GATE EE 2015 Official Paper: Shift 1 (Previous Year Paper)

The given sentence uses the word 'enjoy' which has a positive connotation and is followed by the adverb ‘immensely’, which means to a large extent.

All of the options except option 3 use adjectives with negative connotations - terrible, horrible, terrifying - to describe the subject's experience and and are hence, incorrect.

Option 3 uses the word 'terrific' which has a positive connotation and is hence accurate.

Hence, option 3 is correct.

Total number of possibilities \(= 4 \times 5 = 20\)

Number of possibilities that the sum of two numbers equals 16 = 4

And they are \(= \left{ {\left( {2,14} \right),;\left( {3,13} \right),;\left( {4,12} \right),;\left( {5,11} \right)} \right}\)

Probability \(= \frac{4}{{20}} = 0.2\)

The correct determiner to fit the given blank should be ‘any’. This is because it is used to refer to one or some of a thing or number of things, no matter how much or how many.

‘No paper’ would make the construction of the sentence a double negative since ‘didn’t’ has been used. 

'Much’ is used for uncountable nouns and hence, cannot be used for an object like paper.

‘A few paper’ is also grammatically incorrect since the correct way to use this determiner would be to say "a few sheets of paper".

Hence, option 1 is correct.

Distance between two floors of building = 9 feet

Height of each step = 3/4 (Statement I)

Width of each step = 1 (Statement II)

Using the data give in statement I, we may calculate the number of steps in a set of stairs as

\(n = \frac{9}{3} \times 4 = 12;step\)

So, statement I is sufficient to give the answer.

However, there is no use of width given in statement II to calculate number of required steps in set of stairs. So, width of foot can be anything, and it is not sufficient to obtain the number of steps.

Thus, statement I alone is sufficient, but statement II alone is not sufficient.

The combinations of words in all options except B are synonyms of each other.

‘Wheedle’ means to use flattery to obtain something and ‘roundabout’ means not following a direct route. Thus, these two words are unrelated and do not form a synonymous combination.

Hence, the right answer is option 2.

The phrase ‘take to one's heels’ means 'to start running in order to escape'. Out of the given options, this meaning is best expressed by the phrase ‘took to flight’, which means the same thing.

Hence, option 3 is correct.

Answer: 32 to 32 

Explanation: 

From the pie-chart, we have 

The proportion of male to female students in each department is 5:4. ( male ratio = 5/9 and female ratio = 4/9)

Hence 40 males in Electrical Department, So total students in Electrical Department = 40 *(9/5) = 72 

Total electrical student percentage = 20%

Hence Total number of students in college = (72/20)*100 = 360

Hence number of females in Mechanical Department = 0.1(4/9)(360)** = 16

and number of females in Civil Department = 0.3(4/9)360 = 48

Hence the Difference = 48-16 = 32

Probability of non-pass = 1 - Probability of at least one pass

=1 - 0.75 = 0.25

(1 - m) (1 - p) (1 - c) = 0.25

(1 + mp - m - p) (1 - c) = 0.25

1 + mp - m - p - c - mpc + mc + pc = 0.25

m + p + c - mp - pc - mc + mpc = 0.75 -------Eq(1)

Probability of exactly 2 pass = 0.4

mp(1- c) + pc(1 - m) + mc(1 - p) = 0.4

mp + pc + mc - 3mpc = 0.4

mp + pc + mc - 2mpc = 0.5 ----------Eq(2)

Note: Adding the probability of all passes to probability of exactly 2 pass gives probability of at least 2 passes.

So, mpc = 0.1 = 1/10 ----------Eq(3)

From Eq(2) and Eq(3):

mp + pc+ mc - mpc = 0.6 ----------Eq(4)

And from Eq(1) and Eq(4):

m + p + c = 0.75 + 0.6 = 1.35

m + p + c = 135 / 100 = 27/20

Hence, the correct answer is option D.

Average marks of the class in the examination is \(= \frac{{total;amrks;obtained}}{{total;number;of;student}}\)

\(= \frac{{21 \times 2 + 15 \times 3 + 11 \times 1 + 23 \times 2 + 31 \times 5}}{{44}} = 6.795\)

Explanation:

• The fall in the drop of water levels in the lakes has to be tackled by both courses of actions I and II in order to supply adequate water to the city. There has to be a partial decrease in the water supply by the government so that everyone is supplied with water and nobody faces a total water cut.  Also, the citizens must be made aware of the situation and they should be encouraged to make responsible and minimal use of water in order to conserve it. A total ban on the water supply is illogical since water is a necessity for everyone and thus the third course of action does not follow.

\(\mathop \smallint \limits_{ - \infty }^\infty f\left( x \right)dx = 1\)

So \(\mathop \smallint \limits_0^1 \left( {a + bx} \right)dx = 1\) 

\(a + \frac{b}{2} = 1\),  \(2a + b = 2\) 

Given \(E\left[ X \right] = \frac{2}{3} = \mathop \smallint \limits_0^1 X\left[ {a + bx} \right]dx\)

\(\frac{2}{3} = \frac{a}{2} + \frac{b}{3}\) ,   \(3a + 2b = 4\) 

From above equations, we get      \(a = 0,;b = 2\)

\(P\left[ {X < 0.5} \right] = \mathop \smallint \limits_0^{0.5} f\left( x \right)dx = 2\mathop \smallint \limits_0^{0.5} x;dx = 0.25\)

We know that, (Intermediate value theorem)

If \(f\left( a \right);f\left( b \right) < 0\) then f(x) has at least one root in (a, b)

Since there is no root in [a, b] this implies f(a) and f(b) are of same sign 

i.e. either they both are positive or they both are negative

In both cases \(f\left( a \right)f\left( b \right) > 0\)

Sum of the diagonals elements is \(- 6\) for \(2\times 2\) matrix

The possible Eigen values are

\(\to \begin{array}{*{20}{c}} { - 1, - 5}&{ - 5, - 1,}&{ - 8,2}\ { - 2, - 4}&{ - 4, - 2}&{ - 9,3}\ {0, - 6}&{ - 3, - 3}&{ - 10,4} \end{array} - - - -\) 

Maximum possible value of determinant is \(- 3 \times - 3 = 9\).

Concept:

Given : \(\bar f = \frac{1}{{{r^2}}}\hat r\), 

The Divergence theorem states that:

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

where ∇.D is the divergence of the vector field D.

=\(\mathop{{\int!!!\ !!\int}\mkern-21mu \ \bigcirc} {( \frac{1}{{{r^2}}}\hat r).r^2 Sin\theta.d\theta .d\phi .\hat r}\)

=\(\mathop{{\int!!!\ !!\int}\mkern-21mu \ \bigcirc} { Sin\theta.d\theta .d\phi }\)

=4π

CONCEPT:

Wheatstone Bridge:

It is an arrangement of four resistance that can be used to measure one of them in terms of rest. Here arms AB and BC are called ratio arm and arms AC and BD are called conjugate arms.

The bridge is said to be balanced when deflection in the galvanometer is zero i.e. no current flows through the galvanometer or in other words VB = VD.

In the balanced condition

\(\frac{P}{R} = \frac{Q}{S}\)

Calculation:

Given,

R1 = 50 Ω, R2 = 65 Ω & R3 = 100 ± 5% Ω

R3 = 100 ± (100 × 0.5)

R3 = [95, 105]

For bridge balance condition

R₁ R_x = R₂ R₃

For R₃ = 95 Ω 

⇒ R_x = (65 × 95) / 50 = 123.5 Ω 

For R₃ = 105 Ω 

⇒ R_x = (65 × 105) / 50 = 136.5 Ω 

Therefore the range of R_x will be

⇒ R_x = [123.5 Ω, 136.5Ω]

Concept: 

• Shunts are used for the range extension of ammeters.
• A shunt is a low-value resistance having minimum temperature co-efficient and is made up of manganin. Because manganin has a very low value of temperature coefficient.
• It is connected in parallel with the ammeter whose range is to be extended. The combination is connected in series with the circuit whose current is to be measured.
• Shunt provides a path for extra current as it is connected across (in parallel with) the instrument.
• These shunted instruments can be used to measure currents many times greater than their normal full-scale deflection currents.

In the figure

I is total current flowing in the circuit

Ish is the current through the shunt resistor

Rm is the ammeter resistance

As the shunt and meter are in parallel, therefore, the voltage drop across them will be equal​

Calculation:

Given,

The voltage drop across the ammeter is 0.1 V for full-scale deflection

Current flowing through the meter at full-scale deflection is 50 A

Now the range is extended to 500 A,

Current flowing through shunt I_sh = 500 - 50 = 450 A

As the shunt and meter are in parallel, therefore, the voltage drop across them will be equal

The voltage drop across shunt = Shunt resistance × shunt current

⇒ 0.1V = R_sh × 450 A

⇒ R_sh = 0.22 mΩ

Concept:

Instrumentation Amplifier: An instrumentation amplifier is usually employed to amplify low-level signals, rejecting noise and interference signals.

The circuit of the Instrumentation amplifier can be given as follows

Explanation:

Additional characteristics include very low DC offset, low drift, low noise, very high open-loop gain, very high common-mode rejection ratio, and very high input impedances. Instrumentation amplifiers are used where great accuracy and stability of the circuit both short and long-term are required.

An instrumentation amplifier is usually employed to amplify low-level signals, rejecting noise and interference signals. Therefore, the characteristics of a good Instrumentation amplifier should include:

Finite, Accurate and Stable Gain: Since the instrumentation amplifiers are required to amplify very low-level signals from the transducer device, high and finite gain is the basic requirement. The gain also needs to be accurate and the closed-loop gain must be stable.

High Input Impedance: To avoid the loading of input sources, the input impedance of the instrumentation amplifier must be very high (ideally infinite).

Low Output Impedance: The output impedance of a good instrumentation amplifier must be very low (ideally zero), to avoid loading effect on the immediate next stage.

High CMRR: The output from the transducer usually contains common-mode signals, when transmitted over long wires. A good instrumentation amplifier must amplify only the differential input, completely rejecting common-mode inputs. Thus, the CMRR of the instrumentation amplifier must be ideally infinite.

High Slew Rate: The slew rate of the instrumentation amplifier must be as high as possible to provide maximum undistorted output voltage swing.

Low Power Consumption: A good Instrumentation amplifier have low input offset currents and therefore has very low power consumption which is always a desirable property.

Concept:

Identification of DC-DC Converter:

• Buck converter: Inductor is connected in series with the combination of the capacitor and resistor.
• Boost converter: Inductor is connected in series with the supply voltage source.
• Buck-Boost converter: Inductor is connected in parallel to the supply voltage source.

 

Under steady-state operating conditions:

• The average voltage across the inductor is zero. VL = 0
• The average current across the capacitor is zero. IC = 0

 

For resistive R load,

 

Calculation:

Given DC-DC converter is a Buck converter with a duty cycle ratio δ = 0.4

V₀ = D V_dc = 0.4 × 20 = 8 V

Since the connected load is the battery the output current is given as

\( {I_0} = \frac{{{V_0} - E}}{R} = \frac{{8 - 5}}{3} = 1A\)

Concept:

Definite Integral:

If ∫ f(x) dx = g(x) + C, then \(\rm ∫_a^b f(x)\ dx = [ g(x)]_a^b\) = g(b) - g(a).

Calculation:

\(u\left( \tau \right) = \sin \left( {\omega \tau } \right)\)

\(\begin{array}{l} \omega = 2\pi f = 2\pi .\frac{1}{{2T}} = \frac{\pi }{T}\ \omega T = \pi \ y\left( t \right) = \frac{1}{T}\mathop \smallint \limits_{t - T}^t \sin \left( {\omega \tau } \right)d\tau = \left. {\frac{{\cos \left( {\omega \tau } \right)}}{{\omega T}}} \right|_t^{t - T} = \frac{1}{\pi }\left[ {\cos \omega \left( {t - T} \right) - \cos \omega t} \right]\ = \frac{1}{\pi }\left[ {\cos \omega t\cos \omega T + \sin \omega tsin\omega T - \cos \omega t} \right]\ y\left( t \right) = - \frac{2}{\pi }\cos \omega t = \frac{2}{\pi }\sin \left( {\omega t - 90 } \right)\ x\left( t \right) = \sin \omega t\ \phi = 90^\circ \end{array}\)

Overall impulse response \(= {\rm{;}}g\left( f \right){\rm{*}}g\left( t \right)\) 

\(h\left( f \right){\rm{;}} = {\rm{;}}g\left( f \right){\rm{;*;}}g\left( f \right)\)

From the above graph, the maximum value cascaded response is 1.

\(P = \frac{{V_{emf}^2}}{R}\)

\({V_{emf}} = \frac{{ - d\psi }}{{dt}}\)

\(\psi = \mathop \smallint \limits_s^; B \cdot ds = B \cdot S \cdot = \frac{1}{{800}}sin\omega t\)

\({V_{emf}} = \frac{{ - d\psi }}{{dt}} = \frac{{ - 1}}{8}\pi ;cos\omega t\)

\(P = \frac{{{\pi ^2}}}{{64}}{\cos ^2}\omega t \times \frac{1}{R}\)

\(P = \frac{{{\pi ^2}}}{{0.4 \times 64}}\left[ {1 + \frac{{cos2\omega t}}{2}} \right]\)

\({P_{avg}} = \frac{{{\pi ^2}}}{{2 \times 0.4 \times 64}}+ \frac{{{\pi ^2}}}{{0.4 \times 64 \times 2}}cos2\omega t\)

\({P_{avg}} = \frac{{{\pi ^2}}}{{2 \times 0.4 \times 64}}+ \frac{{{\pi ^2}}}{{0.4 \times 64 \times 2}}cos2\omega t\)

The average value of cos over one complete cycle will be 0. ∴ The average value will be the constant value, i.e.

\({P_{avg}} = \frac{{{\pi ^2}}}{{2 \times 0.4 \times 64}} = 0.192;W\)

Magnetic field intensity at (0, L, 0) due to conductor 1 and conductor 2 is given by

\(\begin{array}{l} H = {H_1} + {H_2}\ = \frac{I}{{2\pi \left( {\frac{L}{2}} \right)}}\left( { - {a_z}} \right) + \frac{I}{{2\pi \left( {\frac{{3L}}{2}} \right)}}\left( { - {a_z}} \right)\ = \frac{I}{{2\pi }}\left( { - {a_z}} \right)\left[ {\frac{2}{L} + \frac{2}{{3L}}} \right]\ = \frac{{4I}}{{3\pi L}}\left( { - {a_z}} \right) \end{array}\)

Magnetic flux density B = μ₀ H

\(B=- \frac{{4{\mu _0}I}}{{3\pi L}}\hat z\)

Concept:

Differential amplifier:

The differential amplifier is to amplify the difference between the two signals received at the input

\({{\rm{V}}_{{\rm{out}}}} = - \frac{{{{\rm{R}}_{\rm{f}}}}}{{{{\rm{R}}_1}}}\left( {{{\rm{V}}_2} - {{\rm{V}}_1}} \right){\rm{;assuming;that;}}{{\rm{R}}_1} = {{\rm{R}}_2}{\rm{;and;}}{{\rm{R}}_{\rm{f}}} = {{\rm{R}}_{\rm{g}}}{\rm{;}}\)

Differential amplifiers are used in

• Inverting amplifiers
• Non-inverting amplifiers
• Summing amplifiers
• Instrumentation amplifiers
• Wheatstone Bridge Differential Amplifier

Virtual ground -

The concept of the virtual ground is stated as if anyone of the i/p terminals is grounded physically the other i/p terminal will also be at ground potential even though, it is not grounded physically.

• One key feature of an Op-Amp is the differential input, and when put together in a circuit, this can form a virtual ground.
• The virtual ground concept is helpful for the analysis of Op Amps. This concept makes Op-Amp circuit analysis much easier.

Calculation:

Given circuit can be re drawn as

using voltage Division Rule:

\(V_{NI} = \frac{R}{R + \frac{1}{Cs}} \times V_2\)

Since, V2 = 0

⇒ VNI = 0 V

From virtual Ground concept:

VI = VNI = 0

Applying KCL at inverting terminal

\(\frac{V_i ∠ 0 - V_I}{\frac{1}{Cs}}= \frac{V_I-V_0 ∠ δ}{R}\)

\(\frac{V_i ∠ 0^\circ}{\frac{1}{Cs}} = \frac{0- V_0 ∠ δ}{R}\)

V0∠δ = -RCs Vi ∠0

V0∠δ = -jRCω Vi < 0

V0∠δ = RCω Vi∠-90°

On comparing

\(\delta = - \frac{-\pi}{2}\)

Concept:

In a transistor, emitter current (Ie)can be calculated as

Ie = Ib + Ic

Where,

Ib = base current

Ic = collector current

Calculation:

Value of emitter current is given by

I_E = I_B + I_C

I_c + I_B = 6/R_c

I_B = 8.3/R_B

β = 75, I_C = βI_B

(76) I_B = 6/R_C , I_B = 8.3/R_B

R_B/R_C = (76 × 8.3)/6 = 105.1

Multiplexers:

• A multiplexer is a combinational circuit.
• A multiplexer is Many to one data selector.
• A multiplexer selects one of the many data available at its input depending on the bits on the select line.
• For 2n inputs, there are n select lines that determine, which input is to be connected to the output.

 

A multiplexer is used for Parallel to serial data conversion with the help of a counter as shown:

• The Q-outputs of the modulus-8 counter is connected to the data select inputs of an 8- multiplexer.
• The first byte (8-bit) of parallel data is applied to the multiplexer inputs.
• As the counter goes through a binary sequence from 0 to 7, each bit beginning with D0 is sequentially selected and passed through the multiplexer to the output line.
• After 8 clock pulses, the data byte is converted to a serial format and sent out on the transmission line.

 

Example:

A 4-input multiplexer is as shown below.

Output, Y = I0S̅1S̅0 + I1S̅1S0 + I2S1S̅0 + I3S1S0

The output of a 4-input multiplexer represents one combinational function of 3-variables each.

Calculation:

Given  

Given select lines, S1 and S0 are A and B

To obtain the given expression:

\(F = A \oplus B = A \bar B + \bar AB\)

Output Y = I0 A̅.B̅ +  I1 A̅.B + I2 A.B̅ + I3 A.B

To get the required output of XOR gate, the required input I3I2I1I0 must be 0110.

\(I = \frac{{{V_1} - {V_2}}}{R}\)

For power to be reversed, polarities of voltages or direction of current must be reversed.

As current is unidirectional due to HVDC connection. So polarities of voltages are changed.

Hence due to unidirectional nature of current, voltage V₁ must be greater than V_2.

\(I = \frac{{{V_2} - {V_1}}}{R}\left( { + {V_e}} \right)\)

V₁ = -485 kV, V₂ = -500 kV, I = 1.5 kA

Baseload power station: Baseload is the minimum level of electricity demand required over a period of 24 hours. It is needed to provide power to components that keep running at all times (also referred to as continuous load). Plants that are running continuously over extended periods of time are said to be baseload power plants.

Example:

• Nuclear power plant
• Coal or thermal power plant
• Run of the river hydroelectric plant
• Geothermal plant
• Biogas plant
• Biomass plant
• Solar plant
• Ocean thermal energy conversion
• Wind turbines

 

Peak load power station: Peak load is the time of high demand. These peaking demands are often for only shorter durations. To cater to the demand peaks, peak load power plants are used. They are started up whenever there is a spike in demand and stopped when the demand recedes.

Example:

• Gas plant
• Solar power plants
• Pumped storage plants
• Diesel generators

Conclusion: Run off power plants and nuclear power plants are base load power plants. Wind farms are also base load plants.

Concept:

Kirchhoff’s Current Law (KCL)

It states that the amount of current flowing into a node or junction is equal to the sum of the currents flowing out of it

Current entering through a load has a positive polarity and exit point has a negative polarity and for source, it will be reversed.

Calculation:

Let current through 3 Ω is I₁ and through 2 Ω is I₂ and current through 5 V is I ampere respectively as shown below

\(I_1 = \frac{{6~V}}{{3{\rm{~\Omega }}}} = 2~A\)

\(I_2 = \frac{{2V}}{{2{\rm{Ω }}}} = 1A\)

Apply KCL at N/W N₂,

I + I2 = I1

I + 1 = 2

I = 1 A

P = 5 × 1 = 5 W

 

Kirchhoff's current law (KCL) is applicable to networks that are:

• Kirchhoff’s law is applicable to both AC and DC circuits. It is not applicable for time-varying magnetic fields.
• Unilateral or bilateral
• Active or passive
• Linear or non-linear
• Lumped network

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 solve Problems with Dependent Sources:

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

To find R_th: 

To calculate Thevenin’s Resistance we should replace all independent current sources by Open circuit and Independent voltage sources by Short circuit (keep dependent sources as it is).

Circuits which have only dependent sources can’t function on their own so V_th and I_sc don’t exists but still, they exhibit resistance, that resistance can be indirectly determined by V/I method by placing an active source across the terminals.

1. Place a voltage source of 1V across the terminal and find the current (I_T) flowing through it. Then,  

\({{\rm{R}}_{{\rm{th}}}} = {{\rm{R}}_{\rm{N}}} = \frac{{1{\rm{V}}}}{{{{\rm{I}}_{\rm{T}}}}}\)

(or) 

2. Place a current source of 1A across the terminals and find the voltage (V_t) across the current source. Then,

 \({{\rm{R}}_{{\rm{th}}}} = {{\rm{R}}_{\rm{N}}} = \frac{{{{\rm{V}}_{\rm{T}}}}}{{1{\rm{A}}}}\)

Calculation:

Applying the KCL

\(\frac{{V - 2}}{1} + \frac{{V - 0}}{1} + \frac{{{V_{th}} - 0}}{2} = 0\) ⋯ (i)

i = V / 1

V_th – V = 20 i ⋯ (ii)

V_th  = 20(V) + V

V_th = 21 V ⋯ (iii)

Substituting the value of the V_th in equation (i)

\(\frac{{V - 2}}{1} + \frac{{V - 0}}{1} + \frac{{21V - 0}}{2} = 0\)

\(V - 2 + V + \frac{{21V}}{2} = 0\)

\(2V + \frac{{21V}}{2} = 2\)

\(\frac{{4V + 21V}}{2} = 2\)

25 V = 4

V = 4 / 25

Substituting this value in equation (iii)

\({V_{th}} = 21 \times \frac{4}{{25}}\)

V_th = 84 / 25

V_th = 3.36 V

Parallel LC circuit:

Resonance in electric circuits is the phenomenon, in which at a certain input voltage frequency the voltage and current drawn by the circuit are in phase.

In a parallel resonant circuit at resonance frequency the impedance is maximum.

For parallel LC circuit

\(Y = j\omega C + \frac{1}{{j{\rm{\omega L}}}};\)

\(Y = j\left( {\omega C - \frac{1}{{\omega L}}} \right)\)

\(Z = \frac{{j\omega L \times \frac{1}{{j\omega C}}}}{{\left( {j\omega L + \frac{1}{{j\omega C}}} \right)}}\)

\(Z = \left[ {\frac{{j\omega L}}{{1 - {\omega ^2}LC}}} \right]\)

At resonance

\(\omega = \frac{1}{{\sqrt {LC;} }}\)

\(Y = j\left( {\frac{1}{{\sqrt {LC;} }}C - \frac{1}{{\frac{1}{{\sqrt {LC;} }}L}}} \right)\)

\(Y = j\left( {\sqrt {\frac{C}{L}} ; - \sqrt {\frac{C}{L}} } \right)\)

Y = 0

Since Y = 0 , Z = ∞

Circuit characteristics with reference to the resonant frequency (f₀)

• At frequency below resonant frequency i.e. f < f₀, X_L >> X_C. Hence the circuit is inductive.
• At frequency above resonant frequency i.e. f > f₀, X_C >> X_L. Hence the circuit is capacitive.
• At resonant frequency i.e. f = f₀, X_L = X_C, the current is minimum and impedance is maximum. In this state, the circuit can act as a rejector circuit.

Impedance response with respect to frequency

At rated speed and rated field current - 

Open circuit voltage Ea₁ = 200 V

At half of rated speed & half field current

∴ E_a ∝ Nϕ

⇒ \(\frac{E_{a_2}}{E_{a_1}}=\frac{N_2\phi_2}{N_1\phi_1}\)

\(\left{N_2=\frac{N_1}{2}, \phi_2=\frac{\phi_1}{2}\right}\)

⇒ \(E_{a_2}=\frac{1}{2}×\frac{1}{2}× 200=50\) V

If capacitor is connected across armature terminal then equivalent circuit is given as,

τ = Rc = 0.1 × 1000 × 10^-6 = 10^-4 sec

voltage across capacitor is given by,

\(V_c(t)=V_C(∞)+\left[V_c(0^-)-V_ c(∞)\right]e^{-t/\tau}\)

Assume uncharged capacitor

i.e., \(V_c(0^-)=V_c(0^+)=0\)

For V_c(∞) capacitor open circuited

⇒ \(V_c(t)=E_2+(0-E_2)e^{-t/Rc}\)

⇒ \(V_c(t)=E_2(1-e^{-t/\tau})\)

⇒ \(25=50(1-e^{-t/10^{-4}})\)

t = 69.3 μ sec

Concept:

Mutual Inductance:

When two coils are placed close to each other, a change in current in the first coil produces a change in magnetic flux, which cuts not only the coil itself but also the second coil as well. The change in the flux induces a voltage in the second coil, this voltage is called induced voltage and the two coils are said to have a mutual inductance.

Consider a pair of coupled inductors with self-inductance L1 and L2, magnetically coupled through coupling coefficient k.

Input and output voltage expressions are given as

V1 = jωL1I1 + jωMI2   ...(1)

V2 = jωL2I2 + jωMI1    ...(2)

Where,

ω = 2πf

\(M = K\sqrt {{L_1}{L_2}} \)

M = Mutual inductance

L1 = Inductance of coil one

L2 = Inductance of coil two

Calculation:

s-domain representation of the circuit diagram is as follows:

Apply KVL in the input loop,

V1(s) = (0.8s) I1(s) + 0.48sI2(s)  ...(1)

Apply KVL in the output loop

(0.6s) I2(s) + 0.48sI1(s) = 0

\(I_2(s)=\dfrac{{-0.48sI_1(s)}}{{0.6s}}\)

Substitute I2(s) in equation(1)

\(V_1(s)=(0.8s)I_1(s)-\dfrac{{0.23s^2I_1(s)}}{{0.6s}}\)

\(\dfrac{{V_1(s)}}{{I_1(s)}}=\dfrac{{0.48s^2-0.23s^2}}{{0.6s}}=\frac{0.2496s^2}{0.6s}\)

Z_in(s) = 0.416s

As resistance is negligible compare the above equation with Z_in = jωL_in

Therefore effective inductance seen from the source is 0.416H or 416 mH

The Primary mmf is least affected by the secondary terminal in current transformer.

Additional Information

Current Transformer (C.T.):

• It is a type of instrument transformer that produces an alternating current in primary winding which is proportional to the current to be measured.
• The primary current of the transformer is dictated by the load current.
• Current transformers reduce high currents to a much lower value and provide a convenient way of safely monitoring the actual electrical current flowing in an AC transmission line using a standard ammeter.
• The principle of operation of a basic current transformer is slightly different from that of an ordinary voltage transformer.
• The current transformer consists of only one or very few turns as its primary winding.
• This primary winding can be of either a single flat turn, a coil of heavy-duty wire wrapped around the core or just a conductor or bus bar placed through a central hole.

Concept:

The standard transfer function of a Bode magnitude plot is:

\(TF = \frac{{K\left( {1 + \frac{s}{{{\omega _1}}}} \right)\left( {1 + \frac{s}{{{\omega _2}}}} \right) \ldots }}{{{s^n}\left( {1 + \frac{s}{{{\omega _3}}}} \right)\left( {1 + \frac{s}{{{\omega _4}}}} \right) \ldots }}\)

Here, ω1, ω2, ω3, ω4, … are the corner frequencies.

n is the number poles at the origin.

In a Bode magnitude plot,

• For a pole at the origin, the initial slope is -20 dB/decade
• For a zero at the origin, the initial slope is 20 dB/decade
• The slope of magnitude plot changes at each corner frequency
• The corner frequency associated with poles causes a slope of -20 dB/decade
• The corner frequency associated with poles causes a slope of -20 dB/decade
• The final slope of Bode magnitude plot = (Z – P) × 20 dB/decade

Where Z is the number of zeros and P is the number of poles

Calculation:

In the given bode plot,

Corner frequencies are 10 and 1000

The slope is zero for frequency less than 10 Hz

The slope changes to - 20 dB for frequency 10 Hz to 1000 Hz (a pole is added)

The slope changes to 0 dB for frequency above 1000 Hz (a zero is added)

\(\therefore G\left( s \right) = \frac{{K.\left( {1 + \frac{1}{{1000}}s} \right)}}{{\left( {1 + \frac{1}{{10}}s} \right)}}\)

Calculation of K:

\({M_{dB}} = 20;dB\) at initial frequency

\(20;log;M; = ;20\) ,    \(20;log;K; = ;20\)

\(\begin{array}{l} K; = ;10\ G\left( s \right) = \frac{{10\left( {s + 1000} \right) \times 10}}{{1000\left( {s + 10} \right) \times 1}} = \frac{{\left( {s + 1000} \right)}}{{10\left( {s + 10} \right)}} \end{array}\)

Concept:

Mason’s gain formula is

\(T = \frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{{\mathop \sum \nolimits_{i = 1}^N {P_i}{{\rm{Δ }}_i}}}{{\rm{Δ }}}\)

Where,

C(s) is the output node

R(s) is the input node

T is the transfer function or gain between R(s) and C(s)

Pi is the ith forward path gain

Δ = 1−(sum of all individual loop gains) + (sum of gain products of all possible two non-touching loops) − (sum of gain products of all possible three non-touching loops) + ........

Δi is obtained from Δ by removing the loops which are touching the ith forward path.

Calculation:

Given signal flow graph,

The number of forward paths from X₂ to Y is 1

P₁Δ₁ = G₂

Number of loops 

L₁ = - G₁G₂, L₂ = G₁

Δ = 1 - (L₁ + L₂) = 1 + G₁ + G₁G₂ 

\(TF = \frac{{{P_1}{{\rm{Δ }}_1}}}{{\rm{Δ }}} = \frac{{{G_2}}}{{1 + {G_1}\left[ {1 + {G_2}} \right]}} \)

To revise the concept, click here.

The characteristic equation of A is  \(\left| {A - XI} \right| = 0\)

\(\begin{array}{l} \Rightarrow f\left( x \right) = {x^3} + 6{x^2} + 11x + 6 + 2a\ = \left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 3} \right) + 2a = 0 \end{array}\)

\(f\left( x \right)\) cannot have all \(3\) real roots (if any) equal

for if \(f\left( x \right) = {\left( {x; - ;k} \right)^3}\), then comparing coefficients, we get

\(6 = - 3k,;3{k^2} = 11\)

No such \(k\) exists

a) Thus \(f\left( x \right) = 0\) has repeated (2) roots (say) \(\alpha ,;\alpha ,;\beta\)

or       b) \(f\left( x \right); = ;0\) has real roots (distance) (say) \(\alpha ,;\beta ,;\delta\)

Now, \(\begin{array}{*{20}{c}} {f'\left( x \right) = 0 \Rightarrow {x_1} = \frac{{ - 6 - \sqrt 3 }}{3} \approx - 2.577a;}&{{x_2} = \frac{{ - 6 + \sqrt 3 }}{3} \approx - 1.422} \end{array}\) 

At \({x_1};f\left( x \right)\) has relative max.

At \({x_2};f\left( x \right)\) has relative min.

The graph of \(f\left( x \right)\) will be as below

    Case (a) repeated roots (α, α, β)

  Case (b) distinct roots

Note that the graph of \(f\left( x \right)\) cannot be like the one given below

Thus in all possible cases we have \(f\left( {{x_2}} \right) \le 0 \Rightarrow \left( {a - \frac{{\sqrt 3 }}{9}} \right) \le 0 \Rightarrow a \le \frac{1}{{3\sqrt 3 }}\)

Given a differential equation,

\(\frac{{{d^2}y}}{{d{t^2}}} + 5\frac{{dy}}{{dt}} + 6y = 0\)

Apply Laplace transform to the above differential equation,

⇒ (s² + 5s + 6).Y(s) = 0

The characteristic equation will be s2 + 5s + 6

The roots of the equation are - 3, - 2

Solution for the differential equation can be written as 

\( y\left( t \right) = {C_2}{e^{ - 3t}} + {C_2}{e^{ - 2t}}\)

Given y(0) = 2

Therefore

\( y\left( 0 \right) = {C_1} + {C_2} = 2\)

Also given,

\(y\left( 1 \right) = - \frac{{1 - 3e}}{{{e^3}}}\)

Therefore 

\( y\left( 1 \right) = - \left( {\frac{{1 - 3e}}{{{e^3}}}} \right) = - {e^{ - 3}} + 3{e^{ - 2}} = {C_1}{e^{ - 3}} + {C_2}{e^{ - 2}} \)

So \({C_1} = - 1,{C_2} = 3\)

So,  \(y\left( t \right) = - {e^{ - 3t}} + 3{e^{ - 2t}}\)

\(\frac{{dy\left( t \right)}}{{dt}} = 3{e^{ - 3t}} - 6{e^{ - 2t}},\frac{{dy\left( 0 \right)}}{{dt}} = 3 - 6 = - 3\)

Concept:

Even Symmetricities:

⇒ FS expansion of an even signal does not contents sin terms

ODD symmetricities:

⇒ FS expansion of an odd signal contains only sin term.

Half wave symmetricities:

\(\begin{array}{*{20}{c}} {x\left( t \right) = - x\left( {t + \frac{{{T_o}}}{2}} \right)}\ {\begin{array}{*{20}{l}} \downarrow \ {{c_n} = - {c_n}.{e^{\frac{{Jn{\omega _0}{T_0}}}{2}}}} \end{array}} \end{array}\)

Time shifting

\(\begin{array}{*{20}{c}} ;\ {x\left( {t - {t_0}} \right){c_n}e} \end{array} - jn{\omega _0}{t_0}\)

\(1 = - {e^{\frac{{Jn{\omega _0}{T_o}}}{2}}},\frac{{{\omega _o}{T_o}}}{2} = \pi \)

1 = -e^Jnπ

1 + (e^jπ)ⁿ = 0

∴ e^πj = -1

1 + (-1)ⁿ = 0      -----(1)

Equation (1) will be satisfied for n = odd – integer

nω₀  = ODD harmonic

∴ F.S expansion of a H.W.S contains only ODD harmonic

Even + H.W.S:

⇒ F.S expansion of even H.W.S signal contains cos terms with ODD harmonic.

Cos terms with ODD harmonic

Odd + H.W.S

⇒ F.S expansion of odd H.W.S. signal contains sin terms with ODD harmonics.

Solution:

we have

\(\rm sgn (x) = \left{ \begin{matrix} \frac{x}{|x|}; & x \ne 0 \\ 0; & x = 0 \end{matrix}\right.\)

put x = cos t

\(\rm sgn (\cos t) = \left{ \begin{matrix} \frac{\cos t}{|\cos t |}; & \cos t \ne 0 \\ 0; & \cos t = 0 \end{matrix}\right.\)

\(\rm sgn (\cos t) = \left{ \begin{matrix} 1; & \cos t > 0 \\ -1; & \cos t < 0 \end{matrix}\right.\)

sgn (cos t) is an Even function, with half wave symmetry.

Thus, the function sgn(cos t) has only cosine terms due to Even symmetry with odd numbered harmonics due to half wave symmetry.

Probability of getting 6 is \(= \frac{6}{{36}} = \frac{1}{6}\)

i.e. Probability of A winning the game \(= \frac{1}{6}\) 

Probability of A not winning the game \(= 1 - \frac{1}{6} = \frac{5}{6}\)

Probability of B winning the game \(= \frac{1}{6}\)

Probability of B not winning the game \(= \frac{5}{6}\)

 If A starts the game, Probability that A will win the game

\(\begin{array}{l} = P\left( A \right) + P\left( {\bar A} \right)P\left( {\bar B} \right)P\left( A \right) + P\left( {\bar A} \right)P\left( {\bar B} \right)P\left( {\bar A} \right)P\left( {\bar B} \right)P\left( A \right) + \ldots \ = \frac{1}{6} + \frac{5}{6}\frac{5}{6}\frac{1}{6} + \frac{5}{6}\frac{5}{6}\frac{5}{6}\frac{5}{6}\frac{1}{6} + \ldots = \frac{1}{6}\left[ {1 + \frac{5}{6}\frac{5}{6} + \frac{5}{6}\frac{5}{6}\frac{5}{6}\frac{5}{6} + \ldots } \right]\ = \frac{1}{6}\left[ {1 + {{\left( {\frac{5}{6}} \right)}^2}+{{\left( {\frac{5}{6}} \right)}^4} + \ldots } \right]\ = \frac{1}{6}\left[ {\frac{1}{{1 - {{\left( {\frac{5}{6}} \right)}^2}}}} \right] = \frac{1}{6} \times \frac{{36}}{{11}} = \frac{6}{{11}} \end{array}\)

Given circuit,

The voltage at node A will be

\({V_A} = \frac{{R\left( {1 + x} \right)E}}{{R\left( {1 + x} \right) + PR}}\)

The voltage at node B will be

\( {V_B} = \frac{{{RE}}}{{R + pR}}\)

V₀ = V_A - V_B 

\({V_0} = \frac{{R\left( {1 + x} \right)E}}{{R\left( {1 + x} \right) + pR}}- \frac{{{RE}}}{{R + pR}}\)

For V₀ to be maximum at p 

Differentiate V₀ w.r.t. to p and equate it to zero

\(\frac{{d{V_0}}}{{dp}} = 0\)

\(\frac{{d{V_0}}}{{dp}}= \frac{d}{dp}(\frac{{R\left( {1 + x} \right)E}}{{R\left( {1 + x} \right) + pR}}- \frac{{{RE}}}{{R + pR}})=0\)

\( \Rightarrow P = \sqrt {1 + x}\)

Therefore for the maximum value of V₀ p should be equal to \(\sqrt {\left( {1 + x} \right)}\)

Given that only one of the two switches is ON. At any instant. For 0.2 ms, S₁ is ON and for 0.3 ms S₂ is ON. Here, we analyse the given circuit for the two cases:

Case 1: For 0.2 ms cycle

For this case, the circuit becomes as

So, the voltage 10 V appears across load resistor R_L for 0.2 ms cycle.

Case 2: For 0.3 ms cycle

So, the voltage 5 V appears across load resistor R_L for 0.3 ms cycle.

Hence, we get average output voltage as

\({V_0} = \frac{{10 \times 0.2 + 5 \times 0.3}}{{0.5}} = 7V;\)

Identification of DC-DC Converter:

• Buck converter: Inductor is connected in series with the combination of the capacitor and resistor.
• Boost converter: Inductor is connected in series with the supply voltage source.
• Buck-Boost converter: Inductor is connected in parallel to the supply voltage source.

 

Under steady-state operating conditions:

• The average voltage across the inductor is zero. V_L = 0
• The average current across the capacitor is zero. I_C = 0

 

 

Given DC-DC converter is a Boost converter with duty cycle ratio δ.

The voltage across the load is the same as the voltage across the capacitor.

\({V_C} = ;{V_0} = {V_{dc}}\left[ {\frac{1}{{1 - δ }}} \right]\)

Concept:

Single-phase full-bridge voltage source inverter (VSI):

The fundamental peak of the output voltage is, Vn1 = m_a × V_dc

Single-phase half-bridge voltage source inverter (VSI):

The fundamental peak of the output voltage is, V_n1 = m_a × V_dc / 2

m_a = modulation index

Calculation:

Given modulation index ma = 0.7, Vdc = 100 V

The fundamental peak of the output voltage of VSI is,

Vn1 = ma × Vdc = 0.7 × 100 = 70 V

The reactance of Inductor for fundamental frequency

\({\mathrm X}_\mathrm L=2\mathrm\pi×50×9.55×10^{-3}=3\mathrm;\Omega\)

The reactance of Capacitor for Fundamental Frequency

\({\mathrm X}_\mathrm C=\frac1{2\mathrm\pi×50×63.66×10^{-6}}=50\mathrm;\Omega\)

Equivalent Impedance at the output

\(\mathrm Z=\frac{5×(-\mathrm j50)}{5-\mathrm j50}=4.975\angle-5.71^\circ\mathrm;\Omega\)

Fundamental voltage amplitude at the output is

\({\mathrm V}_{0\mathrm 1}=70×\frac{4.975\angle-5.71^\circ}{4.975\angle-5.71^\circ+\mathrm j3}=62.76\angle-32.55^\circ\mathrm ;V\)

Concept:

5^th harmonic results in negative phase sequence i.e., Motor will be operating in braking mode.

At 5^th harmonic,

\(\rm N_{s(5)} = \frac{120f}{5p}\)

where,

f = fundamental frequency

p = Number of poles

N_s(5) = synchronous speed due to 5^th harmonic

\(\rm N_{s(5)} = \frac{-N_s}{5}\)

where,

N_s = synchronous speed due to fundamental component

(-)_ve sign is for braking mode as it is operating in reverse phase sequence.

Calculation:

For the 3-phase, 4 pole, 50 Hz induction motor, synchronous speed is given by,

synchronous speed \(\rm N_{s} = \frac{120f}{p}\)

\(\rm ⇒ N_s = \frac{120 \times 50}{4}\)

⇒ N_s = 1500 rpm

operating slip due to 5^th harmonic,

\(\rm s = \frac{N_s - N_r}{N_s} = \frac{N_s - \left( -\frac{N_s}{5} \right)}{N_s}\)

\(⇒ \rm s = \frac{6/5 N_s}{N_s}\)

⇒ s = 1.2

Concept:

\(\rm (a)^n u[n] \overset{Z.T}\longleftrightarrow \frac{z}{z -a}; |z| > |a| \)

\((a)^n u[-n-1] \overset{Z.T}\longleftrightarrow \frac{-Z}{Z -a}; |Z| < |a|\)

Calculations:

we have, 

x[n] = (-0.25)n u[n] + (0.5)n u[-n - 1]

Taking z-transform of above signal,

\(X(Z) = \underset{ROC_1: |Z| > |-0.25|} {\underbrace{{\frac{Z}{Z-(-0.25)}}}} + \underset{ROC_2: |Z| < |0.5|} {\underbrace{{\frac{-Z}{Z-0.25}}}} \)

Hence,

ROC of x(z) = [ROC1] ∩ [ROC2]

= [|Z| > |0.25|] ∩ [|Z| < |0.5|]

= [|Z| > 0.25] ∩ [|Z| < 0.5]

The ROC of Z-transform of x[n] is

0.25 < |Z| < 0.5

Given parallel plate capacitor is

The capacitance of air will be C₁ = Aε₀/d

The capacitance of glass will be C₂ = 4Aε₀/d

Both the dielectrics are in series connection as the area of the plate is the same in both the dielectrics (air and glass)

The equivalent capacitance will be as follows

\(\begin{array}{l} {C_{eq}} = \frac{{{C_1}{C_2}}}{{{C_1} + {C_2}}} = \frac{{4A{\varepsilon _0}}}{{5d}}\ {D_n} = {\rho _s} = \frac{Q}{A} = \frac{{CV}}{A}\ {D_n} = \frac{{4A{\varepsilon _0}}}{{5dA}}V\ {D_n} = \frac{{4{\varepsilon _0}}}{{5d}}V\ {E_1} = \frac{{{D_n}}}{{{\varepsilon _0}}} = \frac{4}{{5d}}V\ 30 \times {10^5} = \frac{{4V}}{{5d}} \Rightarrow V = \frac{{30 \times 5 \times 5 \times {{10}^{ - 3}} \times {{10}^5}}}{4} \end{array}\)

V = 18.75 kV

Analysis:

Given digital circuit 

Let the flipflops be named as 1, 2, and 3 from the left side

Flipflop 2 and 3 are synchronously clocked by Q₀ which is CLK/2

 

Therefore after 3 clock cycles of CLK/2 Q₂Q₁ is attaining 00 again

Given that negative edge triggered J-K flip-flops, 

So, after the 6^th clock cycle of CLK, Q2Q1Q0 = 000

Therefore the number of cycles of CLK is 6

CONCEPT

Maxterm:

A sum (OR) of n Boolean variables or its complement but not both, in a Boolean function of n variables, is called a maxterm

K map:

The K-map is a graphical method that provides a systematic method for simplifying and manipulating the Boolean expressions or to convert a truth table to its corresponding logic circuit in a simple, orderly process.

In an 'n' variable K map, there are 2n cells

For 4 variables there will be 24 = 16 cells as shown:

Analysis:

Given that

f(A,B,C,D) = π M(0,1,3,4,5,7,9,11,12,13,14,15 )

from the K- map it is given by

f (A, B, C, D) = A̅ CD̅ + AB̅D̅

By using Dulaity property

. →  +

+ → -

Q̅ → Q

Q → Q̅ 

we get

(A + C̅ + D)(A̅ + B + D)

​Hence option (1) is correct

Concept:

Virtual ground -

The concept of the virtual ground is stated as if anyone of the i/p terminals is grounded physically the other i/p terminal will also be at ground potential even though, it is not grounded physically.

• One key feature of an Op-Amp is the differential input, and when put together in a circuit, this can form a virtual ground.
• The virtual ground concept is helpful for the analysis of Op Amps. This concept makes Op-Amp circuit analysis much easier.

 

Calculation:

\({V_0} = - X.A = - 1000x\)

By nodal analysis

\(\frac{{x - {V_i}}}{{1k}} + C\frac{{d\left( {x - {V_0}} \right)}}{{dt}} = 0\)

So, \(x = \frac{{{V_0}}}{{1000}}\)

\(\begin{array}{l} - \frac{{{V_0}}}{{1M}} - \frac{{{V_i}}}{{1k}} - \frac{{1\mu \times 1001}}{{1000}};\frac{{d{V_0}}}{{dt}} = 0\ \end{array}\)

\(\frac{{d{V_0}}}{{dt}} + \frac{{1000}}{{1001}}{V_0} = \frac{{ - 1}}{{1\mu \times 1000}}{V_i} \)

By LT on both sides

\(\begin{array}{l} {V_0}\left( {S + \frac{{1000}}{{1001}}} \right) = \frac{{ - 1}}{{S.1\mu \times 1001}}\end{array}\)

\( {V_0} = - \frac{{10^\circ }}{{1001}}\left[ {\frac{1}{{S\left( {S + \frac{{1000}}{{1001}}} \right)}}} \right]\)

\( {V_0} = - \frac{{{{10}^3}}}{{\left( {S + \frac{{1000}}{{1001}}} \right)}} + \frac{{{{10}^3}}}{S} = {10^3}\left( {1 - {e^{\frac{{ - 1000t}}{{1001}}}}} \right)u\left( t \right)\)

\( \tau = \frac{{1001}}{{1000}} = 1001ms \)

Concept:

A successive-approximation ADC is a type of analog-to-digital converter that converts a continuous analog waveform into a discrete digital representation using a binary search through all possible quantization levels before finally converging upon a digital output for each analog voltage conversion

If V_in  ≥ V_DAC ⇒ maintain a loaded bit

V_in ≤  V_DAC⇒ clear the loaded bit

Calculation:

Given,

Reference voltage V_R = 5 V, number of bits n is 8

Resolution \(= \frac{5}{{{2^8} - 1}} \approx 20\ mv\)

Clock pulse 1:

1 st clock = 10000000

Output voltage V₀ = 20 mV × 2⁷ = 2.5 V

As 3.5 V > 2.5 V it maintains the loaded bit

Clock pulse 2:

2 nd clock pulse = 11000000

Output voltage V₀ = 20 mV × (2⁷ + 2⁶) = 3.75 V

As 3.5V < 3.75 V it clears the loaded bit

Clock pulse 3:

So at the end of the 3^rd clock pulse output is = (10100000)₂

Concept:

The condition for optimum operation and economic dispatch is

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

Where,

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

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

Calculation:

\(\begin{array}{l} \frac{{d{C_1}}}{{d{P_1}}} = 2 \times 0.01{P_1} + 30 = 0.02{P_1} + 30\ \frac{{d{C_2}}}{{d{P_2}}} = 2 \times 0.05{P_2} + 10 = 0.10{P_2} + 10\ \frac{{d{C_1}}}{{d{P_1}}} = \frac{{d{C_2}}}{{d{P_2}}}\ ⇒ {P_2} = {\rm{ }}200,{\rm{ }}{P_1} = 0\end{array}\)

But P₂  upper limit is 180 MW.

⇒ P₂ = 180 MW, P₁ = 200 - 180 = 20 MW

Now, P₁ lower limit is 100 MW

Therefore,

⇒ P₁ = 100 MW, P₂ = 200 - 100 = 100 MW

Incremental fuel cost of both the power plant 1

\(\frac{{d{C_1}}}{{d{P_1}}}= 0.02{P_1} + 30=32 Rs/MWh\)

Incremental fuel cost of both the power plant 2

\(\frac{{d{C_2}}}{{d{P_2}}}= 0.10{P_2} + 10=20 Rs/MWhr\)

Power plant 2 has more control over power generation as the increase in the demand can be supported only by power plant 2

Therefore, the incremental cost (in Rs/MWh) of the power plant when it supplies 200 MW is 20 Rs/MWh

• Load flow study determines the operating state of the system for a given loading.
• Load flow solves a set of simultaneous non-linear algebraic power equations for the two unknown variables (|V| and ∠δ) at each node in a system.
• The output of the load flow analysis is the voltage and phase angle, real and reactive power (both sides in each line), line losses, and slack bus power.
• Gauss seidel, Newton Raphson, and Fast decoupled load flow method are the different load flow methods.
• The fast decoupled load flow method gives an approximate load flow solution because it uses fixed number of iterations.
• Accuracy depends on the power mismatch vector tolerance.
• The fast decoupled load flow method is an extension of the Newton-Raphson method formulated in polar coordinates with certain approximations, which results in a fast algorithm for load flow solution.
• The fast decoupled method requires a greater number of iterations than the Newton-Raphson method.

 

​Important Points:

Concept:

The angle variation with respect to time during fault with constant accelerating power(P_a) is given as

\(δ (t) = {δ _o} + \frac{1}{2}\left( {\frac{{{P_a}}}{M}} \right){t^2}\)  ....(1)

Where δ_o = Initial rotor angle

P_a = Accelerating power

M = Angular momentum

M_pu = 2GH/ω_s = H/180f

G = Rating of the generator in pu

H = Inertia constant

Calculation:

Given δ_o = 5°, H = 2 MJ/MVA, t = 0.02 second, f = 50 Hz, P_a = 1pu

Rating of the generator(G) is always asumed as 1 pu, then 

\(M = \frac{{2 × 1}}{{180 × 50}} = \frac{1}{{4500}};pu\)

Substitute all the values in equation(1), we get

δ = 5 + 0.5 × 4500 × 0.02²

δ  = 5.9°

Concept :

As shown in the phasor diagram 

• Current I₂ & I₄ have phase difference of 180°. So that they cancel out each other.
• Magnitude of the current I₁ is more than current I₃. Also I₁ is more lagging than I₃
• So that  fault is located at point Q.

Concept:

Z(s) = R+ jX_L

Where

R = Resistance

X_L = Inductive reactance

Admittance is the inverse of impedance.

\(\begin{array}{l} Y\left( S \right) = \frac{1}{{Z\left( S \right)}} = \frac{1}{{\left( {R + jω L } \right)}}\ Y\left( S \right) = \frac{1}{{\sqrt {R^2 + {(ω L)^2}} }}\angle - ta{n^{ - 1}}\left( ω L/R \right) \end{array}\)

V(t) = I(t) Z(s)

\(I\left( t \right) = \frac{{V\left( t \right)}}{Z(s)}\)

I(t) = V(t) Y(s)

Explanation:

Given 

R = 1Ω 

L = 1H

ω = 1 rad/sec

DC voltage  = 8 V

AC voltage  = 12 sin t

Total input voltage:

V_in(t) = 8 + 12 sin t 

I(t) = V(t) Y(s)

As 8 V is DC voltage so frequency is equal to zero.

\(\begin{array}{l} i\left( t \right) = 8 \cdot \frac{1}{{\sqrt {1 + 0} }}\angle - ta{n^{ - 1}}\left( 0 \right) + \frac{{1 2}}{{\sqrt {1 + 1} }}{\rm{sin}}\left( {t - 45^\circ } \right)\ i\left( t \right) = 8 + \frac{{12}}{{\sqrt 2 }}\left( {sin \cdot \frac{1}{{\sqrt 2 }} - \cos t\frac{1}{{\sqrt 2 }}} \right) \end{array}\)

I(t) = 8 + 6 sin t – 6 cos t

\({I_{rms}} = \sqrt {{8^2} + {{\left( {\frac{6}{{\sqrt 2 }}} \right)}^2} + {{\left( {\frac{6}{{\sqrt 2 }}} \right)}^2}} = 10;A\)

I_rms = 10 A

Concept:

I1 = Y11V1 + Y12V1  ....1

I₂ = Y₂₁V₁ + Y₂₂V₂  .....2

Where, I₁, V₁ are input variables and I₂, V₂ are output variables.

Y11, Y12, Y21, and Y22 are Y matrix elements. 

Calculation:

Case 1: When output port is short circuit so V₂ = 0 so from eqn 2

4 = 10 Y₂₁ ⇒ Y₂₁ = 0.4

Case 2: Now connect 10 V at input terminal and current flow in output terminal is 1.25 A so from eqn 2

1.25 = (0.4)( 5 ) + 1.25 Y₂₂

⇒ Y₂₂ = -0.6

Now input voltage is 3 V so current through 2 Ω connected in output port is

I₂ = 0.4 × 3 – 0.6 (2I₂)

= 1.2 – 1.2 I₂ = 0.545 A

Concept:

KCL:

According to Kirchhoff’s current law (KCL), the algebraic sum of the electric currents meeting at a common point is zero. I.e. the sum of currents entering a node is equal to the sum of currents leaving the node. It is based on the conservation of charge.

By applying KCL in the above diagram,

i1 + i5 = i2 + i3 + i4

The power dissipated through the load(R) is I²/R or V²/R where I is current flow through load R and V is the voltage across that load R

Calculation:

The power dissipated through 2 Ω is 12.5 W

The voltage across  2 Ω  is V_o

\(\frac{(V_o)^2}{R} = 12.5 W\)

Vo = 5 V

Apply the KCL at the junction of 2 Ω, 5 Ω, and 10 Ω in the circuit.

\(\frac{V_o}{2} + kV_o = 5\)    .......1

Now put the value of Vo in the above equation and we get a value of k

k = 0.5 unit

Ohm's law:

At constant temperature, the current flowing through a resistance is directly proportional to the potential difference across its end.

V = R I

Where V is the potential difference, R is resistance and I is current flowing.

N₁ = 1000 rpm

V = 200 V

R_a = 1 Ω

No load Back emf (E_b1) = 200 V

N₂ = 500 rpm

T₃ = 0.5 T_­2

N₃ = 520 rpm

\(\frac{{{E_{b1}}}}{{{E_{b2}}}} = \frac{{1000}}{{500}}\)

⇒ E_b2 = 100 V

E_b2 = V – I_a2

100 = V – I_a2

\(\frac{{{E_{b1}}}}{{{E_{b2}}}} = \frac{{1000}}{{500}}\)

⇒ E_b3 = 104

E_b3 = V – I_a3

104 = V – I_a3

\(\frac{{{T_2}}}{{{T_3}}} \propto \frac{{{I_{a2}}}}{{{I_{a3}}}}\)

\( \Rightarrow \frac{{{T_2}}}{{0.5{T_2}}} = \frac{{{I_{a2}}}}{{{I_{a3}}}}\)

\( \Rightarrow {I_{a3}} = \frac{{{I_{a2}}}}{2} = 8;A\)

Concept:  

Back emf:

• In dc motor also generator action takes place.
• Because of this generator action the rotating conductor's cuts the flux and emf induced in the conductors.
• This induced emf is called back emf. It always opposes the supply voltage.

 

The back emf induced in the dc motor can be given by,

Eb = (ϕZNP) / (60 A)

Where,

ϕ = flux/pole

Z = total number of conductors

A = number of parallel paths

N = speed in RPM

P = number of poles

Calculation:

\(E = \frac{{\phi ZNp}}{{60A}} = \frac{{0.01 \times 666 \times 4 \times 1000}}{{60 \times 2}} = 222 V\)

I_a = (V - E) / R_a = (230 - 222) / 0.267

I_a = 29.96 A

Internal power ( P_d)  \(= ;EI_a = 222; \times ;29.96 = 6651.685 Watt\)

Shaft Power output,

P_out = P_d - Rotational losses

\(\begin{array}{l} {P_{out}} = 6651.685;-;600; = 6051.685\ T = \frac{{{P_{out}}}}{\omega } = \frac{{6051.685}}{{2\pi \times \frac{{1000}}{{60}}}} = 57.78;Nm \end{array}\)

Given information,

two winding 200/400 V transformer supply frequency (f) = 50 Hz

kVA rating = 20 kVA

Auto transformer 200/600 V with 12 Ω load.

Explanation:

using power balance equation,

\(V_LI_K=V_HI_ H\)

⇒ 200 × IL = 600 × IH

⇒ \(I_H=\frac{V_H}{R}=\frac{600}{12}=50\ A\)

Now,

\(I_L=\frac{600\times I_H}{200}=\frac{600\times50}{200}=150\ A\)

Therefore,

\(R_{eq}=\frac{V_L}{I_L}=\frac{200}{150}=1.33\) Ω

Hence, equivalent load resistance seen from L. V. side is 1.33 Ω

Alternate method:

R is transformed to low voltage side, its value is given as,

\(R_{eq}=\frac{R}{K^2}\)

where, K = autotransformer transformation ratio

\(K=\frac{600}{200}=3\)

Therefore,

\(R_{eq}=\frac{12}{(3)^2}=1.33\ \Omega\)

Concept:

For parallel operation of two transformer T₁ and T₂, load shared by transformer T₂ is,

\(S_2=S_L\times\frac{Z_1}{Z_1+Z_2}\)

Explanation:

percentage impedance:

Z₁ = (1 + j6) Ω

Z₂ = (0.8 + j4.8) Ω

load (SL ) = 1000 kVA

Load shared by transformer T₂ is given by,

∴ \(S_2=S_L\times\frac{Z_1}{Z_1+Z_2}\)

⇒ \(S_2=(1000\angle36.87^\circ)\times\frac{(1+j6)}{1+j6+0.8+j4.8}\)

⇒ \(S_2=(1000\angle36.87^\circ)\times\frac{(1+j6)}{1.8+j10.8}\)

⇒ \(S_2=(1000\angle36.87^\circ)\times\frac{(1+j6)}{1.8(1+j6)}\)

⇒ \(S_2=\frac{1000}{8}\angle36.87^\circ\)

∴ S₂ = 555.55 kVA at 0.8 pf lag

From the signal flow graph, we have

\({\dot x_1} = {u_1} + 5{x_1} - 2{x_2}\)

\({\dot x_2} = {u_2} + {u_1} + 2{x_1} + {x_2}\)

\({y_1} = {x_1}\)

\({y_2} = - {x_2} + {x_1}\)

\(\left[ {\begin{array}{*{20}{c}} {{{\dot x}_1}} \ {{{\dot x}_2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 5&{ - 2} \ 2&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}} \ {{x_2}} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 1&0 \ 1&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{u_1}} \ {{u_2}} \end{array}} \right]\)

\(\left[ {\begin{array}{*{20}{c}} {{y_1}} \ {{y_2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0 \ 1&{ - 1} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}} \ {{x_2}} \end{array}} \right]\)

If u₁ is zero, u₂ is only input, and y₁ is only output. Then, the above expression becomes

\(\left[ {\begin{array}{*{20}{c}} {{{\dot x}_1}} \ {{{\dot x}_2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 5&{ - 2} \ 2&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}} \ {{x_2}} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 0 \ 1 \end{array}} \right]\left[ {{u_2}} \right]\)

\({y_1} = \left[ {\begin{array}{*{20}{c}} 1&0 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}} \ {{x_2}} \end{array}} \right]\)

To check the controllability, we form the matrix as

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

Since, determinant of the above matrix is 2(≠0), so its rank will be 2, ie. Equal to the number of variables. Hence, the system is controllable.

Again, to check the observability of the system, we form the matrix

\(\left[ {C:CA} \right] = \left[ {\begin{array}{*{20}{c}} 1&0 \ {54}&{ - 2} \end{array}} \right]\)

Determinant of the above matrix is -2, so its rank will be 2, ie. Equal to the number of variables. Hence, the system is observable. Thus, the given system is controllable and observable when u₂ is only input and y₁ is only output.

Given pole of the root locus is (2 - j3)

The root locus is symmetrical about real axis 

So, there must be another conjugate pair of pole 

Therefore, another conjugate pair of pole is 2 + j3

The pair of poles is (2 ± j3)

Now it is also given that perfectly flat magnitude response of unity 

The magnitude of poles must be equal to the magnitude of zeros 

Therefore the zeros must be (- 2 ± j3)

The location of poles and zeros are as follows in root locus diagram

Concept:

Mason’s gain formula is

\(T = \frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{{\mathop \sum \nolimits_{i = 1}^N {P_i}{{\rm{Δ }}_i}}}{{\rm{Δ }}}\)

Where,

C(s) is the output node

R(s) is the input node

T is the transfer function or gain between R(s) and C(s)

Pi is the ith forward path gain

Δ = 1−(sum of all individual loop gains) + (sum of gain products of all possible two non-touching loops) − (sum of gain products of all possible three non-touching loops) + ........

Δi is obtained from Δ by removing the loops which are touching the ith forward path.

Calculation:

Given block diagram is,

There are two forward paths,

Δ₁P₁ = G₁, Δ₂P₂ = G₂

There are two loops,

• G₁H, - G₂H

Δ = 1 - (- G1H - G2H) = 1 + H (G₁ + G₂)

From mason's gain formula

\( \Rightarrow \frac{Y(s)}{X(s)} = \frac{{{G_1} + {G_2}}}{{1 + H\left( {{G_1} + {G_2}} \right)}} \)

Before adding zero in the system, open loop T.F. of the system is

\(G\left( S \right)H\left( S \right) = \frac{{{K_1}}}{{S\left( {S + 1} \right)\left( {S + 2} \right)}}\)

\(\begin{gathered} 1 + G\left( S \right)H\left( S \right) = S\left( {S + 1} \right)\left( {S + 2} \right) + {K_1} \ \ = {S^3} + 3{S^2} + 2S + {K_1} \ \end{gathered} \)

We check the stability of the system by Routh criteria. The routh array is shown below.

\(\begin{array}{*{20}{c}} {{S^3}}&1&2 \ {{S^2}}&3&{{K_1}} \ S&{\frac{{6 - {K_1}}}{3}}&0 \ {{S^0}}&{{K_1}}&0 \end{array}\)

For the system to be unstable (ie, root locus cross the imaginary axis)

\(\frac{{6 - {K_1}}}{3} \leqslant 3\)

\( \Rightarrow {K_1} \geqslant 6\)

Now adding zero to the system, we have

\(G\left( S \right)H\left( S \right) = \frac{{{K_1}\left( {S + 3} \right)}}{{S\left( {S + 1} \right)\left( {S + 2} \right)}}\)

\(1 + G\left( S \right)H\left( S \right) = {S^3} + 3{S^2} + 2S + {K_1}S + 3{K_1}\)

\( = {S^3} + 3{S^2} + \left( {{K_1} + 2} \right)S + 3{K_1}\)

Routh array for the characteristic equation is obtained as

\(\begin{array}{*{20}{c}} {{S^3}}&1&{{K_1} + 2} \ {{S^2}}&3&{3{K_1}} \ {{S^1}}&2&0 \ {{S^0}}&{3{K_1}}&0 \end{array}\)

In the Routh array, for any value of \({K_1} \geqslant 6\), system does not go in right half cycle of S = 0 plane and in negative feedback system, dc gain is always positive. So, it is always stable. Hence, the root locus of modified system never transits to unstable region.