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

Transcription

1 GATE EE 008 Q. - Q.0 carry one mark each. MCQ. The number of chords in the graph of the given circuit will be SOL. (A) 3 (B) 4 (C) 5 (D) 6 No. of chords is given as l b n+ b " no. of branches n " no. of nodes l " no. of chords b 6, n 4 l Hence (A) is correct option. MCQ. The Thevenin s equivalent of a circuit operation at ω 5 rads/s, has % Voc V and Z0.38 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. Hence (A) is correct option. Impedance Z o.38 j0.667 Ω

2 Page GATE EE 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 θ tan ( ω CR) (Due to capacitor) j Z o R ωc So, ωc and R.38 Ω θ tan. 38 b l # 74.34c [ 5.9c given V oc c So, there is an inductor also connected in the circuit -αt MCQ.3 A signal e sin( ωt) is the input to a real Linear Time Invariant system. Given K -βt and φ are constants, the output of the system will be of the form Ke 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.3 Hence (D) is correct option. Let xt () L Xs () yt () L Ys () ht () L Hs () So output of the system is given as Ys () XsHs () () Now for input xt ( τ) L -sτ e X() s (shifting property) ht ( τ) L sτ e H() s So now output is s s Y'( s ) e X() s $ e H() s -sτ Y'( s ) e X() s H() s -sτ Y'( s ) e Y() s Or y'( t ) yt ( τ) MCQ.4 X is a uniformly distributed random variable that takes values between 0 and. The value of E{ X 3 } will be (A) 0 (B) /8 (C) /4 (D) /

3 Page 3 GATE EE SOL.4 X is uniformly distributed between 0 and So probability density function, 0 < x < fx ( X) ) 0, otherwise So, E{ X 3 3 } # XfX( Xdx ) 0 # 0 4 X : 4 D 3 X () dx 4 Hence (C) is correct option 0 MCQ.5 The characteristic equation of a (3# 3) matrix P is defined as a( λ) λi P λ 3 + λ + λ+ 0 If I denotes identity matrix, then the inverse of matrix P will be (A) ( P + P+ I) (B) ( P + P+ I) (C) ( P + P+ I) (D) ( P + P+ I) SOL.5 According to CAYLEY-HAMILTON Theorem every non-singular square matrix satisfies its own characteristic equation. Characteristic equation a( λ) λi P λ 3 + λ + λ+ 0 Matrix P satisfies above equation 3 P + P + P+ I 0 3 I ( P + P + P) Multiply both sides by P P ( P + P+ I) Hence (D) is correct option. MCQ.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 (C) QQT will be invertible (D) Q T Q will be invertible SOL.6 Rank of a matrix is no. of linearly independent rows and columns of the matrix. Here Rank ρ ( Q) 4

4 Page 4 GATE EE So Q will have 4 linearly independent rows and flour independent columns. Hence (A) is correct option MCQ.7 A function yt () satisfies the following differential equation : dy() t + yt () δ() t dt where δ () t is the delta function. Assuming zero initial condition, and denoting the unit step function by ut (), yt () can be of the form (A) e t (B) e -t t (C) eu() t -t (D) e u() t SOL.7 MCQ.8 Given differential equation for the function dy() t + yt () δ() t dt Taking Laplace on both the sides we have, sy() s + Y() s ( s+ ) Y( s) Ys () s + Taking inverse Laplace of Ys () yt () e t u() t, t > 0 Hence (D) is correct option. 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 V 0 of the circuit will be

5 Page 5 GATE EE SOL.8 Assume the diode is in reverse bias so equivalent circuit is Output voltage V 0 0 sin ωt # 0 5sinωt Due to resistor divider, voltage across diode V D < 0 (always). So it in reverse bias for given input. Output, V 0 5sinωt Hence (A) is correct option. MCQ.9 SOL.9 Two 8-bit ADCs, one of single slope integrating type and other of successive approximate type, take T A and T B times to convert 5 V analog input signal to equivalent digital output. If the input analog signal is reduced to.5 V, the approximate time taken by the two ADCs will respectively, be (A) TA, TB (B) T /, T (C) TA, TB / (D) TA/, TB/ Conversion time does not depend on input voltage so it remains same for both type of ADCs. Hence (A) is correct option A B

6 Page 6 GATE EE MCQ.0 SOL.0 MCQ. SOL. MCQ. An input device is interfaced with Intel 8085A microprocessor as memory mapped I/O. The address of the device is 500H. In order to input data from the device to accumulator, the sequence of instructions will be (A) LXI H, 500H (B) LXI H, 500H MOV A, M MOV M, A (C) LHLD 500H (D) LHLD 500H MOV A, M MOV M, A Hence ( ) is Correct Option 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 Distributed winding and short chording employed in AC machine will result in reduction of emf and harmonics. Hence (D) is correct option. 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 d (C) Y d6 (D) Y d SOL. Transformer connection will be represented by Y d. Hence (B) is correct option. MCQ.3 SOL.3 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 Detent torque/restraining toque: The residual magnetism in the permanent magnetic material produced.

7 Page 7 GATE EE 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.4 A two machine power system is shown below. The Transmission line XY has positive sequence impedance of Z Ω and zero sequence impedance of Z 0 Ω SOL.4 MCQ.5 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 V a Volts and I a amperes, respectively. Then, the impedance measured by the ground distance relay located at the terminal X of line XY will be given by (A) ^Z /h Ω (B) ^Z 0 /hω (C) ( Z0+ Z)/Ω (D) ^Va/ IahΩ Given for X to F section of phase a V a -Phase voltage and I a -phase current. Impedance measured by ground distance, Bus voltage Relay at X Current from phase 'a' V I a Ω a Hence (D) is correct option. An extra high voltage transmission line of length 300 km can be approximate by a lossless line having propagation constant β radians per km. Then the percentage ratio of line length to wavelength will be given by (A) 4.4 % (B). % (C) 9.05 % (D) 6.06 % SOL.5 For EHV line given data is Length of line 300 km and β S rad/km wavelength λ π. π km β So l % λ # # l % λ Hence (D) is correct option. MCQ.6 A-3-phase transmission line is shown in figure :

8 Page 8 GATE EE Voltage drop across the transmission line is given by the following equation : R V R VR V S3Va W SZs Zm Zm WSI aw S3Vb W SZm Zs Zm WSIb W S3V cw SZm Zm Z s WSI cw T X T XT X Shunt capacitance of the line can be neglect. If the has positive sequence impedance of 5 Ω and zero sequence impedance of 48 Ω, then the values of Z s and Z m will be (A) Z 3.5 Ω; Z 6.5 Ω s (B) Z 6 Ω; Z Ω s (C) Z 6.5 Ω; Z 3.5 Ω s (D) Z Ω; Z 6 Ω s m m m m SOL.6 MCQ.7 For three phase transmission line by solving the given equation R ΔV V R S aw ( Xs X ) 0 0 VR I V S m WS aw We get, SΔVb W S 0 ( Xs Xm) 0 WSI bw SΔVc W S 0 0 ( Xs+ Xm) WSI cw T X T XT X Zero sequence Impedance Xs+ Xm 48...() and Positive Sequence Impedance Negative Sequence Impedance ( Xs Xm) 5...() By solving equation () and () Zs or Xs 6 and Zm or Xm Hence (B) is correct option. In the single phase voltage controller circuit shown in the figure, for what range of triggering angle ( α ), the input voltage ( V 0 ) is not controllable? (A) 0c < α < 45c (B) 45c < α < 35c (C) 90c < α < 80c (D) 35c < α < 80c

9 Page 9 GATE EE SOL.7 Hence (A) is correct option. MCQ.8 ` R+ jxl j tan φ ω R L φ 45c so, firing angle α must be higher the 45c, Thus for 0 < α < 45c, V 0 is uncontrollable. A 3-phase voltage source inverter is operated in 80c conduction mode. Which one of the following statements is true? (A) Both pole-voltage and line-voltage will have 3 rd harmonic components (B) Pole-voltage will have 3 rd harmonic component but line-voltage will be free from 3 rd harmonic (C) Line-voltage will have 3 rd harmonic component but pole-voltage will be free from 3 rd harmonic (D) Both pole-voltage and line-voltage will be free from 3 rd harmonic components SOL.8 A 3-φ voltage source inverter is operated in 80c 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 3 rd harmonic components. Hence (D) is correct option. MCQ.9 The impulse response of a causal linear time-invariant system is given as ht. () Now consider the following two statements : Statement (I): Principle of superposition holds Statement (II): ht () 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.9 Since the given system is LTI, So principal of Superposition holds due to linearity. For causal system ht () 0, t < 0 Both statement are correct. Hence (D) is correct option. MCQ.0 It is desired to measure parameters of 30 V/5 V, kva,

10 Page 0 GATE EE single-phase transformer. The following wattmeters are available in a laboratory: W : 50 V, 0 A, Low Power Factor W : 50 V, 5 A, Low Power Factor W 3 : 50 V, 0 A, High Power Factor W 4 : 50 V, 5 A, High Power Factor The Wattmeters used in open circuit test and short circuit test of the transformer will respectively be (A) W and W (B) W and W 4 (C) W and W 4 (D) W and W 3 SOL.0 Given: -φ transformer, 30 V/5 V, kva W : 50 V, 0 A, Low Power Factor W : 50 V, 5 A, Low Power Factor W 3 : 50 V, 0 A, High Power Factor W 4 : 50 V, 5 A, High Power Factor In one circuit test the wattmeter W is used and in short circuit test of transformer W 3 is used. Hence (D) is correct option. Q. to Q.75 carry two marks each. MCQ. The time constant for the given circuit will be (A) /9 s (C) 4 s (B) /4 s (D) 9 s SOL. Time constant of the circuit can be calculated by simplifying the circuit as follows

11 Page GATE EE C eq F 3 Equivalent Resistance MCQ. R eq Ω Time constant τ R eq C eq 6 # 4 sec 3 Hence (C) is correct option The resonant frequency for the given circuit will be (A) rad/s (C) 3 rad/s (B) rad/s (D) 4 rad/s SOL. Impedance of the circuit is jωc R Z jωl+ jωc + R j L R jωcr ω + + jωcr # jωcr R( jωcr) jωl+ + ω CR jωl( + ω C R ) + R jωcr + ω CR R j[ ωl( + ω C R ) ωcr ] + + ω CR + ω CR For resonance Im( Z ) 0 So, ωl( + ω C R ) ωcr

12 Page GATE EE L 0. H, C F, R Ω So, ω# 0.[ + ω ()()] ω()() + ω 0 & ω 9 3 rad/sec Hence (C) is correct option. MCQ.3 Assuming ideal elements in the circuit shown below, the voltage V ab will be (A) 3V (B) 0 V (C) 3 V (D) 5 V SOL.3 By applying KVL in the circuit Vab i+ 5 0 i A, V ab # 5 3 Volt Hence (A) is correct option MCQ.4 A capacitor consists of two metal plates each 500 # 500 mm 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 mm thickness. The relative primitivities of the glass and paper are 8 and respectively. Neglecting the fringing effect, the - capacitance will be (Given that ε # 0 F/m ) (A) pf (B) 475 pf (C) pf (D) pf SOL.4 Here two capacitance C and C are connected in series, so equivalent capacitance is C CC eq C + C C 0 r εε A d # 0 # 8 # 500 # 500 # 0 3 4# # 0 F 6

13 Page 3 GATE EE C C eq 0 r εε A d # 0 # # 500 # 500 # 0 3 # 0. 5 # 0 F # 0 #. 5 # # # # pf Hence (B) is correct option. 6 MCQ.5 SOL.5 MCQ.6 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 mm. The inductance of the coil corresponding to a magnetizing current of 3 A will be -7 (Given that μ0 4π# 0 H/m) (A) μ H (B) 3.04 μh (C) μ H (D).304 μh Hence (B) is correct option. Circumference l 300 mm no. of turns n 300 Cross sectional area A 300 mm μ Inductance of coil L 0n A l 4π 0 (300) # # # # ( 300 # 0 3 ) μh 7 6 In the circuit shown in the figure, the value of the current i will be given by (A) 0.3 A (C).75 A (B).5 A (D).5 A SOL.6 By writing node equations at node A and B Va V 5 + a 0 0 V a 5 0

14 Page 4 GATE EE & V a.5 V Similarly Vb V V 3 4 ab b Vb 4 ( Va Vb) + Vb 0 3 Vb 4(.5 Vb) + 3Vb 0 8V b 0 0 & V b.5 V Current i Vb.5 A Hence (B) is correct option. MCQ.7 Two point charges Q 0 μc and Q 0 mc are placed at coordinates (,,0) and (,, 0) respectively. The total electric flux passing through a plane z 0 will be (A) 7.5 μ C (B) 3.5 μc (C) 5.0 μ C (D).5 μc SOL.7 Hence ( ) is Correct Option MCQ.8 Given a sequence xn, [ ] to generate the sequence yn [ ] x[ 3 4n], which one of the following procedures would be correct? (A) First delay xn ( ) by 3 samples to generate z [ n], then pick every 4 th sample of z [ n] to generate z [ n], and than finally time reverse z [ n] to obtain yn. [ ] (B) First advance xn [ ] by 3 samples to generate z [ n], then pick every 4 th sample of z [ n] to generate z [ n], and then finally time reverse z [ n] to obtain yn [ ] (C) First pick every fourth sample of xn [ ] to generate v [ n], time-reverse v [ n] to obtain v [ n], and finally advance v [ n] by 3 samples to obtain yn [ ] (D) First pick every fourth sample of xn [ ] to generate v [ n], time-reverse v [ n] to obtain v [ n], and finally delay v [ n] by 3 samples to obtain yn [ ] SOL.8 In option (A) z [ n] xn [ 3] z [ n] z[4 n] x[4n 3] yn [ ] z [ n] x[ 4n 3] Y x[3 4 n] In option (B) z [ n] xn [ + 3] z [ n] z[4 n] x[4n+ 3] yn [ ] z [ n] x[ 4n+ 3] In option (C) v [ n] x[ 4n]

15 Page 5 GATE EE v [ n] v[ n] x[ 4 n] yn [ ] v[ n+ 3] x[ 4( n+ 3)] Y x[3 4 n] In option (D) v [ n] x[ 4n] v [ n] v[ n] x[ 4 n] yn [ ] v[ n 3] x[ 4( n 3)] Y x[3 4 n] Hence (B) is correct option. MCQ.9 A system with xt () and output yt () is defined by the input-output relation : -t yt () # xtd () τ -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.9 Input-output relation - t yt () x() τ d τ 3 # - Causality : Since yt () depends on x( t), So it is non-causal. Time-variance : -t yt () x( τ τ0) dτ Y y( t τ 0 ) # -3 So this is time-variant. Stability : Output yt () is unbounded for an bounded input. For example Let x() τ e -τ ( bounded) -t -τ -t yt () -τ e d τ e # 8 $ Unbounded B- Hence (D) is correct option sin( x) x MCQ.30 A signal xt () sinc( αt) where α is a real constant ^sinc() x h is the input to a Linear Time Invariant system whose impulse response ht () sinc( βt), where β is a real constant. If min ( αβ, ) denotes the minimum of α and β and similarly, max ( αβ, ) 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( γ t) where γ min( α, β) (B) It will be of the form Ksinc( γ t) where γ max( α, β) (C) It will be of the form Ksinc( αt) (D) It can not be a sinc type of signal π π

16 Page 6 GATE EE SOL.30 Output yt () of the given system is yt () xt ()) ht () Or Yjω ( ) Xj ( ω) Hj ( ω) given that xt () sinc( αt) ht () sinc( βt) Fourier transform of xt () and ht () are Xjω ( ) F[ xt ( )] π rect ω, < < α ` α ω α α j Hjω ( ) F[ ht ( )] π rect ω, < < β ` β ω β β j Yjω ( ) π rect ω rect ω αβ `α j `β j So, Yjω ( ) K rect ω `γ j Where γ min( αβ, ) And yt () Ksinc( γt) Hence (A) is correct option. MCQ.3 Let xt () be a periodic signal with time period T, Let yt ( ) xt ( t0) + xt ( + t0) for some t 0. The Fourier Series coefficients of yt () are denoted by b k. If bk 0 for all odd k, then t 0 can be equal to (A) T/ 8 (B) T/ 4 (C) T/ (D) T SOL.3 Let a k is the Fourier series coefficient of signal xt () Given yt () xt ( t0) + xt ( + t0) Fourier series coefficient of yt () -jkωt0 jkωt0 b k e ak + e ak b k ak coskωt0 b k 0 (for all odd k ) kω t 0 π, k " odd k π t T 0 π

17 Page 7 GATE EE For k, t 0 T 4 Hence (B) is correct option. MCQ.3 Hz () is a transfer function of a real system. When a signal xn [ ] ( + ) is the input to such a system, the output is zero. Further, the Region of convergence (ROC) of ^ z h H(z) is the entire Z-plane (except z 0). It can then be inferred that Hz () can have a minimum of (A) one pole and one zero (B) one pole and two zeros (C) two poles and one zero (D) two poles and two zeros SOL.3 Hence ( ) is correct option. MCQ.33 Given Xz () z with z > a, the residue of Xzz () n - at z a ( z a) for n $ 0 will be (A) a n - (B) a n SOL.33 (C) na n (D) na n - Hence (D) is correct option. Given that Xz () z, z > a ( z a) Residue of Xzz () n - at z a is d ( z a ) X ( z ) z n - dz d ( z a ) z dz ( z a ) d z n dz z a - nan - nz n z a z a z n - z a MCQ.34 Consider function fx () ( x 4) where x is a real number. Then the function has (A) only one minimum (B) only tow minima (C) three minima (D) three maxima j n SOL.34 Given function fx () ( x 4) f'( x ) ( x 4) x To obtain minima and maxima f'( x ) 0 4xx ( 4) 0 x 0, x 4 0 & x! So, x 0, +, f''( x ) 4x( x) + 4( x 4) x 6

18 Page 8 GATE EE For x 0,''() f 0 () < 0 (Maxima) x +,''() f () 6 3 > 0 (Minima) x,''( f ) ( ) 6 3 > 0 (Minima) So fx () has only two minima Hence (B) is correct option. MCQ.35 Equation e x 0 is required to be solved using Newton s method with an initial guess x0. Then, after one step of Newton s method, estimate x of the solution will be given by (A) (B) (C) (D) SOL.35 An iterative sequence in Newton-Raphson method can obtain by following expression fx ( n) x n+ xn f'( x ) We have to calculate x, so n 0 n x So, x fx ( 0) x0, Given x0 f'( x ) fx ( 0 ) e x 0 e f'( x 0 ) e x 0 e (. ) 0 63 ( ) Hence (A) is correct option 0 MCQ.36 A is m# n full rank matrix with m > n and I is identity matrix. Let matrix A' ( A T - A) A T, Then, which one of the following statement is FALSE? (A) AA' A A (B) ( AA') (C) AA ' I (D) AA' A A' SOL.36 Hence (D) is correct option T T A ' ( A A) A A ( A T ) A A I Put A' A I in all option. T option (A) AA' A A AA A A

19 Page 9 GATE EE A A (true) option (B) ( AA') ( AA I) () I I I I (true) option (C) option (D) AA ' A IA I AA' A AA IA I I I (true) A' A Y A' (false) t MCQ.37 A differential equation dx/ dt e - u( t), has to be solved using trapezoidal rule of integration with a step size h 00. s. Function ut () indicates a unit step function. - If x( 0) 0, then value of x at t 00. s will be given by (A) (B) (C) (D) SOL.37 Hence (C) is correct option dx e u() t dt t # x e t u() t dt # t x e dt 0 x # ftdt (), 0 t.0 s From trapezoid rule t0+ nh # ftdt () h f () 0 + f (. 0) t0 # 0 ftdt (). 0 0 e e. 0 6 h MCQ.38 Let P be a # real orthogonal matrix and x is a real vector [ x,x ] T with length / x ( x + x ). Then, which one of the following statements is correct? (A) Px # x where at least one vector satisfies Px < x (B) Px (C) Px # x for all vector x $ x where at least one vector satisfies Px > x (D) No relationship can be established between x and Px SOL.38 P is an orthogonal matrix So PP T I

20 Page 0 GATE EE Let assume P PX cos θ sin θ > sin θ cos θ H cos θ sin θ > x sin θ cos θ H8 x T B cos θ sin θ x > sin θ cos θ H> x H xcos θ xsin θ > x sin θ+ x cos θ H PX ( xcos θ xsin θ) + ( xsin θ+ xcos θ) x + x PX X Hence (B) is correct option. MCQ.39 Let xt () rect^t h (where rect() x for # x # and zero otherwise. If sin( x) sinc() x π πx, then the Fourier Transform of xt () + x( t) will be given by (A) sinc ` ω π j (B) sinc ω ` π j SOL.39 (C) sinc ω cos ω ` π j ` j (D) sinc ω sin ω ` π j ` j Given signal xt () rect t ` j, # t # or 0 # t # So, xt () * 0, elsewhere Similarly x( t) rect t ` j, # t # or # t # 0 x( t) * 0, elsewhere # 3 -jωt # jωt -jωt # # 0 - -jωt F [ xt ( ) + x( t)] xte () dt+ x( te ) dt () e dt+ () e dt -jωt -jωt e e ; jω E + ; 0 jω E - j ( e ) j ( e ) jω - ω + ω jω 0

21 Page GATE EE jω/ Hence (C) is correct option. jω/ e jω/ j / ( e e ) e j / j / ( e e ) jω - ω + ω jω - ω jω/ -jω/ -jω/ jω/ ( e e )( e + e ) jω sin ω $ cos ω ω `j `j cos ω sinc ω ` π j MCQ.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 (C) 4.3 ma (B) 3.6 ma (D) 5.7 ma SOL.40 Hence (C) is correct option. This is a current mirror circuit. Since β is high so IC IC, IB I V B ( ) 4.3 volt Diode D is forward biased. So, current I is, I IC IC 0 ( 4. 3) 4.3 ma B MCQ.4 In the voltage doubler circuit shown in the figure, the switch S is closed at t 0

22 Page GATE EE Assuming diodes D and D to be ideal, load resistance to be infinite and initial capacitor voltages to be zero. The steady state voltage across capacitor C and C will be (A) V 0 V, V 5 V (B) V 0 V, V 5 V c c c c (C) V 5 V, V 0V (D) V 5 V, V 0V c c c c SOL.4 In positive half cycle of input, diode D is in forward bias and D is off, the equivalent circuit is Capacitor C will charge upto + 5 volt. V C + 5 volt In negative halt cycle diode D is off and D is on. Now capacitor V C will charge upto 0 volt in opposite direction. Hence (D) is correct option MCQ.4 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

23 Page 3 GATE EE SOL.4 Let input V in is a sine wave shown below According to given transfer characteristics of rectifiers output of rectifier P is.

24 Page 4 GATE EE Similarly output of rectifier Q is Output of a full wave rectifier is given as MCQ.43 To get output V 0 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 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 FABC (,, ) 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)

25 Page 5 GATE EE SOL.43 In SOP form, F is written as F Σm(,3,5,6) X Y Z + X YZ + XY Z + XYZ Solving from K- map F XZ+ YZ+ XYZ In POS form F ( Y+ Z)( X+ Z)( X+ Y+ Z) since all outputs are active low so each input in above expression is complemented F ( Y+ Z)( X+ Z)( X+ Y+ Z) Hence (B) is correct option. MCQ.44 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 Q and Q are T ON and T ON respectively. The frequency and the duty cycle of the signal at Q will respectively be (A) f, D T + T 5 ON ON (B) f, T + T TON D T + T ON ON ON ON (C) f TON, D T T + T ON ON ON

26 Page 6 GATE EE SOL.44 MCQ.45 SOL.45 (D) f TON, D TON T ON + T ON In this case f T + T and, D T TON + T Hence (B) is correct option. ON ON ON ON The content of some of the memory location in an 8085 accumulator based system are given below Address Content g g 6FE 00 6FF g g The content of stack (SP), program counter (PC) and (H,L) are 700 H, 00 H and 0000 H respectively. When the following sequence of instruction are executed. 00 H: DAD SP 0 H: PCHL the content of (SP) and (PC) at the end of execution will be (A) PC 0 H, SP 700 H (B) PC 700 H, SP 700 H (C) PC 800 H, SP 6FE H (D) PC A0 H, SP 70 H Given that SP 700 H PC 00 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 H HL 700 H PCHL & Load program counter with HL contents PC HL 700 H So after execution contents are, PC 700 H, HL 700 H

27 Page 7 GATE EE Hence (B) is correct option. MCQ.46 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 V i is made + 5. V, the voltage waveform at point P will become SOL.46 MCQ.47 Hence ( ) is correct option. A 30 V, 50 Hz, 4-pole, single-phase induction motor is rotating in the clockwise (forward) direction at a speed of 45 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 (A) Ω (B) 4 Ω (C) 78 Ω (D) 56 Ω SOL.47 Given: 30 V, 50 Hz, 4-Pole, -φ induction motor is rotating in clock-wise(forward) direction N s 45 rpm Rotar resistance at stand still(r ) 78Ω. So

28 Page 8 GATE EE N s 0 # Slip(S ) Resistance in backward branch r b Ω Hence (B) is correct option. R S MCQ.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.5 kw and 900 W respectively. The friction and windage losses are 050 W and the core losses are 00 W. The air-gap power of the motor will be (A) 3.06 kw (B) 4. kw (C) 5.0 kw (D) 6. kw SOL.48 MCQ.49 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.5 kw and 900 W fraction and windage losses 050 W Core losses 00 W. kw So, Input power in stator 3 # 400 # 50 # kw Air gap power kw Hence (C) is correct option. The core of a two-winding transformer is subjected to a magnetic flux variation as indicated in the figure. The induced emf ( ) e rs in the secondary winding as a function of time will be of the form

29 Page 9 GATE EE SOL.49 Hence (A) is correct option. dφ Induced emf in secondary N dt During 0 < t < dφ E ( 00) V dt E and E are in opposition E E 4V During time < t < During < t <. 5 dφ dt 0, then E E 0 MCQ.50 Then dφ E (00) 4 V dt E 0 48 V Voltages phasors at the two terminals of a transmission line of length 70 km have a magnitude of.0 per unit but are 80 degree out of phase. Assuming that the maximum load current in the line is /5 th 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- set to 80% of the line impedance.

30 Page 30 GATE EE (B) Directional over current protection set to pick up at.5 times the maximum load current (C) Pilot relaying system with directional comparison scheme (D) Pilot relaying system with segregated phase comparison scheme SOL.50 MCQ.5 SOL.5 Hence ( ) is Correct Option A loss less transmission line having Surge Impedance Loading (SIL) of 80 MW is provided with a uniformly distributed series capacitive compensation of 30%. Then, SIL of the compensated transmission line will be (A) 835 MW (B) 80 MW (C) 75 MW (D) 357 MW SIL has no effect of compensation So SIL 80 MW Hence (B) is correct option. MCQ.5 A loss less power system has to serve a load of 50 MW. There are tow generation ( G and G ) in the system with cost curves C and C respectively defined as follows ; C( PG) PG # PG C( PG) 3PG # PG Where P G and P G are the MW injections from generator G and G respectively. Thus, the minimum cost dispatch will be (A) P 50 MW; P 0 MW (B) P 50 MW; P 00 MW G G (C) P 00 MW; P 50 MW (D) P 0 MW; P 50 MW G G G G G G SOL.5 Hence (C) is correct option Given PG + PG 50 MW...() and C( PG) PG PG 4 C( PG) 3PG PG...() from equation () dc dpg + 0.P G...(3a) and dc dpg P G...(3b) Since the system is loss-less Therefore dc dc dpg dpg So from equations (3a) and (3b) We have 0.P G 0.06P G...(4) Now solving equation () and (4), we get P G 00 MW

31 Page 3 GATE EE P G 50 MW MCQ.53 A loss less single machine infinite bus power system is shown below : The synchronous generator transfers.0 per unit of power to the infinite bus. The critical clearing time of circuit breaker is 0.8 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 0.4 s (B) reduce but will be more than 0.4 s (C) remain constant at 0.8 s (D) increase beyond 0.8 s SOL.53 MCQ.54 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 0.8 s but will not be less than 0.4 s for transient stability purpose. Hence (B) is correct option. Single line diagram of a 4-bus single source distribution system is shown below. Branches e, e, e3 and e 4 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) e (B) e (C) e 3 (D) e 4

32 Page 3 GATE EE SOL.54 MCQ.55 Given that the each section has equal impedance. Let it be R or Z, then by using the formula line losses / IR On removing ( e ); losses ( ) R+ ( + ) R+ ( + + 5) R R+ 9R+ 64R 74R Similarly, On removing e ;losses 5 R+ ( 5+ ) R+ ( 5+ + ) R 38R lossess on removing e 3 () R+ ( ) R+ ( 5+ ) R R+ 4R+ 49R 54R on removing e 4 lossless () R+ ( + ) R+ 5 R 4R+ 9R+ 5R 38R So, minimum losses are gained by removing e 4 branch. Hence (D) is correct option 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) SOL.55 Given α 30c, in a -φ fully bridge converter we know that, Power factor Distortion factor # cos α D.f. (Distortion factor) Is(fundamental) / Is 0.9 power factor 09. # cos30c 078. Hence (B) is correct option MCQ.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 ( ) V 0 waveform of the converter will become

33 Page 33 GATE EE SOL.56 Output of this Here the inductor makes T and T 3 in ON because current passing through T and

34 Page 34 GATE EE T 3 is more than the holding current. Hence (A) is correct option MCQ.57 A 0 V0 A, 000 rpm, separately excited dc motor has an armature resistance of.5 Ω. The motor is controlled by a step down chopper with a frequency of khz. The input dc voltage to the chopper is 50 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) 0.58 (B) (C) 0.85 (D) 0.90 SOL.57 Hence ( ) is Correct Option MCQ.58 A 0 V, 400 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 0 V (rms). The approximate firing angles of the dual converter for motoring operating at 50% of rated torque and 000 rpm will be (A) 43c, 37c (B) 43c, 47c (C) 39c, 4c (D) 39c, 5c SOL.58 MCQ.59 Hence ( ) is Correct Option A single phase source inverter is feeding a purely inductive load as shown in the figure The inverter is operated at 50 Hz in 80c square wave mode. Assume that the load current does not have any dc component. The peak value of the inductor current i 0 will be (A) 6.37 A (C) 0 A (B) 0 A (D) 40 A SOL.59 Input is given as Here load current does not have any dc component

35 Page 35 GATE EE ` Peak current occur at ( π/ ω) ` V s L di dt Here 00. di 0# dt di So di ( max ) Hence (C) is correct option π aπkb 50 l # 00 # 0 0 A. MCQ.60 SOL.60 A 400 V, 50 Hz, 4-pole, 400 rpm, star connected squirrel cage induction motor has the following parameters referred to the stator: ' R r.0 Ω, X s X' r.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 Vf / control. The stator line-to-line voltage(rms) and frequency to obtain the maximum torque at starting will be : (A) 0.6 V,.7 Hz (B) 33.3 V, 6.7 Hz (C) 66.6 V, 33.3 Hz (D) 33.3 V, 40.3 Hz Given 400 V, 50 Hz, 4-Pole, 400 rpm star connected squirrel cage induction motor. R.00 Ω, Xs Xlr.5 Ω So, for max. torque slip S Rl r m Xsm + Xlrm for starting torque Sm Then Xsm + Xl rm Rl r πfmls+ 0. πfmll r Frequency at max. torque f m π( Ls+ Llr) L Xs. s 5 π# 50 π# 50 Ll. r 5 π # 50 f m f m 6.7 Hz In const Vf / control method

36 Page 36 GATE EE MCQ.6 V f a V 8 f So V f # # 8 V 33.3 V Hence (B) is correct option. 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 I 0 0 A will be (A) 44c (B) 5c (C) 9c (D) 36c SOL.6 Here for continuous conduction mode, by Kirchoff s voltage law, average load current V Ia V + 50 ` 0 A, So I V m π cos I a V 30 V α 30 # # 30 cos α 30c π

37 Page 37 GATE EE α 9c Hence (C) is correct option. MCQ.6 SOL.6 A three phase fully controlled bridge converter is feeding a load drawing a constant and ripple free load current of 0 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) 3% and 6.8 A (B) 3% and 7.8 A (C) 66% and 6.8 A (D) 66% and 7.8 A Hence (B) is correct option. Total rms current I a A 3 # Fundamental current I a 0.78 # A where ` THD DF THD DF Ia # Ia 0 86 # 0 b 3% l MCQ.63 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) 0 V, A (B) 0 V, 8 A (C) 40 V A (D) 40 V, 8 A SOL.63 Hence (C) is correct option.

38 Page 38 GATE EE In the given diagram when switch S is open I0 IL 4 A, Vs 0 V when switch S is closed ID 0, V0 0 V Duty cycle 05. so average voltage is V s δ Average current amp Average voltage V MCQ.64 SOL.64 The transfer function of a linear time invariant system is given as Gs () s + 3s + The steady state value of the output of the system for a unit impulse input applied at time instant t will be (A) 0 (B) 0.5 (C) (D) Given transfer function Gs () s + 3s + Input rt () δ( t ) Rs () L[ δ( t )] e Output is given by Ys () RsGs () () s Ys () e s + 3s + Steady state value of output lim yt () lim sy() s t " 3 s " 0 lim s " 0 s Hence (A) is correct option. s se s + s MCQ.65 The transfer functions of two compensators are given below : 0( s + ) C, C s 0 + ( s + 0) 0( s + ) Which one of the following statements is correct? (A) C is lead compensator and C is a lag compensator (B) C is a lag compensator and C is a lead compensator (C) Both C and C are lead compensator (D) Both C and C are lag compensator SOL.65 For C Phase is given by

39 Page 39 GATE EE MCQ.66 θ C tan ( ω) tan ω a0k Jω ω N θ C tan K 0 O K + ω O L 0 P θ C tan 9ω c > ω m (Phase lead) Similarly for C, phase is θ C tan ω a tan ( ω) 0k J ω ωn tan K0 O K + ω O L 0 P tan 9ω c < ω m (Phase lag) Hence (A) is correct option 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 (C) Two poles and two zero (B) Two poles and one zero (D) One pole and two zeros SOL.66 From the given bode plot we can analyze that: Slope 40 db/decade" poles Slope 0 db/decade (Slope changes by + 0 db/decade)" Zero Slope 0 db/decade (Slope changes by + 0 db/decade)" Zero So there are poles and zeroes in the transfer function. Hence (C) is correct option. MCQ.67 Figure shows a feedback system where K > 0

40 Page 40 GATE EE 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 (D) K > 390 SOL.67 MCQ.68 SOL.68 Characteristic equation for the system + K 0 ss ( + 3)( s+ 0) ss ( + 3)( s+ 0) + K 0 s 3 + 3s + 30s+ K 0 Applying Routh s stability criteria s 3 30 s 3 K s ( 3 # 30) K 3 s 0 K 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 The transfer function of a system is given as 00 s + 0s+ 00 The system is (A) An over damped system (C) A critically damped system Given transfer function is Hs) () 00 s + 0s + 00 Characteristic equation of the system is given by s + 0s ω n 00 & ω n 0 rad/sec. ξω n 0 or ξ 0 # 0 (B) An under damped system (D) An unstable system

41 Page 4 GATE EE ( ξ ) so system is critically damped. Hence (C) is correct option. MCQ.69 Two sinusoidal signals p( ω, t) Asinωt and q( ω 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( ω t) will be represented as (A) q( ω t) Asin ω t, ω ω (B) q( ω t) Asin ω t, ω ω / (C) q( ω t) Acos ω t, ω ω (D) q( ω t) Acos ω t, ω ω / SOL.69 Hence (D) is correct option. Frequency ratio f f Y X f f Y X meeting points of horizontal tangents meeting points of vertical tangents 4 f Y ( f ) X ω ω / Since the Lissajous figures are ellipse, so there is a phase difference of 90c exists between vertical and horizontal inputs. So q( ω t) Acos t, ω ω / ω MCQ.70 The ac bridge shown in the figure is used to measure the impedance Z.

42 Page 4 GATE EE If the bridge is balanced for oscillator frequency f khz, then the impedance Z will be (A) (60 + j0) Ω (B) (0 + j00) Ω (C) (60 j00) Ω (D) (60 + j00) Ω SOL.70 Impedance of different branches is given as Z AB 500 Ω Z BC j# π# # 0 # μf + Ω - ( 00j + 300) Ω 3 Z AD j # π# # 0 # 5.9 mh Ω - ( 00j + 300) Ω To balance the bridge ZABZCD ZADZBC 500Z ( 00j+ 300)( 00j+ 300) 500Z Z (60 + j0) Ω Hence (A) is correct option. Common Data for Questions 7, 7 and 73: Consider a power system shown below: Given that: Vs Vs + j0p.u; The positive sequence impedance are Zs Zs j0.0 p.u and ZL j0.06 p.u 3-phase Base MVA 00 voltage base 400 kv(line to Line) Nominal system frequency 50 Hz.

43 Page 43 GATE EE The reference voltage for phase a is defined as Vt () 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 S to point F equals 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.7 The instant ( t 0 ) of the fault will be (A) 4.68 ms (C) ms (B) ms (D) ms SOL.7 Given Vt () Vm cos( ωt) For symmetrical 3 φ fault, current after the fault ( RLt / ) it () Ae Vm + cos( ωt α) Z At the instant of fault i.e t t 0, the total current it () 0 ( ` 0 Ae RLt / ) 0 Vm + cos( ωt0 α) Z V m cos( ωt0 α) Z Ae ( RLt0 / ) Maximum value of the dc offset current Ae ( RLt0 / ) V m cos( ωt0 α) Z For this to be negative max. ( ωt 0 α) 0 or t 0 α ω...() and Z j0. 04 Z Z + α c ` α 84.9cor.47 rad. From equation () t sec ( π # 50) t ms Hence (A) is correct option. MCQ.7 The rms value of the component of fault current ( I f ) will be (A) 3.59 ka (B) 5.07 ka (C) 7.8 ka (D) 0.5 ka

44 Page 44 GATE EE SOL.7 Since the fault F is at mid point of the system, therefore impedance seen is same from both sides. Z c Z (Positive sequence) Z c also Z Z Z0 (for 3-φ fault) So magnitude ` If ( pu) + 0c Z I f (p.u.) c c ` Fault current I f # 3 # ka Hence (C) is correct option. MCQ.73 SOL.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 ( I x ) for phase a will be (A) 4.97 p.u (B) 7.0 p.u (C) 4.93 p.u (D) 9.85 p.u If fault is LG in phase a Z Z c Z Z c and Z 0 3Z c

45 Page 45 GATE EE Then Ia / 3 I I I a a a0 ` Ia ( pu) c Z + Z + Z and 0 I a pu ( ) Fault Current I f I 3I pu So Fault current I f # 3 # ka Hence (A) is correct option. a a 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. MCQ.74 SOL.74 The motor is coupled to a 0 V, separately excited d.c generator feeding power to fixed resistance of 0 Ω. Two-wattmeter method is used to measure the input power to induction motor. The variable resistance is adjusted such the motor runs at 40 rpm and the following readings were recorded W 800 W, W 00 W. 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) 500 rpm in the direction of rotation (D) 500 rpm in the opposite direction of rotation Given 3-φ, 440 V, 50 Hz, 4-Pole slip ring motor

46 Page 46 GATE EE Motor is coupled to 0 V N 40 rpm, W 800 W, W 00 W So, 0f N s P 0 # rpm 4 Relative speed rpm in the direction of rotation. Hence (A) is correct option. MCQ.75 SOL.75 Neglecting all losses of both the machines, the dc generator power output and the current through resistance ( R ex ) will respectively be (A) 96 W, 3.0 A (B) 0 W, 3.46 A (C) 504 W,.6 A (D) 880 W, 3.7 A Neglecting losses of both machines Slip(S ) Ns N N s total power input to induction motor is P in W Output power of induction motor P out ( SP ) in ( 0. 06) W Losses are neglected so dc generator input power output power 504 W So, IR 504 I Hence (C) is correct option. A Linked Answer Questions: Q.76 to Q.85 carry two marks each. Statement for Linked Answer Questions 76 and 77: The current it () sketched in the figure flows through a initially uncharged 0.3 nf

47 Page 47 GATE EE capacitor. MCQ.76 The charge stored in the capacitor at t 5 μs, will be (A) 8 nc (B) 0 nc (C) 3 nc (D) 6 nc SOL.76 Charge stored at t 5 μ sec Q # itdt () 0 5 area under the curve Q Area OABCDO Area (OAD)+Area(AEB)+Area(EBCD) # # + # # + # nc Hence (C) is correct option

48 Page 48 GATE EE MCQ.77 SOL.77 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 μ s will approximately be (A) 8.8 V (B) 3.5 V (C) 3.5 V (D) 30.6 V Initial voltage across capacitor Q V o 0 3 C. nc 03 nf Volt When capacitor is connected across an inductor it will give sinusoidal esponse as vc () t Vocos ωot where ω o LC # 0 # 0. 6 # 0.35 # 0 6 rad/sec at t μ sec 6 6 So, vc () t cos(.35 # 0 # # 0 ) # ( 0.70) V Hence (D) is correct option. MCQ.78 SOL.78 Statement for Linked Answer Question 78 and 79. 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 0 A, B, C [ 0] 0 G G The transfer function G(s) of this system will be (A) s (B) s + ( s + ) ss ( ) (C) s (D) ( s ) ss ( + ) State space equation of the system is given by, X o AX+ Bu 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) BU( s) ` Y() s CX() s

49 Page 49 GATE EE So Y () s CsI ( A) BU( s) Y() s T.F CsI ( A) B U() s s ( si A) 0 0 > 0 s H > 0 H ( si A) s + ss ( + ) > 0 s H R V S W s ss ( ) S + W S 0 W S ( s + ) W T X Transfer function R V S W s ss ( ) 0 Gs () CsI [ A] B 8 0 S + W BS 0 W > H S ( s + ) W T X ss ( + ) Hence (D) is correct option. s > 0 s + H R V S W ss ( ) 0 S + W 8 BS W S ( s + ) W T X MCQ.79 A unity feedback is provided to the above system Gs () to make it a closed loop system as shown in figure. For a unit step input rt, () the steady state error in the input will be (A) 0 (B) (C) (D) 3 SOL.79 Steady state error is given by, sr() s e ss lim s " 0 + GsHs ( ) ( ) G Here Rs () L[ rt ( )] s (Unit step input) Gs () ss ( + ) Hs () (Unity feed back) R V S sb s l W So, e ss lims W s " 0S + W S ss ( + ) W T ss ( ) X lim + s 0 ss ( + ) + G "

50 Page 50 GATE EE Hence (A) is correct option. Statement for Linked Answer Questions 80 and 8. A general filter circuit is shown in the figure : MCQ.80 If R R R A and R 3 R 4 R B, the circuit acts as a (A) all pass filter (B) band pass filter (C) high pass filter (D) low pass filter SOL.80 For low frequencies, ω " 0, so " 3 ωc Equivalent circuit is, By applying node equation at positive and negative input terminals of op-amp. va vi + va vo 0 R R v A vi+ vo, a R R R A Similarly, va vi + va 0 0 R R 3 4

51 Page 5 GATE EE v A, a R 3 R 4 R B v in So, v o 0 It will stop low frequency signals. For high frequencies, ω " 3, then " 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.8 The output of the filter in Q.80 is given to the circuit in figure : The gain v/s frequency characteristic of the output ( v o ) will be SOL.8 In Q.80 cutoff frequency of high pass filter is given by, ω h πr C A

52 Page 5 GATE EE Here given circuit is a low pass filter with cutoff frequency, ω L RA C RA π π C ωl ωh When both the circuits are connected together, equivalent circuit is, So this is is Band pass filter, amplitude response is given by. Hence (D) is correct option. Statement for Linked Answer Question 8 and 83. MCQ.8 A 40 V, dc shunt motor draws 5 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) 34 V (C) 40 V (D) 474 V SOL.8 Given: V 40 V, dc shunt motor I 5 A Rated load at a speed 80 rad/s Armature Resistance 05Ω. Field winding Resistance 80 Ω So, E 40 # 0.5 E 34 V plugging V+ E V Hence (D) is correct option. MCQ.83 The external resistance to be added in the armature circuit to limit the armature current to 5% of its rated value is (A) 3. Ω (B) 3.9 Ω (C) 5. Ω (D) 5.9 Ω]

53 Page 53 GATE EE SOL.83 External Resistance to be added in the armature circuit to limit the armature current to 5%. So I a # R + R Ra + Rexternal 3. 6 R external 3. Ω Hence (A) is correct option a external Statement for Linked Answer Question 84 and 85. A synchronous motor is connected to an infinite bus at.0 pu voltage and draws 0.6 pu current at unity power factor. Its synchronous reactance is.0 pu resistance is negligible. MCQ.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).7 pu and 30.96c lead (D).7 pu and 30.96c lag SOL.84 A synchronous motor is connected to an infinite bus at.0 p.u. voltage and 0.6 p.u. current at unity power factor. Reactance is.0 p.u. and resistance is negligible. So, V + 0c p.u. I a c p.u. Z s Ra+ jxs 0+ j + 90c p.u. V E+ δ + I a Z s + 0c c# + 90c E+δ c p.u. Excitation voltage 7. p.u. Load angle ( δ ) c(lagging) Hence (D) is correct option. MCQ.85 Keeping the excitation voltage same, the load on the motor is increased such that the motor current increases by 0%. The operating power factor will become (A) lagging (B) leading (C) 0.79 lagging (D) leading SOL.85 Hence (D) is correct option.