General Aptitude. Q. No. 1 5 Carry One Mark Each

Transcription

1 EE GATE-6-PAPER- General Aptitude Q. No. 5 arry One Mark Each. The chairman requested the aggrieved shareholders to him. (A) bare with (B) bore with () bear with (D) bare (). Identify the correct spelling out of the given options: (A) Managable (B) Manageable () Manageble (D) Managible (B). Pick the odd one out in the following,,, 4, 5 (A) (B) () 4 (D) 5 (B) Given numbers are,,, 4, 5. All the numbers have second digits as. If We sum of the digits of each number we get 4, 5, 6, 7, 8. All given numbers are irrational numbers except which is rotation. So odd one out is. 4. RD is a robot. RD can repair aeroplanes. No other robot can repair aeroplanes. Which of the following can be logically inferred from the above statements? (A) RD is a robot which can only repair aeroplanes. (B) RDis the only robot which can repair aeroplanes. () RD is a robot which can repair only aeroplanes. (D) Only RD is a robot. (B) 5. If 9y 6, then y 4y is. (A) (B) () (D) undefined () 9Y 6 Possibility (A): 9y 9 y Possibility (B): 9y y When y= 4y 4() 4 y IP Intensive lassroom Program egate-live Internet Based lasses DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ enters across India

2 When y EE GATE-6-PAPER Q. No. 6 arry Two Mark Each 6. The following graph represents the installed capacity for cement production (in tones) and the actual production (in tones) of nine cement plants of a cement company apacity utilization of a plant is defined as ratio of actual production of cement to installed capacity. A plant with installed capacity of at least tonnes is called a large plant and a plant with lesser capacity is called a small plant. The difference between total production of large plants and small plants, in tones is Installedapacity Actual Pr oduction 5 apacity production 5 tonnes Plant Number Largent plant Installed apacity 5 Actual production Plant number Total production of larger plants = =77 tonnes Smaller Plants Installed apacity Actual production 5 6 Plant number Total production of smallest plants = = 65tonnes Difference = = tonnes IP Intensive lassroom Program egate-live Internet Based lasses DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ enters across India

3 EE GATE-6-PAPER A poll of students appearing for masters in engineering indicated that 6% of the students believed that mechanical engineering is a profession unsuitable for women. A research study on women with master or higher degrees in mechanical engineering found that 99% of such women were successful in their professions. Which of the following can be logically inferred from the above paragraph? (A) Many students have isconceptions regarding various engineering disciplines (B) Men with advanced degrees in mechanical engineering believe women are well suited to be mechanical engineers. () Mechanical engineering is a profession well suited for women with masters or higher degrees in mechanical engineering. (D) The number of women pursuing high degrees in mechanical engineering is small. () 8. Sourya committee had proposed the establishment of Sourya Institutes of Technology (SITs) in line with Indian Institutes of Technology (IITs) to cater to the technological and industrial needs of a developing country. Which of the following can be logically inferred from the above sentence? Based on the proposal, (i) In the initial years, SIT students will get degrees from IIT. (ii) SITs will have a distinct national objective (iii) SIT like institutions can only be established in consultation with IIT. (iv) SITs will serve technological needs of a developing country. (A) (iii) and (iv) only () (ii) and (iv) only () (B) (i) and (iv) only (D) (ii) and (iii) only 9. Shaquille O Neal is a 6% career free throw shooter, meaning that he successfully makes 6 free throws out of attempts on average. What is the probability that he will successfully make exactly 6 free throws in attempts? (A).58 (B).86 ().94 (D).6 (A). The numeral in the units position of (7) is. IP Intensive lassroom Program egate-live Internet Based lasses DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ enters across India

4 EE GATE-6-PAPER- Electrical Engineering Q. No. 5 arry One Mark Each. The output expression for the Karnaugh map shown below is (A) A B B A (B) A () A (D) A (B) A B B B B F A,B, A A. The circuit shown below is an example of a R (A) low pass filter (B) band pass filter 5V () high pass filter (D) notch filter V in R 5V V out (A) R R Z R cs i, Z R R S c Z R TF Z R S R s WhenS R R ; TF non zero value When S ; TF Given circuit is an example of low pass filter IP Intensive lassroom Program egate-live Internet Based lasses DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ enters across India 4

5 EE GATE-6-PAPER- The following figure shows the connection of an ideal transformer with primary to secondary turns ratio of :. The applied primary voltage is V (rms), 5Hz, A. The rms value of the current I, in ampere, is. X L I : R 8k V ~ I X 4k () Where Z secondary 8 j4 ZL 8 4i n I 6.86 j Z J 8 j4 8 j6 L Since all the calculation done with respect to RMS, I also in RMS 4. onsider a causal LTI system characterized by differential equation dy t y t x t. The dt 6 response of the system to input (A) (D) 9e t t u t t x t e u t, where u(t) denotes the unit step function, is 6 6 () 9e ut 6e ut (D) 54e ut 54e ut 6 Hs s 6 S ys s Similarly X s s ys s s s s 6 6 t 6 t Thus y t 54 e u t 54e u t (B) 9e t 6 u t t t IP Intensive lassroom Program egate-live Internet Based lasses DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ enters across India 5

6 EE GATE-6-PAPER Suppose the maximum frequency in a band-limited signal x(t) is 5kHz. Then, the maximum 6 frequency in x(t) cos t, in khz, is. Maximum frequency of x t 5kHz Maximum frequency of cos t khz x tcos t Gives convolution between their respective spectrums in frequency domain max frequency of xtcos t 6kHz 6. onsider the function f(z) z z* where z is a complex variable and z* denotes its complex conjugate. Which one of the following is TRUE? (A) f(z) is both continuous and analytic (B) f(z) is both continuous but not analytic () f(z) is not continuous but is analytic (D) f(z) is neither continuous not analytic (B) f z z z x iy x iy z is conjugate of z x i f z x is continues but not analytic, since R equations will not satisfy 7. A matrix P is such that, (A),, - (B),.5 j.866,.5 j.866 (),.5 j.866,.5 j.866 (D),, - (D) If is an Eigen value of p then value of p. p p P P. Then the eigenvalues of P are is an Eigen ; ; IP Intensive lassroom Program egate-live Internet Based lasses DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ enters across India 6

7 EE GATE-6-PAPER- t, y t y' t y t with initial " 8. The solution of the differential equation, for conditions (A) () t te u t y = and y', is ut denotes the unit step function), t (B) e t t e t te ut (D) e ut (A) The operator form of the of given D.E is D D y The A.E is D D t yt e t D D,. Given y & y' i.e, t ; y y' from (s); dy dt dy At t ; y' dt t t From (), e t e From () t t y(t) e (t) te u(t) 9. The value of the line integral xy dx x ydy dz c IP Intensive lassroom Program egate-live Internet Based lasses DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ enters across India t te u t along a path joining the origin (,,) and the point (,,) is (A) (B) () 4 (D) 6 (B) Let f xy i x yj k i j k curl f f x y z f is irrotational xy x y onsider a straight line passing through,, &,, x y z i.e, t x y z t dx dy dz dt 4 xy dx x ydy dz t dt t dt dt t t 4t dt t t 7

8 EE GATE-6-PAPER- Let f(x) be a real, periodic function satisfying f x f x. The general form of its Fourier series representation would be (A) f x a a cos kx (B) f x b sin kx k k () f x a a cos kx k (D) k k k k f x a sin k x (B) We know that a periodic function f(x) defined in (-c, c) can be represented by the poisoins series a nx nx c c o f x ancos bnsin n n If a periodic function f(x) is odd, its Fourier expression contains only sine terms. A resistance and a coil are connected in series and supplied from a single phase, V, 5Hz ac source as shown in the figure below. The rms values of plausible voltage across the resistance V and coil R V respectively, in volts, are k V R ~ VS V (A) 65, 5 (B) 5, 5 () 6, 9 (D) 6, 8 (). The voltage (v) and current (A) across a load are as follows., it sin t 6 O sin t sin 5 t v t sin t The average power consumed by the load, in W, is. 5 The instantaneous power of load is p t V i t sin tsin( t 6 sin t sin t avg sin t 5sin5 t T since P P t dt, in the above expression Only st term will result non zero answer Remaining terms will be. so directly consider P t sin t sin t 6 IP Intensive lassroom Program egate-live Internet Based lasses DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ enters across India 8

9 EE GATE-6-PAPER- cos6 P V I cos avg rms rms V I 5 watt. A power system with two generators is shown in the figure below. The system (generators, buses and transmission lines) is protected by six overcurrent relays R to R 6. Assuming a mix of directional and nondirectional relays at appropriate locations, the remote backup relays for R 4 are S ~ R R R R 4 S R R 5 6 ~ (A) R, R (B) R, R 6 () R, R 5 (D) R, R 6 (D) 4. A power system has buses including generator buses. For the load flow analysis using Newton-Raphson method in polar coordinates, the size of the Jacobian is (A) (B) () 9 9 (D) 8 8 (A) Total no of Buses = generator buses = = 9 (n) load buses = - = 9(m) Jacobeam matrix size = m nm n The inductance and capacitance of a 4kV. Three-phase. 5Hz lossless transmission line are.6 mh/km/phase and nf km phase respectively. The sending end voltage is maintained at 4kV. To maintain a voltage of 4kV at the receiving end, when the line is delivering MW load, the shunt compensation required is (A) apacitive (B) Inductive () Resistive (D) Zero (B) XL jl j4.6 j.54 j j X j Since X X L, the shunt compensation is inductive 6. A parallel plate capacitor field with two dielectrics is shown in the figure below. If the electric field in the region A is 4kV/cm, the electric field in the region B, in kv/cm, is A r B 4 r cm IP Intensive lassroom Program egate-live Internet Based lasses DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ enters across India 9

10 EE GATE-6-PAPER- (A) (B) () 4 (D) 6 () Since the voltage & distance bw the two plates are same for both the regions. The electric field is same for both the regions V E d Electric field in region B = 4kV cm 7. A 5MVA, kv, 5Hz, star-connected, unloaded three-phase alternator has a synchronous reactance of p.u. and a sub-transient reactance of. p.u. If a -phase short circuit occurs close to the generator terminals, the ratio of initial and final values of the sinusoidal component of the short circuit current is. 5 Vprefault IS X P.U S IS initial 5 I cfinal. 8. onsider a liner time-invariant system transfer function Hs. If the input is cos(t) and s the steady state output is Acos t, then the value of A is..77 H tan O H 45 So when input is cost then O/P yt O cost 45 So A A three-phase diode bridge rectifier is feeding a constant D current of A to a highly inductive load. If three-phase, 45V, 5Hz A source is supplying to this bridge rectifier then the rms value of the current in each diode, in ampere, is I A O RMS, diode current = 57.7A IP Intensive lassroom Program egate-live Internet Based lasses DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ enters across India

11 EE GATE-6-PAPER- A buck-bost D-D converter shown in the figure below, is used to convert 4 V battery voltage to 6 V D voltage to feed a load of 7 W. It is operated at khz with an inductor of mh and output capacitor of F. All devices are considered to be ideal. The peak voltage across the solid-state switch (S), in volt, is. 6. For the network shown in the figure below, the frequency phase lag occurs is.. 9 4V S mh in rad s at which the maximum Load 6V in F.6 The given circuit is standard lag compensator Whose Transfer function G s s s s a s s s So, the frequency at which maximum phase lag happen m.6 rad sec. The direction of rotation of a single-phase capacitor run induction motor is reversed by (A) interchanging the terminals of the A supply (B) interchanging the terminals of the capacitor () interchanging the terminals of the auxiliary winding. (D) interchanging the terminals of both the windings (). In the circuit shown below, the voltage and current are ideal. The voltage (V out ) across the current source, in volts, is V 5A V outs IP Intensive lassroom Program egate-live Internet Based lasses DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ enters across India

12 (D) EE GATE-6-PAPER- (A) (B) 5 () (D) Writing KVL V O V V 5A V O V O V 4. The graph associated with an electrical network has 7 branches and 5 nodes. The number of independent KL equations and the number of independent KVL equations, respectively, are (A) and 5 (B) 5 and () and 4 (D) 4 and (D) No of branches = 7 Nodes = 5 No of KL equations = No of Modal equations = n = 5 = 4 No of KVL equations = No of Mesh equations = b-(n ) = 7-4 = Since no information given regarding how many simple & principal node, if we assume all principal nodes then the answer for nodal is 5 = 4 5. The electrodes, whose cross-sectional view is shown in the figure below, are at the same potential The maximum electric field will be at the point A D B (A) A (B) B () (D) D (A) At A Fields are additive F F At Fields are subtractive F F At D field is due to one electrode F At B field make an angle F F FF cos So maximum electric field is at A IP Intensive lassroom Program egate-live Internet Based lasses DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ enters across India

13 EE GATE-6-PAPER- Q. No. 6 5 arry Two Mark Each 6. The Boolean expression a b c d b c simplifies to (A) (B) a.b () a.b (D) (D) F = a b c d b c = a d b b c c = a d = = 7. For the circuit shown below, taking the opamp is ideal, the output voltage V out in terms of the input voltages V, V and V is 9 V V V x V x V out 4 V (A).8V 7.V V (B) V 8V 9V () 7.V.8V V (D) 8V V 9V (D) V V V V 4V V ; Vx 4 5 x x Vx V Vx Vout 9 4V V 4V V out V V V V 5V out 4V V 5V 9 6V 9V 45V 4V V 5V out 4V V 45V Vout 5 V 8V V 9V out IP Intensive lassroom Program egate-live Internet Based lasses DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ enters across India

14 EE GATE-6-PAPER Let x t X and x t X be two signals whose Fourier Transforms are as shown in the figure below. In the figure h(t) = t e denotes the impulse response. X X B B B B B B x t h t e t yt For the system shown above, the minimum sampling rate required to sample y(t), so that y(t) can be uniquely reconstructed from its samples, is (A) (B) x B (B) B B () 4B B yt x tx t * ht In frequency duration y X *X H Max. Frequency of X B Max, frequency of X B Max frequency of X *X B B Max frequency of H Thus max, freq of y B B Max frequency t Nyquist frequency = B B (D) sin t 9. The value of the integral dt t is equal to (A) (B).5 () (D) (D) IP Intensive lassroom Program egate-live Internet Based lasses DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ enters across India 4

15 EE GATE-6-PAPER- sin t sin t sin t dt dt is an even function t t t.t sin t 4 e.dt t By the defining of L.T; we have sin t 4L ; where S t 4 sin t L ; wheres t Putting s = ; than 4 4 tan 4 sin at a tan ; where s L tan s t s sin t dt t. Let y(x) be the solution of the differential equation d y dy y and. Then the value of y () is. dx x 7.98 The operate form of given D.E is D 4D 4 y The A.E is D 4D 4 D D, D D, The solution is y e x x Given that y & dx y' i.e x y i.e at x, y' dy 4 4y with initial conditions dx from (); from x from y e.x y xe x y e e y e or y 7.89 y e x c x x IP Intensive lassroom Program egate-live Internet Based lasses DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ enters across India 5

16 EE GATE-6-PAPER- ˆ ˆ ˆ. The line integral of the vector field F = (,,) parametrized by t,t,t is F 5xzi x y j x z k x t; y t ; z t dx dt d t dt; dz dt The line integral of the vector field is F.dr 5xzdx x y dy x z dz 5t dt t dt t dt t t 5t dt t dt 4 t t ` 5xzi x y j x zk along a path from (,,) to. Let P =. onsider the set S of all vector x y such than a x a b where p b y. Then S is (A) A circle of radius (B) a circle of radius = () an ellipse with major axis along (D) an ellipse with minor axis along (D) P a x a x P b y b y a b x y x y x y xy a ; b ; h 6 x y a x y b IP Intensive lassroom Program egate-live Internet Based lasses DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ enters across India 6

17 EE GATE-6-PAPER- It represents ellipse 4 ab h r a b r The length of semi-axes is 4 64r r r or r 4 6 r values are positive, so it represents ellipse Both r or r 4 Length of major axis = r Length of minor axis = r 4 Equation of the major axis is a x hy r 4 x 6y x y Equation of the minor axis is a x hy r 6 x 6y y x Major axis exists along y = -x and minor axis exists along y = x. The vector Lies on the line y = x. Let the probability density function of random variable, X, be given as: x 4x fx x e u x ae u x where u(x) is the unit step function. Then the value of a and ProbX, respectively, are (A) (A), (B) we have f x dx x x x f x dx f x dx 4x x a e dx e dx u x for x, other wise 4, (), 4 (D) 4, 4 IP Intensive lassroom Program egate-live Internet Based lasses DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ enters across India 7

18 u x for x EE GATE-6-PAPER- otherwise 4x x a e dx e dx a 4 Prob 4x x e e a 4 a a a 4 4 4x 4x x fx x dx a.e dx a e dx 4x e y 4 4. The driving point input impedance seen from the source V S of the circuit shown below, in, is. I S V V S 4V 4 The Driving point impedance is nothing but the ratio of voltage to current from the defined port. In this case it is V I S S Writing KL at node x Vx Vx IS 4V 6 Substituting these in Eq() VS Is VS Is IS 8Is 6 VS IS S V I V 6 I S S S V S I S V 4V V x I V x 4 V 6 IP Intensive lassroom Program egate-live Internet Based lasses DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ enters across India 8

19 EE GATE-6-PAPER The z-parameters of the two port network shown in the figure are Z 4, Z 6, Z 8, Z. The average power delivered to RL, in watts, is. I I V V Z V R L 5.55 In the given terminated port network the Z matrix is known and for load of we want to find power on the load. The get it assuming R L as load let first obtain the thevenin equivalent of port Thevenin equivalent means V th & R th V V i.e., O. voltage of port th I IS I V i.e., s.c current of port, R in V I th sc. Evaluation of V. th The Z matrix equation is V 4I 6I V 8I I In the above two equations if I then V 4I () V 8I () From the input side we can say V I I 4I I A Then equation becomes 5 V 8 I 8 V 5 so Vth V Evaluation of I S In the Z matrix equation if we put V then V 4I 6I (5) 8I I (4) IP Intensive lassroom Program egate-live Internet Based lasses DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ enters across India 9

20 EE GATE-6-PAPER- I I & V I I 8 8 Using these V & I in equation 4 I I 6I I 4I 48I IS I 8A 6 I I 8A V in Rin 4 Isc 8 Now the ckt is from port is P I watt Vth Rth 4 RL 6. In the balanced -phase, 5Hz, circuit shown below, the value of inductance (L) is mh. The value of the capacitance () for which all the line current are zero, in millifarads, is. L L L.4 I ph L jl j X L.X c Zph X j L j L X c L b c 4.4 mf L I L L L IP Intensive lassroom Program egate-live Internet Based lasses DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ enters across India

21 EE GATE-6-PAPER In the circuit shown below, the initial capacitor voltage is 4V. Switch S is closed at t =. The charge in lost by the capacitor from t 5 s to t s is. 4V 5F S V 4 It is given R 5, 54f so 4 R Since it is a source free network we can say V t5sec t V t V e ; t 4e 4 t We are asked to find the charge last by capacitor From t 5s to s We know in a capacitor Q V or Q V Q V 5 sec V sec 4t 4e V 4e.47 tsec 6 4 V 4e.7 Q s 8. The single line diagram of a balanced power system is shown in the figure. The voltage magnitude at the generator internal bus is constant and. p.u. the p.u. reactances of different components in the system are also shown in the figure. The infinite bus voltage magnitude is.p.u. A three phase fault occurs at the middle of line. The ratio of the maximum real power that can be transferred during the pre-fault condition to the maximum real power that can be transferred under the faulted condition is Generator int ernal bus j. j. ~ j. j.5 Line j.5 Line inf inite bus IP Intensive lassroom Program egate-live Internet Based lasses DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ enters across India

22 EE GATE-6-PAPER- Before fault During fault ~ ~ Xeq..5 P.U EV Pe max EV X eq ~. b b xb.6 xc xe.5.5 c b a onverting Y a X ac X ac Pe max EV Pe Prefault Pe during fault.5 9. The open loop transfer function of a unity feedback control system is given by Ks G s, K, T. s Ts s The closed loop system will be stable if (A) () 4 K T K T K T (B) (D) IP Intensive lassroom Program egate-live Internet Based lasses DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ enters across India 4 T K T 8K T K () To comment closed b system stability we need the characteristic equation. Here it is given that it is a unity feedback system. Unity feedback system So the characteristic equation is S TS S KS

23 EE GATE-6-PAPER- S TS S KS K S S TS TS KS K S T S T S k k T k k T T T s s s for stability using Rt criterion T K K T T T K K T T K K ` K T T T K K T T 4. At no load condition a -phase, 5Hz, lossless power transmission line has sending end and receiving-end voltage of 4 kv and 4kV respectively. Assuming the velocity of traveling wave to be the velocity of light, the length of the line, in km, is VS Vr km 8 4. The power consumption of industry is 5kVA, at.8 p.f. lagging. A synchronous motor is added to raise the power factor of the industry to unity. If the power intake of the motor is kw. The p.f. of the motor is..6 cos P cos S P 4; P motor Q P tan P P tan 4 tan tan kw S j motor cos.6 m 6.86 IP Intensive lassroom Program egate-live Internet Based lasses DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ enters across India

24 4. The flux linkage EE GATE-6-PAPER- and current (i) relation for an electromagnetic system is i g. When i = A and g air gap length cm, the magnitude of mechanical force on the moving part, in N, is. 86 to 9 4. The starting line current of a 45V. -phase, delta connected induction motor is A, when the rated voltage is applied to its stator winding. The starting line current at a reduced voltage of V, in ampere is. to 44. A single-phase, kva, V transformer is reconnected as an auto-transformer such that its 6 ` kva rating is maximum. The new rating in kva, is. A V V A A V max KVA rating 6kVA 45. A full-bridge converter supplying in RLE load is shown in figure. The firing angle of the bridge converter is O. The supply voltage t sin t V, R, E 8V. The m inductor L is large enough to make the output current I L a smooth dc current. Switches are lossless. The real power fed back to the source. In kw is. Load I L T T L ` ~ V in Bridge R T T 4 E 8V 6 Vm V cos cos V IP Intensive lassroom Program egate-live Internet Based lasses DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ enters across India 4

25 I EE GATE-6-PAPER- E V 8 R A Pfedback V I 6kW. 46. A three phase Voltage Source Inverter (VSI) as shown in the figure is feeding a delta connected resistive load of phase. If it is fed from a 6V battery, with 8 O conduction of solid-state devices, the power consumed by the load, in kw, is. 6V 4 Vph Vdc 6 V R R ph ph Vph PLoad 4kW R 47. A D-D boost converter, as shown in the figure below, is used to boost 6V to 4V, at a power of kw. All devices are ideal. onsidering continuous inductor current, the rms current in the solid state switch (S), in ampere, is. mh 6V S mf Load 4 V to A single-phase bi-directional voltage source converter (VS) is shown in the figure below. All devices are ideal. It is used to charge a battery at 4V with power of 5kW from a source Vs V(rms),5Hz sinusoidal A mains at unity p.f. If its ac side interfacing inductor is 5mH and the switches are operated at khz, then the phase shift and fundamental A rms VS voltage V, in degree, is. between A mains voltage V S IP Intensive lassroom Program egate-live Internet Based lasses DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ enters across India 5

26 EE GATE-6-PAPER- VA ~ I S 5mH X S mh 4V I S V V S IX 9. to onsider a linear time invariant system x Ax, with initial condition x() at t =. Suppose and are eigenvectors of matrix A corresponding to distinct eigenvalues and respectively. Then the response (A) xt of the system due to initial condition x () = is e t (B) e t () e t t (D) e e (A) Eigen values are nothing but pole location Here with respect to the pole is wrt Pole is The section should be of form e t e t but we are asking w.r.t Initial condition x only so the response should be IP Intensive lassroom Program egate-live Internet Based lasses DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ enters across India e t 5. A second-order real system has the following properties: (a) the damping ratio.5 and undamped natural frequency n rad s, (b) the steady state value of the output, to a unit step input, is.. The transfer function of the system is. (A) (B) () s 5s s s s s (B) The standard nd order T/F is it is given that G s K s s.5 & n K s s Now to satisfy the steady state O/P. n n n y t K. K. s s s s. s s s s Gs t (D) s 5s 6

27 EE GATE-6-PAPER Three single-phase transformers are connected to form a delta-star three-phase transformer of kv kv. The transformer supplies at kv a load of 8MW at.8 p.f. lagging to a nearby plant. Neglect the transformer losses. The ratio of phase current in delta side to star side is (A) : (B) : () : (D) : (A) N : N : : I N I N I I. i I I 5. The gain at the breakaway point of the root locus of a unity feedback system with open loop Ks transfer function Gs is s s 4 (A) (B) () 5 (D) 9 (A) Gs Ks s s 4 To find Break away point We need to find the root of dk ds where s s 4 s 5s 4 K s s d d s s 5s 4 s 5s 4 s dk ds ds ds s S S 5 S 5S 4 S 5S S 5S 4 S 4 S From the pole zero plot it is clean that Break away point must be S as it is in between poles Now to find gain at this point use magnitude condition KS s s 4 s KS K 4 IP Intensive lassroom Program egate-live Internet Based lasses DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ enters across India 7

28 EE GATE-6-PAPER Two identical unloaded generators are connected in parallel as shown in the figure. Both the generators are having positive, negative and zero sequence impedance of j.4p.u, j.p.u and terminals of the generators, the fault current, in p.u., is. ~ ~ 6.4. Z.5 P.U ; Z. P.U ; Z.5P.U Vprefault If 6p.u Z Z Z.5..5 O 54. An energy meter, having meter constant of revolutions kwh, makes revolutions in seconds for a constant load. The load, in kw is revolutions K re v kwh PkW hr 6 P kw z 55. A rotating conductor of m length is placed in a radially outward (about the z-axis) magnetic flux density (B) of Tesla as shown in figure below. onductor is parallel to and at m distance from the z-axis. The speed of the conductor in r.p,m. required to induce a voltage of V across it, should be Voltage = B Voltage Velocity m s B i.e, m takes m Takes r r for rotation in minute = 6 Velocity = rpm B m m IP Intensive lassroom Program egate-live Internet Based lasses DLP TarGATE-All India Test Series Leaders in GATE Preparation 65+ enters across India 8