GATE EE 2008 With Solutions

GATE EE 2008

Q.1 - Q.20 carry one mark each. MCQ 1.1

The number of chords in the graph of the given circuit will be

(A) 3

(B) 4

(C) 5

(D) 6

SOL 1.1

No. of chords is given as l = b−n+1 b " no. of branches n " no. of nodes l " no. of chords b = 6, n = 4 l = 6 − 4 + 1= 3 Hence (A) is correct option.

MCQ 1.2

The Thevenin’s equivalent of a circuit operation at ω = 5 rads/s, has Voc = 3.71+ − 15.9% V and Z0 = 2.38 − j0.667 Ω . At this frequency, the minimal realization of the Thevenin’s impedance will have a (A) resistor and a capacitor and an inductor (B) resistor and a capacitor (C) resistor and an inductor (D) capacitor and an inductor

SOL 1.2

Hence (A) is correct option. Impedance Zo = 2.38 − j0.667 Ω

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Constant term in impedance indicates that there is a resistance in the circuit. Assume that only a resistance and capacitor are in the circuit, phase difference in thevenin voltage is given as (Due to capacitor) θ =− tan− 1 (ωCR) j Zo = R − ωC 1 = 0.667 So, ωC and

R = 2.38 Ω θ =− tan− 1 b 1 # 2.38 l 0.667

=− 74.34c =− [ 15.9c given Voc = 3.71+ − 15.9c So, there is an inductor also connected in the circuit MCQ 1.3

A signal e - αt sin (ωt) is the input to a real Linear Time Invariant system. Given K and φ are constants, the output of the system will be of the form Ke - βt sin (vt + φ) where (A) β need not be equal to α but v equal to ω (B) v need not be equal to ω but β equal to α (C) β equal to α and v equal to ω (D) β need not be equal to α and v need not be equal to ω

SOL 1.3

Hence (D) is correct option. L Let x (t) L y (t) L h (t)

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

So output of the system is given as Y (s) = X (s) H (s) L Now for input x (t − τ) e - sτ X (s) (shifting property) L e− sτ H (s) h (t − τ) So now output is Y' (s) = e - sτ X (s) $ e - τs H (s) Y' (s) = e - 2sτ X (s) H (s) Y' (s) = e - 2sτ Y (s) Or y' (t) = y (t − 2τ) MCQ 1.4

X is a uniformly distributed random variable that takes values between 0 and 1. The value of E {X3} will be (A) 0 (B) 1/8 (C) 1/4


(D) 1/2


Page 3 SOL 1.4

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X is uniformly distributed between 0 and 1 So probability density function 1, 0 < x < 1 fX (X) = ) 0, otherwise So, 1 E {X3} = # X3 fX (X) dx 0

=

#0

1

X3 (1) dx 4 1

= :X D 4 0 =1 4 Hence (C) is correct option MCQ 1.5

The characteristic equation of a (3 # 3 ) matrix P is defined as a (λ) = λI − P = λ3 + λ2 + 2λ + 1 = 0 If I denotes identity matrix, then the inverse of matrix P will be (A) (P2 + P + 2I) (B) (P2 + P + I) (C) − (P2 + P + I)

SOL 1.5

(D) − (P2 + P + 2I)

According to CAYLEY-HAMILTON Theorem every non-singular square matrix satisfies its own characteristic equation. Characteristic equation a (λ) = λI − P = λ3 + λ2 + 2λ + 1 = 0 Matrix P satisfies above equation P 3 + P 2 + 2P + I = 0 I =− (P3 + P2 + 2P) Multiply both sides by P− 1 P− 1 =− (P2 + P + 2I) Hence (D) is correct option.

MCQ 1.6

If the rank of a (5 # 6) matrix Q is 4, then which one of the following statement is correct ? (A) Q will have four linearly independent rows and four linearly independent columns (B) Q will have four linearly independent rows and five linearly independent columns T QQ (C) will be invertible (D) QT Q will be invertible

SOL 1.6

Rank of a matrix is no. of linearly independent rows and columns of the matrix. Here Rank ρ (Q) = 4



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So Q will have 4 linearly independent rows and flour independent columns. Hence (A) is correct option MCQ 1.7

A function y (t) satisfies the following differential equation : dy (t) + y (t) = δ (t) dt where δ (t) is the delta function. Assuming zero initial condition, and denoting the unit step function by u (t), y (t) can be of the form (A) et (B) e - t (C) et u (t)

SOL 1.7

(D) e - t u (t)

Given differential equation for the function dy (t) + y (t) = δ (t) dt Taking Laplace on both the sides we have, sY (s) + Y (s) = 1 (s + 1) Y (s) = 1 Y (s) = 1 s+1 Taking inverse Laplace of Y (s) y (t) = e− t u (t), t > 0 Hence (D) is correct option.

MCQ 1.8

The equivalent circuits of a diode, during forward biased and reverse biased conditions, are shown in the figure. (I) (II)

If such a diode is used in clipper circuit of figure given above, the output voltage V0 of the circuit will be



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SOL 1.8

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Assume the diode is in reverse bias so equivalent circuit is

Output voltage

V0 = 10 sin ωt # 10 = 5 sin ωt 10 + 10

Due to resistor divider, voltage across diode VD < 0 (always). So it in reverse bias for given input. Output, V0 = 5 sin ωt Hence (A) is correct option. MCQ 1.9

Two 8-bit ADCs, one of single slope integrating type and other of successive approximate type, take TA and TB times to convert 5 V analog input signal to equivalent digital output. If the input analog signal is reduced to 2.5 V, the approximate time taken by the two ADCs will respectively, be (A) TA, TB (B) TA /2, TB (C) TA, TB /2

SOL 1.9

(D) TA /2, TB /2

Conversion time does not depend on input voltage so it remains same for both type of ADCs. Hence (A) is correct option



Page 6 MCQ 1.10

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An input device is interfaced with Intel 8085A microprocessor as memory mapped I/O. The address of the device is 2500H. In order to input data from the device to accumulator, the sequence of instructions will be (A) LXI H, 2500H (B) LXI H, 2500H MOV (C) LHLD MOV

A, M 2500H

MOV (D) LHLD

A, M

MOV

M, A 2500H M, A

SOL 1.10

Hence ( ) is Correct Option

MCQ 1.11

Distributed winding and short chording employed in AC machines will result in (A) increase in emf and reduction in harmonics (B) reduction in emf and increase in harmonics (C) increase in both emf and harmonics (D) reduction in both emf and harmonics

SOL 1.11

Distributed winding and short chording employed in AC machine will result in reduction of emf and harmonics. Hence (D) is correct option.

MCQ 1.12

Three single-phase transformer are connected to form a 3-phase transformer bank. The transformers are connected in the following manner :

The transformer connecting will be represented by (A) Y d0 (B) Y d1 (C) Y d6

(D) Y d11

SOL 1.12

Transformer connection will be represented by Y d1. Hence (B) is correct option.

MCQ 1.13

In a stepper motor, the detent torque means (A) minimum of the static torque with the phase winding excited (B) maximum of the static torque with the phase winding excited (C) minimum of the static torque with the phase winding unexcited (D) maximum of the static torque with the phase winding unexcited

SOL 1.13

Detent torque/Restraining toque: The residual magnetism in the permanent magnetic material produced.



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The detent torque is defined as the maximum load torque that can be applied to the shaft of an unexcited motor without causing continuous rotation. In case the motor is unexcited. Hence (D) is correct option. MCQ 1.14

A two machine power system is shown below. The Transmission line XY has positive sequence impedance of Z1 Ω and zero sequence impedance of Z0 Ω

An ‘a’ phase to ground fault with zero fault impedance occurs at the centre of the transmission line. Bus voltage at X and line current from X to F for the phase ‘a’, are given by Va Volts and Ia amperes, respectively. Then, the impedance measured by the ground distance relay located at the terminal X of line XY will be given by (A) ^Z1 /2h Ω (B) ^Z0 /2h Ω (C) (Z0 + Z1) /2 Ω SOL 1.14

(D) ^Va /Ia h Ω

Given for X to F section of phase ‘a’ Va -Phase voltage and Ia -phase current. Impedance measured by ground distance, Bus voltage Relay at X = Current from phase 'a' = Va Ω Ia Hence (D) is correct option.

MCQ 1.15

An extra high voltage transmission line of length 300 km can be approximate by a lossless line having propagation constant β = 0.00127 radians per km. Then the percentage ratio of line length to wavelength will be given by (A) 24.24 % (B) 12.12 % (C) 19.05 %

SOL 1.15

(D) 6.06 %

For EHV line given data is Length of line = 300 km and β = 0.00127 S rad/km wavelength λ = 2π = 2π = 4947.39 km 0.00127 β l % = 300 So 100 = 0.06063 # 100 4947.39 # λ l % = 6.063 λ Hence (D) is correct option.

MCQ 1.16

A-3-phase transmission line is shown in figure :



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Voltage drop across the transmission line is given by the following equation : VR V V R R S3 Va W S Zs Zm Zm WSIa W S 3 Vb W = SZm Zs Zm WSIb W SS 3 V WW SSZ Z Z WWSSI WW c s m m c XT X X T T Shunt capacitance of the line can be neglect. If the has positive sequence impedance of 15 Ω and zero sequence impedance of 48 Ω, then the values of Zs and Zm will be (A) Zs = 31.5 Ω; Zm = 16.5 Ω (B) Zs = 26 Ω; Zm = 11 Ω (C) Zs = 16.5 Ω; Zm = 31.5 Ω (D) Zs = 11 Ω; Zm = 26 Ω SOL 1.16

MCQ 1.17

For three phase transmission line by solving the given equation VRI V RΔV V R(X − X ) 0 0 m WS aW S aW S s We get, 0 (Xs − Xm) 0 WSIbW SΔVbW = S SSΔV WW SS 0 0 (Xs + 2Xm)WWSSIcWW c XT X X T T Zero sequence Impedance = Xs + 2Xm = 48 and Positive Sequence Impedance = Negative Sequence Impedance = (Xs − Xm) = 15 By solving equation (1) and (2) Zs or Xs = 26 and Zm or Xm = 11 Hence (B) is correct option.

...(1)

...(2)

In the single phase voltage controller circuit shown in the figure, for what range of triggering angle (α), the input voltage (V0) is not controllable ?

(A) 0c < α < 45c

(B) 45c < α < 135c

(C) 90c < α < 180c

(D) 135c < α < 180c



Page 9 SOL 1.17

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Hence (A) is correct option.

R + jXL = 50 + 50j tan φ = ωL = 50 = 1 50 R

`

φ = 45c so, firing angle ‘α’ must be higher the 45c, Thus for 0 < α < 45c, V0 is uncontrollable. MCQ 1.18

A 3-phase voltage source inverter is operated in 180c conduction mode. Which one of the following statements is true ? (A) Both pole-voltage and line-voltage will have 3rd harmonic components (B) Pole-voltage will have 3rd harmonic component but line-voltage will be free from 3rd harmonic (C) Line-voltage will have 3rd harmonic component but pole-voltage will be free from 3rd harmonic (D) Both pole-voltage and line-voltage will be free from 3rd harmonic components

SOL 1.18

A 3-φ voltage source inverter is operated in 180c mode in that case third harmonics are absent in pole voltage and line voltage due to the factor cos (nπ/6). so both are free from 3rd harmonic components. Hence (D) is correct option.

MCQ 1.19

The impulse response of a causal linear time-invariant system is given as h (t). Now consider the following two statements : Statement (I): Principle of superposition holds Statement (II): h (t) = 0 for t < 0 Which one of the following statements is correct ? (A) Statements (I) is correct and statement (II) is wrong (B) Statements (II) is correct and statement (I) is wrong (C) Both Statement (I) and Statement (II) are wrong (D) Both Statement (I) and Statement (II) are correct

SOL 1.19

Since the given system is LTI, So principal of Superposition holds due to linearity. For causal system h (t) = 0 , t < 0 Both statement are correct. Hence (D) is correct option.

MCQ 1.20

It

is

desired

to

measure


parameters

of

230

V/115

V,

2

kVA,


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single-phase transformer. The following wattmeters are available in a laboratory: W1 : 250 V, 10 A, Low Power Factor W2 : 250 V, 5 A, Low Power Factor W3 : 150 V, 10 A, High Power Factor W4 : 150 V, 5 A, High Power Factor The Wattmeters used in open circuit test and short circuit test of the transformer will respectively be (A) W1 and W2 (B) W2 and W4 (C) W1 and W4 SOL 1.20

(D) W2 and W3

Given: 1-φ transformer, 230 V/115 V, 2 kVA W1 : 250 V, 10 A, Low Power Factor W2 : 250 V, 5 A, Low Power Factor W3 : 150 V, 10 A, High Power Factor W4 : 150 V, 5 A, High Power Factor In one circuit test the wattmeter W2 is used and in short circuit test of transformer W3 is used. Hence (D) is correct option.

Q.21 to Q.75 carry two marks each. MCQ 1.21

SOL 1.21

The time constant for the given circuit will be

(A) 1/9 s

(B) 1/4 s

(C) 4 s

(D) 9 s

Time constant of the circuit can be calculated by simplifying the circuit as follows



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Ceq = 2 F 3 Equivalent Resistance

Time constant

Req = 3 + 3 = 6 Ω τ = Req Ceq = 6 # 2 = 4 sec 3

Hence (C) is correct option MCQ 1.22

SOL 1.22

The resonant frequency for the given circuit will be

(A) 1 rad/s

(B) 2 rad/s

(C) 3 rad/s

(D) 4 rad/s

Impedance of the circuit is 1 jωC 1 j ωC

R +R 1 − jωCR R = jω L + 1 + jωCR # 1 − jωCR R (1 − jωCR) = jω L + 1 + ω2 C2 R2 jωL (1 + ω2 C2 R2) + R − jωCR2 = 1 + ω2 C2 R2 j [ωL (1 + ω2 C2 R2) − ωCR2] R = + 1 + ω2 C2 R2 1 + ω2 C2 R2

Z = jω L +

For resonance Im (Z) = 0 So, ωL (1 + ω2 C2 R2) = ωCR2 Brought to you by: Nodia and Company PUBLISHING FOR GATE


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L = 0.1 H, C = 1 F, R = 1 Ω So, ω # 0.1 [1 + ω2 (1) (1)] = ω (1) (1) 2 1 + ω2 = 10 & ω = 9 = 3 rad/sec Hence (C) is correct option. MCQ 1.23

Assuming ideal elements in the circuit shown below, the voltage Vab will be

(A) − 3 V

(B) 0 V

(C) 3 V

(D) 5 V

SOL 1.23

By applying KVL in the circuit Vab − 2i + 5 = 0 i = 1 A, Vab = 2 # 1 − 5 =− 3 Volt Hence (A) is correct option

MCQ 1.24

A capacitor consists of two metal plates each 500 # 500 mm2 and spaced 6 mm apart. The space between the metal plates is filled with a glass plate of 4 mm thickness and a layer of paper of 2 mm thickness. The relative primitivities of the glass and paper are 8 and 2 respectively. Neglecting the fringing effect, the capacitance will be (Given that ε0 = 8.85 # 10 - 12 F/m ) (A) 983.3 pF (B) 1475 pF (C) 637.7 pF

SOL 1.24

(D) 9956.25 pF

Here two capacitance C1 and C2 are connected in series, so equivalent capacitance is Ceq = C1 C2 C1 + C 2

C1 = ε0 εr1 A d1 − 12 −6 # 500 # 10 = 8.85 # 10 # 8 # 500 4 # 10− 3 = 442.5 # 10− 11 F



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C2 = ε0 εr2 A d2 − 12 −6 # 500 # 10 = 8.85 # 10 # 2 # 500 −3 2 # 10 = 221.25 # 10− 11 F − 11 − 11 Ceq = 442.5 # 10 − 11 # 221.25 # 10− 11 442.5 # 10 + 221.25 # 10 = 147.6 # 10− 11 - 1476 pF Hence (B) is correct option.

MCQ 1.25

A coil of 300 turns is wound on a non-magnetic core having a mean circumference of 300 mm and a cross-sectional area of 300 mm2. The inductance of the coil corresponding to a magnetizing current of 3 A will be (Given that μ0 = 4π # 10 - 7 H/m) (B) 113.04 μH (A) 37.68 μH (C) 3.768 μH

SOL 1.25

(D) 1.1304 μH

Hence (B) is correct option. Circumference l = 300 mm no. of turns n = 300 Cross sectional area A = 300 mm2 μ n2 A Inductance of coil L = 0 l 4π # 10− 7 # (300) 2 # 300 # 10− 6 (300 # 10− 3) = 113.04 μH =

MCQ 1.26

SOL 1.26

In the circuit shown in the figure, the value of the current i will be given by

(A) 0.31 A

(B) 1.25 A

(C) 1.75 A

(D) 2.5 A

By writing node equations at node A and B Va − 5 + Va − 0 = 0 1 1 2Va − 5 = 0



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& Va = 2.5 V Similarly Vb − 4Vab ++Vb − 0 = 0 3 1 Vb − 4 (Va − Vb) + Vb = 0 3 Vb − 4 (2.5 − Vb) + 3Vb = 0 8Vb − 10 = 0 & Vb = 1.25 V Current i = Vb = 1.25 A 1 Hence (B) is correct option. MCQ 1.27

Two point charges Q1 = 10 μC and Q2 = 20 mC are placed at coordinates (1,1,0) and (− 1, − 1, 0) respectively. The total electric flux passing through a plane z = 20 will be (A) 7.5 μC (B) 13.5 μC (C) 15.0 μC

(D) 22.5 μC

SOL 1.27


MCQ 1.28

Given a sequence x [n], to generate the sequence y [n] = x [3 − 4n], which one of the following procedures would be correct ? (A) First delay x (n) by 3 samples to generate z1 [n], then pick every 4th sample of z1 [n] to generate z2 [n], and than finally time reverse z2 [n] to obtain y [n]. (B) First advance x [n] by 3 samples to generate z1 [n], then pick every 4th sample of z1 [n] to generate z2 [n], and then finally time reverse z2 [n] to obtain y [n] (C) First pick every fourth sample of x [n] to generate v1 [n], time-reverse v1 [n] to obtain v2 [n], and finally advance v2 [n] by 3 samples to obtain y [n] (D) First pick every fourth sample of x [n] to generate v1 [n], time-reverse v1 [n] to obtain v2 [n], and finally delay v2 [n] by 3 samples to obtain y [n]

SOL 1.28

In option (A) z1 [n] = x [n − 3] z2 [n] = z1 [4n] = x [4n − 3] y [n] = z2 [− n] = x [− 4n − 3] = Y x [3 − 4n] In option (B) z1 [n] = x [n + 3] z2 [n] = z1 [4n] = x [4n + 3] y [n] = z2 [− n] = x [− 4n + 3] In option (C) v1 [n] = x [4n]



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v2 [n] = v1 [− n] = x [− 4n] y [n] = v2 [n + 3] = x [− 4 (n + 3)] = Y x [3 − 4n] In option (D) v1 [n] = x [4n] v2 [n] = v1 [− n] = x [− 4n] y [n] = v2 [n − 3] = x [− 4 (n − 3)] = Y x [3 − 4n] Hence (B) is correct option. MCQ 1.29

A system with x (t) and output y (t) is defined by the input-output relation :

#

- 2t

y (t) = x (t) dτ -3 The system will be (A) Casual, time-invariant and unstable (B) Casual, time-invariant and stable (C) non-casual, time-invariant and unstable (D) non-casual, time-variant and unstable SOL 1.29

Input-output relation y (t) =

#- 3x (τ) dτ - 2t

Causality : Since y (t) depends on x (− 2t), So it is non-causal. Time-variance : y (t) =

#- 3x (τ − τ0) dτ =Y y (t − τ0) - 2t

So this is time-variant. Stability : Output y (t) is unbounded for an bounded input. For example Let x (τ) = e - τ (bounded) - τ - 2t - 2t y (t) = e - τ dτ = 8 e B $ Unbounded − 1 -3 -3

#

Hence (D) is correct option. MCQ 1.30

A signal x (t) = sinc (αt) where α is a real constant ^sinc (x) = πx h is the input to a Linear Time Invariant system whose impulse response h (t) = sinc (βt), where β is a real constant. If min (α, β) denotes the minimum of α and β and similarly, max (α, β) denotes the maximum of α and β, and K is a constant, which one of the following statements is true about the output of the system ? (A) It will be of the form Ksinc (γt) where γ = min (α, β) sin (πx)

(B) It will be of the form Ksinc (γt) where γ = max (α, β) (C) It will be of the form Ksinc (αt) (D) It can not be a sinc type of signal Brought to you by: Nodia and Company PUBLISHING FOR GATE


Page 16 SOL 1.30

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Output y (t) of the given system is y (t) = x (t) ) h (t) Or Y (jω) = X (jω) H (jω) given that x (t) = sinc (αt) h (t) = sinc (βt) Fourier transform of x (t) and h (t) are X (jω) = F [x (t)] = π rect` ω j, − α < ω < α α 2α H (jω) = F [h (t)] = π rect` ω j, − β < ω < β β 2β 2 Y (jω) = π rect` ω j rect` ω j αβ 2α 2β

So,

Y (jω) = K rect ` ω j 2γ

Where γ = min (α, β) And y (t) = K sinc (γt) Hence (A) is correct option. MCQ 1.31

Let x (t) be a periodic signal with time period T , Let y (t) = x (t − t0) + x (t + t0) for some t0 . The Fourier Series coefficients of y (t) are denoted by bk . If bk = 0 for all odd k , then t0 can be equal to (B) T/4 (A) T/8 (C) T/2

SOL 1.31

(D) 2T

Let ak is the Fourier series coefficient of signal x (t) Given y (t) = x (t − t0) + x (t + t0) Fourier series coefficient of y (t) bk = e - jkωt ak + e jkωt ak bk = 2ak cos kωt0 bk = 0 (for all odd k ) kωt0 = π , k " odd 2 k 2π t0 = π 2 T 0


0


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For k = 1,

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t0 = T 4

Hence (B) is correct option. MCQ 1.32

H (z) is a transfer function of a real system. When a signal x [n] = (1 + j) n is the input to such a system, the output is zero. Further, the Region of convergence (ROC) of ^1 − 12 z - 1h H(z) is the entire Z-plane (except z = 0 ). It can then be inferred that H (z) can have a minimum of (A) one pole and one zero (B) one pole and two zeros (C) two poles and one zero

SOL 1.32 MCQ 1.33

Hence ( ) is correct option. z Given X (z) = with z > a , the residue of X (z) zn - 1 at z = a 2 (z − a) for n $ 0 will be (B) an (A) an - 1 (D) nan - 1

(C) nan SOL 1.33

(D) two poles and two zeros

Hence (D) is correct option. z , z >a Given that X (z) = (z − a) 2 Residue of X (z) zn - 1 at z = a is = d (z − a) 2 X (z) zn - 1 z = a dz z = d (z − a) 2 zn - 1 dz (z − a) 2 z=a n d z = dz z = a = nzn - 1 z = a = nan - 1

MCQ 1.34

Consider function f (x) = (x2 − 4) 2 where x is a real number. Then the function has (A) only one minimum (B) only tow minima (C) three minima

SOL 1.34

(D) three maxima

Given function f (x) = (x2 − 4) 2 f' (x) = 2 (x2 − 4) 2x To obtain minima and maxima f' (x) = 0 2 4x (x − 4) = 0 x = 0, x2 − 4 = 0 & x = ! 2 So, x = 0, + 2, − 2 f'' (x) = 4x (2x) + 4 (x2 − 4) = 12x2 − 16



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MCQ 1.35

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For x = 0, f'' (0) = 12 (0) 2 − 16 =− 16 < 0 x =+ 2, f'' (2) = 12 (2) 2 − 16 = 32 > 0 x =− 2, f'' (− 2) = 12 (− 2) 2 − 16 = 32 > 0 So f (x) has only two minima Hence (B) is correct option.

(Maxima) (Minima) (Minima)

Equation ex − 1 = 0 is required to be solved using Newton’s method with an initial guess x0 =− 1. Then, after one step of Newton’s method, estimate x1 of the solution will be given by (A) 0.71828 (B) 0.36784 (C) 0.20587

SOL 1.35

(D) 0.00000

An iterative sequence in Newton-Raphson method can obtain by following expression f (xn) xn + 1 = xn − f' (xn) We have to calculate x1 , so n = 0 x1 = x 0 −

f (x 0) , Given x 0 =− 1 f' (x 0)

f (x 0) = ex − 1 = e− 1 − 1 =− 0.63212 f' (x 0) = ex = e− 1 = 0.36787 0

0

So,

x1 =− 1 −

(− 0.63212) (0.36787)

=− 1 + 1.71832 = 0.71832 Hence (A) is correct option MCQ 1.36

A is m # n full rank matrix with m > n and I is identity matrix. Let matrix A' = (AT A) - 1 AT , Then, which one of the following statement is FALSE ? (A) AA'A = A (B) (AA') 2 (C) A'A = I

SOL 1.36

(D) AA'A = A'

Hence (D) is correct option A' = (AT A) − 1 AT = A− 1 (AT ) − 1 AT = A− 1 I Put A' = A− 1 I in all option. option (A)

AA'A = A AA− 1 A = A



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A = A (true) option (B)

(AA') 2 = I (AA− 1 I) 2 = I (I) 2 = I (true)

option (C)

A'A = I A IA = I I = I (true) −1

option (D) MCQ 1.37

AA'A = A' AA− 1 IA = A = Y A' (false)

A differential equation dx/dt = e - 2t u (t), has to be solved using trapezoidal rule of integration with a step size h = 0.01 s. Function u (t) indicates a unit step function. If x (0 -) = 0 , then value of x at t = 0.01 s will be given by (A) 0.00099 (B) 0.00495 (C) 0.0099

SOL 1.37

(D) 0.0198

Hence (C) is correct option dx = e− 2t u (t) dt x = # e− 2t u (t) dt x =

1

# e− 2t dt 0

x =

1

# f (t) dt , 0

t = .01 s From trapezoid rule t + nh f (t) dt = h 6f (0) + f (.01)@ # 2 t 1 e0 + e− .02@, h = .01 #0 f (t) dt = .01 2 6 0

0

= .0099 MCQ 1.38

Let P be a 2 # 2 real orthogonal matrix and x is a real vector [x1, x2] T with length x = (x12 + x22) 1/2 . Then, which one of the following statements is correct ? (A) Px # x where at least one vector satisfies Px < x (B) Px # x for all vector x (C) Px $ x where at least one vector satisfies Px > x (D) No relationship can be established between x and Px

SOL 1.38

P is an orthogonal matrix So PPT = I



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cos θ − sin θ Let assume P = > sin θ cos θ H cos θ − sin θ x x T PX = > sin θ cos θ H8 1 2B cos θ − sin θ x1 => sin θ cos θ H>x2H x1 cos θ − x2 sin θ => x1 sin θ + x2 cos θH = (x1 cos θ − x2 sin θ) 2 + (x1 sin θ + x2 cos θ) 2 = x 12 + x 22 PX = X Hence (B) is correct option. PX

MCQ 1.39

SOL 1.39

Let x (t) = rect^t − 12 h (where rect (x) = 1 for − 12 # x # 12 and zero otherwise. If sin (πx) sinc (x) = πx , then the Fourier Transform of x (t) + x (− t) will be given by (B) 2 sinc` ω j (A) sinc` ω j 2π 2π (C) 2 sinc` ω j cos ` ω j 2π 2 Given signal

(D) sinc` ω j sin ` ω j 2π 2

x (t) = rect `t − 1 j 2 1, − 1 # t − 1 # 1 or 0 # t # 1 2 2 2 x (t) = * 0, elsewhere

So, Similarly

x (− t) = rect`− t − 1 j 2 1, − 1 # − t − 1 # 1 or − 1 # t # 0 2 2 2 x (− t) = * 0, elsewhere F [x (t) + x (− t)] = =

#- 3 x (t) e- jωt dt + #- 3 x (− t) e- jωt dt 3

3

#0 (1) e- jωt dt + #- 1 (1) e- jωt dt 0

1

- jω t - jω t = ; e E +; e E − jω 0 − jω - 1 = 1 (1 − e - jω) + 1 (e jω − 1) jω jω 1


0


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- j ω /2 j ω /2 =e (e jω/2 − e - jω/2) + e (e jω/2 − e - jω/2) jω jω

(e jω/2 − e - jω/2) (e - jω/2 + e jω/2) jω = 2 sin ` ω j $ 2 cos ` ω j ω 2 2 = 2 cos ω sinc` ω j 2 2π

=

Hence (C) is correct option. MCQ 1.40

SOL 1.40

Two perfectly matched silicon transistor are connected as shown in the figure assuming the β of the transistors to be very high and the forward voltage drop in diodes to be 0.7 V, the value of current I is

(A) 0 mA

(B) 3.6 mA

(C) 4.3 mA

(D) 5.7 mA

Hence (C) is correct option.

This is a current mirror circuit. Since β is high so IC1 = IC2, IB1 = IB2 VB = (− 5 + 0.7) =− 4.3 volt Diode D1 is forward biased. So, current I is, I = IC2 = IC1 0 − (− 4.3) = = 4.3 mA 1 MCQ 1.41

In the voltage doubler circuit shown in the figure, the switch ‘S ’ is closed at t = 0



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. Assuming diodes D1 and D2 to be ideal, load resistance to be infinite and initial capacitor voltages to be zero. The steady state voltage across capacitor C1 and C2 will be

SOL 1.41

(A) Vc1 = 10 V,Vc2 = 5 V

(B) Vc1 = 10 V,Vc2 =− 5 V

(C) Vc1 = 5 V,Vc2 = 10 V

(D) Vc1 = 5 V,Vc2 =− 10 V

In positive half cycle of input, diode D1 is in forward bias and D2 is off, the equivalent circuit is

Capacitor C1 will charge upto + 5 volt. VC1 =+ 5 volt In negative halt cycle diode D1 is off and D2 is on.

Now capacitor VC2 will charge upto − 10 volt in opposite direction. Hence (D) is correct option MCQ 1.42

The block diagrams of two of half wave rectifiers are shown in the figure. The transfer characteristics of the rectifiers are also shown within the block.

It is desired to make full wave rectifier using above two half-wave rectifiers. The resultants circuit will be Brought to you by: Nodia and Company PUBLISHING FOR GATE


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SOL 1.42

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Let input Vin is a sine wave shown below

According to given transfer characteristics of rectifiers output of rectifier P is.



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Similarly output of rectifier Q is

Output of a full wave rectifier is given as

To get output V0 V0 = K (− VP + VQ) K − gain of op-amp So, P should connected at inverting terminal of op-amp and Q with non-inverting terminal Hence (D) is correct Option MCQ 1.43

A 3-line to 8-line decoder, with active low outputs, is used to implement a 3-variable Boolean function as shown in the figure

The simplified form of Boolean function F (A, B, C) implemented in ‘Product of Sum’ form will be (A) (X + Z) (X + Y + Z ) (Y + Z) (B) (X + Z ) (X + Y + Z) (Y + Z ) (C) (X + Y + Z) (X + Y + Z) (X + Y + Z) (X + Y + Z ) (D) (X + Y + Z) (X + Y + Z ) (X + Y + Z) (X + Y + Z ) Brought to you by: Nodia and Company PUBLISHING FOR GATE


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In SOP form, F is written as F = Σm (1, 3, 5, 6) = X Y Z + X YZ + XY Z + XYZ Solving from K- map

F = X Z + Y Z + XYZ In POS form F since all outputs are F Hence (B) is correct MCQ 1.44

= (Y + Z) (X + Z) (X + Y + Z ) active low so each input in above expression is complemented = (Y + Z ) (X + Z ) (X + Y + Z) option.

The truth table of monoshot shown in the figure is given in the table below :

Two monoshots, one positive edge triggered and other negative edge triggered, are connected shown in the figure, The pulse widths of the two monoshot outputs Q1 and Q2 are TON and TON respectively. 1

2

The frequency and the duty cycle of the signal at Q1 will respectively be 1 (A) f = , D= 1 TON + TON 5 1

(B) f =

2

TON 1 , D= + TON TON + TON 2

TON

1

2

1

TON (C) f = 1 , D = TON TON + TON

2

1

1

1


2


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TON (D) f = 1 , D = TON TON + TON In this case 1 f = TON1 + TON 2 TON 2 and, D = TON1 + TON 2 1

2

SOL 1.44

1

2


The content of some of the memory location in an 8085 accumulator based system are given below Address

Content

g

g

26FE

00

26FF

01

2700

02

2701

03

2702

04

g g The content of stack (SP), program counter (PC) and (H,L) are 2700 H, 2100 H and 0000 H respectively. When the following sequence of instruction are executed. 2100 H: DAD SP 2101 H: PCHL the content of (SP) and (PC) at the end of execution will be (A) PC = 2102 H, SP = 2700 H (B) PC = 2700 H, SP = 2700 H (C) PC = 2800 H, SP = 26FE H SOL 1.45

(D) PC = 2A02 H, SP = 2702 H

Given that SP = 2700 H PC = 2100 H HL = 0000 H Executing given instruction set in following steps, DAD SP &Add register pair (SP) to HL register HL = HL + SP HL = 0000 H + 2700 H HL = 2700 H PCHL & Load program counter with HL contents PC = HL = 2700 H So after execution contents are, PC = 2700 H, HL = 2700 H



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A waveform generator circuit using OPAMPs is shown in the figure. It produces a triangular wave at point ‘P’ with a peak to peak voltage of 5 V for Vi = 0 V .

If the voltage Vi is made + 2.5 V, the voltage waveform at point ‘P’ will become

SOL 1.46

Hence ( ) is correct option.

MCQ 1.47

A 230 V, 50 Hz, 4-pole, single-phase induction motor is rotating in the clockwise (forward) direction at a speed of 1425 rpm. If the rotor resistance at standstill is 7.8 Ω, then the effective rotor resistance in the backward branch of the equivalent circuit will be (B) 4 Ω (A) 2 Ω (C) 78 Ω

SOL 1.47

(D) 156 Ω

Given: 230 V, 50 Hz, 4-Pole, 1-φ clock-wise(forward) direction Ns = 1425 rpm Rotar resistance at stand still(R2 ) = 7.8 Ω So


induction

motor

is

rotating


in

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Ns = 120 # 50 = 1500 4 Slip(S ) = 1500 − 1425 = 0.05 1500 Resistance in backward branch rb = R2 2−S = 7.8 2 − 0.05 =4Ω Hence (B) is correct option. MCQ 1.48

A 400 V, 50 Hz 30 hp, three-phase induction motor is drawing 50 A current at 0.8 power factor lagging. The stator and rotor copper losses are 1.5 kW and 900 W respectively. The friction and windage losses are 1050 W and the core losses are 1200 W. The air-gap power of the motor will be (A) 23.06 kW (B) 24.11 kW (C) 25.01 kW

(D) 26.21 kW

SOL 1.48

Given: a 400 V, 50 Hz, 30 hp, 3-φ induction motor Current = 50 A at 0.8 p.f. lagging Stator and rotor copper losses are 1.5 kW and 900 W fraction and windage losses = 1050 W Core losses = 1200 W = 1.2 kW So, Input power in stator = 3 # 400 # 50 # 0.8 = 27.71 kW Air gap power = 27.71 − 1.5 − 1.2 = 25.01 kW Hence (C) is correct option.

MCQ 1.49

The core of a two-winding transformer is subjected to a magnetic flux variation as indicated in the figure.

The induced emf (ers) in the secondary winding as a function of time will be of the form



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SOL 1.49

GATE EE 2008

Hence (A) is correct option. Induced emf in secondary =− N2 During

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dφ dt

−0 < t < 1 E1 =− (100)

dφ =− 12 V dt

E1 and E2 are in opposition E2 = 2E1 = 24 V During time 1 < t < 2 dφ = 0 , then E1 = E2 = 0 dt During 2 < t < 2.5 E1 =− (100)

dφ =− 24 V dt

Then E2 =− 0 − 48 V MCQ 1.50

Voltages phasors at the two terminals of a transmission line of length 70 km have a magnitude of 1.0 per unit but are 180 degree out of phase. Assuming that the maximum load current in the line is 1/5th of minimum 3-phase fault current. Which one of the following transmission line protection schemes will not pick up for this condition ? (A) Distance protection using ohm relay with zoen-1 set to 80% of the line impedance.



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(B) Directional over current protection set to pick up at 1.25 times the maximum load current (C) Pilot relaying system with directional comparison scheme (D) Pilot relaying system with segregated phase comparison scheme SOL 1.50


MCQ 1.51

A loss less transmission line having Surge Impedance Loading (SIL) of 2280 MW is provided with a uniformly distributed series capacitive compensation of 30%. Then, SIL of the compensated transmission line will be (A) 1835 MW (B) 2280 MW (C) 2725 MW

(D) 3257 MW

SOL 1.51

SIL has no effect of compensation So SIL = 2280 MW Hence (B) is correct option.

MCQ 1.52

A loss less power system has to serve a load of 250 MW. There are tow generation ( G 1 and G 2 ) in the system with cost curves C1 and C2 respectively defined as follows ; 2 C1 (PG1) = PG1 + 0.055 # PG1 2 C2 (PG2) = 3PG2 + 0.03 # PG2 Where PG1 and PG2 are the MW injections from generator G 1 and G 2 respectively. Thus, the minimum cost dispatch will be (A) PG1 = 250 MW; PG2 = 0 MW (B) PG1 = 150 MW; PG2 = 100 MW (C) PG1 = 100 MW; PG2 = 150 MW

SOL 1.52

Hence (C) is correct option Given PG1 + PG2 C1 (PG1) and C2 (PG2) from equation (2) dC1 dPG1 dC2 and dPG2

(D) PG1 = 0 MW; PG2 = 250 MW

= 250 MW = PG1 + 0.055PG12 = 3PG2 + 0.03PG22

...(1) 4

...(2)

= 1 + 0.11PG1

...(3a)

= 3 + 0.06PG2

...(3b)

Since the system is loss-less dC1 = dC2 Therefore dPG1 dPG2 So from equations (3a) and (3b) We have 0.11PG1 − 0.06PG2 = 2 Now solving equation (1) and (4), we get PG1 = 100 MW Brought to you by: Nodia and Company PUBLISHING FOR GATE

...(4)


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PG2 = 150 MW MCQ 1.53

A loss less single machine infinite bus power system is shown below :

The synchronous generator transfers 1.0 per unit of power to the infinite bus. The critical clearing time of circuit breaker is 0.28 s. If another identical synchronous generator is connected in parallel to the existing generator and each generator is scheduled to supply 0.5 per unit of power, then the critical clearing time of the circuit breaker will (A) reduce to 0.14 s (B) reduce but will be more than 0.14 s (C) remain constant at 0.28 s

(D) increase beyond 0.28 s

SOL 1.53

After connecting both the generators in parallel and scheduled to supply 0.5 Pu of power results the increase in the current. ` Critical clearing time will reduced from 0.28 s but will not be less than 0.14 s for transient stability purpose. Hence (B) is correct option.

MCQ 1.54

Single line diagram of a 4-bus single source distribution system is shown below. Branches e1, e2, e3 and e4 have equal impedances. The load current values indicated in the figure are in per unit.

Distribution company’s policy requires radial system operation with minimum loss. This can be achieved by opening of the branch (A) e1 (B) e2 (C) e3


(D) e4


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SOL 1.54

Given that the each section has equal impedance. Let it be R or Z , then by using the formula line losses = / I2 R On removing (e1); losses = (1) 2 R + (1 + 2) 2 R + (1 + 2 + 5) 2 R = R + 9R + 64R = 74R Similarly, On removing e2 ;losses = 52 R + (5 + 2) 2 R + (5 + 2 + 1) 2 R = 138R lossess on removing e 3 = (1) 2 R + (2) 2 R + (5 + 2) 2 R = 1R + 4R + 49R = 54R on removing e 4 lossless = (2) 2 R + (2 + 1) 2 R + 52 R = 4R + 9R + 25R = 38R So, minimum losses are gained by removing e 4 branch. Hence (D) is correct option

MCQ 1.55

A single phase fully controlled bridge converter supplies a load drawing constant and ripple free load current, if the triggering angle is 30c, the input power factor will be (A) 0.65 (B) 0.78 (C) 0.85

(D) 0.866

SOL 1.55

Given α = 30c, in a 1-φ fully bridge converter we know that, Power factor = Distortion factor # cos α D.f. (Distortion factor) = Is(fundamental) /Is = 0.9 power factor = 0.9 # cos 30c = 0.78 Hence (B) is correct option

MCQ 1.56

A single-phase half controlled converter shown in the figure feeding power to highly inductive load. The converter is operating at a firing angle of 60c.

If the firing pulses are suddenly removed, the steady state voltage (V0) waveform of the converter will become



Page 33

SOL 1.56

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Output of this

Here the inductor makes T1 and T3 in ON because current passing through T1 and Brought to you by: Nodia and Company PUBLISHING FOR GATE


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T3 is more than the holding current. Hence (A) is correct option MCQ 1.57

A 220 V 20 A, 1000 rpm, separately excited dc motor has an armature resistance of 2.5 Ω. The motor is controlled by a step down chopper with a frequency of 1 kHz. The input dc voltage to the chopper is 250 V. The duty cycle of the chopper for the motor to operate at a speed of 600 rpm delivering the rated torque will be (A) 0.518 (B) 0.608 (C) 0.852

(D) 0.902

SOL 1.57


MCQ 1.58

A 220 V, 1400 rpm, 40 A separately excited dc motor has an armature resistance of 0.4 Ω. The motor is fed from a single phase circulating current dual converter with an input ac line voltage of 220 V (rms). The approximate firing angles of the dual converter for motoring operating at 50% of rated torque and 1000 rpm will be (A) 43c, 137c (B) 43c, 47c (C) 39c, 141c

(D) 39c, 51c

SOL 1.58


MCQ 1.59

A single phase source inverter is feeding a purely inductive load as shown in the figure The inverter is operated at 50 Hz in 180c square wave mode. Assume that the load current does not have any dc component. The peak value of the inductor current i0 will be

SOL 1.59

(A) 6.37 A

(B) 10 A

(C) 20 A

(D) 40 A

Input is given as

Here load current does not have any dc component Brought to you by: Nodia and Company PUBLISHING FOR GATE


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(π/ω)

`Peak current occur at `

Here So

Vs = L di dt 200 = 0.1 # di dt di = a π kb 1 l = 1 2π 50 100 di(max) = 200 # 1 # 1 = 20 A 100 0.1

Hence (C) is correct option MCQ 1.60

A 400 V, 50 Hz, 4-pole, 1400 rpm, star connected squirrel cage induction motor has the following parameters referred to the stator: R'r = 1.0 Ω, Xs = X'r = 1.5 Ω Neglect stator resistance and core and rotational losses of the motor. The motor is controlled from a 3-phase voltage source inverter with constant V/f control. The stator line-to-line voltage(rms) and frequency to obtain the maximum torque at starting will be : (A) 20.6 V, 2.7 Hz (B) 133.3 V, 16.7 Hz (C) 266.6 V, 33.3 Hz

SOL 1.60

(D) 323.3 V, 40.3 Hz

Given 400 V, 50 Hz, 4-Pole, 1400 rpm star connected squirrel cage induction motor. R = 1.00 Ω, Xs = Xlr = 1.5 Ω So, for max. torque slip Rlr Sm = Xsm + Xlrm for starting torque Sm = 1 Then Xsm + Xlrm = Rlr 2πfm Ls + 0.2πfm Llr = 1 Frequency at max. torque 1 2π (Ls + Llr ) Xs Ls = = 1.5 2π # 50 2π # 50 Llr = 1.5 2π # 50 1 fm = = 50 1. 5 + 1. 5 3 50 50 fm =

fm = 16.7 Hz In const V/f control method Brought to you by: Nodia and Company PUBLISHING FOR GATE


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So

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V1 = 400 = 8 50 f1 a V2 = 8 f1

V2 = f2 # 8 = 16.7 # 8 V2 = 133.3 V Hence (B) is correct option. MCQ 1.61

A single phase fully controlled converter bridge is used for electrical braking of a separately excited dc motor. The dc motor load is represented by an equivalent circuit as shown in the figure.

Assume that the load inductance is sufficient to ensure continuous and ripple free load current. The firing angle of the bridge for a load current of I0 = 10 A will be (A) 44c (B) 51c (C) 129c SOL 1.61

(D) 136c

Here for continuous conduction mode, by Kirchoff’s voltage law, average load current

V − 2Ia + 150 = 0 Ia = V + 150 2 ` I1 = 10 A, So V =− 130 V 2Vm cos α =− 130 π 2#

2 # 230 cos α =− 130c π



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α = 129c Hence (C) is correct option. MCQ 1.62

A three phase fully controlled bridge converter is feeding a load drawing a constant and ripple free load current of 10 A at a firing angle of 30c. The approximate Total harmonic Distortion (%THD) and the rms value of fundamental component of input current will respectively be (A) 31% and 6.8 A (B) 31% and 7.8 A (C) 66% and 6.8 A

SOL 1.62

(D) 66% and 7.8 A

Hence (B) is correct option. Total rms current Ia =

2 10 = 8.16 A 3#

Fundamental current Ia1 = 0.78 # 10 = 7.8 A THD = where

1 −1 DF2

DF = Ia1 = 0.78 # 10 = 0.955 0.816 # 10 Ia ` MCQ 1.63

THD =

1 2 b 0.955 l − 1 = 31%

In the circuit shown in the figure, the switch is operated at a duty cycle of 0.5. A large capacitor is connected across the load. The inductor current is assumed to be continuous.

The average voltage across the load and the average current through the diode will respectively be (A) 10 V, 2 A (B) 10 V, 8 A (C) 40 V 2 A SOL 1.63

(D) 40 V, 8 A




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In the given diagram when switch S is open I 0 = IL = 4 A, Vs = 20 V when switch S is closed ID = 0, V0 = 0 V Duty cycle = 0.5 so average voltage is Vs 1−δ Average current = 0 + 4 = 2 amp 2 Average voltage = MCQ 1.64

20 = 40 V 1 − 0.5

The transfer function of a linear time invariant system is given as G (s) = 2 1 s + 3s + 2 The steady state value of the output of the system for a unit impulse input applied at time instant t = 1 will be (A) 0 (B) 0.5 (C) 1

SOL 1.64

(D) 2

Given transfer function 1 s2 + 3s + 2 Input r (t) = δ (t − 1) R (s) = L [δ (t − 1)] = e− s Output is given by Y (s) = R (s) G (s) −s Y (s) = 2 e s + 3s + 2 Steady state value of output lim y (t) = lim sY (s) G (s) =

t"3

s"0

se− s =0 s " 0 s + 3s + 2 Hence (A) is correct option. = lim

MCQ 1.65

2

The transfer functions of two compensators are given below : 10 (s + 1) C1 = , C2 = s + 10 (s + 10) 10 (s + 1) Which one of the following statements is correct ? (A) C1 is lead compensator and C2 is a lag compensator (B) C1 is a lag compensator and C2 is a lead compensator (C) Both C1 and C2 are lead compensator (D) Both C1 and C2 are lag compensator

SOL 1.65

For C1 Phase is given by



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θC = tan− 1 (ω) − tan− 1 a ω k 10 Jω − ω N −1 K 10 O θC1 = tan 2 KK O ω 1+ O 10 L P 9 ω −1 θC1 = tan c > 0 (Phase lead) 10 + ω2 m Similarly for C2 , phase is θC2 = tan− 1 a ω k − tan− 1 (ω) 10 J ω − ωN − 1 K 10 O = tan 2 KK O ω 1+ O 10 L P 9 − ω −1 < 0 (Phase lag) = tan c 10 + ω2 m Hence (A) is correct option 1

MCQ 1.66

The asymptotic Bode magnitude plot of a minimum phase transfer function is shown in the figure :

This transfer function has (A) Three poles and one zero

(B) Two poles and one zero

(C) Two poles and two zero

(D) One pole and two zeros

SOL 1.66

From the given bode plot we can analyze that: Slope − 40 dB/decade"2 poles Slope − 20 dB/decade (Slope changes by + 20 dB/decade)"1 Zero Slope 0 dB/decade (Slope changes by + 20 dB/decade)"1 Zero So there are 2 poles and 2 zeroes in the transfer function. Hence (C) is correct option.

MCQ 1.67

Figure shows a feedback system where K > 0



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The range of K for which the system is stable will be given by (A) 0 < K < 30 (B) 0 < K < 39 (C) 0 < K < 390 SOL 1.67

(D) K > 390

Characteristic equation for the system K =0 1+ s (s + 3) (s + 10) s (s + 3) (s + 10) + K = 0 s3 + 13s2 + 30s + K = 0 Applying Routh’s stability criteria s3

1

30

s2

13

K

s1

(13 # 30) − K 13 K

s0 For stability there should be no sign change in first column So, 390 − K > 0 & K < 390 K >0 0 < K < 90 Hence (C) is correct option MCQ 1.68

The transfer function of a system is given as 100 2 s + 20s + 100 The system is (A) An over damped system (B) An under damped system (C) A critically damped system

SOL 1.68

(D) An unstable system

Given transfer function is 100 s2 + 20s + 100 Characteristic equation of the system is given by s2 + 20s + 100 = 0 ωn2 = 100 & ωn = 10 rad/sec. 2ξωn = 20 or ξ = 20 = 1 2 # 10 H (s)) =



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(ξ = 1) so system is critically damped. Hence (C) is correct option. MCQ 1.69

SOL 1.69

Two sinusoidal signals p (ω1, t) = A sin ω1 t and q (ω2 t) are applied to X and Y inputs of a dual channel CRO. The Lissajous figure displayed on the screen shown below : The signal q (ω2 t) will be represented as

(A) q (ω2 t) = A sin ω2 t, ω2 = 2ω1

(B) q (ω2 t) = A sin ω2 t, ω2 = ω1 /2

(C) q (ω2 t) = A cos ω2 t, ω2 = 2ω1

(D) q (ω2 t) = A cos ω2 t, ω2 = ω1 /2

Hence (D) is correct option.

Frequency ratio

meeting points of horizontal tangents fY = fX meeting points of vertical tangents fY =2 4 fX fY = 1 (fX ) 2

ω2 = ω1 /2 Since the Lissajous figures are ellipse, so there is a phase difference of 90c exists between vertical and horizontal inputs. So q (ω2 t) = A cos ω2 t, ω2 = ω1 /2 MCQ 1.70

The ac bridge shown in the figure is used to measure the impedance Z .



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If the bridge is balanced for oscillator frequency f = 2 kHz, then the impedance Z will be (A) (260 + j0) Ω (B) (0 + j200) Ω (C) (260 − j200) Ω SOL 1.70

(D) (260 + j200) Ω

Impedance of different branches is given as ZAB = 500 Ω 1 + 300 Ω j # 2π # 2 # 103 # 0.398 μF - (− 200j + 300) Ω = j # 2π # 2 # 103 # 15.91 mH + 300 Ω

ZBC = ZAD

- (200j + 300) Ω To balance the bridge ZAB ZCD = ZAD ZBC 500Z = (200j + 300) (− 200j + 300) 500Z = 130000 Z = (260 + j0) Ω Hence (A) is correct option.

Common Data for Questions 71, 72 and 73: Consider a power system shown below:

Given that: Vs1 = Vs2 = 1 + j0 p.u ; The positive sequence impedance are Zs1 = Zs2 = 0.001 + j0.01 p.u and ZL = 0.006 + j0.06 p.u 3-phase Base MVA = 100 voltage base = 400 kV(Line to Line) Nominal system frequency = 50 Hz. Brought to you by: Nodia and Company PUBLISHING FOR GATE


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The reference voltage for phase ‘a’ is defined as V (t) = Vm cos (ωt). A symmetrical three phase fault occurs at centre of the line, i.e. point ‘F’ at time ‘t 0 ’. The positive sequence impedance from source S1 to point ‘F’ equals 0.004 + j0.04 p.u. The wave form corresponding to phase ‘a’ fault current from bus X reveals that decaying d.c. offset current is negative and in magnitude at its maximum initial value, Assume that the negative sequence impedances are equal to postive sequence impedance and the zero sequence impedances are three times positive sequence impedances. MCQ 1.71

SOL 1.71

The instant (t0) of the fault will be (A) 4.682 ms

(B) 9.667 ms

(C) 14.667 ms

(D) 19.667 ms

Given V (t) = Vm cos (ωt) For symmetrical 3 − φ fault, current after the fault i (t) = Ae− (R/L) t + 2 Vm cos (ωt − α) Z At the instant of fault i.e t = t 0 , the total current i (t) = 0 ` 0 = Ae− (R/L) t + 2 Vm cos (ωt 0 − α) Z 0

Ae− (R/L) t =− 2 Vm cos (ωt 0 − α) Z 0

Maximum value of the dc offset current Ae− (R/L) t =− 2 Vm cos (ωt 0 − α) Z 0

For this to be negative max. (ωt 0 − α) = 0 or t0 = α ω and ` From equation (1)

...(1)

Z = 0.004 + j0.04 Z = Z +α = 0.0401995+84.29c α = 84.29cor 1.471 rad. t0 =

1.471 = 0.00468 sec (2π # 50)

t 0 = 4.682 ms Hence (A) is correct option. MCQ 1.72

The rms value of the component of fault current (If ) will be (A) 3.59 kA (B) 5.07 kA (C) 7.18 kA


(D) 10.15 kA Visit us at: www.nodia.co.in

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SOL 1.72

Since the fault ‘F’ is at mid point of the system, therefore impedance seen is same from both sides.

Z = 0.0201+84.29c 2 Z1 (Positive sequence) = Z = 0.0201+84.29c 2 also Z1 = Z2 = Z 0 (for 3-φ fault) 1+0c ` I f (pu) = 1+0c = Z1 0.0201+84.29c So magnitude

If

(p.u.)

= 49.8

I f = 49.8 #

` Fault current

100 3 # 400

= 7.18 kA Hence (C) is correct option. MCQ 1.73

Instead of the three phase fault, if a single line to ground fault occurs on phase ‘a’ at point ‘F’ with zero fault impedance, then the rms of the ac component of fault current (Ix) for phase ‘a’ will be (A) 4.97 p.u (B) 7.0 p.u (C) 14.93 p.u

SOL 1.73

(D) 29.85 p.u

If fault is LG in phase ‘a’

Z1 = Z = 0.0201+84.29c 2 and

Z2 = Z1 = 0.0201+84.29c Z 0 = 3Z1 = 0.0603+84.29c



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Then Ia /3 = Ia1 = Ia2 = Ia0 ` and

1.0+0c Z1 + Z 2 + Z 0 1. 0 = = 9.95 pu (0.0201 + 0.0201 + 0.0603)

Ia1 (pu) = Ia1

Fault Current I f = Ia = 3Ia1 = 29.85 pu So

Fault current I f = 29.85 #

100 3 # 400

= 4.97 kA Hence (A) is correct option.

Common Data for Questions 74 and 75: A 3-phase, 440 V, 50 Hz, 4-pole slip ring induction motor is feed from the rotor side through an auto-transformer and the stator is connected to a variable resistance as shown in the figure.

The motor is coupled to a 220 V, separately excited d.c generator feeding power to fixed resistance of 10 Ω. Two-wattmeter method is used to measure the input power to induction motor. The variable resistance is adjusted such the motor runs at 1410 rpm and the following readings were recorded W1 = 1800 W, W2 =− 200 W. MCQ 1.74

The speed of rotation of stator magnetic field with respect to rotor structure will be (A) 90 rpm in the direction of rotation (B) 90 rpm in the opposite direction of rotation (C) 1500 rpm in the direction of rotation (D) 1500 rpm in the opposite direction of rotation

SOL 1.74

Given 3-φ, 440 V, 50 Hz, 4-Pole slip ring motor



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Motor is coupled to 220 V N = 1410 rpm, W1 = 1800 W, W2 = 200 W So, 120f Ns = P = 120 # 50 = 1500 rpm 4 Relative speed = 1500 − 1410 = 90 rpm in the direction of rotation. Hence (A) is correct option. MCQ 1.75

Neglecting all losses of both the machines, the dc generator power output and the current through resistance (Rex) will respectively be (A) 96 W, 3.10 A (B) 120 W, 3.46 A (C) 1504 W, 12.26 A

SOL 1.75

(D) 1880 W, 13.71 A

Neglecting losses of both machines Slip(S ) = Ns − N Ns = 1500 − 1410 = 0.06 1500 total power input to induction motor is Pin = 1800 − 200 = 1600 W Output power of induction motor Pout = (1 − S) Pin = (1 − 0.06) 1600 = 1504 W Losses are neglected so dc generator input power = output power = 1504 W So, I2 R = 1504 I =

1504 = 12.26 A 10


Linked Answer Questions: Q.76 to Q.85 carry two marks each.

Statement for Linked Answer Questions 76 and 77: The current i (t) sketched in the figure flows through a initially uncharged 0.3 nF Brought to you by: Nodia and Company PUBLISHING FOR GATE


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capacitor.

MCQ 1.76

The charge stored in the capacitor at t = 5 μs , will be (A) 8 nC (B) 10 nC (C) 13 nC

SOL 1.76

(D) 16 nC

Charge stored at t = 5 μ sec 5

Q =

# i (t) dt 0

=area under the curve

Q =Area OABCDO =Area (OAD)+Area(AEB)+Area(EBCD) = 1#2#4+1#2#3+3#2 2 2 = 4+3+6 = 13 nC Hence (C) is correct option



Page 48 MCQ 1.77

GATE EE 2008

The capacitor charged upto 5 ms, as per the current profile given in the figure, is connected across an inductor of 0.6 mH. Then the value of voltage across the capacitor after 1 μs will approximately be (A) 18.8 V (B) 23.5 V (C) − 23.5 V

SOL 1.77

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(D) − 30.6 V

Initial voltage across capacitor Q V0 = o = 13 nC 0.3 nF C = 43.33 Volt When capacitor is connected across an inductor it will give sinusoidal esponse as vc (t) = Vo cos ωo t where ωo = 1 LC 1 = −9 0.3 # 10 # 0.6 # 10− 3 = 2.35 # 106 rad/sec at t = 1 μ sec So, vc (t) = 43.33 cos (2.35 # 106 # 1 # 10− 6) = 43.33 # (− 0.70) =− 30.44 V Hence (D) is correct option.

Statement for Linked Answer Question 78 and 79.

MCQ 1.78

The state space equation of a system is described by Xo = AX + Bu,Y = CX where X is state vector, u is input, Y is output and 0 1 0 A == , B = = G, C = [1 0] G 0 −2 1 The transfer function G(s) of this system will be s (A) (B) s + 1 s (s − 2) (s + 2) s 1 (D) (s − 2) s (s + 2) State space equation of the system is given by, o = AX + Bu X Y = CX Taking Laplace transform on both sides of the equations. sX (s) = AX (s) + BU (s) (sI − A) X (s) = BU (s) X (s) = (sI − A) − 1 BU (s) ` Y (s) = CX (s) (C)

SOL 1.78



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Y (s) = C (sI − A) − 1 BU (s)

So

Y (s) = C (sI − A) − 1 B U (s) s 0 0 1 s −1 => (sI − A) = > H − > H 0 s 0 −2 0 s + 2H

T.F =

1 >s + 2 1H s (s + 2) 0 s R V 1 W S1 s s (s + 2)W = SS 1 W 0 S (s + 2) W T X Transfer function V V R R 1 W S 1 W S1 s s (s + 2)W 0 Ss (s + 2)W G (s) = C [sI − A] − 1 B = 81 0BSS >1H = 81 0BS 1 W W 1 S (s + 2) W S0 (s + 2) W X X T T 1 = s (s + 2) (sI − A) − 1 =

Hence (D) is correct option. MCQ 1.79

A unity feedback is provided to the above system G (s) to make it a closed loop system as shown in figure.

For a unit step input r (t), the steady state error in the input will be (A) 0 (B) 1 (C) 2 SOL 1.79

(D) 3

Steady state error is given by, sR (s) ess = lim = G s " 0 1 + G (s) H (s) Here R (s) = L [r (t)] = 1 (Unit step input) s 1 G (s) = s (s + 2)

So,

H (s) = 1 (Unity feed back) V R sb 1 l W S s W ess = lim S 1 s"0S W + 1 S s (s + 2) W T s (s + 2) X = lim = G s " 0 s (s + 2) + 1



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=0 Hence (A) is correct option.

Statement for Linked Answer Questions 80 and 81. A general filter circuit is shown in the figure :

MCQ 1.80

If R1 = R2 = RA and R3 = R4 = RB , the circuit acts as a (A) all pass filter (B) band pass filter (C) high pass filter

SOL 1.80

(D) low pass filter

For low frequencies, ω " 0 , so 1 " 3 ωC Equivalent circuit is,

By applying node equation at positive and negative input terminals of op-amp. vA − vi + vA − vo = 0 R1 R2 2vA = vi + vo ,

a R1 = R 2 = R A Similarly, vA − vi + vA − 0 = 0 R3 R4 Brought to you by: Nodia and Company PUBLISHING FOR GATE


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2vA = vin ,

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a R 3 = R 4 = RB

So, vo = 0 It will stop low frequency signals. For high frequencies, ω " 3 , then 1 " 0 ωC Equivalent circuit is,

Output, vo = vi So it will pass high frequency signal. This is a high pass filter. Hence (C) is correct option MCQ 1.81

The output of the filter in Q.80 is given to the circuit in figure : The gain v/s frequency characteristic of the output (vo) will be

SOL 1.81

In Q.80 cutoff frequency of high pass filter is given by, 1 ωh = 2πRA C



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Here given circuit is a low pass filter with cutoff frequency, 1 2 ωL = = R 2 R π AC 2π A C 2 ωL = 2ωh When both the circuits are connected together, equivalent circuit is,

So this is is Band pass filter, amplitude response is given by.


Statement for Linked Answer Question 82 and 83. MCQ 1.82

A 240 V, dc shunt motor draws 15 A while supplying the rated load at a speed of 80 rad/s. The armature resistance is 0.5 Ω and the field winding resistance is 80 Ω. The net voltage across the armature resistance at the time of plugging will be (A) 6 V (B) 234 V (C) 240 V

(D) 474 V

SOL 1.82

Given: V = 240 V , dc shunt motor I = 15 A Rated load at a speed = 80 rad/s Armature Resistance = 0.5 Ω Field winding Resistance = 80 Ω So, E = 240 − 12 # 0.5 E = 234 Vplugging = V + E = 240 + 234 = 474 V Hence (D) is correct option.

MCQ 1.83

The external resistance to be added in the armature circuit to limit the armature current to 125% of its rated value is (A) 31.1 Ω (B) 31.9 Ω (C) 15.1 Ω


(D) 15.9 Ω] Visit us at: www.nodia.co.in

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SOL 1.83

External Resistance to be added in the armature circuit to limit the armature current to 125%. 474 So Ia = 12 # 1.25 = Ra + R external Ra + R external = 31.6 R external = 31.1 Ω Hence (A) is correct option

Statement for Linked Answer Question 84 and 85. A synchronous motor is connected to an infinite bus at 1.0 pu voltage and draws 0.6 pu current at unity power factor. Its synchronous reactance is 1.0 pu resistance is negligible. MCQ 1.84

The excitation voltage (E ) and load angle (δ) will respectively be (A) 0.8 pu and 36.86c lag (B) 0.8 pu and 36.86c lead (C) 1.17 pu and 30.96c lead

(D) 1.17 pu and 30.96c lag

SOL 1.84

A synchronous motor is connected to an infinite bus at 1.0 p.u. voltage and 0.6 p.u. current at unity power factor. Reactance is 1.0 p.u. and resistance is negligible. So, V = 1+0c p.u. Ia = 0.6+0c p.u. Zs = Ra + jXs = 0 + j1 = 1+90c p.u. V = E+δ + Ia Zs = 1+0c − 0.6+0c # 1+90c E+δ = 1.166+ − 30.96c p.u. Excitation voltage = 1.17 p.u. Load angle (δ) = 30.96c(lagging) Hence (D) is correct option.

MCQ 1.85

Keeping the excitation voltage same, the load on the motor is increased such that the motor current increases by 20%. The operating power factor will become (A) 0.995 lagging (B) 0.995 leading (C) 0.791 lagging

SOL 1.85

(D) 0.848 leading




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Answer Sheet

1.

(A)

19.

(D)

37.

(C)

55.

(B)

73.

(A)

2.

(A)

20.

(D)

38.

(B)

56.

(A)

74.

(A)

3.

(D)

21.

(C)

39.

(C)

57.

(*)

75.

(C)

4.

(C)

22.

(C)

40.

(C)

58.

(*)

76.

(C)

5.

(D)

23.

(A)

41.

(D)

59.

(C)

77.

(D)

6.

(A)

24.

(B)

42.

(D)

60.

(B)

78.

(D)

7.

(D)

25.

(B)

43.

(B)

61.

(C)

79.

(A)

8.

(A)

26.

(B)

44.

(B)

62.

(B)

80.

(C)

9.

(A)

27.

(*)

45.

(B)

63.

(C)

81.

(D)

10.

(*)

28.

(B)

46.

(*)

64.

(A)

82.

(D)

11.

(D)

29.

(D)

47.

(B)

65.

(A)

83.

(A)

12.

(B)

30.

(A)

48.

(C)

66.

(C)

84.

(D)

13.

(D)

31.

(B)

49.

(A)

67.

(C)

85.

(D)

14.

(D)

32.

(*)

50.

(*)

68.

(C)

15.

(D)

33.

(D)

51.

(B)

69.

(D)

16.

(B)

34.

(B)

52.

(C)

70.

(A)

17.

(A)

35.

(A)

53.

(B)

71.

(A)

18.

(D)

36.

(D)

54.

72. (D) **********

(C)



GATE EE 2008 With Solutions

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