GATE EE 2014 Official Paper: Shift 2 (Previous Year Paper)

The correct answer is Option 1 i.e. it was a former British colony.

Key Points

The sentence given above ends with 'because' and we know that 'because' is a subordinating conjunction. It shows the cause, a conjunction connects two sentences or phrases. As per the context of the first phrase, the second phrase must be a cause. A  country is called postcolonial if it came into existence after the colonies of the British and the Europeans were abolished and the countries then under their rule were declared independent.

India was under the British colonial rule till 1947, i.e. it was a former British colony and thus is called a postcolonial country. All the other options do not seem to be a valid cause, Option 2 is an independent statement, whereas other options are irrelevant.

So, the** complete sentence will be**- India is a post-colonial country because it was a former British colony.

The correct answer is Option 2) i.e. ​did you say

We usually form wh-questions with wh- + an auxiliary verb (be, do or have) + subject + main verb or with wh- + a modal verb + subject + main verb. For eg- When are you leaving?

Among all the options, Option 2 is the appropriate choice as it concurs with the context of the sentence.

• Option 1 is incorrect because the auxiliary verb is missing.
• Option 3 is incorrect because 'that' doesn't fit as per the context of the sentence.
• Option 4 is incorrect because we need simple past tense, and in the given option past perfect has been used.

The correct answer is Option 1) i.e. ​1 : S, 2 : P, 3 : Q, 4 : R

Data:

Range: 2 to 198

Multiples of 10 are 10, 20, 30 ..... 190

Since the numbers are in Arithmetic progression:

First term = a = 10, 

Last term = tn= 190

Difference = d = 20 - 10 = 10

number of terms = n

Sum = S_n

Formula:

tn = a + (n - 1)d

Sn = $\frac{n(a + t_n)}{2}$

Calculation:

190 = 10 + (n - 1) 10

n = 19

Sn = $\frac{19(10 + 190)}{2} = 1900$

Average = $\frac{1900}{19} = 100$ 

Tips and Tricks:

$Average = \frac{a + t_n}{2} = \frac{10 + 190}{2} = 100$

Let y = $\sqrt {12 + \sqrt {12 + \sqrt {12 + \ldots } } }$

∴ y = $\sqrt {12 +y }$

Squaring on both sides

y²  = 12 + y

y² - y - 12 = 0

y2 - 4y + 3y - 12 = 0

y(y - 4) + 3(y - 4) = 0

(y - 4)(y + 3) = 0

∴ y = 4 or y = -3

The value of $\sqrt {12 + \sqrt {12 + \sqrt {12 + \ldots } } }$ is 4.000

The correct answer is Option 2 i.e Kaliningrad is a part of Russia despite it not being contiguous with the rest of Russia

 

 

 Explanation

From the passage, we find that:

• Koenigsberg became Russian Kaliningrad after the war.
• It had a German majority population before the war.
• Poland is in its South and West and Lithuania to the east.

From the above points, we find that

• Option 1 is wrong as Kaliningrad was historically German with the German population.
• Option 2 concurs perfectly to the above points and is the correct answer.
• Option 3 is wrong as Kaliningrad was not the original name as it was Koenigsberg.
• Option 4 is wrong as Poland and Lithuania are on two different sides of the city.

 

Hence we find that Option 2 is the only correct option, and therefore the correct answer.

The correct answer is option 4 i.e. the number of people with malarial fever (also contracted from mosquito bites) has increased this year

Explanation

Option 4 is the correct answer, as it states both Dengue fever and malarial fever is caused by Mosquito bite; dengue fever had increased because of some other reasons which are not mentioned by the municipal authorities. Since Municipal authorities have concluded that measures to control the mosquito population have failed in this region so as its consequences we can easily say "The number of people with malarial fever (also contracted from mosquito bites) has increased this year"

Let’s see the given points-

• Statement (1) negates or contradicts the conclusion of the municipal authorities, as it concluded that measures taken to control the mosquito population had failed; Hence 1st option is incorrect.
• As there is no data mentioned in the passage related to the reporting of cases and the efficiency of the administrative capabilities; it concludes that the 2nd option is also incorrect
• Statement 3rd also contradicts the conclusion of the municipal authorities mentioned in the paragraph; hence this option will be incorrect.

|𝑥2 − 2𝑥 + 3| = 11

𝑥2 − 2𝑥 + 3 = 11 or 𝑥2 − 2𝑥 + 3 = -11

By taking: 

𝑥2 − 2𝑥 + 3 = 11

𝑥2 − 2𝑥 − 8 = 0

𝑥2 − 4𝑥 + 2𝑥 − 8 = 0

𝑥(𝑥 - 4) + 2(𝑥 − 4) = 0

(𝑥 + 2)(𝑥 − 4) = 0

𝑥 = −2 or 𝑥 = 4

Substitute:  𝑥 = -2

|− 𝑥3 + 𝑥2 − 𝑥|  = | − (−2)3 + (−2)2 − (−2)|  = |14 | = 14

Substitute:  𝑥 = 4

|− 𝑥3 + 𝑥2 − 𝑥|  = | − 43 + 42 − 4|  = |-52 | = 52

The possible values of | − 𝑥3 + 𝑥2 − 𝑥| include 14 and 52

1 Rotation of clock consists of 360^o

In 1 hour, minute hand covers 360^o

60 minutes → 360^o

1 minute → 6o

Therefore, the minute hand covers 6o in 1 minute.

In 12 hour, hours hand covers 360o

1 hour → 30o

60 minutes → 30o

1 minute → 0.50

Therefore, the hour hand covers 0.5^o in 1 minute.

Option 1:

22 minutes will move 110 of an hour hand

2 minutes will move 120 of a minutes hand

Angle covered:

(4 to 5) + (5 to 6) + (6 to 7)

(30 - 2× 6)o + 30o + (22 × 0.5)o

18o + 30o + 11⁰ = 59o

This is closest to 60o

Explanation:

$f\left( x \right) = 10{\left( {x - 1} \right)^{\frac23}}$

Now, given that f(x) is a real-valued function.

If (x – 1) is negative, then the factional power of the negative number is imaginary.

So, (x – 1) cannot be negative.

Therefore, (x – 1) should be ≥ 0 and x ≥ 1.

f(x) is an increasing function. So, f(x) will be minimum at the minimum value of x.

Hence, f(x) will be minimum at x = 1.

In terms of complex function, 1 can be written as

1 = cos (2m π) + i sin (2m π)

Where m = integer

$\Rightarrow 1 = {e^{i\left( {2m\pi } \right)}}$ 

${1^i} = {\left( {{e^{i\left( {2m\pi } \right)}}} \right)^i}$

Given that, $i = \sqrt { - 1} \Rightarrow {i^2} = - 1$

Now, (1)^I = e^-2mπ

It is a positive value.

Therefore, all values of 1ⁱ are a non-negative and real number.

Given the differential equation is,

${x^2}\frac{{{d^2}y}}{{d{x^2}}} + x\frac{{dy}}{{dx}} - y = 0$

Put, x = e^t

$\Rightarrow t = ln;x,;\frac{{dt}}{{dx}} = \frac{1}{x}$ 

Now, $\frac{{dy}}{{dx}} = \frac{{dy}}{{dt}} \cdot \frac{{dt}}{{dx}} = \frac{{dy}}{{dt}} \cdot \left( {\frac{1}{x}} \right)$

$\Rightarrow x\frac{{dy}}{{dx}} = \frac{{dy}}{{dt}}$ 

$\frac{{{d^2}y}}{{d{x^2}}} = \frac{1}{x}\left( {\frac{{{d^2}y}}{{d{t^2}}} \cdot \frac{{dt}}{{dx}}} \right) - \frac{1}{{{x^2}}}\frac{{dy}}{{dt}}$ 

$= \frac{1}{{{x^2}}}\frac{{{d^2}y}}{{d{t^2}}} - \frac{1}{{{x^2}}}\frac{{dy}}{{dt}}$ 

$\Rightarrow {x^2}\frac{{{d^2}y}}{{d{x^2}}} = \frac{{{d^2}y}}{{d{t^2}}} - \frac{{dy}}{{dt}}$

Now the given differential becomes,

$\frac{{{d^2}y}}{{d{t^2}}} - \frac{{dy}}{{dt}} + \frac{{dy}}{{dt}} - y = 0$ 

$\Rightarrow \frac{{{d^2}y}}{{d{t^2}}} - y = 0$ 

Characteristic equation is: (m² - 1)y = 0

⇒ m = ±1

Roots of the characteristic equation are distinct real, so there are two possible solutions.

y = e^t and y = e^-t

since, x = e^t

$\Rightarrow y = x;;or;;y = \frac{1}{x}$  are two possible solutions.

Concept:

I) Inductor do NOT allow sudden change in current i.e. iL(0-) = iL(0+)

II) Capacitor do NOT allow sudden change in voltage i.e. VL(0-) = VL(0+)

Considering the initial conditions, Laplace transform of capacitor and inductor will be replaced as follows 

Calculation:

Given circuit diagram 

When the switch is at position 1 for a long time (time t < 0):

The capacitor will be completely charged to V₀₁ and it will become an open circuit

When the position of switch is changed from position 1 to position 2 at time t = 0⁺

Apply KVL in the loop and let I(s) be the current flowing in the circuit

$\frac{{ - {V_{02}}}}{S} + 2I\left( s \right)R + \left[ {\frac{1}{{SC}}} \right]\left[ {I\left( s \right)} \right] + \frac{{{V_{01}}}}{S} = 0$

$I\left( s \right) = \frac{{\frac{{{V_{01}}}}{S} - \frac{{{V_{02}}}}{S}}}{{2R + \frac{1}{{SC}}}} = \frac{{{V_{01}} - {V_{02}}}}{{S\left( {2R + \frac{1}{{SC}}} \right)}}\$

The voltage across the capacitor considering the initial conditions will be V_c(s)

${V_c} = \frac{{ - {V_{01}}}}{S} + \frac{1}{{SC}}I\left( s \right)$

Substitute the value of I(s) which is obtained above

$= \frac{{ - {V_{01}}}}{S} + \frac{1}{{SC}}\left[ {\frac{{{V_{01}} - {V_{02}}}}{{SR\left( {2 + \frac{1}{{SRC}}} \right)}}} \right]$

$\begin{array}{l} \frac{{ - {V_{01}}}}{S} + \frac{1}{{SC}}\left[ {\frac{{\left( {{V_{01}} - {V_{02}}} \right)}}{{SR\frac{{\left( {2RCS + 1} \right)}}{{SRC}}}}} \right]\ = \frac{{ - {V_{01}}}}{S} + \frac{1}{{SC}}\left[ {\frac{{{V_{01}} - {V_{02}}}}{{2RCS + 1}}} \right] \end{array}$

$\begin{array}{l} = \frac{{ - {V_{01}}}}{S} + \frac{{{V_{01}} - {V_{02}}}}{{2RC}}\left[ {\frac{1}{{S\left( {S + \frac{1}{{RC}}} \right)}}} \right]\ = \frac{{ - {V_{01}}}}{S} + \frac{{{V_{01}} - {V_{02}}}}{{2RC}}\left[ {\frac{1}{S} - \frac{1}{{S + \frac{1}{{2RC}}}}} \right]2RC\ = - {v_{01}} + \left( {{V_{01}} - {V_{02}}} \right)\left[ {1 - {e^{ - \frac{t}{{2RC}}}}} \right] \end{array}$

Additional Information

Q = CV

C = (εA)/d

$\frac{{{\varepsilon _1}}}{{{\varepsilon _2}}} = \frac{{{V_2}}}{{{V_1}}} \Rightarrow {V_2} = \frac{{{\varepsilon _1}}}{{{\varepsilon _1} + {\varepsilon _2}}}\left( V \right)$

$\Rightarrow 6 = \frac{{{\varepsilon _1}}}{{{\varepsilon _1} + {\varepsilon _2}}}\left( {10} \right)$

$\Rightarrow \frac{{{\varepsilon _1}}}{{{\varepsilon _2}}} = \frac{3}{2}$

Concept:

Response to sinusoidal/cosine input

Then, the steady state response to u(t) = cos(ω₀t) is given by,

$\rm y_{ss} (t) = |H(jω )|_{ω = ω_0} u(t + ∠ H (j ω_0))$....eq (1)

Calculation:

we have:

output, y(t) = A sin(3t + α)

input, u(t) = cos(3t)

Transfer function,

$\rm H(s) = \frac{Y(s)}{U(s) } = \frac{1}{s(s + 4)}$

Here, ω = ω₀ = 3 rad/sec

From equation (1)

$\rm y (t) = |H(jω )|_{ω = ω_0} u(t + ∠ H (j ω_0))$

$\rm |H(jω)|_{ω = 3}= \frac{1}{ω_0 \sqrt{ ω_0^2 + 4^2}}= \frac{1}{3 \sqrt {3^2 + 4^2}} = \frac{1}{15}$

$\rm ∠ H(j ω_0 )= -90^\circ - \tan^{-1} \left( \frac{ω_0}{4} \right)-90 ^\circ- tan^{-1} \left( \frac{3}{4} \right)$

∠H(jω₀) = -126.86°

so,

$\rm y(t) = \frac{1}{15} \cos (3t - 126.86^\circ)$

or,

$\rm y(t) = \frac{1}{15} \sin (3t - 126.89^\circ + 90^\circ)$

$\rm y(t) = \frac{1}{15} \sin (3t - 36.86^\circ)$

Now comparing above expression with given output expression y(t) = A sin(3t + α)

we get,

A = $\frac{1}{15}$

Given:

Impulse response, h(t) = e^-5t u(t)

output , Y(t) = e^-3t u(t) – e^-5t u(t)

Taking Laplace transform of h(t),

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

Taking Laplace transform of y(t),

$Y\left( s \right) = \frac{1}{{s + 3}} - \frac{1}{{s + 5}} = \frac{2}{{\left( {s + 3} \right)\left( {s + 5} \right)}}$

Now,

Y(s) = X (s) H(s)

$X\left( s \right) = \frac{{Y\left( s \right)}}{{H\left( s \right)}} = \frac{2}{{s + 3}}$

Taking inverse Laplace,

x(t) = 2e^-3t u(t)

Concept:

According to Thevenin’s theorem, any linear circuit across a load can be replaced by an equivalent circuit consisting of a voltage source Vth in series with a resistor Rth as shown:

Vth = Open circuit Voltage at a – b (by removing the load), i.e.

If a linear circuit contains dependent sources only, i.e., there is no independent source present in the network, then the open-circuit voltage or Thevenin voltage will simply be zero. (Since there is no excitation present)

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:

In the given circuit diagram Thevenin's resistance referred from the x-y side is nothing but the impedance of the transformer from the secondary side

​

Thevenin's resistance, 

Z_th = a2 Z1 = (2)² × 1Ω 

Zth = 4 Ω 

Thevenin's voltage is nothing but voltage from the secondary side 

V_xy = V_th

$\frac{{{N_2}}}{{{N_1}}} = \frac{{{V_2}}}{{{V_1}}}$

V₂ = V₁ × (N₂ / N₁)

⇒ V₂ = V_th = sin ωt × (2 / 1) = 2 sin ωt

Therefore the Thevenin's voltage equivalent voltage and impedance as seen from the terminals x and y for the circuit is 2 sin (ωt), 4 Ω

For a 3-ϕ, 4 poles self-excited Induction generator with frequency f₁, if the load is partially removed, frequency is f₂ and speed of generator maintained at 1500 rpm.

The speed vs torque curve of the induction generator is given below.

As the load decreases, the speed increases.

The relative speed between stator flux and rotor gets reduced which decreases the slip consequently, emf induced in the rotor and current also decreases. Hence frequency decreases.

f₂ > f₁ and f_1, f₂ < 50 Hz.

Concept:

Under no-load conditions or light load conditions, medium and long transmission lines may operate at the leading power factor due to the capacitance effect.

So that receiving end voltage becomes greater than sending end voltage.

In this case, shunt reactors are needed to bring down receiving end voltage at light loads.

The leading power factor can be changed to a lagging power factor by using a shunt reactor. By using a shunt reactor, it will compensate for the effect of capacitance and changes the power factor.

Note:

• The shunt capacitor is used to improve the power factor.
• A series reactor smoothens the wave shape.
• A Series capacitor reduces the net reactance in a line.
• The shunt inductor reduces the Ferranti effect by limiting overvoltages at the load side under lightly loaded conditions.

Concept:

State transition matrix:

It is defined as inverse Laplace transform of |sI - A|-1

⇒ L-1 |sI - A|-1 = eAt = ϕ(t)

General state equation: 

$\dot x\left( t \right) = A;x\left( t \right) + BU\left( t \right)$

Also, $\dot x\left( t \right) = \frac{{dx}}{{dt}}$

\(\left[ {\begin{array}{{20}{c}} {{{\dot x}_1}\left( t \right)}\ {{{\dot x}_2}\left( t \right)}\ \vdots \ \vdots \ {{{\dot x}_n}\left( t \right)} \end{array}} \right] = \left[ A \right]\left[ {\begin{array}{{20}{c}} {{x_1}}\ {{x_2}}\ \vdots \ \vdots \ {{x_n}} \end{array}} \right] + \left[ B \right]\left[ {\begin{array}{*{20}{c}} {{U_1}}\ {{U_2}}\ \vdots \ {{U_n}} \end{array}} \right]\)

Where, x1, x2, x3 …. xn are state variables

A is state matrix

B is the input matrix

Calculation:

Given state equation

\(\left[ {\begin{array}{{20}{c}} {{{\dot x}_1}}\ {{{\dot x}_2}} \end{array}} \right] = \left[ {\begin{array}{{20}{c}} 1&0\ 1&1 \end{array}} \right];\left[ {\begin{array}{{20}{c}} {{x_1}}\ {{x_2}} \end{array}} \right] + \left[ {\begin{array}{{20}{c}} 1\ 1 \end{array}} \right]u\)

\(A = \left[ {\begin{array}{{20}{c}} 1&0\ 0&1 \end{array}} \right],;B = \left[ {\begin{array}{{20}{c}} 1\ 1 \end{array}} \right]\)

State transition matrix.

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

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

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

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

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

\({L^{ - 1}};{\left[ {sI - A} \right]^{ - 1}} = \left[ {\begin{array}{*{20}{c}} {{e^t}}&0\ {t{e^t}}&{{e^t}} \end{array}} \right]\)

Concept:

• 8421 is also known as BCD code
• BCD is a weighted code.
• In weighted codes, each successive digit from right to left represents weights equal to some specified value, and to get the equivalent decimal number to add the products of the weights by the corresponding binary digit.

 

The following represents the 4-bit binary representation of decimal values: 

Explanation:

• A number with an 'n' decimal digit will require 4k bits in BCD.
• Ex- Decimal 396 is represented in BCD with 12 bits as 0011 1001 0110, with each group of 4 bits representing one decimal digit.
• As the decimal digits relating to 1010, 1011, 1100, 1101, 1110, and 1111 do not exist, these are invalid.

In BCD valid states are from 0 to 9. So 1100 is an invalid state.

Concept:

(V / F) control or frequency control method: 

It is basically frequency control but to maintain Bmax constant

The frequency variations should be done by keeping the (V / f) ratio constant.

The synchronous speed of the induction motor in rpm is

${N_S} = \frac{{120f}}{P}$ rpm 

NS ∝ f

NS = Synchronous speed

f = Frequency

P = No. of poles

Torque produced in the motor is

$T = \frac{{3 \times 60}}{{2\pi {N_S}}} \times \frac{{sE_2^2{R_2}}}{{R_2^2 + {{\left( {s{X_2}} \right)}^2}}}$

E2 = Rotor induced emf

R2 = Rotor resistance

X2 = Rotor reactance

s = slip      

${E_{ph}} = 4.44\phi nf;;$

$\begin{array}{l} E ∝ V\ \phi ∝ {B_{max}} ∝ \frac{V}{f} \end{array}$

The Torque - Speed characteristics:

Explanation:

$\frac{V}{f}$ is maintained constant from f = 0  to Base value in V/f control of IM.

Now, at low frequency, the effect of resistance can not be neglected as compared to reactance and must be compensated by increasing the stator voltage.

V/f = constant which implies that Torque will be constant. But at low frequency, magnitude of air gap flux decreases due to resistive effect. As a result, the stator flux decreases from its rated value, as at low frequency, flux density increases, which results in saturation of stator. Hence to avoid saturated, stator flux should decrease.

Given integral \(\mathop \smallint \limits_0^8 \left( {\mathop \smallint \limits_{\frac{y}{2}}^{\frac{y}{2} + 1} \left( {\frac{{2x - y}}{2}} \right)dx} \right)dy\)

$u = \frac{{2x - u}}{2}$

$\Rightarrow u = x - \frac{y}{2}$

⇒ du = dx

$v = \frac{y}{2}$

⇒ 2dv = dy

Limits of u:

at x = y/2, u = 0

at $x = \frac{y}{2} + 1,;;u; = \frac{{2\left( {\frac{y}{2} + 1} \right) - y}}{2} = 1$

 Limits of v:

At y = 0, v = 0

At y = 8, v = 4

Therefore integral becomes:

\(\mathop \smallint \limits_0^4 \left( {\mathop \smallint \limits_0^1 2;u;du} \right)dv\)

Concept:

The method of finding Maxima and Minima of y = f(x)

• Find f’(x) and f”(x) for given function y = f(x)
• Equate f’(x) to zero to obtain stationary points x = a
• Calculate f”(x) at each stationary points x = a (i.e f”(a))

The following three conditions are obtained:

1. If f”(a) > 0 then f(x) has a minimum at x = a and minimum value will be f(a)
2. If f”(a) < 0 then f(x) has a maximum at x = a and maximum value will be f(a)
3. If f”(a) = 0 then f(x) may or may not have a maximum or a minimum at x = a

 

Calculation:

f(x) = x³ – 3x² – 24x + 100

to find minimum value of function f(x) on [-3, 3]

Now, the given function is a polynomial and so continuous everywhere. Now to find derivative of the function to find the critical points:

f’(x) = 3x² – 6x – 24

Put f’(x) = 0 for critical point –

3x² – 6x – 24 = 0

x² – 2x – 8 = 0

⇒ (x - 4)(x + 2) = 0

x = 4, x = -2

We consider the critical point that actually fall in the given interval [-3, 3]. So only take x = -2. Now, we find the function value at the critical point x = -2 and end points x = -3, x = 3

Now, f(x = -2) (-2)³ – 3(-2)² – 24(-2) + 100

f(-2) = 128

f(x = -3) = (-3)³ – 3(-3)² – 24(-3) + 100

f(-3) = 118

f(x = 3) = (3)³ – 3(3)² – 24(3) + 100

f(3) = 28

from the above values we can see that the absolute minimum is 28 which occurs at x = 3

Concept:

Dual Clipper: Circuit which removes that part of i/p signal which lies above greater reference voltage and also which lies before smaller resistance.

Explanation:

Assume both the diodes are OFF.

${V_o} = {V_i}\left( {\frac{{10}}{{10 + 10}}} \right) = \frac{{{V_i}}}{2}$

For proper clipping operation, both the diode cannot be forward biased together.

When both diodes are OFF, ${V_o} = \frac{{{V_i}}}{2}$

Let V_a1, V_c1 are voltage of anode and cathode of diode D₁ and V_a2, V_c2 are voltage of anode and cathode of diode D₂.

So,

${V_o} = {V_{a1}} = \frac{{{V_i}}}{2}$

${V_o} = {V_{c2}} = \frac{{{V_i}}}{2}$

To make both diodes forward biased,

V_a1 > -1 V and V_c2 < -2 V

$\Rightarrow \frac{{{V_i}}}{2} > - 1;V;and\frac{{{V_i}}}{2} < - 2;V$

⇒ V_i = -2 to -4 V range does not allow proper clipping operation.

Concept:

Superposition Theorem:

• It is stated that in any linear, active, bilateral network having more than one source, the response across any element is the Algebraic sum of the response obtained from each source considered separately and all other sources are replaced by their internal resistance.
• The principle of the superposition theorem is based on Linearity.
• Voltage Source  →   short
• Current source   →  open
• Do not disturb the dependent source present in the network.

 

Step to solving Network by superposition theorem

• Step 1 – Take only one independent source of voltage or current and deactivate the other sources.
• Step 2 – If there is a voltage source then short circuit it and if there is a current source then just open-circuit it.
• Step 3 – Thus, by activating one source and deactivating the other source find the current in each branch of the network.
• Step 4 – Now to determine the net branch current utilizing the superposition theorem, add the currents obtained from each individual source for each branch.
• Step 5 – If the current obtained by each branch is in the same direction then add them and if it is in the opposite direction, subtract them to obtain the net current in each branch.

 

Calculation:

By using the superposition theorem,

The voltage across the capacitor ${V_{{C_1}}}\left( t \right)$ when 20 sin 10t voltage source acting, current source will be replaced with the open circuit

There won't be any current through the inductor and the voltage will be applied across the resistor and the capacitor 

$H\left( {j\omega } \right) = \frac{{\frac{1}{{j\omega c}}}}{{R + \frac{1}{{j\omega c}}}} = \frac{1}{{\left( {10j + 1} \right)}}$

${V_{{C_1}}} = \frac{1}{{\sqrt {101} }}20\sin \left( {10t - {{\tan }^{ - 1}}\left( {10} \right)} \right)$

The voltage across the capacitor ${V_{{C_2}}}\left( t \right)$ when 10 sin 5t current source is acting, voltage source will be replaced with the short circuit

${V_{{C_2}}}\left( t \right) = \frac{{10\angle 0}}{{1 - 0.2j}} \times \left( { - 0.2j} \right)$

${V_{{C_2}}}\left( t \right) = \frac{2}{{\sqrt {1 + {{\left( {0.2} \right)}^2}} }}{\text{sin}}\left( {5t - {\theta _2}} \right)$

${V_{{C_2}}}\left( t \right) = 1.98;{\text{sin}}\left( {5t - {\theta _2}} \right)$

Therefore from the superposition theorem 

${V_{{C}}}\left( t \right) = {V_{{C_1}}}\left( t \right) + {V_{{C_2}}}\left( t \right)$

V_C(t) = 2 sin (10t – θ₁) + 1.98 sin (5t – θ₂)

By comparing with given expression 

A₁ = 2, A₂ = 1.98

The given vectors are along the x-axis and y-axis respectively.

The point where H̅ is to be found in (2, 1, 0) which is on the x-y plane 

Magnetic field intensity of point P due to 4a_x is ${H_x} = \frac{1}{{2π \rho }}a\phi = \frac{4}{{2π \left( 1 \right)}}\left( {{a_x} × {a_y}} \right) = \frac{2}{π }{a_z}$

Magnetic field intensity of point P due to 2a_y is ${H_y} = \frac{1}{{2π \rho }}a\phi = \frac{2}{{2π \left( 2 \right)}}\left( {{a_y} × {a_x}} \right) = \frac{{ - 1}}{{2π }}{a_z}$

H = H_x + H_y

_{$H = \frac{1}{π }\left( {2 - \frac{1}{2}} \right) = \frac{3}{{2π }}{a_z}$}

Concept:

The poles of the system are the roots of the characteristic equation.

The characteristic equation is: |zI – A| = 0

Calculation:

From the given representation, the matrix A

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

\(zI - A = \left[ {\begin{array}{{20}{c}} z&0\ 0&z \end{array}} \right] - \left[ {\begin{array}{{20}{c}} a&{a - 1}\ {a + 1}&a \end{array}} \right]\)

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

For a = 1,

Characteristic equation: |zI – A| = 0

⇒ (z – 1)² = 0

⇒ z = 1 + j0

The roots of characteristic equation gives the system poles.

Concept:

Final value theorem,

$Y(∞) = \displaystyle\lim _{Z \rightarrow 1} ( 1 - Z^{-1}) Y(Z)$

Calculation:

We have:

$\frac{Y(z)}{X(z)} = \frac{1}{N} \left( \frac{1 - Z^{-N}}{1 - Z^{-1}} \right)$....(1)

and x(t) = 2 + 5sin(100πt)

sampling frequency, f_s = 400 Hz

put t = nT_s, the output of the sampling process is,

x(nT_s) = 2 + sin(100πnT_s)

x(nT_s) = 2 + 5 sin$\left( 100 π n \times \frac{1}{400} \right)$

x(nT_s) = 2 + 5$\sin \left(\dfrac{n π}{4} \right)$

where,

ω₀ = π/4

$N = \frac{2\pi}{\omega_0} = \frac{2\pi}{\frac{\pi}{4}} =8$

The z-transform of x(n) is,

$\rm ZT[x(n)] = ZT \left[ 2 + 5 \sin \left( \frac{\pi n}{4} \right) \right]$

$\rm X(Z) = 2 + \frac{5Z \sin \left( \frac{\pi} {4} \right)}{Z^2 - 2Z \cos \left( \frac{\pi}{4} \right) + 1}$

$\rm X(Z) = 2 + \frac{\frac{5Z}{\sqrt 2}}{Z^2 - \frac{2z}{\sqrt Z } + 1}$

$X(Z) = 2+ \frac{2.5 \sqrt 2 Z}{Z^2 - \sqrt 2 Z + 1}$

From equation (i)

$\frac{Y(z)}{X(z)} = \frac{1}{N} \left( \frac{1 - Z^{-N}}{1 - Z^{-1}} \right)$

$\rm Y(Z) = \frac{1}{8} \left( \frac{1 - Z^{-8}}{1 - Z^{-1}} \right)X(Z)$

$\rm Y(Z) = \frac{1}{8} \left( \frac{1 - Z^{-8}}{1 - Z^{-1}} \right) \left[ 2+ \frac{2.5 \sqrt 2 Z}{Z^2 - \sqrt 2 Z + 1} \right]$

using final value theorem,

$Y(∞) = \displaystyle\lim _{Z \rightarrow 1} ( 1 - Z^{-1}) Y(Z)$

$Y(∞) = \displaystyle\lim _{Z \rightarrow 1} ( 1 - Z^{-1}) \frac{1}{8} \left( \frac{1 - Z^{-N}}{1 - Z^{-1}} \right) \left[ 2+ \frac{2.5 \sqrt 2 Z}{Z^2 - \sqrt 2 Z + 1} \right]$

$Y(∞) = \displaystyle\lim _{Z \rightarrow 1} \frac{1}{8}( 1 - Z^{-8}) \left[ 2+ \frac{2.5 \sqrt 2 Z}{Z^2 - \sqrt 2 Z + 1} \right]$

y(∞) = 0

The given signal is 10 kHz even symmetrical square wave.

Since the bandpass filter is centered at 30 kHz it follows half-wave symmetry. Hence only odd harmonics exist.

The frequency components present in this wave: 10 kHz, 30 kHz, 50 kHz, 70 kHz.

Now, 30 kHz component will pass through filter output is nearly perfect cosine wave at 10 kHz cosine as the signal is even signal.

Concept: 

When the slip rings are shorted, the induced current in rotor oppose the relative motion between rotor conductor & stator field according to the Lenz's law. So, the rotor also starts rotating in the same direction as that of stator field. As the rotor rotates in clockwise direction, the direction of stator field is also in clockwise direction.

When the slip rings are opened, then rotor is rotate in the Anticlockwise direction.

Explanation: 

Rotor is made to rotate with speed

$N_r = \frac{120 f_r}{p}$ (in ACW direction) when slip ring are opened.

Relative motion between rotor & stator field

= N_s + N_r

Therefore frequency of induced EMF in rotor winding

$\rm f_1 = \frac{p}{120} \times ( Relative \ motion)$

$\rm f_1 = \frac{p}{120} \times (N_s + N_r)$

$\rm f_1 = \frac{N_sP}{120} + \frac{N_rP}{120}$

f_{1 = f + fr}

At the same time, the frequency of voltage across commutator brushes is same as stator frequency.

⇒ f₂ = f

Distribution factor:

• The Distribution Factor is also known as the Breadth Factor or Belt factor or Spread factor.
• It is defined as the ratio of the actual voltage obtained to the possible voltage if all the coils of a polar group were concentrated in a single slot.
• It is denoted by Kd

                   Kd = A / B

Where,

           A is the vector sum of induced EMF

           B is the arithmetic sum of induced EMF

Distribution factor Kd

                              ${K_d} = \frac{{\sin \frac{{mβ }}{2}}}{{m\sin \frac{β }{2};;;}}$

Where, 

           m is slots per pole per phase

           β angular displacement of the slots

 

Single layer winding: One coil side occupies one slot completely, Number of coils (C) is equal to the half of the number of slots (S)

Double layer winding: Number of coils (C) is equal to the number of slots (S)

Integral slot winding: Number of slots per pole per phase is an integer

Fractional slot winding: Number of slots per pole per phase is not an integer

Full pitched winding: Coil span is equal to slots per pole (S/P)

Short pitched winding: Coil span is less than slots per pole (S/P)

Calculation:

Given that, poles of alternator = 20

Stator slots = 180

Conductors in each slot = 6

For single-phase winding

No. of coils in phase, $n = \frac{{180}}{{20 \times 1}} = 9$ 

Slots span in electrical degrees, β =180/(no. of slots per pole)

$\Rightarrow \beta = \frac{{180}}{{180/20}} = 20^\circ$

Distribution factor ${k_d} = \frac{{\sin n\beta /2}}{{n\sin \beta /2}}$ 

$= \frac{{\sin 9 \times 20/2}}{{9\sin 20/2}}$

k_d1 = 0.6398

No. of turns N₁ = No. of slots × (conductors/2)

$= 180 \times \frac{6}{2} = 540$

For three-phase winding

Slot span, β = 20°

No. of coils in a phase $= \frac{{180}}{{20 \times 3}} = 3 = n$

Distribution factor ${k_d} = \frac{{\sin 3 \times 20/2}}{{3\sin 20/2}}$

k_d2 = 0.9597

No. of turns N₂ = no. of slots × (conductors/2) × (1/number of phases)

$= 180 \times \frac{6}{2} \times \frac{1}{3}$

N₂ = 180

Now, the RMS value of generated emf for single phase is

V₁ = 4.44 k_d1 ϕ N₁

For 3-phase, the RMS value of generated emf is

V₂ = 4.44 k_d2 ϕ N₂

Now, $\frac{{{V_1}}}{{{V_2}}} = \frac{{{k_{d1}} \cdot {N_1}}}{{{k_{d2}} \cdot {N_2}}} = \frac{{0.6398 \times 540}}{{0.9597 \times 180}}$ 

$\Rightarrow \frac{{{V_1}}}{{{V_2}}} \approx 2$

Concept:

​Hysteresis loss:

It is due to the reversal of magnetization of the transformer core whenever it is subjected to the alternating nature of the magnetizing force.

The power consumed by the magnitude domains to change their orientation after every half cycle whenever core is subjected to alternating nature of magnetizing force is called as hysteresis loss.

​Hysteresis loss can be determined by using the Steinmetz formula  given by

${W_h} = \eta B_{max}^2fV$

Where

x is the Steinmetz constant, Bm = maximum flux density

f = frequency of magnetization or supply frequency, V = volume of the core

Eddy current losses: 

Eddy current loss is basically I2R loss present in the core due to the production of eddy currents in the core, because of its conductivity.

Eddy current losses are directly proportional to the conductivity of the core.

Eddy current loss ${W_e} = K{f^2}B_m^2{t^2}$

Where

K - coefficient of eddy current., Bm - maximum value of flux density 

t - thickness of lamination in meters, f - frequency of eddy current

If (V / f) ratio is constant:

As we know,

Bmax ∝ (V / f)

⇒ Bmax = constant

⇒ Hysteresis loss Wh ∝ f

And, eddy current loss We ∝ f2

If V/f is not constant:

Hysteresis losses

$\Rightarrow {W_h} \propto V_1^{1.6}{f^{ - 0.6}}$

Eddy current losses

$\Rightarrow {W_e} \propto V_1^2$

Explanation:

Supply frequency (f) and voltage (V) increased by 10%.

${B_m} \propto \frac{V}{f}$

Hysteresis losses:

P_h ∝ f

Therefore, change in hysteresis loss

⇒ ΔP_h = Δf = 10%

Eddy current losses:

P_e ∝ f²

$\Rightarrow \frac{{{p_{e1}}}}{{{p_{e2}}}} = \frac{{f_1^2}}{{f_2^2}}$

Now, f₂ = 1.1 f₁

$\Rightarrow \frac{{{p_{e1}}}}{{{p_{e2}}}} = \frac{1}{{1.1}} \Rightarrow {p_{e2}} = \left( {1.21} \right){p_{e1}}$

$\Rightarrow {\rm{\Delta }}{p_e} = \frac{{{p_{e2}} - {p_{e1}}}}{{{p_{e1}}}} = \frac{{1.21 - 1}}{1} \times 100\%$

⇒ Δp_e = 21%

Concept:

The capacitance of line:

The capacitance of each conductor to neutral is given by,

$C_{an}=\frac{2\pi\epsilon_{0}}{ln(\frac{d}{r})}$

Where,

d = distance between the conductors 

r = radius of conductors

Calculation:

Diameter of conductor = 1.6 m

Radius, r = 0.8 m

Distance d = 6m

The capacitance of each line os

$C = \frac{{2\pi {\varepsilon _o}}}{{\ln \left( {\frac{d}{r}} \right)}} = \frac{{2\pi \times 8.85 \times {{10}^{ - 12}}}}{{\ln \left[ {\frac{6}{{0.8 \times {{10}^{ - 2}}}}} \right]}}$

= 8.4 × 10^-12

The capacitance per km will be

$= \frac{C}{{km}} = 8.4 \times {10^{ - 9}}F$

Concept:

Bode plot transfer function is represented in standard time constant form as

 $T\left( s \right) = \frac{{k\left( {\frac{s}{{{\omega _{{c_1}}}}} + 1} \right) \ldots }}{{\left( {\frac{s}{{{\omega _{{c_2}}}}} + 1} \right)\left( {\frac{s}{{{\omega _{{c_3}}}}} + 1} \right) \ldots }}$

ωc1, ωc2, … are corner frequencies.

k is the constant gain term

Calculation:

Given transfer function is 

$G\left( s \right) = \frac{{5\left( {s + 4} \right)}}{{s\left( {s + 0.25} \right)\left( {{s^2} + 10s + 25} \right)}}$

$= \frac{{5 \times 4\left( {1 + \frac{s}{4}} \right)}}{{0.25 \times 25 \times s \times \left( {1 + \frac{s}{{0.25}}} \right)\left( {1 + \frac{{10s}}{{25}} + \frac{{{s^2}}}{{25}}} \right)}}$

$= \frac{{3.2\left( {1 + \frac{s}{4}} \right)}}{{s\left( {1 + \frac{s}{{0.25}}} \right)\left( {1 + \frac{{10s}}{{25}} + \frac{{{s^2}}}{{25}}} \right)}}$

Constant gain = 3.2

Corner frequencies = 0.25, 4, 5

Highest corner frequency = 5 rad/sec

Concept:

Controllability:

It is the internal states of the system are changed from one value to another value in a finite time by a finite input, then we can say the system is controllable otherwise it is not controllable.

To check Controllability, consider the controllability matrix (ϕc)

\(\left[ {{\phi c}} \right] = {\left[ {{A^0}B;;;{A^1}B} \right]{2 \times 2}};;;;\left[ {{\phi c}} \right] = {\left[ {{A^0}B;;;;;{A^1}B;;;;{A^2}B} \right]{3 \times 3}}\)

If |ϕc| = 0; then the system is uncontrollable

If |ϕc| ≠ 0, then the system is controllable.

Observability:

If the internal states of the system can be evaluated from the output of the system of any time, then we can say, the system is observable otherwise it is not observable.

To check observability, consider the observability matrix

\(\left[ {{\phi 0}} \right] = {\left[ {{C^T};;;;A{C^T}} \right]{2 \times 2}}; \to {2^{nd}};order:::;:::\left[ {{\phi 0}} \right] = {\left[ {{C^T};;;;;A{C^T};;;;;{A^2}{C^T}} \right]{3 \times 3}}\)

If |ϕo| = 0; Then system is not-observable

If |ϕo| ≠ 0; then the system is observable.

Calculation:

Controllability matrix, C = [Q PQ]

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

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

|C| = -1 ≠ 0

System is controllable.

Observability matrix, \(O = \left[ {\begin{array}{*{20}{c}} R\ {RP} \end{array}} \right]\)

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

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

|O| = 0

System is not observable.

R1 = 300 ± 1% = [297, 303]

R2 = 200 ± 1% = [198, 202]

Actual Reff $= \frac{{{R_1} \times {R_2}}}{{{R_1} + {R_2}}} = \frac{{200 \times 300}}{{500}} = 120$

R1 = 297, R2 = 198

Reff = 118.8

R1 = 303, R2 = 202

Reff = 121.2

⇒Reff = [118.8, 121.2]

= 120 ± 1.2 Ω 

= 120 ± 1 %

Tolerance = (118.8 - 120) / 120 = - 0.01

Tolerance = (121.2 - 120) / 120 = 0.01

⇒ Tolerance = 1 %

Concept:

The SOP representation of the circuit is:

F = Σm (minterms)

The POS representation of the circuit:

F = ΠM (max terms)

Calculation:

Given,

The SOP (sum of products) form of a Boolean function is ∑(0,1,3,7,11)

But the options are in POS (Product Of Sums) form 

Therefore minimize in POS form using k-map as follows:

Minimized expression

$= \left( {\bar B + C} \right)\left( {\bar A + C} \right)\left( {\bar A + \bar B} \right)\left( {\bar C + D} \right)$

Analysis 

Step 1:

The characteristic table of the required flipflop i.e JK flipflop is:

 

Step-2:

The excitation table of the given flip-flop i.e. T flipflop is:

 

Step-3:

Finding the boolean expression

 

\(T = {̅ Q_n}J + QK\)

= JQ̅_n (1 + K) + KQ_n (1 + J)

= JQ̅_n + JQ̅_nK + JKQ_n + KQ_n

= JQ̅_n + KQ_n + JK(Q_n + Q̅_n)

= (J + Q_n) (Q̅_n + K)

Logic diagram of the above expression will be 

2001 H XRA A; A ← 00H

2001 H    MVI B, 04 H; B ← 04 H

2003 H     MVI A, 03 H; A ← 03 H

2005 H   RAR (Rotate Accumulator Right with carry)

2006 H     DCR B; B ← 03 H

2007 H     JNZ 2005; Jump to 2005 H till B does not come to zero.

The below table shows the values of A for each iteration.

 

The program halts after B comes to 00 H.

Therefore, at the end of the program, register A contains 60H.

Given,

V_s = 230 V, 50 Hz

R = 5 Ω, L = 16 mH

The maximum firing angle at which the voltage across the device becomes zero is the angle at which device trigger i.e., maximum firing angle to converter

α = ϕ = tan^-1 (ωL/R)

$= {\tan ^{ - 1}}\left( {\frac{{2π × 50 × 16 × {{10}^{ - 3}}}}{5}} \right) = 45^\circ$

Current flowing through SCR is max at this angle

α = ϕ,  γ = π

\(\begin{array}{l} {I_{Trms}} = {\left( {\frac{1}{{2π }}\mathop \smallint \limits_\alpha ^{π + \alpha } {{\left( {\frac{{{V_m}}}{2}\sin \left( {ω t - \alpha } \right)} \right)}^2}.dω t} \right)^{\frac{1}{2}}}\ {I_{Trms}} = \frac{{{V_m}}}{{2z}} = \frac{{√ 2 × 230}}{{2 × √ {{5^2} + 5.042} }}\ = {\rm{ }}23{\rm{ }}A \end{array}\)

Alternate Method:

Each for half of the cycle, Thyristor T₁ & T₂ conducts alternate pattern in order to carry load current (I₀)

So, RMS load voltage (V_or) = V_s = 230 V

$\rm \Rightarrow I_{or} = \frac{V_{or}}{|Z|}$

where,

load impedance (Z) = R + jωL

Z = 5 + j(2π × 50 × 16 × 10^-3)

$|Z| = √ {5^2 + (2 \pi \times 50 \times 16 \times 10^{-3})^2}$

|Z| = 5√2 Ω

$\rm \Rightarrow I_{or} = \frac{230}{5\sqrt 2} \ A$

Thyristor rms current,

$I_{T(rms)} = \frac{I_{or}}{\sqrt 2} = \frac{(230/5\sqrt 2)}{\sqrt 2}$

I_T(rms) = 23 A