GATE EE 2011 Official Paper (Previous Year Paper)

Explanation:

Let; Total number of voters = X

Number of voters P was supposed to get = 0.4X

Number of voters Q was supposed to get =  0.6X

According to the given information;

Number of voters P got = 85/100 (0.4X) + 25/100 (0.6X)

Number of voters Q got = 75/100 (0.6X) + 15/100 (0.4X)

Q - P = 2

⇒ 75/100 (0.6X) + 15/100 (0.4X) - 85/100 (0.4X) - 25/100 (0.6X) = 2

⇒ 50/100 (0.6X) - 70/100 (0.4X) = 2

⇒ 30X - 28X = 200

⇒ 2X = 200

⇒ X = 100

Hence, total number of voters are 100.

The correct answer is 'exacerbated'.

Key Points

• Let's explore options;
• identified: establish or indicate who or what (someone or something) is.
• Example: Even the smallest baby can identify its mother by her voice.
• ascertained: find (something) out for certain; make sure of.
• Example: The police have so far been unable to ascertain the cause of the explosion.
• exacerbated: make (a problem, bad situation, or negative feeling) worse.
• Example: This attack will exacerbate the already tense relations between the two communities.
• analysed: discover or reveal (something) through detailed examination.
• Example: Researchers analysed the purchases of 6,300 households.
• Thus, from above, we can conclude that 'exacerbated' will be used in the blank because of the keyword 'counterproductive' which suggests technocrats must have done something to make the situation worse.
• Therefore, the correct answer is option 3.

The correct answer is 'rarity'.

Key Points

• Frequency: the fact of being frequent or happening often.
• Example: Complaints about the frequency of buses rose in the last year.
• Rarity: the state or quality of being rare.
• Example: Snow in Florida is a rarity.
• Thus, the correct answer is option 2.

Additional Information

• Let's explore the other options:
• periodicity: the quality or character of being periodic; the tendency to recur at intervals.
• gradualness: Occurring or developing slowly or by small increments
• persistency: the continued or prolonged existence of something.

The correct answer is 'available'.

Key Points

• Let's explore options:
• similar: a person or thing similar to another.
• Example: My father and I have similar views on politics.
• most: a large number of.
• Example: What's the most you've ever won at poker?
• uncommon: out of the ordinary; unusual.
• Example: Accidents due to failure of safety equipment are uncommon nowadays.
• available: able to be used or obtained.
• Example: Is this dress available in a larger size?
• Thus, from above, we can conclude that the correct answer is option 4.

Therefore, the correct sentence is: Under ethical guidelines recently adopted by the Indian Medical Association, human genes are to be manipulated only to correct diseases for which available treatments are unsatisfactory.

The Gladiator fights in the Arena to win battles. 

Similarly; 

A lawyer fights in the court to win cases.

Hence, "Lawyer ∶ Courtroom" is the correct answer.

Additional Information1) Dancer ∶ Stage →

Dancer dances on the Stage.

2) Commuter ∶ Train →

Commuter travels in train or Traveller travels in Train. (commuter means traveler)

3) Teacher ∶ Classroom →

Teacher teaches in Classroom.

Calculation:

Consumption (km/liter) means distance covered by the car in 1 liter of fuel

⇒ Fuel consumption per liter = $\frac{1}{Consumption\ (km/liter)}$

From the above table, fuel consumption per km was least during the lap Q.

∴ The correct answer is Q.

Let the total number of toffees in bowl be x

R took 1/3 of toffees and returned 4 to the bowl

∴ Number of toffees with R = x/3 - 4

Remaining of toffees in bowl = 2x/3 + 4

Number of toffees with $S = \frac{1}{4}[\frac{2}{3}x+4]-3$

Remaining toffees in bowl = $\frac{3}{4}[\frac{2}{3}x+4]+3$

Number of toffees with T = $\frac{1}{2}[\frac{3}{4}(\frac{2}{3}x+4)+3]-2$

Remaining toffees in bowl = $\frac{1}{2}[\frac{3}{4}(\frac{2}{3}x+4)+3]+2$

Given: $\frac{1}{2}[\frac{3}{4}(\frac{2}{3}x+4)+3]+2=17$

⇒ x = 48

∴ There were 48 toffees in the bowl.

Given

f(y) = |y|/y

f(q) = |q|/q      ----(1)

f(-q) = |-q|/(-q)

f(-q) = |q|/-(q) 

f(-q) = - |q|/q     ----(2)

From (1) and (2)

|f(q) - f(-q)| = 2|q|/q = 2

∴ The correct answer is 2.

Let S = 4(1 + 11 + 111 + ...) = $\frac{4}{9}(9 + 99 + 999 + ...)$

$\Rightarrow\frac{4}{9}[(10 - 1) + (10^2-1)+(10^3-1)+...]$

$\Rightarrow\frac{4}{9}[(10 + 10^2 +...10^n)-n]=\frac{4}{9}[10\frac{(10^n-1)}{9}-n]$

$\Rightarrow \frac{4}{81}[10^{n+1}-9n-10]$

The correct answer is 'generally quite immune to diseases'.

Key Points

• Let's refer to the following lines of the passage:
• Horses were injected with toxins of diseases until their blood built up immunities. Then a serum was made from their blood.
• From above it can be inferred that the horses are quite immune to diseases as they themselves build the immunities in their blood to fight the toxins.
• Thus, the correct answer is option 2.

Applying KCL in the given circuit,

I_s = I_RL + I_c

or, I_c = Is - I_RL

or, I_c = √2 ∠π/4 - √2 ∠-π/4 = j2 A

Given Block Diagram:

Given block diagram can be reduced by,

Further, It can be reduced by,

Hence,

$\frac{Y(S)}{R(S)}=\frac{1}{S^2+S(K+1)+1}$ .... (1)

The standard equation of 2nd order system can be written as,

TF = $\frac{ω_n^2}{S^2+2ζω_nS+ω_n^2}$ .... (2)

From equation (1) & (2),

ω_n = ±1

2ζω_n = (K + 1)

or, 2ζ = (K + 1)

or, $\zeta=\frac{K+1}{2}$

Peak over shoot (M_p) = ${\large e}^{(\frac{-\zeta π}{\sqrt{1-\zeta^2}})}$

We have,

Now,

$\frac{Y(S)}{U(S)}=\frac{1}{S(S+1+K)}$

Here, $ϕ = -\frac{π}{2}-tan^{-1}\frac{ω}{K+1}$

At: ω → 0 ⇒ ϕ = -(π/2)

At: ω → ∞ ⇒ ϕ = -(π)

Hence, K will only affct the Peak overshoot.

Concept:

The inductive reactance for an inductor operating in a frequency 'f' is given by:

X_L = jωL

L = inductance

Similarly, the capacitive reactance of a capacitor operating in a frequency 'f' is given by:

$X_c=\frac{1}{j\omega C}$

C = Capacitance

Also, for a sinusoidal signal, the RMS value for the given maximum amplitude is given by:

RMS value = 0.707 × Maximum value

Calculation:

Given supply voltage ω = 1 rad/sec

The inductive reactance of the inductor 1 H will be:

X_L = jωL = j Ω

And the capacitive reactance of the capacitor 1 F will be:

X_C = 1/jωC = -j Ω

Now, the given circuit is redrawn as:

The series combination of inductive and capacitive reactance will be:

j + (-j) = 0 Ω

The circuit diagram is redrawn as:

i(t) = v(t)/1 = sin t

From the above current equation:

Maximum value of current i(t) = 1

For sinusoidal waveform, RMS value = 0.707 × maximum value

RMS value of current i(t) = 0.707 A = 1/√2 A

\(f\left( t \right) = {a_0} + \mathop \sum \limits_{n = 1}^\infty ({a_n}\cos n\omega t + {b_n}\sin n\omega t)\)

The given function f(t) is an even function, therefore b_n = 0

f(t) is nonzero average value function, so it will have a nonzero value of a₀

\({a_0} = \frac{1}{{\frac{T}{2}}}\mathop \smallint \limits_0^{\frac{T}{2}} f\left( t \right)\ dt\) (average value of f(t)

a_n is zero for all even values of n and nonzero for odd n

\({a_n} = \frac{2}{T}\mathop \smallint \limits_0^T f\left( t \right)\) cos(nωt) d(ωt)

So Fourier expansion of f(t) will have a₀ and a_n, n=1, 3, 5, …..∞.

A  4 point starter is used to start and control the speed of a Dc shunt motor with field weakening control.

High speed protection of DC shunt motor is not provided by 4 point starter.

Four-point starter:

The four-point starter works as a current controlling device in the deficiency of back EMF while it starts running off the DC motor. A four-point starter also works as a protecting device. The main difference between a 4-point starter compared to a 3-point starter is, the holding coil is detached from the shunt-field circuit.

The 4-point starter uses four terminals for speeding up the motor. These four terminals namely, armature terminal (A), field terminal (F), and the line terminal (L).

• NVC (No Volt Coil): The connection of a four-point starter can be done in parallel with the field coil
• The line terminal (L) is connected to a positive supply
• The armature terminal (A) is connected to the winding of an armature
• The field terminal (F) is connected to the field winding
• It is provided as not to affect the current flowing through 'Hold on' coil even when the field current changes.

Important:

A resistor is used in the starter for a DC shunt motor

• When the connected dc motor is to be started, the lever is turned gradually to the right.
• When the lever touches point 1, the field winding gets directly connected across the supply, and the armature winding gets connected with resistances R1 to R5 in series.
• During starting, full resistance is added in series with the armature winding.
• Then, as the lever is moved further, the resistance is gradually is cut out from the armature circuit.
• Now, as the lever reaches to position 6, all the resistance is cut out from the armature circuit and armature gets directly connected across the supply.

A 3-phase synchronous motor, connected to infinite bus, is operating at no load at normal excitation. The field excitation of the motor is first decreased to zero and then increased in the reverse direction. The armature current of the synchronous motor will increase first and then decrease.

$SIL = \frac{{{V^2}}}{{{Z_C}}} = \frac{{400 \times 400}}{{400}} = 400MW$

The carry 500 MW a double circuit 400 kV line is used.

We say that 2 vectors are orthogonal if they are perpendicular to each other. i.e. the dot product of the two vectors is zero.

We have two vectors [1, 1, 1] and [1, a,a2]

Dot product = [1, 1, 1] ∙ [1, a,a2] = (1 + a + a2)

We have,

a = $\left(-\frac{1}{2}+j\frac{\sqrt{3}}{2}\right)$

(1 + a + a2) = 1 + $\left(-\frac{1}{2}+j\frac{\sqrt{3}}{2}\right)$ + $\left(-\frac{1}{2}+j\frac{\sqrt{3}}{2}\right)^2$

or, (1 + a + a2) = 1+ $\left(-\frac{1}{2}+j\frac{\sqrt{3}}{2}\right)$ + $(\frac{1}{4}-\frac{3}{4}-j\frac{\sqrt3}{2})$

or, (1 + a + a2) = 0

Hence, the given vector is orthogonal.

Concept:

The steady-state error for a system is defined as:

${e_{ss}} = \mathop {\lim }\limits_{s \to 0} \frac{{sR\left( s \right)}}{{1 + G\left( s \right)H\left( s \right)}}$

R(s) = Input

G(s) = open loop transfer function

H(s) = feedback gain = 1 for unity feedback system

Analysis:

For a unit step input, we have:

$R\left( s \right) = \frac{1}{s}$

The steady-state error becomes:

${e_{ss}} = \mathop {\lim }\limits_{s \to 0} \frac{{s\left( {\frac{1}{s}} \right)}}{{1 + G\left( s \right)}} = 0.1$

1 + G(0) = 10

G(0) = 9

Again we have the input as:

r(t) = 10 [u(t) – u(t - 1)]

$R\left( s \right) = 10\left[ {\frac{1}{s} - \frac{1}{s}{e^{ - s}}} \right]$

$= 10\left( {\frac{{1 - {e^{ - s}}}}{s}} \right)$

∴ The steady-state error for the input will be:

$e_{ss}' = \mathop {\lim }\limits_{s \to 0} \frac{{s \times 10\frac{{\left( {1 - {e^{ - s}}} \right)}}{s}}}{{1 + G\left( s \right)}}$

$= \frac{{10\left( {1 - {e^0}} \right)}}{{1 + 9}}$

$e_{ss}' = 0$

Features of LPF wattmeter:

The circuit of the LPF wattmeter is shown below.

• The pressure coil must be connected to the load side.
• Series multiplier resistance (R_p) slightly reduced.
• To eliminate the effect of inductive reactance(L_p) of the pressure coil, a capacitor must be connected in parallel with R_p.
• A compensating coil is placed over the current coil in order to cancel the field produced by the pressure coil, which means it compensates for the effect of the impedance of the voltage coil circuit.

Points to remember:

The value of the capacitor connected in parallel to R_p is

C = 0.41 (L_p / R_p²)

L_p = Inductance of the pressure coil

The frequency spectrum for a low pass filter with a cut-off frequency of 30 Hz can be drawn as:

Similarly, the frequency spectrum for a high-pass filter with a cut-off frequency of 20 Hz is as shown:

The cascaded system will have a response which will be the convolution of the two impulse responses.

In the frequency domain, the frequency response will be the multiplication of the two responses, i.e. the frequency response of the cascaded filter will be:

∴ It will act as a band-pass filter.

Concept:

When the symbol represents the input ↑, then the arrow part represents the potential of V_i, and the bottom part is the 0 V or ground.

The output of inverting OP-AMP is given by:

$V_{o1} = V_i({-R_2 \over R_1})$

V_o1 = -V_i

The second circuit is a Schmitt trigger circuit.

Positive feedback in the OP-AMP forms a Schmitt trigger circuit.

The output of the Schmitt trigger circuit is always either ±V_sat.

Case 1: When V_I < V_NI

V_o = +V_sat

Case 2: When VI > VNI

Vo = -V_sat

The input at V_NI is either V_UTP or V_LTP

${V_{UTP}} = +V_{sat}\left( {\frac{R}{{2R}}} \right)$

VUTP = 6V

${V_{LTP}} = -V_{sat}\left( {\frac{R}{{2R}}} \right)$

V_LTP = -6V

Calculation:

Case 1: When -Vi < 6V (or V_i > 6V)

V_O = +V_sat = 12

Case 2: When -Vi > 6V (or Vi < 6V)

VO = -Vsat = -12

The transfer characteristic curve is:

It has to carry unidirectional current and has to block bidirectional voltage.

Only option A allow bi direction power flow from source to the drive

$\bar Z$ is |Z| = 0 where θ is around 45° or so

$\therefore \bar Z$ = |Z| with angle 45° where |Z| <1

$\begin{array}{l} \bar Y = \frac{1}{z} = \frac{1}{{\left| Z \right|angle\left( {45^\circ } \right)}} = \frac{1}{{\left| Z \right|}}angle\left( { - 45^\circ } \right)\ \left| {\bar Y} \right| > 1(because\left| Z \right| < 1) \end{array}$

So Y will be out of unity circle.

$\begin{array}{l} V\left( t \right) = 100\sqrt 2 \cos \left( {100\pi t} \right)V\ i\left( t \right) = 10\sqrt 2 \sin \left( {100\pi t + \frac{\pi }{4}} \right)A \end{array}$

Taking the voltage as the reference phasor,

Voltage phasor, $\vec V = 100\sqrt 2 {e^{j0}}V$

For $i\left( t \right) = 10\sqrt 2 \sin \left( {100\pi t + \frac{\pi }{4}} \right)A$

$\begin{array}{l} = 10\sqrt 2 \cos \left( {100\ \pi t + \pi /4} \right) + 3\pi /2\ = 10\sqrt 2 \cos \left( {100\ \pi t + 2\pi - \pi /4} \right)\ = 10\sqrt 2 \cos \left( {100\ \pi t - \pi /4} \right) \end{array}$

∴ Current phasor, $\vec I$ with $\vec V$ as the reference

$= 10 {e^{ - j\pi /4}} = 10\ \angle - \pi /4$

For maximum power transfer to R_L,

R should be zero so that maximum current flows through the load resistance. and hence maximum power is transferred to the load resistance.

Common Mistake:

Maximum power theorem states that for maximum power to be transferred to the load resistance RL, RL must equal the Thevenin Equivalent resistance, i.e.

RL = Rth

But here, we are asked to find the value of R, and not RL that will result in the maximum power to be transferred to load RL. So we cannot go by the standard procedure of equating RL with the Thevenin equivalent resistance.

Concept:

Some important convolution results are

• u(t) * u(t) = t u(t)
• e­^-at u(t) * e­^-at u(t) = t e­^-at u(t)
• e­^-t u(t) * u(t) = (1- e­^-t) u(t)
• e­^-at u(t) * e­^-bt u(t) $= \frac{1}{{\left( {b - a} \right)}}{e^{ - at}} - {e^{ - bt}};u\left( t \right)$
• $t;{\rm{u}}\left( {\rm{t}} \right){\rm{*u}}\left( {\rm{t}} \right) = \frac{{{{\rm{t}}^2}}}{2}{\rm{u}}\left( {\rm{t}} \right)$

Calculation:

x(t) = e^-t and y(t) = e^-2t

Now convolution

z(t) = x(t) * y(t)

z(t) = e^-t – e^-2t

∵ air cored

∵ 

∴ i = 

Negative sequence relay:

• It protects generators from the unbalanced load by detecting negative sequence current.
• A negative sequence current may cause a dangerous situation for the machine.
• Phase to phase fault mainly occurs because of the negative sequence component.
• The negative sequence relay has earthing which protects from phase-to-earth fault but not from phase-to-phase fault.

Note: 

$P = \frac{{{V^2}}}{X}$

Series capacitive will reduce the reactance. So power transmission will improve.

Concept:

Minimum phase system: It is a system in which poles and zeros will not lie on the right side of the s-plane.  In particular, zeros will not lie on the right side of the s-plane.

For a minimum phase system,

$\mathop {\lim }\limits_{\omega \to \infty } \angle G\left( s \right)H\left( s \right) = \left( {P - Z} \right)\left( { - 90^\circ } \right)$

Where P & Z are finite no. of poles and zeros of G(s)H(s)

Non-minimum phase system: It is a system in which some of the poles and zeros may lie on the right side of the s-plane. In particular, zeros lie on the right side of the s-plane.

Stable system: A system is said to be stable if all the poles lie on the left side of the s-plane.

Application:

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

As one zero lies in the right side of the s-plane, it is a non-minimum phase transfer function.

As there no poles on the right side of the s-plane, it is a stable system.

Given bridge is Maxwell inductance – Capacitance Bridge and it is suitable for the measurement of medium ‘Q’ inductor.

Important Points

Dual Trace CRO:

• In a dual-trace oscilloscope, a single electron beam generates 2 traces, that undergoes deflection by two independent sources.
• In order to produce two separate traces, basically, 2 methods are used, known as an alternate and chopped mode.
• Alternate mode of Dual Trace Oscilloscope:
• Whenever we activate the alternate mode, then it permits the connection between both the channels alternately.
• The control input of the multiplexer used in the circuit is fed with a signal having a frequency equal to the frequency of the time base (sweep) oscillator.
• This alternation or switching between channels A and B takes place at the beginning of each upcoming sweep.
• Also, there exists synchronization between the switching rate and the sweep rate.
• This leads to the spotting of traces of each channel on one sweep. Like in the first sweep traces of channel A will be spotted, then in the next sweep traces of channel B will be considered by the CRT.
• In this way, the alternate connection of the two-channel input with the vertical amplifier is performed.
• The change in the electronic switch from one channel to the other occurs at flyback sweep duration. At the flyback period, the electron beam will be invisible and so the changeover from one channel to other.
• Hence a complete sweep signal from one vertical channel will be displayed on the screen. While for the next sweep, the signal from another vertical channel will be displayed.
  

The figure below represents the waveform the oscilloscope output operating in alternate mode:

XOR GATE

Symbol:

Truth Table:

Output Equation: $Y = {\bf{A}} \oplus {\bf{B}} = \bar AB + A \bar B$

Key Points: 

1) If B is always High, the output is the inverted value of the other input A, i.e. A̅.

1) The output is low when both the inputs are the same. 

2) The output is high when both the inputs are different.

Explanation:

$Y = {\bf{\bar X}} \oplus {\bf{X}} = \bar{\bar X} X+\bar X \bar X$

$Y = XX+\bar X \bar X$

$Y = X+\bar X$

Y = 1

It is defined as the time during which a reverse voltage is applied across the thyristor during its commutation process.

We know that matrix A is equal to the product of lower triangular matrix L and upper triangular matrix U,

A = [L][U]

Let's check option wise,

Option 1:

L = ​$\begin{bmatrix}1&0\\ 4&-1\end{bmatrix}$and U = $\begin{bmatrix}1&1\\ 0&-2\end{bmatrix}$

A = [L][U] = $\begin{bmatrix}1&1\\ 4&6\end{bmatrix}$

Option 2:

L = $\begin{bmatrix}2&0\\ 4&-1\end{bmatrix}$ U = $\begin{bmatrix}1&1\\ 0&1\end{bmatrix}$

A = [L][U] = ​ $\begin{bmatrix}2&2\\ 4&3\end{bmatrix}$

Option 3:

L = $\begin{bmatrix}1&0\\ 4&1\end{bmatrix}$ U = $\begin{bmatrix}2&1\\ 0&-1\end{bmatrix}$

A = [L][U] = ​$\begin{bmatrix}2&1\\ 8&3\end{bmatrix}$

Option: 4

L =$\begin{bmatrix}2&0\\ 4&-3\end{bmatrix}$ U = $\begin{bmatrix}1&.5\\ 0&1\end{bmatrix}$

A = [L][U] = ​$\begin{bmatrix}2&1\\ 4&-1\end{bmatrix}$

f(x) = 2x - x² +3

f'(x) = 0

⇒ 2 - 2x = 0 ⇒ x = 1

f''(x) = -2

⇒f''(x) < 0

So the equation f(x) having only maxima at x = 1.

C_p = 0.102 μF

R_p = 1.25 MΩ

$\tan \delta = \frac{{\left| {{I_R}} \right|}}{{\left| {{I_C}} \right|}}$

$\tan \delta = \frac{{V/{R_p}}}{{V \cdot \omega {C_p}}}$ 

$\tan \delta = \frac{1}{{\omega {C_p}{R_p}}}$

$\tan \delta = \frac{1}{{\left( {2\pi \times 50} \right)\left( {0.102;\mu F} \right)\left( {1.25;M{\rm{\Omega }}} \right)}}$

tan δ = 0.0249

Power loss = I² R

= V² / R

= V² ω C tan δ

= 5000 × 5000 × 2 × π × 50 × 0.102 × 0.025 × 10^-6

P = 20.027 W

\(\begin{array}{l} f\left( t \right)\mathop \leftrightarrow \limits^Z {F_1}\left( s \right)\ f\left( {t - \tau } \right)\mathop \leftrightarrow \limits^Z {e^{ - s\tau }}{F_1}\left( s \right) = {F_2}\left( s \right)\ G\left( s \right) = \frac{{{F_2}\left( s \right){F_1}\left( s \right)}}{{{{\left| {{F_1}\left( s \right)} \right|}^2}}}\ = \frac{{{e^{ - s\tau }}{F_1}\left( s \right){F_1}\left( s \right)}}{{{{\left| {{F_1}\left( s \right)} \right|}^2}}}\ = \frac{{{e^{ - s\tau }}|{F_1}\left( s \right){|^2}}}{{|{F_1}\left( s \right){|^2}}}\ \left( {because\ {F_1}\left( S \right)F_1^{*}\left( S \right) = |{F_1}\left( S \right){|^2}} \right) \end{array}\)

= e ^–sτ

Taking inverse Laplace,

g(t) = L^-1 [e^-sτ ] = δ (t – τ)

Let a signal p(x) is uniformly distributed between limits – a to + a

Variance, \(\bar \sigma = \mathop \smallint \limits_{ - a}^a {x^2}p\left( x \right)dx\)

\(\begin{array}{l} = \mathop \smallint \limits_{ - a}^a {x^2}.\frac{1}{{2a}}.dx\ = \frac{1}{{2a}}{\left[ {\frac{{{x^3}}}{3}} \right]^{a}}\ = \frac{{2{a^3}}}{{6a}} = \frac{{{a^2}}}{3} \end{array}\)

Its mean square value is equal to its variance

\(\begin{array}{l} {P^2}rms = {\sigma p} = \frac{{{a^2}}}{3}\ {p{rms}} = \frac{a}{{\sqrt 3 }} \end{array}\)

${T_2} = {T_1}$           ${\phi _2} = 0.9{\phi _1}$            ${\phi _1} \times {I_1} = {\phi _2} \times {I_2}$

$\begin{array}{l} {I_2} = 11.11A\ {E_1} = 220 - 10 \times 1 = 210\ V\ {E_2} = 220 - 11.1 \times \left( {1 + R} \right)\ \frac{{{N_2}}}{{{N_1}}} = \frac{{{E_2}}}{{{E_1}}} \times \frac{{{\phi _1}}}{{{\phi _2}}}\ I = \frac{{220 - 11.1\left( {1 + R} \right)}}{{210}} \times \frac{{{\phi _1}}}{{0.9{\phi _1}}} \end{array}$

R = 1.79 Ω

Power delivers to load center = 120 MW

The load is Connected through 25 km & 75 km long transmission lines by G₁ (100 kW) and G₂, G₃ (100 MW each) respectively.

For minimum loss, higher power should come from a shorter distance transmission line connected to the generator.

Here, P₁ comes from a 25 km transmission line, and P₂, P₃ is from a 75 km line.

So that G1 supplied power equal to the three times of the power supplied by the combination of G2 & G3.

Therefore, from option P1 = 90 MW, P2 = 15 MW & P3 = 15 MW. 

∵ P1 = 3 (P2 + P3).

LXI D, DISP

LP: CALL SUB

LP + 3

When CALL SUB is executed LP +3 the value is pushed in the stack

POP H ⇒ HL = LP + 3

DAD D ⇒ HL = HL + DE

                LP + 3 + DE

PUSH H ⇒ The last two values of the stack will be HL value

Concept:

Centroid: It is the intersection of the asymptotes and always lies on the real axis. It is denoted by σ.

$\sigma = \frac{{\sum {P_i} - \sum {Z_i}}}{{\left| {P - Z} \right|}}$

ΣPi is the sum of real parts of finite poles of G(s)H(s)

ΣZi is the sum of real parts of finite zeros of G(s)H(s)

The Angle of asymptotes: ${θ _l} = \frac{{\left( {2l + 1} \right)180^\circ }}{{P - Z}}$

l = 0, 1, 2, … |P – Z| – 1

Application:

We have,

Centroid = $\frac{-2-\frac{-2}{3}}{3-1}=\frac{-2}{3}$

Asymptotes = $\frac{2\pm 1\times 180^\circ}{2}$

Hence,

θ₁ = 90°

θ₂ = 270°

Concept

The circuit of the NPN transistor is:

For an NPN transistor to operate in:

1.) Active mode

$V_{BE}\space > 0.7$

and $V_{CB}=V_{CE}-V_{BE}\space > 0$

2.) Saturation mode

$V_{BE}\space > 0.7$

and $V_{CB}=V_{CE}-V_{BE}\space < 0$

3.) Cut-off mode

$V_{BE}\space < 0.7$

and $V_{CB}=V_{CE}-V_{BE}\space < 0$

Calculation

The value of V_BE is calculated by:

$V_{BE}=V_B-V_E$

The base terminal of the transistor is connected to the diode whose cut-in voltage is 0.7 V

$V_{BE}=0.7-(-12)$

$V_{BE}=12.7$ that is greater than 0.7 V

Hence, the diode can either be in saturation or active mode.

Assuming that the transistor is operating in the active region then, applying KVL to the base-emitter loop, we get

-5 + (1× I_B) + 0.7 + 0.7 - 12 = 0

I_B = 15.6 mA

IC = βIB 

I_C = 0.468 A

Applying KVL to the collector-emitter loop, we get

• 0 + 2.2I_C + V_CE - 12 = 0

V­_CE = 10.97 V

$V_{CB}=V_{CE}-V_{BE}\space$

$V_{CB}=10.97-12.7$

$V_{CB}=-1.72 \space V\space <0$

Hence, our assumption is wrong.

∴ the transistor is operating in the saturation region.

V_+(sat) = 0.2 V

Again applying KVL,

-0 + 2.2 I_C + 0.2 - 12 = 0

I_C = 5.36 mA

$\begin{array}{l} {I_{{T_M}}} = {I_o} + {V_s}\sqrt[{}]{{\frac{C}{L}}}\ = 10 + 200\sqrt {\frac{{0.1 \times {{10}^{ - 6}}}}{{1 \times {{10}^{ - 3}}}}} = 12A \end{array}$

∴Maximum current through auxiliary thyristor

= I_o = 10 A

We have given,

Function 1 (f1) = 10x2 sin x1 - 0.8 = 0

Function 2 (f2)  = $\rm 10x_2^2-10x_2\cos x_1-0.6=0$

Jacobian Matrix is given by,

J = $\begin{bmatrix}\frac{\delta f_1}{\delta x_1}&\frac{\delta f_1}{\delta x_2}\\ \frac{\delta f_2}{\delta x_1}&\frac{\delta f_2}{\delta x_2}\end{bmatrix}$ = $\begin{bmatrix}10x_2cosx_1&10sinx_1\\ 10x_2sinx_1&(20x_2-10cosx_1)\end{bmatrix}$

For, x1 = 0.0 and x2 = 1.0

J = $\begin{bmatrix}10&0\\ 0&10\end{bmatrix}$

Gain margin $= \frac{1}{{\left| {G\left( {j{\omega _{pc}}} \right)} \right|}}$

ωpc is phase cross over frequency

ωpc occurs at ∠G(jω) = -180°

From the given table, at ∠G(jωpc) = -180°,

|G(jωpc)| = 0.5

Gain margin $= \frac{1}{0.5} = 2$ = 20 log(2) = 6 dB

At | G(jω) | = 1, phase angle ∠ G(jω) = – 150°

∴ P.M = 180 + (– 150) = 30°

Concept:

In an induction motor,

• The stator is stationary and the stator field rotates with synchronous speed.
• The rotor field rotates with synchronous speed and the rotor rotates at a speed less than the synchronous speed.

Synchronous speed in rpm of a three-phase induction motor is given by,

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

Where f is the frequency in Hz

P is the number of poles

A rotor of a three-phase induction motor rotates at a speed close to the synchronous speed but not equal to the synchronous speed.

The difference between the synchronous speed and the rotor speed is known as slip speed. It is given by

Slip speed = Ns - Nr

Slip is defined as the ratio of slip speed to the rotor speed.

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

⇒ Nr = Ns (1 – s)

Calculation:

Given that,

Number of poles (P) = 6

Frequency (f) = 50 Hz

Synchronous speed, ${N_s} = \frac{{120 \times 50}}{6} = 1000;rpm$

Slip at no load, s = 5 %

Rotor speed N = Ns (1 – s) = 1000(1 – 0.05) = 950 rpm

The speed of stator magnetic field = speed of rotor magnetic field = 1000 rpm

The speed of the stator magnetic field with respect to the rotor magnetic field = 1000 – 1000 = 0

The speed of the rotor with respect to the stator magnetic field = 950 – 1000 = -50 rpm

Area of the rectangular plates,

A = 20 × 10^-2 × 40 × 10^-2 = 0.08 m²

ε = ε_oε_r = (2.26) × 8.85 × 10^-12

Breakdown strength of the dielectric = 50 kV/cm for a distance of separation d meter: Maximum voltage that can be applied

= 5 × 10⁶ d volts.

Maximum charge in the capacitor,

Q_m = CVm

$\Rightarrow {Q_m} = \frac{{{\rm{\varepsilon }}A}}{d} \times 5 \times {10^6} \times d$

= 2.26 × 8.85 × 10^-12 × 20 × 10^-2 × 40 × 10^-2 × 5 × 10⁶

= 8 μC

h (t) = e^-t + e^-2t

Taking Laplace transform,

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

For unit step input, r (t) = u (t)

∴ R (s) = 1/s

$Y\left( s \right) = R\left( s \right)H\left( s \right) = \frac{1}{s}\left[ {\frac{1}{{s + 1}} + \frac{1}{{s + 2}}} \right]$

$Y\left( s \right) = \frac{A}{s}+\frac{B}{s+1}+\frac{C}{s}+\frac{D}{s+2}$

A = 1, B = -1, C = 0.5, D = -0.5

Now, Y(s) becomes, 

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

Taking inverse Laplace,

Y (t) = u (t) [ 1.5 – e^-t – 0.5 e ^–2t ]

$\begin{array}{l} \tan {\rm{\Psi }} = \frac{{Vsin\phi + {I_a}{X_q}}}{{Vcos\phi + {I_a}{R_a}}} = \frac{{0 + 1{\rm{x}}1}}{{1{\rm{x}}1 + 0}}\ \therefore {\rm{\Psi }} = {45^\circ} \end{array}$

Maximum count with 4(1/2) digits display = 19999

Given, Full-scale reading = 200 V

Hence, 1 count = $\frac{200}{19999}$

For, 10 count it will be, $10\times \frac{200}{19999}\approx ±0.1$ V

And, error corresponded to 0.2% of reading (for 100 V) = $0.2\times \frac{100}{100}=±0.2$ V

Hence, total error will be the sum of error due to 10 counts and error corresponded to 0.2% of reading,

or, Total error = ± 0.1 V + ± 0.2 V = ± 0.3 V

y₁₁ = -j10

y₁₂ = -j5

y₁₃ = j 12.5

y₃₃ = -j10

y₁₁ = -j15         y₁₂ = j5                          y₁₃ = 0

y₂₁ = j5                        y₂₂ = j 7.5          y₂₃ = -j12.5

y₃₁ = 0             y₃₂ = -j 12.5       y₃₃ = j 2.5

Given a differential equation,

$\frac{dy}{dx}=e^{-3x}$

or, dy = e^-3x dx

Integrated on both side,

∫dy = ∫e-3x dx

or, y = e-3x × $\frac{-1}{3}$ + K

From the above table we observe that before the start of the next clock pulse, the inputs are:

J = 1, K = 0, T = 1

The output after the first clock pulse will be:

QA QB = 11

The inputs are now:

J = 0, K = 1, T = 1

The output after the second clock pulse will be:

QA QB = 00

The inputs are now:

J = 1, K = 0, T = 0

The output after the third clock pulse will be:

QA QB = 10

Concept:

For normal diode:

If V_p > V_n; diode is ON

If V_p < V_n; Diode is OFF

Where V_p and V_n are the voltages applied at the p-side and n-side respectively.

For Zener diode:

If |V_pn| > V_z ; Zener diode is ON

If |V_pn| < V_z ; Zener diode is OFF

Application:

Case I: V_i < – 0.7 V

For V_i < -0.7 V, the PN diode will be OFF and the Zener diode will be forward bias, behaving like a normal diode.

The equivalent circuit is drawn as:

∴ The output voltage V₀ will be:

V₀ + 0.7 = 0

V₀ = - 0.7 V

Case II: -0.7 ≤ V_i ≤ 5.7

The circuit is drawn as:

The output voltage will be:

V₀ = V_i

Case III: 5.7 < V_i < 10

For this range, Diode D will be ON and the Zener will be OFF. The equivalent circuit is as shown:

The output voltage is therefore:

V₀ – 0.7 – 5 = 0

V₀ = 5.7

From the result obtained above, the transfer characteristic will be:

Input power factor

\(\begin{array}{l} = \frac{{{V_s}{I_{s1}}\cos {\phi 1}}}{{{V_s}{I_s}}}\ = \frac{{{I{{s_1}}}}}{{{I_s}}}cos{\phi _1}\ {I_s} = \sqrt {{{10}^2} + {5^2} + {2^2}} = 11.35A\ pf = \frac{{10}}{{11.35}} \times \cos 60 = 0.44 \end{array}\)

Active power draws = Power drawing by fundamental

Is1 = RMS value of fundamental current, V_s = RMS value of source voltage

$\begin{array}{l} = {V_s}{I_{s1}}\cos {\phi _1}\ = 100 \times 10 \times \cos 60 \end{array}$