ESE (Prelims) - Offline Test Series Test - 22 ELECTRICAL ENGINEERING Mock-1 (Paper - II) - SOLUTIONS

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1 TEST D: ESE- 9 (Prelim) - Offline Tet Serie Tet - ELECTCAL ENGNEENG Mock- (Paper - ) - SOLUTONS. An: (c) The input line current of a three-phae fullwave un-controlled rectifier upplying an LE load contain only odd harmonic but no triplen harmonic with zero dc component. An: (d) Second breakdown failure mode doe exhibit only in BJT.. An: (a) The input voltage = 4in 4t When E load i connected acro ingle pule converter the range of the firing on the i to 4in = = ange of firing angle = to 5 4. An: (b) The nature of load current depend on both type of load and firing angle delay. 5. An: (d) ml 6 o co 6 6. An: (c) 7. An: (d) 8. An: (d) o ml ml = co n S the output current depend on value of load parameter. f load i pure inductive the current wave form will be a triangular. 9. An: (b) Step-down chopper drive Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru Chennai ijayawada izag Tirupati Kukatpally Kolkata

2 : : Electrical Engineering Separately excited d.c motor Motor rating:, rpm, A E a = k m W m + a a Neglect armature reitance drop km 6 k m 6 volt ec rad n tep-down chopper, = T on.5 T =.5 = volt E a = k m W m 6 W m N. An: (a) Cae: W m rad/ ec rpm 6 - half wave converter Speed of drive = rpm Firing angle () = 45 For a - half wave converter m Output voltage ( ) = ( co ) At N Cae: For - emi converter m Output voltage co So, N N = rpm. An: (a) Semi converter work alway a a rectifier in both ingle phae and three phae mode of operation. Becaue the current direction i alway poitive and voltage will never be negative in any intant for a emi converter becaue of the free wheeling action of diode.. An: (a) Thermal run away i a condition where temperature of device increae it reitance decreae and current through the device increae and caue temperature rie and further damage the witch. Thi i not poible in MOSFET becaue a temperature increae the mobility of device decreae and reitance increae. Thu MOSFET i a poitive temperature coefficient device. Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru Chennai ijayawada izag Tirupati Kukatpally Kolkata Ahmedabad

3 : : ESE - 9 (Prelim) Offline Tet Serie. An: (b) 6. An: (b) k TF= For any value of and the ytem i table for k = to k = k A mg -plane e O t i alway table 4. An: (d) Phae and gain cro over frequencie doe not exit GM = and PM = 5. An: (b) M j tan j M + N (, ) = tan 8 AO co k O 7. An: (c) Apply KL at node A A A C Apply KL at node B B B = A ( +) B = + A = B (By virtual ground concept) O Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru Chennai ijayawada izag Tirupati Kukatpally Kolkata Ahmedabad

4 : 4 : Electrical Engineering Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru Chennai ijayawada izag Tirupati Kukatpally Kolkata Ahmedabad

5 8. An: (d) ( ) (+) () 4 AE = AE 4 4 AE = = Zero row Zero row : 5 : ESE - 9 (Prelim) Offline Tet Serie 9. An: (c) From fig.(i) Number of individual loop = (ad, be, cf) Number of forward path = (abc) Tranfer function (T.F) abc ad be cf adbe adcf becf adbecf From fig.(ii) Number of individual loop = (ad, be, cf) Number of forward path = (abc) Tranfer function (T.F) = abc ad be cf adcf 5 4 +ve ve ve +ve +ve ve AE root: 4 +4 = 4 + = =, j + Symmetrical root in HP. An: (c) C. E i + (+T) +T+ = outh Tabulation T +T ( T)T T For Stability, + T> and T T T T > (+T) T > T + T > (T+6) (T 5) > T + 6 > and T 5 > Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru Chennai ijayawada izag Tirupati Kukatpally Kolkata Ahmedabad

6 : 6 : Electrical Engineering T > 6 T > 5 T > 5 ( the ytem i table). An: (b) T.F = = 8 ( ) 8( a) 8 ( ) ( ) n = 8a+ 4 4 = 8a + 8a =. An: (b) a = 8 (8a ) 6 dx Given that, 4 +x = f(t) dt Applying LT on both ide with Zero initial condition 4 X() + X() = F() X() F() Y() = 4 X() F().5 y(t) =.5e t/ Time contant T= ec.. An: (a) Unit ramp Unit tep r(t) = t u(t) t u(t) =, t = t < =, t < Lr t Lu t Unit impule t t Unit doublet= (t) L[ (t)] = L = t t 4. An: (a) From the magnitude equation [ A] 9 9 Adj[ A] TF = TF CAdj[ A]B A 9 [ ] [ ] = ( 9) 9 = 9 CE = 5. An: (c) To get radiu of circle, we required break point. Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru Chennai ijayawada izag Tirupati Kukatpally Kolkata Ahmedabad

7 : 7 : ESE - 9 (Prelim) Offline Tet Serie dk Break point are obtained by d CE G 5 K K 5 K dk d [ ] = [ + + 9] = (+)(+9) = Break point are =, 9 9 Centre 5, (mid point of Break point) Centre ( 5, ) adiu i ditance between centre to break point. adiu = 4 6. An: (c) BP 9 Given that e i the difference between and meaured voltage. 5 r =4 r =4 e = meaured value BP m meaured voltage = e e e 7. An:(c) m e m m m Actual energy per minute 4.8 E 6 e m m m. kwh Speed of the dic in rpm = Energy upplied per minute meter contant in rev/kwh 8. An: (d) =. 6 = 9. rpm n Kelvin double bridge two et of reading are taken when meauring low reitance, one with current in one direction and the other with the direction of current revered. Thi i done to eliminate the effect of thermoelectric emf Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru Chennai ijayawada izag Tirupati Kukatpally Kolkata Ahmedabad

8 : 8 : Electrical Engineering 9. An: (c) n -, power meaurement by two wattmeter method, W = co( ) W = co ( + ) Pure capacitance i connected a a load load p.f = ZPF lead Load power factor angle, = 9 W = co [ ( 9)] = W co[ ( 9)] Statement (iii) &(iv) are correct. Unknown reitance,. An: (a) =.67 n reitive hygrometer ome hygrocopic ort exhibit a change in reitivity with humidity example lithium chloride.. An: (c) 5 oltmeter v=k. An:(c) At bridge i balanced a = b b a =.75.5A.75.75A 5 oltage drop acro =.75 a b 5 7.5A = 6.5 eq 5k = 4 k 5 Current flowing through 4 i = [ +] = 45 Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru Chennai ijayawada izag Tirupati Kukatpally Kolkata Ahmedabad = 6 =.5 k

9 : 9 : ESE - 9 (Prelim) Offline Tet Serie. An: (b) A B S C P D Q 4 a = =.666 nm 4. An: (b) Statement & 4 are correct 5. An: (c) We Know p C p = i C i 6. An: (a) 9 M C p = M 45 pf C P M 45pF 9M = 5 pf ariable capacitance device ued for preure meaurement. Orifice meter ued for flow meaurement. Thyritor are ued for temperature meaurement. 7. An: (b) Given data pacing (,, ) nter planner ditance 4 = a a h a k a ; =.5 nm 8. An: (c) f the radiu ratio X =.44, a more table configuration i poible with ix anion bonding with a cation. Thi configuration called the octahedral configuration i table for.44 < X <.7. Here the cation occupie the void created by ix anion forming an octahedral tructure. 9. An: (b) Carbon- Agriculture odine- Medical therapy Krypton-5 ndutry Ceium-7 Calibrate equipment 4. An: (a) High humidity allow conduction and high temperature i the favourable condition chemical reaction. 4. An: (d) The critical magnetic field, H c for type- uper conductor i of the order of. Tela Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru Chennai ijayawada izag Tirupati Kukatpally Kolkata Ahmedabad

10 : : Electrical Engineering (or) le o. High magnetic field cannot be produced by type- uper conductor. catalyt uch a iron. Thi method produced tube with open end. 4. An: (d) Structural formula for ferrite i AOB O or AB O An: (b) Ueful uperconducting material have very low critical temperature. 44. An: (b) The reduction in the particle ize in the cae of emiconductor reult in the increaed in the band-gap which reult in the hift of the light aborption toward in the highenergy region (blue hift). n addition, the band edge poition of valance and the conduction band are tabilized and detabilized repectively. The rate of recombination of photo excited electron hole pair i greatly reduced. 45. An: (c) Chemical apour Depoition Method involve decompoing a hydrocarbon ga uch a methane (CH 4 ) at C. A the ga decompoe, carbon atom are produced. Carbon atom then condene on a cooler ubtrate that contain variou 46. An: (c) At t = 5m, i, but the voltage uddenly drop to zero value i.e., the element act a a hort circuit. A the voltage acro a capacitor cannot change intantaneouly, the element i not a capacitor. And alo the voltage i not proportional to the current; there fore the element i not a reitance. We conclude that the element i an inductor. di From the figure, dt 5 L 47. An: (c) network N L A / S and 5 di / dt Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru Chennai ijayawada izag Tirupati Kukatpally Kolkata Ahmedabad oc c di dt From the given equation 5 oc i obtained when i = 6 oc = 8 oc = 5v t i obtained when = A B 5mH

11 : : ESE - 9 (Prelim) Offline Tet Serie = 8 c c = 4 N 48. An: (c) oc c 5 8 Number of poible tree= det [A r ][A r ] T Where [A r ]= reduced incidence matrix. Determinant = (6 ) + ( ) () 49. An: (a) i(t) = co (t)a (t) z 6 = rad/ec e = 5 4 = 8 co(t 45) j A j e Z z + z mpednance Z 6 e j e j45 6 e j45 6 co45 jin 45 For inductive load 6 j Z jl j(). =, in erie with L =. H 5. An: (c) 5. An: (a) i = A in t = = 5 in( t + ) p.f. = unity 5. An: (d) =+j A capacitor act like a open circuit to d.c. and voltage cannot change abruptly. A nductor act like a hort circuit to d.c. and current cannot change abruptly. Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru Chennai ijayawada izag Tirupati Kukatpally Kolkata Ahmedabad

12 5. An: (d) B B : : Electrical Engineering At =, Eq. () and () become = + Power acro Bulb (P ) = 4 = Power acro Bulb (P ) = = = + = 4 = 4 = 8 = 6 For Bulb, reitance value i P (8) When bulb, alone i connected acro ; Power conumed by it P. 54. An: (a) i W + 4 = + ( + ) = +.. () = + ( + ) = +.. () = And = =.5 h = h = =.5 =.5 At =, Eq. () and () become = and = h = h = =.5 =.5 h h.5.5 h h An: (a) C F i(t) t t i idt idt di i i dt di 5i dt 6 F Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru Chennai ijayawada izag Tirupati Kukatpally Kolkata Ahmedabad

13 : : ESE - 9 (Prelim) Offline Tet Serie i(t) = 5 t Ae 6 At t = +, i(t) = i(t) = W = e 6 5 t t e 6 dt 6 = e ( ) = 6 Joule t 6 Since thi i independent on, at = 5, energy = 6 J. 56. An: (b) At t = : NW i in teady tate (i) t = + i L ( ) = A = i L ( + ) C ( ) = = C ( + ) i + i i + + C( ) i L ( ) + i + i + L ( + ) + At t= + : NW i in Tranient tate By KCL 5 i.i + = i = 5A By KL 5 i L ( + ) = L ( + ) = 5 5 = 5 di L(t) L dt di L (t) L dt 57. An: (d) t t = 5 A/ L = co (t+7 o ) = co (t+7 8 o ) = co (t o ) = 4in (t o ) = 4co (t 9 o ) = 4co (t o ) Hence the phae angle between and i = o 58. An: (c) Time contant i found by opening the A ource. The reultant circuit i hown in Fig. F F F Fig. Time contant = C = ( + ) = 6 = 4 ec. Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru Chennai ijayawada izag Tirupati Kukatpally Kolkata Ahmedabad

14 : 4 : Electrical Engineering 59. An: (c) Statement and : T d = K a. The developed torque at tarting a well a during running proportional to armature current. A large armature current (about.5 rated value) at tarting increae the tarting torque, and accelerate the motor to it final peed fater. Hence tator i uually deigned for a tarting current greater than rated current. (The current decreae a peed increae). Statement i wrong. Statement i correct.. The motor in the ward-leonard ytem receive an armature voltage which can be increaed moothly by field control of the generator of the ytem. Hence no external tarter i neceary. Statement i correct. 4. n plugging, the armature current nearly double while the peed fall. With large motor, the rated current are large, and when doubled, they may damage the machine and alo the wiring of the electric upply. Hence plugging i not commonly ued to brake large rated dc motor. Statement 4 i wrong. 6. An: (c). Star-delta tarting can be ued for a - phae induction motor only if the motor i delta connected for normal operation. Let thi be atified, and let L be the rated line voltage, and L the rated line current. The correponding phae value are L and L /. f now the motor i tarted with phae in tar, taring current = r L x where r and x are the equivalent circuit value. Thi i likely to be le than L, the maximum permiible value, preventing over heating of winding. correct. Statement i. n auto-tranformer tarting, the applied voltage to the tator phae i reduced at tarting; thereby reducing the tarting current t. But tarting torque = r t ; and the relationhip between tarting torque and tarting line current i nonlinear irrepective of the method of tarting, tatement i wrong.. n tator reactor tarting alo, tarting phae voltage i reduced due to drop in reactance. But relation between tarting torque and tarting line current i nonlinear. Statement i correct. Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru Chennai ijayawada izag Tirupati Kukatpally Kolkata Ahmedabad

15 : 5 : ESE - 9 (Prelim) Offline Tet Serie Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru Chennai ijayawada izag Tirupati Kukatpally Kolkata Ahmedabad

16 : 6 : Electrical Engineering 6. An: (c) Given data: fl =.5. c fl = 6 and T t = T fl We know that in auto tranformer, T T t fl x SC fl fl = x (6).5 x = 6.5 Line current at tarting, t = x SC = 6 fl 6.5 =. fl 6. An: (b) = contant and f < f rated ratio i not contant f Wh f.6.6 A frequency i decreaed, Hyterei lo W h i increaed and eddy current lo i depend only on applied voltage in thi cae which i contant. But W h i due to core lo component of current w, core lo component of current will increae. correponding magnetizing component of current will increae. 6. An: (b) For a dc hunt motor T T % change in torque = T Since T 4 % For a dc erie motor, % change in torque = (ince T in a dc erie motor) 4 = = % % atio = % We know that, B max f, A frequency i decreaed, the required flux denity of the core will increae, therefore Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru Chennai ijayawada izag Tirupati Kukatpally Kolkata Ahmedabad

17 : 7 : ESE - 9 (Prelim) Offline Tet Serie 64. An: (a) High permeability and low reluctance to the flow of flux. x = 8 = 6 Magnetization Curve: (B) e = contant CGO teel Silicon teel Cat teeel Cat ron Mild Steel e (H) With CGO teel higher flux denitie can be achieved, o that ize and weight of tranformer can be reduced. 65. An: (b) Capacity of open delta bank i 57.7% of capacity of cloed delta bank without c over loading of the winding % eduction in load 66. An: (c) = % Critical reitance peed 6 x An: (a), N = N E g zn 6 E g N E g = N E g E g =E g 68. An: (a) P A N Armature reitance Eoc 6 and c = A Z E oc = 6 c X = 69. An: (c) = Ω Z = a a = Ω () () =.7 Ω The tarting torque depend on ine of the angle between the current in the two winding. Therefore erie capacitor value depend on tarting torque. Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru Chennai ijayawada izag Tirupati Kukatpally Kolkata Ahmedabad

18 : 8 : Electrical Engineering 7. An: (c) A the load on the ynchronou motor i uddenly increaed, the motor become hunt i.e, rotor peed fluctuate around ynchronou peed and finally reache to ynchronou peed. 7. An: (a) + Excluding field input (which are ued up a field copper loe), motor electrical input = t m Generator electrical output = t g Loe of both machine = t ( m g ). Armature copper loe of both machine = r a Stray loe of each machine t = 7. An: (b) m g r Given u log x y u t x m g a px & v tan y x fm g m fg r a ra M G x y v m & y g px p x y Uing C- equation u x = v y x y px p x y x On comparing, we get p = 7. An: (b) Given X X + = O X (X X + ) = O X + X = O X = X X = X 74. An: (a) in x ainx Lt b x x By L'Hopital rule, cox acox Lt b x x + a = ( b i finite) a = By L'Hopital rule, Lt x 4in x in x b 6x By L'Hopital rule, 8cox cox Lt b x 6 b = a = & b = Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru Chennai ijayawada izag Tirupati Kukatpally Kolkata Ahmedabad

19 : 9 : ESE - 9 (Prelim) Offline Tet Serie 75. An: (b) f x x F(x) = x a i continuou every where, and F(x) = x a i differentiable every where except at x = a f(x) = x + i not differentiable at x = and [, ] f(x) i continuou in the given interval, but not differentiable. 76. An: (c) Let f (x) and g(x) x x f (x) x Conider 77. An: (d) and g (x) x f (c) f (b) f (a) g (c) g(b) g(a) c b b a a ab c (a,b) a b Let = xz xy 4x 7 = grad i j k x y z At (,, ) z y 4i x j 4xzk grad = 7 i j + 8k = a The equation of the tangent plane i (r r ). a = r r = (xi+yj+zk) (i j+k) = (x )i+(y+)j+(z )k (r r ). a = [(x )i+(y+)j+(z )k].[7i j+8k] = 7(x ) (y+) +8(z ) = 7x y + 8z 7 6 = 7x y + 8z 6 = 7x y + 8z = An: (b) To evaluate F.dr along the curve C y = x from (, ) to (, ) F.dr xydx y dy c (,) c Y We have y = x and dy = 4x dx y = x (,) X C i y = x dy = 4xdx Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru Chennai ijayawada izag Tirupati Kukatpally Kolkata Ahmedabad

20 : : Electrical Engineering F.dr 4 xx dx 4x 4x dx c 79. An: (c) c 5 6x dx 6x dx 6x x 6 Let A = Event of failing in Paper- 6 B = Event of failing in Paper- given that, P(A) =., P(B) =., and P(A B) =.6 equired probability = P(A B) 8. An: (c) Given = P(A B). P(B) = (.6). (.) =. dy x y dx, y() = dy x dx dy y y dx x dy dx y x ntegrating both ide dy dx dx x log y log x c xy c at x = and y = c = 4 The olution i xy = 4 8. An: (b) At z = a, numerator cot(z) = cot(a) z = a i a pole of order alo called a double pole. 8. An: (b) Output y(n) x(n) h(n) y(n) [ke j( n ) ] [ (n 4) (n 4)] [ x(n) (n n) x(n n)] y(n) k[e y(n) k[e y(n) ke j( (n4) ) jn jn y(n) ke 8. An: (a) e e j4 j [e j( n ) e j j4 e e e j( (n4) ) e e jn j4 j4 co(4) A ytem i aid to be caual, if OC i out of the outermot circle. A ytem i aid to ] ] j be table, if OC include unit circle. f OC i z > then ytem i caual, but untable. ] Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru Chennai ijayawada izag Tirupati Kukatpally Kolkata Ahmedabad

21 : : ESE - 9 (Prelim) Offline Tet Serie z then ytem i non-caual and table. Then the correponding difference equation i y(n) y(n ).y(n ).5x(n).x(n ) z < untable. 84. An: (d) Given then ytem i non-caual and y (n) k a k y(n k) bx(n) Apply z-tranform on both ide Y (z) H (z) k a k k z Y(z) bx(z) Y(z) X(z) b H( ) b = 85. An: (a) Given y(n) b k a k z k y(n ).y(n ).5x(n).x(n ) 86. An: (b) Given h (n) & h (n) are connected in cacade. So, the reultant impule repone i h(n) = h (n) h (n) h(n) = (n) [(n) (n )] h(n) = (n) (n ) Given x(n) = u(n), then y(n) = x(n) h(n) 87. An: (b) x (t) y(n) = u(n) [(n) (n )] y(n) = u(n) u(n ) y(n) = {, } y(n) = (n) + (n ) Sin ( t) Sinc( t) rect t X() Apply DTFT on both ide H (e H(e j Y(e ) X(e j( ) j j ).5.e j ) e.e.5.e ) j( ) e.e j( ) j j j( ) h (t) Sin( t) Sinc( t) rect t H() H (e j( ).5.e ) j e.e j j / Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru Chennai ijayawada izag Tirupati Kukatpally Kolkata Ahmedabad

22 : : Electrical Engineering Y( ) X( )H( ) y(t) Sinc(pt), hence p = min(, ) 88. An: (a) Given j k 5 Y(k) X(k)e, N = 5 Apply.D.F.T y(n) = x((n+)) 5 x() = y() = x() = = 89. An: (c) There i a potential problem for frequency ampling realization of the F linear phae filter. The frequency ampling realization of the F filter introduce pole and zero at equally paced point on the unit circle. 9. An: (d) B. n y (n) + y(n) = x (n) Non-linear becaue of quared term, y (n) Time variable becaue of multiplication by n min(, ) Y() min(, ) Memory le becaue of y(n) depending upon x(n) (5) D. y(n + ) y(n) = 4 x(n) Non-linear becaue of product term y(n+) y(n) Time invariant (no multiplication by function of n) Dynamic becaue of y(n + ) depending upon the previou value, y(n) (4) C. y(n + ) + n y(n) = 4 n x(n) Linear becaue of the abence of Non-linear term Time variable becaue of multiplication by n Dynamic becaue of y(n+) depending on the previou value y(n) () A. y(n + ) + y(n + ) + y(n) = x(n +) + x(n) Linear becaue of the abence of non-linear term. Time-invariant becaue of abence of Multiplication by function on n. Dynamic becaue of dependence of y(n +) on y(n) etc. () 9. An: (b) tatement, and are correct. Statement 4 i not correct, becaue MUX i converting parallel to erial. Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru Chennai ijayawada izag Tirupati Kukatpally Kolkata Ahmedabad

23 : : ESE - 9 (Prelim) Offline Tet Serie 9. An: (c) ( 56) 56 in complement repreentation uing 9 bit 56 in complement repreentation uing 6 bit So the number of in complement repreentation uing 6 bit i An: (c) ncluding XO gate, it i becoming T flip Clock Q flop. A T =, the tate Q i complementing at every clock edge T Q = T f Q T 94. An: (b) T T f M 5MHz D D D D Q Q Q Q By oberving the above table, we can ay that D and D input of DAC were interchanged. 95. An: (c). Baed addreing mode: n thi addreing mode, the offet addre of the operand i given by the um of content of BX/BP regiter and 8-bit/6-bit diplacement. Eg: MO DX, [BX+4]. Direct addreing mode: The addreing mode in which the effective addre of the memory location i written directly in the intruction. Eg: MO [4H], Ax Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru Chennai ijayawada izag Tirupati Kukatpally Kolkata Ahmedabad

24 : 4 : Electrical Engineering. mmediate addreing mode: The addreing mode in which the data operand i a part of the intruction itelf i known a immediate addreing mode. Eg: MO AX, 4H 4. ndexed addreing mode: n thi addreing mode, the operand offet addre i found by adding the content of S or D regiter and 8-bit/6-bit diplacement. Eg: Add AL, [S] 96. An: (a) A = 44 H = +B = 56 H = + (AC generated here. So we have to add to lower nibble) An: (a) = H AC mean add the immediate to the accumulator with carry. 7 = +56 = carry = 8E H 98. An: (b) N A = 6 /cm N D = 6 /cm N A > N D p-type p = N A N D for N A N D >> n i p = ( 6 6 ) atom/cm = 6 atom/cm 99. An: (b) C o r A d d. An: (d) =.6 F/m F/m KT N AN D ln ln(doping) q n i. An: (d) AqD AqD p n T T Pno npo[e ] [e ] Lp Ln Where, AqD AqD p n Pno n po Lp Ln AqD L p p n N i D AqD L n n n N i A Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru Chennai ijayawada izag Tirupati Kukatpally Kolkata Ahmedabad

25 : 5 : ESE - 9 (Prelim) Offline Tet Serie. An: (d) C B.6 6 C ma ma 49.mA 5. An: (d) = ( in ) = in in in in F in F in in (tranreitance) in F in F F. An: (c) DC rm m, m m.4 Turn ratio n n p rm rm DC : : Primary( n ) and econdary( n ) turn are, 4. An: (b) p C phae hift : Minimum C ection Crytal Wein bridge Hartley : i : High tability : AF range : F range i D F k 6. An: (c) in = be + i e E = i b r + (+) i b E i in in r b 7. An: (c) E The gain Bandwidth product i alway contant. i.e., G (B.w) = G (B.w) B.w 8. An: (b) t in + G B.w G = 6kHz i b in E k (.) (.4) (.5) Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru Chennai ijayawada izag Tirupati Kukatpally Kolkata Ahmedabad

26 : 6 : Electrical Engineering t =. 45 P P t c t.45 Pt P t =.445 W 9. An: (a) A c [+] = 5 = max A c [] = 4 = min A. An: (d) c max min 76 8 Pre emphai and de emphai ued in FM receiver. Balanced modulator ued to generate DSB SC ignal Companding i ued in PCM to uppre higher amplitude ignal Envelope detector i ued to demodulate AM ignal. An: (d) For a tranmiion line, at any point impedance Z(x) where x i the ditance from the receiving end i given a, (x) x = Z(x) And (x) = L coh(x) + L X inh(x) (x) = L coh(x) + Z(x) = Z Z Z L Z L inh(x) coh( x) Z coh x) Z L inh x) inh x if line i terminated with it characteritic impedance then, Z L = Z Z(x) = Z. An: (b) cae(i): When conductor are horizontally paced. a b c nductance/phae = Capacitance/phae = n thi cae GMD= GMD = d GM =.7788 r d =.6 d d GMD n H/m GM GMD n r F/m d d dd d d Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru Chennai ijayawada izag Tirupati Kukatpally Kolkata Ahmedabad

27 : 7 : ESE - 9 (Prelim) Offline Tet Serie Cae (ii): When conductor are placed in equilateral triangle form. a d d d b c n thi cae GMD = d GM =.7788 r Since GM & r are contant in both cae But GMD for equilateral pacing i low. So, capacitance will increae and inductance will decreae by converting flat pacing into equilateral pacing.. An: (d) ka demand of load at.8 Power factor = in6.86 =.47 ka. [ = co.8= 6.86] ka demand of load at unity power factor = in = ka to be upplied by capacitor bank connected at load =.47 =.47 ka 4. An: (a) Fault current in a fault i Z pu L = = pu. Fault current in a SLG fault i Z Z Z Z...5 X A SLG and -fault current magnitude are ame =.5 X n =.5 + X n =. X n =.5 X n =.66pu 5. An: (d) n =.66 pu All tatement are correct. 6. An: (d) n n bu ytem mot of the element in Y Bu are zero (95% - 98%). So that the memory requirement to tore the data in leer. Further Y-bu i a ymmetric matrix Y-bu called a pare matrix. n Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru Chennai ijayawada izag Tirupati Kukatpally Kolkata Ahmedabad

28 : 8 : Electrical Engineering Z bu i a full matrix modifying Y Bu i eaier 8. An: (c) than Z Bu modification during network change. 7. An: (c) Stator winding Given data: C = (.P + 8.) per MWh C = (.5P +.) per MWh Total load = P + P = MW C = C.P + 8 =.5P +.5P.P = 5.. () Given load P + P =.. () Solving equation () and ().P +.P =.P.P.5P P = 5 5 P = 6 MW 5 5 P = 6 = 4 MW There are two et of current tranformer one CT i connected to line ide and other i connected to the neutral ide of generator in each current in both CT hould match and current flowing through relay i will be zero and relay doe not operate. There during normal operation current at both end (line ide and neutral ide) of the phae winding are equal. 9. An: (a) mho relay i ued for protection of long tranmiion line mpedance relay i ued for protection of medium tranmiion line eactance relay i ued for protection of hort tranmiion line Buchholz relay i ued for protection of inter current fault on tranformer. Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru Chennai ijayawada izag Tirupati Kukatpally Kolkata Ahmedabad

29 : 9 : ESE - 9 (Prelim) Offline Tet Serie. An: (b). An: (d) Short-circuit capacity of a bu: t i defined a the product of prefault voltage and pot fault current. Since trength of a bu i directly related to hort-circuit capacity, the higher the hort-circuit capacity, the more voltage table it i in cae of fault on any other bu. Higher the hort-circuit capacity, lower will be the equivalent impedance. An infinite bu i almot unaffected by the fault on other bu, hence it ha infinite hort-circuit capacity and zero equivalent impedance. hort-circuit capacity MA Zeq bae bc ca 8 o p.u. Zero equence current in loop, ab =. An: (b) 4. An: (d) ab bc ca o = = Electric field intenity at x due to pherical charge ditribution and point charge i given by E = E + E () x= Q / x= () Q x= E? x. An: (d) a = b =8 c = ab b o a c o a = p.u. ab ca bc From Gau' law E Area Qenc r 4 Q 4 Q E E Q 8 E 4 Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru Chennai ijayawada izag Tirupati Kukatpally Kolkata Ahmedabad

30 : : Electrical Engineering E E Q 4 Q 9 Q E 8 7Q E 7 9 Q 5. An: (c) Given A x yzâ A Conider A x x x A y y zâ y A z x z. (x y z)â (x yz) (x z) (x y z) x y z = 6xyz +. A 6xyz, hence A i not olenoidal conider â x â x A x x yz â y y x z â z z (x y z) x x â x y x z â x z x z A (A null vector) y Therefore the vector field A i aid to be conervative (or) irrotational z z 6. An: (a). From magnetic boundary condition acro current-free interface B Ht H t & B t t Tangential component of magnetic field intenity are continuou and magnetic flux denity are dicontinuou. Work done around any cloed path i zero W qe. d L L E. d From Stoke' theorem E. ds E = Therefore an electrotatic field i irrotational (or) conervative.. From faraday' experiment, Diplacement flux denity i independent of permittivity of medium and electric field intenity i dependent on permittivity of medium Q D 4r Q E 4r â C/ m r â / m r tatement '' i incorrect. Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru Chennai ijayawada izag Tirupati Kukatpally Kolkata Ahmedabad

31 : : ESE - 9 (Prelim) Offline Tet Serie 4. From ampere' law B H J & J Aume the medium i homogeneou, B J f the medium i current-free, then B and hence magnetic flux denity i irrotational 5. From electric boundary condition 7. An: (b) D n D n = & E n E n = f = (charge-free) Then D n = D n & E n = E n Therefore acro charge-free interface, normal component of electric flux denity are continuou and electric field intenity are dicontinuou. Statement,, 4 and 5 are correct. n fully aociative mapping, main memory addre format i Tag 8. An: (c) f cache i full then it i capacity mi or ele it will be conflict mi. x Byte offet 6 bit 6 bit bit 9. An: (d) Suppoe n -addre intruction upported are -addre intruction: 6 opcode add add 4 bit 6 bit 6 bit max opcode = 6 ued opcode = n unued opcode = 6 n addre intruction 6 opcode add bit 6 bit max opcode = (6 n) * 6 = 9 n =. An: (a) n C, lock variable of a function will have block cope o the innermot printf( %d, i) will print value i and econd printf( %d, i) will print value i and outer printf( %d, i) will refer to global variable o it will print. The output i,,. Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru Chennai ijayawada izag Tirupati Kukatpally Kolkata Ahmedabad

32 : : Electrical Engineering. An: (d) for loop block ha variable 'x'. t ha no effect on the outide the for block on tatic variable x. x = x + tatement ue tatic variable x and variable x of for block cope complete and not available after the loop termination. x = x + = + = So, output will be produced by the given code.. An: (c) For the above reaon, the block ize hould be larger. While deigning the cache memory.. An: (d) Tak are aigned prioritie in TS 4. An: Effective acce time = H(t TLB + t mm ) + ( H) (t TLB + t mm ) =.9 (+) +. (+) = 8+ = nec 5. An: (a) For a magnetic bar (or) current carrying conductor, a iolated magnetic charge doe not exit and hence alway magnetic flux line are continuou cloed loop and they cloe upon themelve. A the number of magnetic flux line entering equal to number of flux line leaving and hence the net magnetic flux leaving a cloed urface i equal to zero. B.dS net (or) B.dS, thi i generally known a law of conervation of magnetic flux (or) Gau' law of magnetic field. therefore tatement & are individually correct and tatement i the correct explanation for tatement 6. An: (a) The rectifier voltage i given by X Lr r d n co d () The inverter voltage i given by d n N Li co X i d where n = number of bridge S Cloed urface.. () Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru Chennai ijayawada izag Tirupati Kukatpally Kolkata Ahmedabad

33 : : ESE - 9 (Prelim) Offline Tet Serie Lr Li = line to line AC voltage at the rectifier and inverter repectively. X cr, X ci = commutation reactance at the rectifier and inverter, repectively. n HDC ytem, DC voltage hould be a high a poible becaue rectifier control angle i low (equation ()). From equation () and (), control of DC voltage i exercied by the rectifier and inverter control angle and repectively. 7. An: (b) Sampling in one domain make the ignal to be periodic in the other domain. So, tatement () i correct. According to multiplication in time domain property, multiplication in one domain i the convolution in the other domain. 8. An: (c) De-emphai i performed after demodulation. So, tatement () i fale. 9. An: (b) Pule width can be altered by changing and C value of monotable multivibrator. Monotable multivibrator ha a ingle table tate. But both tatement are correct but not related. 4. An: (d) An aynchronou equential circuit may be regarded a a combinational circuit with Feedback. So tatement () i wrong. Statement () i correct. Becaue of the feedback among logic gate, an aynchronou equential circuit may become untable at time. 4. An: (a) z a n u(n), z a z a a n 4. An: (d) u( n ) z z a, z a Applying baic electromagnetic principle to a cro-ection of the machine, it can be hown a follow: When the phae current (with the machine acting a a generator or delivering current) lag the phae induced emf by 9 o, the armature mmf axi coincide with the field mmf axi, the armature mmf oppoing the field mmf. (Thi i called demagnetizing action of the armature mmf). n practice, the angle between a phae induced emf and correponding phae current i determined by the load, and can be any value between -9 o and 9 o. Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru Chennai ijayawada izag Tirupati Kukatpally Kolkata Ahmedabad

34 : 4 : Electrical Engineering Fig. how the phae current lagging the induced emf by ome angle t alo how the field mmf, and mmf due to component of phae current in phae with E and lagging E by 9 o. t i conventional to call the component (lagging wrong. E ph by 9 o ) a d. Statement- i Alo from the above analyi, i clear that tatement- i correct. 4. An: (c) i) Equivalent of the r reitance in the N circuit, referred to the N circuit, i r. N N F f ph F Statement- i correct. ii) Let the current in N circuit be A. Then voltage drop acro r = r. F, : component of ph E ph (induced emf) F and F : correponding mmf F f : Field mmf Fig. Correponding current in the N - circuit N =. N oltage drop acro the equivalent = N N r N N N N r Statement- i wrong. [Statement- correct, Statement- i wrong] 44. An: (b) Statement (): Nyquit path (or) contour: t encloe the entire right ide of -plane and it hould not pa through pole of G()H() which are on the j axi and at the origin. Statement (): Nyquit tability criteria (NSC):Nyquit plot will encircle (, j) critical point a many no. of time a the difference between, the number of right ide pole and zero of F()=+G()H()=. N = P Z N = Number of encirclement of (-, j) by the Nyquit plot. P = Number of right ide pole of F() (or) right ide pole of G()H() Z = Number of right ide zero of F() (or) pole of cloed loop TF Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru Chennai ijayawada izag Tirupati Kukatpally Kolkata Ahmedabad

35 : 5 : ESE - 9 (Prelim) Offline Tet Serie There i no relation between tatement () and tatement (). 45. An:(b) Both the tatement are correct 46. An: (d) The gauge factor of a train gauge i the ratio of per unit change in reitance to the train (per unit change in length). GF Poion ratio i a meaure of Poion effect, in which a material tend to expand in direction perpendicular to the direction of compreion converely, if the material i tretched rather than compreed. t uually tend to contract in the direction tranvere to the direction of tretching. So tatement - i wrong. Statement - i right. 47. An: (b) Unit cell i a baic building block in a lattice, by repetition in dimenion lattice i created. A non crytalline material poee more hardne and brittlene due to random orientation of atom. 48. An: (b) The colour of a natural crytal i becaue of preure of defect in it. Ex: F-centre defect. 49. An: (d) Ferroelectric material have already domain with permanent dipole. On application of electric field, they are aligned in the direction of electric field hence tatement () i wrong. Statement () i correct. 5. An: (c) (Cae ) Both current entering the dotted terminal M a b By KL d Md.. () dt dt L d dt Md. () dt L d M d dt L dt L L ' d ' Subtitute value in equation () dt Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru Chennai ijayawada izag Tirupati Kukatpally Kolkata Ahmedabad

36 : 6 : Electrical Engineering d M d M dt L dt L M L d dt L L eff M L L M Left i L rrepective at the L polarity of the coil. So option C i correct. (Cae: ) One current entering the dotted terminal and other current leaving the dotted terminal M a b By KL L L d Md.. () dt dt L d dt Md. () dt L d M d dt L dt ' d ' Subtitute value in equation () dt d M d M dt L dt L M L d dt L L eff M L L M L eff i L rrepective at the L polarity of the coil. So option c i correct. Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru Chennai ijayawada izag Tirupati Kukatpally Kolkata Ahmedabad

37 : 7 : ESE - 9 (Prelim) Offline Tet Serie Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru Chennai ijayawada izag Tirupati Kukatpally Kolkata Ahmedabad