: 2 : EE-Conventional Test-10 (Solutions)

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2 .(a) Sol: The hunt field current At no load, armature current ounter e.m.f, E a I f I a. A 5 A 9.4 : : EE-onventional Tet- (Solution) otational loe = E al I a watt At full load, armature current I a I L I f 5 5A ounter e.m.f E a 5. Here (no load flux) = (full load flux), becaue the field current i contant at A and effect of A i neglected E E n a a n n n E E (5) 9.4 a a 45 r.p.m Shaft power, P h = Electromagnetic power otational loe = = watt P Shaft torque h m Nm 45.(b) Sol: When torque-lip relation i liner, Full-load torque, From () T e. fl 75.9 Te ().r 74.4 Nm r and r Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

3 : : ESE - Offline Tet-8 4 Or or No-load peed of the motor = (.87) = 987. r.p.m. (c) Sol: y (t)= S(t) co(f c t+) y (t) =A m(t) cof c t co(f c t + ) y A A t mtco4 f t mtco It i paed through LPF,then output i c P out /P in / y A t mtco /4 /4 / /4 P m power of S t A m t P out powerof y A 4 t m tco P P out m co.(d) Sol: Energy lo during turn-on t on = i. t on E dt IS = t tdt 5 t on 4 = t t t on = 4 t t =.5 watt-ec dt dt Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

4 : 4 : EE-onventional Tet- (Solution) Energy lo during turn-off t off 4 = 4 t tdt 75 t off = 4 4 t4 tdt =.44 watt-ec Total energy lo in one cycle = =.94 W-ec Available witching frequency = f = 57.7 Hz 5.94.(e) Sol: Given data Supply voltage = Internal reitance = A voltage of upplying power to, through a ingle diode, i hown in Fig. (a). Waveform for the ource voltage, load current i and diode voltage D are hown in Fig.(b). m t D + D m. i + = i m D I om / t (a) D i D - m t (b) Fig. (a) circuit diagram and (b) waveform Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

5 (i) Dc diode voltage, I in t dt D o D : 5 : ESE - Offline Tet-8 m = Io D.5.5 (ii) At no load, load voltage, on = m.5 At given load,load voltage, o on o oltage regulation on =.9%.49.(a) Sol: (i) Equivalent circuit at the intant of witch S i cloed L L i = I e I = di dt S t / S max L and S I. L L S L dv dt =.8 A/ec 5 = max di dt max = 5.8 = 8.4 /ec Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

6 : : EE-onventional Tet- (Solution) (ii) X L i much lower than = 4, o current i primarily limited by 4 5 I max = A 4 For firing angle delay of 9, I TA = 5 =. A For firing angle delay of 5, I TA = =.4 A MS current rating of the thyritor i 5 =.4 A for any conduction angle. (iii) oltage rating of S i = (.5 to ) time the peak working voltage = (.5 to ) = 8.7 to (b)(i) Sol: We know that P elec mech. Therefore, pole hoe ubtend an angle 4 = electrical Flux per pole = (air gap area under one pole hoe) (uniform flux denity) rl..5 =...4Wb P 8 Total armature turn = 4 = 48. Thi give total armature conductor, Z=48 = 9 Generated e.m.f at No-load, E a ZNP A Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

7 : 7 : ESE - Offline Tet-8 Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

8 : 8 : EE-onventional Tet- (Solution).(b)(ii) 48 Sol: Number of armature turn per path 8 eitance of one path = 8. =.4 For -pole lap-wound D. generator, there are A = parallel path. e iance of one path.4 eitance of armature circuit. 4 Number of parallel path For D. generator, terminal voltage at Full load, t = E a I a r a t = = 79.4.(c)(i) Sol: Pre-emphai: In FM, the noie ha a greater effect on the higher modulating frequencie thi effect can be reduced by increaing the value of modulation index () for higher modulating frequencie (f m ). Thi can be done by increaing the deviation (f) and f can be increaed by increaing the amplitude of modulating ignal at higher modulating frequencie. Thu, if we boot the amplitude of higher frequency modulating ignal artificially then it will be poible to improve the noie immunity at higher modulating frequencie. The artificial booting of higher modulating frequencie i called Pre-emphai. ircuit diagram: Modulating AF ignal FM Modulator Pre-emphaized FM output Pre-emphai circuit Pre emphai circuit baically i a high pa filter. The modulating AF ignal i paed through a high pa filter, before applying it to the FM modulator. A f m increae, reactance of decreae Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

9 : 9 : ESE - Offline Tet-8 and modulating voltage applied to FM modulator goe on increaing. The amount of pre-emphai in FM tranmiion and ound tranmiion in T ha been tandardized at 75 ec. The frequency for the high pa network i Hz haracteritic: Output Slope =db/octave db db Hz KHz f m De-emphai: The proce of pre-emphai i nullified or compenated at the receiver by a proce called Deemphai. The artificially booted high frequency ignal are brought to their original amplitude uing the de-emphai circuit. The de-emphai circuit act a low pa filter. The 75ec deemphai correpond to a frequency repone curve that i db down at a frequency whoe time contant i 75ec. f Hz 75 ircuit diagram: FM input FM demodulator De-emphai ircuit AF output The demodulated FM i applied to the of goe on decreaing. De emphai circuit with increae in f m the reactance haracteritic: Output db -db Hz f m Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

10 : : EE-onventional Tet- (Solution).(c)(ii) Sol: m(t) = inc(4t) Then M(f) repreent a rectangular function a hown in the figure below. M(f) /4 f Then maximum frequency of meage ignal i (Hz) f () f max max f max = maximum frequency deviation = f max f max = = (Hz) () Since = (>) the modulated ignal repreent a WBFM ignal. Power of modulated ignal = power of carrier ignal Power 5kW A () BW of modulated ignal = ( + ) f m = (+) =.8(kHz).(a) Sol: (i) f USB = khz + khz = khz f LSB = khz khz = 59 khz (ii) Modulation efficiency A A m c 4.4 A min = A max = 44 Percentage of Modulation =.4 = 4.% (iii) A max = A c + A m = + 4 = 44 Fig: Output envelope Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

11 : : ESE - Offline Tet-8 A min = A c A m = 4 = (iv) Strength of lower ideband in the pectrum Strength of upper ideband in the pectrum Output pectrum A c =.4995 A c khz. (b) Sol: (i) i() = I ; co = dc ac = di L dt idt Apply Laplace Tranform on both ide dc dc I() = = LI() LI = co dc L dc I() L dc L c L co co I() O LI LI L LI. L L L.I L( ) dc = I + dc L T i + o Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

12 : : EE-onventional Tet- (Solution) i(t) = if co = i(t) = i(t) = dc dc in t I cot L dc in t I cot L in t I cot L A = A B in t L =. I in t dc B I peak = L 5 dc. I 5 = 98.7 A oltage acro capacitor t = idt co L t dc co = in t I cot)dt c L I dc co t t = ( cot) in t c c = ( dc co )(cot)+i L int +co c = L = I in t ( )cot dc c dc co co I If co = L in t (dc c )cot dc c = I L in t dc cot dc Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

13 : : ESE - Offline Tet-8 c max = L I dc =.48 (ii) onduction Time of S = Tan I L / / L ec = 49.8.(c) Sol: Terminal voltage i taken a the reference phaor. omplex power of plant = 4 j4 tan (co.8) Power input to induction motor A = 4 j ka 4.9 kw omplex power requirement of induction motor 4 4 j tan co B = j 5.54 ka omplex power requirement of ynchronou motor 4 4 j tan.9.9 co j.ka Net requirement of plant = A B + = 4 j j j. New p.f. = co.5 =.85 lag = 4 j45.4 ka = ka Original line current = 4.8 t 5 ka t Here, t = line voltage New line current = ka t Percentage reduction in line current.7% 5 Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

14 : 4 : EE-onventional Tet- (Solution) 4.(a) Sol: - upply B BA A AB A B t = T 5 T T T 5 T D D D 4 D D 4 t I SA /+/ + t I I 45 5 t =9 T T 5 T T T 5 D 5 D D 4 D 4 t B A BA I SA 5 4 t I I t (i) When / ( = ) / m t in d t m in t d t m m cot co co co co 4 m co co t Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

15 : 5 : ESE - Offline Tet-8 When > / ( = 9) 4 / m in t dt / m m (ii) I / 4 co co I co = co I co MS value of ource current I, when < / I I / = when / I I I = I I I co DF= = = co I I ( i.e., = ) = I co I = co ( > i.e., = 9) (iii) IPF = DF DPF = = = co co co / ( = ) co co ( ) co ( = 9) ( ) Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

16 : : EE-onventional Tet- (Solution) 4.(b) Sol: Here / 95., t E f 4 / 9.5 Z.4 5 = 5. tan.4/ 5) 4.57 Per phae power input i given by P = Ef t in( ) Z Z t r a or in ( ) =.4 or = =.88 or We know that, for leading power factor load I Z I a E E co / f t f t in( ) co. / az 88 I = 84.4A Now t I co =, W, co Here E f co + I a r a co = 9.5 co = > t (=95.) Therefore, motor i operating at a p.f, co =.948 leading.4 4.(c) (i) Sol: Given fig i hown in fig(a) Applying KL at node x x p I I 4 I S p - + Fig(a) Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

17 : 7 : ESE - Offline Tet-8 Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata p =..() KL at node P p 4 p p 4 p = 4 4 p = p..() Put () in () = = S S = 4 4.(c)(ii) Sol: Applying linearity and homogeneity principle to the above circuit we get =.5 k.k.k.5k.5k k.k = = 4.5

18 : 8 : EE-onventional Tet- (Solution) 5. (a)(i) Sol: x z D A T B A B clock From the logic circuit D xb x B T A x Z A B PS NS Z A B x = x = x = X 5. (a)(ii) Sol: Y m(,5,,7 ) y ab Y = (, 7) + (5, 7) + (, 7) = ab + yb + ya = ab + y(a+b) y a b Y Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

19 : 9 : ESE - Offline Tet-8 Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

20 : : EE-onventional Tet- (Solution) 5.(b) Sol: o + t i o t The output voltage and current expreion of a - full Bridge Inverter are o n,, n in t volt i o n,, n.z n in nt n Where Z n = nl n n nl tan n rad Given that = = 8 L =.4H f = 5 Hz L Z 5. 4 =.5 L = 4.89 Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

21 : : ESE - Offline Tet-8 tan L tan = 57.5 o 4 in f t 4 in 4t = 8. in4t I 4 in ft Z in 4t = 8.8in(4t 57.5 o ) o 5.(c) Sol:. PUSH intruction Syntax: PUSH r Operation: ontent of bit regiter given in intruction i moved into tack memory (a given below) and tack pointer content decremented by. ((SP) ) (rh) ((SP) ) (rl) (SP) i decremented by Ex: PUSH B PUSH PSW PUSH D PUSH H. POP intruction Syntax: POP r Operation: ontent of memory location from Top of tack i retrieved into pecified bit regiter and tack pointer content will be incremented by (rl) ((SP)) Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

22 : : EE-onventional Tet- (Solution) (rh) ((SP) + ) (SP) i incremented by Ex: POP PSW POP D POP H POP B. XTHL intruction Syntax: XTHL Operation: ontent of memory location from Top of tack exchanged with content of HL pair. The content of tack pointer will be unchanged. Ex: XTHL 5.(d) Sol: For tranformer of ka,. j.p. u Z e For tranformer of 5 ka,.5 j.8p. u hooe a bae of ka. Z e Z e Ze e Z e = pu. j (.5 j.8) Z.7 j Now S Ze Z Z e e. S Total ka load, Similarly, S 5 S ka ka On per un it bae, Z e (=.5) baed on it own ka rating i le than Z e (=.88) baed on it own ka rating. Therefore, a the load on the parallel combination i increaed, tranformer reache it load of ka firt. So tranformer hare a load of = ka Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

23 : : ESE - Offline Tet-8 Larget poible load on two tranformer i ka Method-II (Iz ) fl Larget ka Load on both the tranformer i (ka) T = (ka) + e (ka ) (Iz ) Larget ka Load on both the tranformer i ka ka T e fl (e) Sol: From hort circuit data Pc. 5 I reh c 4 Z eh c 5 I 4 c 5.5. x Z r eh eh eh For the.v ide, the parameter are r x el el Full load econdary current,, 4 I L 5A Now E ILreL co ILxeL in = (5) (.5) (.8) + 5 (.544) (.) =. For = 4, E =4 +. = 4. The voltage applied to the primary i = Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

24 : 4 : EE-onventional Tet- (Solution) Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

25 .(a) Sol: eil i going to round the value to it integer value plu one. eil (.95) = output A B D S S S eil (.) = S (A,B,,D) = m ( 5) S (A,B,,D) = m (4 9) S (A,B,,D) = m (,,, 7, 8, 9,, 4, 5) : 5 : ESE - Offline Tet-8 S D AB AB D S = (,, 4, 5)+(,, 4, 5) S = A+ AB S = (4, 5,,7) + (8, 9) S = AB A B AB D Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

26 S,, 7,5 8,9,5 4,5 : : EE-onventional Tet- (Solution) A BD A B BD AB ABD AB ( A B)D + (A B) + BD+ AB,,,7 8,9,5 4,5 S Or A BD A B AD AB ABD AB,, 7,5 8,9 9, 4,5 S A BD A B BD A B AD AB We will implement the logic circuit with S A B D S S S. (b) Sol: ipple current in inductor (I L ): Since current i varying in a linear fahion and having a lope of w D dc L dc i L + Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

27 : 7 : ESE - Offline Tet-8 I L lope time I L dc L L dc T.DT ON ipple voltage of capacitor ( ): Q Where Q = charge either accumulated or dicharge acro capacitor. I.DT I DT i L I Lmax I Lmin I L I L v D ( + dc ) v L i c t v c = v Given data: dc = = 5 I =.5A F = khz L = 5H Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata t

28 : 8 : EE-onventional Tet- (Solution) = 44F We know that D 5. D 5 = D 5 D 5 7 D D dc ipple current of inductor I ipple voltage of capacitor i I DT.5 44 = L dc. DT L A.(c) Sol: During blocked rotor tet, power input to a -phae induction motor appear almot entirely a I lo in both tator and rotor winding. Stator I lo + rotor I lo = kw It i given that tator ohmic lo = rotor ohmic lo during blocked rotor tet Total rotor I lo (i) Synchronou peed, Starting torque, I r 4f P kw 45 rad/ ec, Te. t.ir Nm (ii) With phae winding in tar, tarting phae current i reduced to / time the tarting current when in delta. Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

29 Total rotor ohmic lo, I r when in tar = / kw : 9 : ESE - Offline Tet-8 Starting torque, T e.t I r, 8.Nm 7.(a) Sol: L a = 5 mh ; = 9 ; E b = 8 + T T T 4 T = in E m = 8 in = 5.5 S t E b i G G G + v t i a + + t When T T pair i ON at t = di La Eb min t dt di in t E dt L m b a Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

30 : : EE-onventional Tet- (Solution) di dt at t = () hence i i maximum E m b di in tdt dt La La L m b i = a E cot t K L initial condition i i = at t = = / a = K = m E L L a a Eb L b co a K Equation for E E L L L m b b i = co t t a a a Maximum value of i i at t = () o co8 5.5 i max = = 5.5 A 7(b): Sol: Note that core lo remain fixed at 4W, wherea ohmic loe vary a the quare of the ka load 7 A.M top.m.,kaload 5kA. 5 5 Ohmic loe for hour = W P.M to P.M ; ka load.5ka.8 Omhic loe for 5 hour = 5W Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

31 : : ESE - Offline Tet-8 P.M to A.M ; ka load ka.9 5 Omhic loe for 7 hour = 77.8W A.M to 7A.M; Ohmic loe = Daily energy lot a ohmic loe = kwh 4 4 Daily energy lot a core lo.9 kwh Total kwh lot = =.9 kwh Daily kwh output = ( ) = 7 kwh Daily loe in kwh All day efficiency = (Daily output Daily loe) in kwh.9 = 7.9 = % (7) 7.(c) (i) Sol: a. The DA ha a -bit input and a.- F.S. output. Thu, the number of total poible tep i -=, and o the tep ize i. m Thi mean that AX increae in tep of m a the counter count up from zero. Since A =.78 and T =. m, AX ha to reach.78 or more before the comparator witche LOW. Thi will require tep m At the end of the converion, the counter will hold the binary equivalent of 7, which i. Thi i the deired digital equivalent of A =.78, a produced by thi AD. Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

32 : : EE-onventional Tet- (Solution) b. Thee hundred eventy-three tep were required to complete the converion. Thu, 7 clock pule occurred at the rate of one per microecond. Thi give a total converion time of 7. c. The reolution of thi converter i equal to tep ize of the DA, which i m. In percent it i / %.%. 7.(c) (ii) Sol: 8-bit ing counter A B Q Q Q Q Q 4 Q 5 Q Q 7 Q clk tart -bit Up counter clear A B A B 8 Decoder Q Q Q Q 4 Q 5 Q EN Q 7 Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

33 : : ESE - Offline Tet-8 T Q Q lock T Q Q Q A B 8 Decoder Q Q Q Q Q Q 4 Q 5 T Q Q EN Q Q 7 Q 8.(a)(i) Sol : (AB. AE) Binary form ( A B. A E ) (. ) (i) Octal form (Bae-8) (. ) ( ) 8 (ii) Bae-4 form (. ) (. ) 4 (iii) 5 complement form of (AB. AE) Subtract each digit from F(5) Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

34 : 4 : EE-onventional Tet- (Solution) FFF. FFF AB.AE 54.5 (iv) complement form = 5 complement form (54.5) (v) Binary form of (AB. AE) i (. ) complement form (Invert all the bit) (. ) (vi) complement form = complement form +. (. ) 8.(a)(ii) Sol : Quadratic equation x -x+ = Let the root be and =, = 7 Product of the root = Sum of the root + = +7 = Let the bae be b () b +(7) b =() b (b +b ) + (b +7b ) = b +b b++b+7 = b+ b + = b+ b = 9 The bae of the number ytem i 9. Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

35 8. (b) Sol: Given that - SI in mode, = 5, dc =, f = 5 Hz. The Fourier erie expreion for line voltage ab ab (t) n dc in nt,5,7, n : 5 : ESE - Offline Tet-8 ( t) in t in 5 t in 7 t in t 5 7 ab (t)=9.in 5 7 t +4.9in 5t +.7in 7 t +9.9 in t The Fourier erie expreion for phae voltage an dc n (t) co in nt n,5,7, n an (t) co in t co in t (t).8in an co in 5t co in 7t 5 7 t 5.in 5 5t 8.5in In tar connection line current i equal to phae current i a an(t) (t).8 in t in 5t 5 7 7t.5 in t in 7t in t 5 5 (t) i a 5 5. in t 5.7 in 5t (i) MS value of phae voltage dc a n r = = 9.89 MS value of line voltage dc ab r = 7. in 7t. in t =. A Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

36 : : EE-onventional Tet- (Solution) (ii) MS value of fundamental phae voltage an()r dc co = co = 89. MS value of fundamental line voltage dc ab = ( )r = 55. (iii) MS value of all harmonic line voltage. oh abr abr =. 55. = 48.7 % THD of line voltage = oh 48.7 = 55. ab =.8% (iv) Load power P I r 5. I or = 8.8 A 5.7. P = 8. 5 = 5 W. / Average ource current P 5 I =. A dc Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

37 : 7 : ESE - Offline Tet-8 (v) Average value of witch current IT A I TA = dc dc = 7.7 A 5 dc i T MS value of witch current I dc T r = dc IT r =.8 A 5 8 (c) Ire Sol: Percentage full load voltage drop in reitance E Ix e Percentage full load voltage drop in leakage reactance 4 E Ire (i) From above,. I E Full load ohmic lo =. ated A = Iron lo (given) ated A ateda I r Iron lo e = ated A ated A. rated A. rated A 9.54% 4 (ii) Maximum voltage drop mean maximum voltage regulation. Therefore, full-load p.f at which voltage drop i maximum Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

38 : 8 : EE-onventional Tet- (Solution) r Z e e Percentage r Percentage Z e e lagging (iii) Load p.f at which voltage drop i zero x Z e e Percentage x Percentage Z e e leading Hyderabad Delhi Bhopal Pune Bhubanewar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata