

Let's consider I_μ as magnetizing current and primary current as I₁. 

Then the phasor diagram for the given condition can be given by,

From the above phasor diagram by the use of parallelogram law, we can write,

\({{\rm{I}}_1} = \sqrt {{{\left( {{{{\rm{I'}}}1}} \right)}^2} + {{\left( {{{\rm{I}}{\rm{μ }}}} \right)}^2} + 2{\rm{I}}{{\rm{'}}1}{{\rm{I}}{\rm{μ }}}\cos θ } \)

Where θ is the angle between I_μ and I₁^' and from the phasor diagram it equals to θ = 90⁰.

⇒ ${{\rm{I}}_1} = \sqrt {{{\left( 2 \right)}^2} + {{\left( 1 \right)}^2} + 2 \times 2 \times 1\cos {{90}^0}}$

⇒ ${{\bf{I}}_1} = \sqrt {4 + 1} = \sqrt 5 = 2.24; A$

From the above system of HVDC link, we can observe that

The power is transferring from system A to system B.

We know that power always flows from a higher voltage side to the lower voltage side.

From that, we can conclude that system A voltage V_AB is greater than system B voltage V_CD (V_AB > V_CD).

And the output of the rectifier circuit V_AB is positive, V_AB > 0.

And the input of the inverter circuit V_CD is also positive, V_CD > 0.

Given that

Line voltage = 11 kV

Frequency = 50 Hz

k = shunt capacitance between link and ground / self-capacitance of insulator = 1/5 = .2

And we know that,

V_ph = e₁ + e₂

(11/√3) kV = e1 + e₂      ---------------  (1)

e₁ = (1 + k) e₂      

e1 = 1.2 e2               -------------- (2)

From (1) and (2)

e1 = 3.46 kV, e2 = 2.89 kV

The reactor is initially open circuit

V₂ = V + V₁ = 1.0 + 1.0 = 2.0 P.u.

Where V₁ is reflected voltage

V₂ is switched voltage.

P = Line impedance angle, δ = 30°, α = 0°, A = D = 1.0, β = 5∠90^o 

$\begin{array}{l} {P_{sending}} = \left| {\frac{D}{B}} \right|{\left| {{V_S}} \right|^2}\cos \left( {β - \alpha } \right) - \frac{{\left| {{V_S}} \right|\left| {{V_r}} \right|}}{{\left| B \right|}}\cos \left( {β + \delta } \right)\ {Q_{sending}} = \left| {\frac{D}{B}} \right|{\left| {{V_S}} \right|^2}\sin \left( {β - \alpha } \right) - \frac{{\left| {{V_S}} \right|\left| {{V_r}} \right|}}{{\left| B \right|}}\cos \left( {β + \delta } \right)\ \therefore {P_{sending}} = \left| {\frac{{1.0}}{5}} \right|{100^2}\cos \left( {90 - 0} \right) - \frac{{100 \times 100}}{{\left| 5 \right|}}\cos \left( {90 + 30} \right) \end{array}$

= 1000 W

${Q_{sending}} = \left| {\frac{{1.0}}{5}} \right|{100^2}\sin \left( {90 - 0} \right) - \frac{{100 \times 100}}{{\left| 5 \right|}}\sin \left( {90 + 30} \right)$

= 268 VAR

Concept:

Breaking current: It expresses the highest number of short-circuit currents that the breakers are capable of breaking under specified conditions of transient recovery voltage and power frequency voltage. It is expressed in kA (RMS) at contact separation. The breaking capacities are divided into two types.

• Symmetrical breaking capacity of a circuit breaker
• Asymmetrical breaking capacity of a circuit breaker

 

Rated symmetrical breaking current ${I_B} = \frac{S}{{\sqrt 3 {V_L}}}$

Making current of a circuit breaker is the peak value of the maximum current loop during sub transient condition including the DC component when the breaker closes.

Symmetrical making current = 2.55 × symmetrical breaking current

Calculation:

Rated current = 1200 A

MVA rating = 2000 MVA

Rated symmetrical breaking current $= \frac{{2000 \times {{10}^6}}}{{\sqrt 3 \times 33 \times {{10}^3}}} = 35;kA$

∴ current through the operating coil = 3.125 – 2.75 = 0.375 A

Concept:

Zero sequence networks for different 3-phase loads are shown below.

The zero sequence equivalent circuits of 3-phase transformers can be drawn by using the following general circuit.

Z is the zero sequence impedance of the windings of the transformer. These are two series and two shunt switches. One series and one shunt switches are for both sides separately.

The series switch of a particular side is closed if it is star grounded and the shunt switch is closed if that side is delta connected, otherwise they are left open.

Solution:

zero-sequence circuit

The given system is

Ẋ = AX + Bu

Where,

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

We determine stability using characteristic equation (i.e. poles or eigenvalues of the system).

|sI - A| = 0

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

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

(s + 1)(s - 2) = 0

s = -1 and s = 2

i.e. the system has one pole in right half of the s-plane.

Hence, the system is unstable.

Now, the controllability matrix is given by:

C_M = [B AB]

Where

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

So,

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

|C_M| = -2 ≠ 0

Hence, the system is Controllable.

Concept:

The condition for an ideal diode to be ON/OFF are:

If VP > Vp, Diode will be ON (Short-circuit)

If Vp < Vn, Diode will be OFF (open-circuit)

Where Vp and Vn are the voltages at the p and side respectively

Analysis:

Checking the given circuit for the above condition, we conclude that D2 is always OFF and D1 is ON.

The equivalent circuit is drawn as:

Using voltage division, we get:

${V_0} = 10 \times \frac{{10;k}}{{10k + 10k}}$

V0 = 5 V

DC to DC converters (DC Choppers):

• A dc chopper converts dc input voltage to a controllable dc output voltage.
• For lower power circuits, thyristors are replaced by power transistors.
• Choppers find wide applications in dc drives, subway cars, trolley trucks, battery-driven vehicles, etc.

 

The load voltage or output voltage is controlled by duty cycle of the circuit. The output voltage can be varied by varying the duty cycle.

 

Important Points

• In both buck-boost converter and Cuk converter, the polarity of the output voltage is the opposite of input voltage.
• In buck-boost converter, energy transfer is associated with the inductor, whereas, in Cuk converter, energy transfer is associated with the capacitor.
• In the Buck-boost converter, the input current is discontinuous, whereas, in the Cuk converter, the input current is continuous.

 

Both the power transistor and the diode are ideal.

When voltage V is positive, both transistor and diode will be ON and making voltage across them zero and current I will be flowing.

⇒ at V = 0, I = I_maximum (∞)

When voltage V is negative, then both will be OFF and offering infinite resistance and hence voltage is maximum, and the current is zero.

⇒ at V = V_max (∞), I = 0

Now, the V-I characteristics are as shown below.

Concept:

The average output voltage of a single-phase thyristor converter with RL load is given by 

V₀ = $\frac{2V_m}{π} cos α$

Where,

V_m = Peak supply voltage

α = firing angle

Calculation:

Case 1:

α = 0° 

V₀ = 300 V

⇒ 300 = $\frac{2V_m}{π} cos 0$

⇒ V_m = 471.24 V

Case 2:

α = 60° 

V₀ = (2 × 471.24/π) cos 60° 

V₀ = 150 V

Concept:

The point of maxima or minima is obtained by solving for the derivative of the function and equating to zero.

Then, to check if the point is a point of maxima ‘or’ minima we check the second derivative at that point.

This is explained with the help of the following graph:

If $\frac{{{d^2}f}}{{d{x^2}}} < 0$; the point will be a point of maxima

If $\frac{{{d^2}f}}{{d{x^2}}} > 0$, the point will be a point of minima.

Calculation:

Method I:

$f\left( t \right) = \frac{{\sin t}}{t} = \frac{{t - \frac{{{t^3}}}{{3!}} + \frac{{{t^5}}}{{5!}} \ldots }}{t} = 1 - \frac{{{t^2}}}{{3!}} + \frac{{{t^4}}}{{5!}}$  

$f'\left( t \right) = \frac{{ - 2t}}{6} + \frac{{4{t^3}}}{{5!}} - \frac{{6{t^5}}}{{7!}}$

$f''\left( t \right) = \frac{{ - 1}}{3} + \frac{{12{t^2}}}{{5!}} - \frac{{30{t^4}}}{{7!}}$

Critical point ⇒ f’(t) = 0

⇒ t = 0 is critical pt.

$\therefore f''\left( 0 \right) = \frac{{ - 1}}{3} < 0$

So, t = 0 is point of maxima.

∴ Maximum value $f\left( 0 \right) = \mathop {\lim }\limits_{t \to 0} \left( {\frac{{\sin t}}{t}} \right)$ = 1

Method II:

Using graph:

From graph: Maximum value = f(0) = 1

Given that,

Number of white balls = 4

Number of red balls = 3

Required probability would be,

P (B/A) = Probability of drawing a red ball in the second draw given that a white ball has already been drawn in the first draw.

Since 6 balls are left after drawing a white ball in the first draw and out of these 6 balls 3 are red, so

P(B/A) = 3/6 = 1/2.

Explanation:

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

Given matrix is upper triangular matrix, so eigen values = 1, 2, 3

Eigen vector corresponding to eigen value λ = 3

For λ = 3, 

(A - λI)X = 0

Where X is the eigen vector

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

\(\left[ {\begin{array}{{20}{c}} -2x+y\ -y+2z\ 0 \end{array}} \right] = \left[ {\begin{array}{{20}{c}} 0\ 0\ 0 \end{array}} \right]\)

⇒ The required eigenvector is [ 1 2 1]T

$\frac{{{d^2}x}}{{d{t^2}}} + \frac{{6dx}}{{dt}} + 8x = 0$

Taking laplace transform (with initial condition) on both sides

S² X(s) – sx(o) –x^’(o) + 6 [sX (s) – x (o) ] +8X(s) =0

S²X (s) – s(1)  - 0 + 6 [s X(s) -1] +8X(s) =0

X(s) [s²+6s+8] –s-6=0

$X\left( s \right) = \frac{{s + 6}}{{{s^2} + 6s + 8}}$

By partial fraction,

$X\left( s \right) = \frac{2}{{s + 2}} - \frac{1}{{s + 4}}$

Taking inverse laplace transform,

x(t) = (2e^-2t – e^-4t)

Given, system of equation

x₁ + 2x₂ + x₃ + 4x₄ = 2

3x₁ + 6x₂ + 3x₃ + 12x₄ = 6

Which is the non-homogeneous system of equations .

The augmented matrix is

Applying R₂ → R₂ – 3R₁

Here, Rank of [A : B] = 1 & Rank of A = 1

i.e. Rank of [A : B] = Rank of A

So, the given system of equation is consistent. But the rank of A is less than the number of unknowns. So, the system of non-homogeneous equation will have infinitely many solutions. SO multiple, non-trivial solution exist.

Concept:

Parseval's theorem

• It relates the energy of a signal x(t) and its spectral component X(ω)
• Parseval’s Theorem can be written as:

$\underset{-\infty }{\overset{\infty }{\mathop \int }}|,{{x}^{2}}\left( t \right)|dt=\frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\mathop \int }},{{\left| X\left( jω \right) \right|}^{2}}dω$

Calculation:

From Concept,

\(\frac{1}{{2\pi }}\mathop \smallint \limits_{ - \infty }^\infty {\left| {X\left( \omega \right)} \right|^2}d\omega = \mathop \smallint \limits_{ - \infty }^\infty {x^2}\left( t \right)\ dt\)

\(\mathop \smallint \limits_{ - \infty }^\infty {\left| {X\left( \omega \right)} \right|^2}d\omega = 2\pi \times 2\)

= 4π

 

• \(\mathop \smallint \limits_{ - \infty }^\infty {x^2}\left( t \right) =Area;under;curve\)
• $\underset{-\infty }{\overset{\infty }{\mathop \int }}|,{{x}^{2}}\left( t \right)|dt=\underset{-\infty }{\overset{\infty }{\mathop \int }},{{\left| X\left( f \right) \right|}^{2}}df =\frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\mathop \int }},{{\left| X\left( jω \right) \right|}^{2}}dω$

\(\begin{array}{l} x\left[ n \right] = \left{ {1, - 1} \right},0 \le n \le 1\ y\left[ n \right] = \left{ {1,0,0,0, - 1} \right},0 \le n \le 4 \end{array}\)

If impulse response is h[n] then y[n] = h[n] * x[n]

Length of convolution ( y[n] ) is 0 to 4, x[n] is of length 0 to 1, so length of h[n] will be 0 to 3.

Let

Convolution

By comparing,

a = 1

• a + b = 0 ⇒ b = a = 1
• b + c = 0 ⇒ c = b = 1
• c + d = 0 ⇒ d = c = 1

So, h[n] = {1, 1, 1, 1}

The circuit is

Current in R Ω resistor is

i = 2 – 1 = 1A

Voltage across 12 Ω resistor is,

V_A = 1 × 12 = 12 V

$i = \frac{{{V_A} - 6}}{R} = \frac{{12 - 6}}{1} = 6{\rm{\ \Omega }}$

V₁ = Z₁₁ I₁ + Z₁₂ I₂

V₂ = Z₂₁ I₁ + Z₂₂ I₂

$\begin{array}{l} V_1' = Z_{11}'I_1' + Z_{12}'I_2'\ V_2' = Z_{21}'I_1' + Z_{22}'I_2' \end{array}$

When R = 1 Ω is connected

$\begin{array}{l} V_1' = {V_1} + \left( {I_1' \times 1} \right)\ V_1' = {V_1} + {I_1}\ V_1' = {Z_{11}}{I_1} + {Z_{12}}{I_2} + {I_1}\ V_1' = \left( {{Z_{11}} + 1} \right){I_1} + {Z_{12}}{I_2}\ V_1' = \left( {{Z_{11}} + 1} \right)I_1' + {Z_{12}}I_2'\ Z_{11}' = {Z_{11}} + 1\ Z_{12}' = {Z_{12}} \end{array}$

Similarly for output port

$\begin{array}{l} V_2' = Z_{21}'I_1' + Z_{22}'I_2'\ V_2' = Z_{21}'{I_1} + Z_{22}'{I_2} \end{array}$

So, $Z_{21}' + {Z_{21}},Z_{22}' + {Z_{22}}$

Concept:

Bridge balance condition is Z₁ Z₄ = Z₂ Z₃

Calculation:

Given Z₁ = R + jωL, Z₂ = R₂, Z₃ = R₃

Z₄ = R₄ // (-j/ωC₄)

${Z_4} = \left( {\frac{{{R_4} \times \frac{{ - j}}{{ω {C_4}}}}}{{{R_4} + ω {C_4}}}} \right)$

${Z_4} = \left( {\frac{{ - j{R_4}}}{{ω {R_4}{C_4} - j}}} \right)$

Applying the bridge balancing condition

Z1 Z4 = Z2 Z3

$\left( {R + jω L} \right)\left( {\frac{{ - j{R_4}}}{{ω {R_4}{C_4} - j}}} \right) = {R_2}{R_3}$

⇒ -j R R₄ + ω R₄ L = ω R₂ R₃ R₄ C₄ - j R₂ R₃

Comparing the real terms on both sides we get

⇒ L = R2 R3 C4

Comparing the imaginary terms on both sides we get

⇒ R = R2 R3 / R₄

The given frequency response is

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

$\begin{array}{l} G\left( {jω } \right) = \frac{1}{{jω \left( {jω + 1} \right)\left( {jω + 2} \right)}}\ \left| {G\left( {jω } \right)} \right| = \frac{1}{{ω \left( {\sqrt {{ω ^2} + 1} } \right)\left( {\sqrt {{ω ^2} + 4} } \right)}}\ ∠ G\left( {jω } \right) = - 90 - {\tan ^{ - 1}}ω - {\tan ^{ - 1}}0.5ω \end{array}$

In Anti-clockwise the phase order is 0, 90, 180, 270, 360 or 0.

In clockwise the phase angle order is 0, -90, -180, -270, -360 or 0.

Here, 90^∘  = -270^∘  and 270^∘ = -90^∘ 

At ω = ω_c (Cut-off frequency), ∠ G(jω) = -180^∘ 

 $- 90 - {\tan ^{ - 1}}ω_c - {\tan ^{ - 1}}0.5ω_c=-180$

${\tan ^{ - 1}}ω_c + {\tan ^{ - 1}}0.5ω_c=90$

${\tan ^{ - 1}}\frac{{{ω _c} + 0.5{ω _c}}}{{1 - 0.5{ω ^2}}}$ = 90° 

1 - 0.5ω_c² = 0

ω_c = 1.414 rad/sec

At ω = ω_c 

|G(jω)| = (1/6) 

The polar plot for given transfer function is

Apply KCL at inverting terminal,

(0 - 2) / R = (2 - V₀) / 2R

• 4 = 2 - V₀

V₀ = 6 V

The characteristic equation of the given closed-loop system is:

s(s + 1)(s + 3) + K(s + 2) = 0 ; K > 0

$1 + \frac{{K\left( {s + 2} \right)}}{{s\left( {s + 1} \right)\left( {s + 3} \right)}} = 0$

So, the open-loop transfer function is given as:

$G\left( s \right)H\left( s \right) = \frac{{K\left( {s + 2} \right)}}{{s\left( {s + 1} \right)\left( {s + 3} \right)}}$

Therefore, we have the open-loop poles and zeros as:

Poles: s = 0, s = -1, s = -3

Zero: s = -2

So, we obtain the pole-zero plot for the system as:

For the pole-zero location, we obtain the following characteristic of root locus

Number of asymptotes: P – Z = 3 – 1 = 2

Angles of asymptotes: ${\phi _a} = \frac{{\left( {2q + 1} \right)180^\circ }}{{P - Z}};;P - Z = 2,;q = 0,;1$ 

ϕa = 90° and 270°

Centroid: ${\sigma _A} = \frac{{Sum;of;Re\left[ P \right] - Sum;of;Re\left[ Z \right]}}{{P - Z}}$  

$= \frac{{\left( {0 - 1 - 3} \right) - \left( { - 2} \right)}}{{3 - 1}} = - \frac{3}{2}$

= -1

Thus, from the above analysis, we sketch the root locus as:

For the root locus, we conclude the following points

1. The breakaway point lies in the range,

-1 < Re[s] < 0

2. Two of its roots tend to infinite along the asymptotes Re[s] = -1.
3. Root locus lies only in the left half of the s-plane.

A long transmission line under no-load conditions behaves as the capacitive load. The effect of armature current is purely magnetization. When the alternator is disconnected, there is no magnetizing effect. So the terminal voltage decreases with the same field current.

Explanation:

When the generator is connected to an open-circuit transmission line, the line draws a charging current, therefore V_t > E_g.

But, when the generator is disconnected from the line, no charging current is delivered by the generator, i.e.,

I_c = 0 ⇒ V_t = E_g.

So, the terminal voltage decreases.

Concept:

Synchronous Speed:

The synchronous speed of a three-phase induction motor is given by:

${N_s} = \frac{{120f}}{P}$

Where Ns = Synchronous speed in rpm

f = Supply frequency

P = Number of poles

Slip:

The slip in an induction motor is the difference between the main flux speed and their rotor speed.

Fractional slip (s) is given by,

$s = \frac{{{N_s} - {N_r}}}{{{N_s}}}$

Where N_r is rotor speed.

Note: If slip (s) is negative then the induction machine acts as a generator.

Calculation:

Synchronize speed of the induction machine is

${N_s} = \frac{{120{\rm{\ x}}\ 50}}{4} = 1500{\rm{\ rpm}}$

N_r = 1600 rpm

s = (1500 - 1600) / 1500

s = - 1/15 = - 0.066

Since slip is negative, the induction machine acts as an induction generator and dc machine as a dc motor.

The thing to be noted here is that while transforming resistance the copper loss should remain the same

$\begin{array}{l} I_1^2{R_{{q_1}}} = I_2^2{R_q}_2\ {\left( {\frac{{\frac{{110}}{{\sqrt 3 }}}}{{{R_1}}}} \right)^2}{R_1} = {\left( {\frac{{\frac{{220}}{{\sqrt 3 }}}}{{{R_2}}}} \right)^2}{R_2} \end{array}$

R₂ = 4 Ω

$\therefore R_1^1 = 1\Omega$

 = 1 + j0 Ω

The approximate equivalent circuit of the induction motor per phase, when referred to stator, is

The magnitude of starting current 

I_st = $\rm \frac{V}{\sqrt{(r_1 + r_2)^2 + (x_1 + x_2)^2}}$ ( during starting slip s = 1)

C₁ = 0.2μF

C₂ = 0.4 μF

Capacitance per phase,

C_eq = 3C₁ + C₂ = 3 × 0.2 + 0.4 = 1 μF

Charging current, ${I_C} = \frac{V}{{{X_C}}} = V\omega C$

$= \frac{{11 \times {{10}^3}}}{{\sqrt 3 }} \times 2\pi \times 50 \times 1 \times {10^{ - 6}} = 2;A$

In the electrical equivalent circuit of the above circuit shown below,

A and C are capacitances and B and D are resistances.

• The water columns h₁ and h₂ are analogous to voltages across capacitances. Hence A and C of equivalent circuit is capacitances.
• Water loses its energy (due to friction) while moving from left to right h₁ to h₂ and then beyond h₂. hence the elements B and D of equivalent circuit is resistances.

Apply KVL from 10 V to group through B – E terminals

10 - 10⁴ I_B - V_BE – 100 I_E = 0

10 – 0.7 = 10⁴ I_B + 100 I_C

$because \ \beta = 100 \ and \ I_C = I_E (approximately)$

9.3 = 100 I_C + 100 I_C = 200 I_C

 

V_o = 100 I_C = 4.65 V

Concept:

The Divergence theorem states that:

$\mathop{{\int!!!\ !!\int}\mkern-21mu \ \bigcirc} {D.ds = \iiint_V {\left( {∇ .D} \right);dV}}$