1. TF(1)

Transformers 01. Two transformers with leakage impedance Z1=0.1+j0.4 pu, and Z2=0.05+j0.2 pu are connected in parallel. The ratio of loads shared will be (EPDCL-10) (a) equal (b) 1:2 (c) 2:1 (d) 2:3 01. Ans: (b) Sol: If two transformers having equal voltage ratings, the equivalent circuit diagram of two transformers connected in parallel is shown in below figure. ZA



RA

XA

IA

RB

XB

IB

E2A = E2B

ZB

IL V2

ZL

0.05  j 0.2 0.05  j 0.2  0.1  j.4 0.1  j 0.4  SL  0.05  j 0.2  0.1  j 0.4

SA  SL 

the ratio of SA & SB is S A 0.05  j 0.2  SB 0.1  j 0.4 5  20 j

= 10  40 j 1 4 j

= 28 j SA  SB

17 1 = 68 2

 SA : SB = 1 : 2 02. In a transformer, all-day efficiency (EPDCL-10) (a) is higher than the full-load efficiency (b) is maximum efficiency (c) is lower than the full-load efficiency (d) None of these 02. Ans: (c) ACE Engg. Publications

03. A transformer is working at its full load and its efficiency is also maximum. The iron loss is 1000 watts. Then, is copper loss at half of full load will be: (SSC-JE-10) (a) 250Watt (b) 300 Watt (c) 400Watt (d) 500Watt 03. Ans: (a) Sol: At max efficiency, full load Cu. Loss= Iron loss  full load Cu. Loss = 1000W At half load, Cu. losses =

1  1000 4

= 250 W.

Load sharing equation of two transformers ZB SA  SL  ZA  ZB ZA SB  SL  ZA  ZB Given data: ZA = 0.1 + j.4, ZB = 0.05 + j 0.2

SB

Sol:  All – day efficiency is depends on load cycle.  All – day efficiency is less than full – load efficiency.

04. A 2 kVA transformer has iron loss of 150W and full load copper loss of 250W. The maximum efficiency of the transformer will occur when the total loss is: (SSC-JE-10) (a) 500W (b) 400W (c) 300W (d) 275W 04. Ans: (c) Sol: Iron loss = 150 W Full load Cu. Loss = 250W Maximum efficiency occurs when Cu. Loss = iron loss = 150W  Total losses = 150 + 150 = 300W 05. In a transformer, the core loss is found to be 46 W at 50 Hz and is 80 W at 70 Hz, both losses being measured at the same peak flux density. The hysteresis loss and eddy current loss at 60 Hz is (SSC-JE-11) (a) 11 W, 20 W (b) 30 W, 45 W (b) 16 W, 30 W (d) 22 W, 40 W 05. Ans: (d) Sol: Core loss = 46 W at 50 Hz = 80 W at 70 Hz Bmax = constant  Wn  f We  f2 2  Af1 + B f1 = 46 W  50 A+ 2500 B = 46 2 Af2 + B f 2 = 80 W  70 A + 4900 B = 80

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:2: Electrical Machines

By solving these equations A = 0.362 B = 0.0111 At f = 60 Hz, hysteresis loss = A(60) = 22 W eddy current loss = B(60)2 = 40 W

08. Ans: (b) Sol:

125A + 200 V

06. A 100 kVA single phase transformer exhibits maximum efficiency at 80 % of full load and the total loss in the transformer under this condition is 1000 W. The ohmic losses at full load will be (SSC-JE-11) (a) 781.25 watt (b) 1250 watt (c) 1562.5 watt (d) 12500 watt 06. Ans: (a) Sol: At maximum efficiency, x2(full load loss) = core loss  x2(full load loss) = 500 W 500  full load ohmic loss = 0.64 = 781.25 W

Full – Load kVA rating of Autotransformer = V2 I2 = 2200  125 = 275 kVA

07. A 40 kVA transformer has a core loss of 400 W and full load copper loss of 800 W. The fraction of rated load at maximum efficiency is (SSC-JE-11) (a) 50 % (b) 62.3 % (c) 70.7 % (d) 100 % 07. Ans: (c) Sol: Fraction of FL corresponding to max efficiency

(a) 0.52 (b) 0.42 (c) 0.62 (d) 0.36 09. Ans: (a) Sol: pu VR = pu R cos 2  pu X sin2 “+”  for lagging p.f “–” for leading p.f Pu VR = 0.2  0.8 + 0.6 × 0.6 = 0.52

=

Iron loss Full load Cu  loss

400  800

+ 2000 V

200 V –

09. A single phase transformer has resistance and reactance of 0.2 pu and 0.6 pu respectively. Its pu voltage regulation at 0.8 pf lagging would be (APGenco12)

3

08. A 25 kVA, 2000/200V, two winding transformer is connected as shown in figure. The full load kVA of connection is (APGenco12)

th

4

of full load. The ratio of its iron loss

(pi) and full load copper loss (pc), is (APGenco12) 3 4 9 (c) 16

16 9 3 (d) 4

(a)

(b)

10. Ans: (c) Sol: “x” of full load corresponding to maximum

125 A

Load

2000 V

efficiency =

iron losses [ Pi ] F.L Cupper losses  PC 

Given data : - x = (a) 125 (c) 375 ACE Engg. Publications

LOAD

12.5A

10. A transformer has maximum efficiency at

1  70.7% 2

2200 V

–

137. 5A

3 4

(b) 275 (d) 175 Hyderabad|Delhi|Bhopal|Pune|Bhubaneswar|Bengaluru|Lucknow|Patna|Chennai|Vijayawada|Vizag | Tirupati |Guntur| Kukatpally(Hyd)

:3: APPSC Previous Questions

Pi Pi 9  = Pc Pc 16

3  4

Vc2N =

v 120  180 2

Vector Diagram: Vc N 1

Va N 2

11. The following connection of three single phase transformer bank results in A

a1

a2 b1

b2 c1

c2

(a) 3-phase to 2-phase conversion (b) 3-phase to 3-phase (c) 2-phase to 3-phase conversion (d) 3-phase to 6-phase conversion 11. Ans: (d) Sol: B – +

V0

2

0

V 2

  180

V 2

V.R

ZPF lead p.f

Va2N = Vb1N = Vb2N = Vc1N =

V

 120 2  120  180

v 0 2 v   180 2 v   120 2 v   120  180 2 v 120 2

ACE Engg. Publications

0.2 0.4 0.6 0.8

0.8 0.6 0.4 0.2

ZPF lag p.f

C

Secondary phase voltages: Va1N =

12. A transformer can have regulation closer to zero (APTranso-12) (a) On lagging power factor (b) On leading power factor (c) On zero power factor (d) On unity power factor 12. Ans: (b)

– V–120

a2 b 1

a1

Vc N 2

Sol:

N

V

Va N 1

From the vector diagram, we can say that these transformer connections are used to 3 -  to 6 -  conversion.

(APGenco-12)

A +

60

60

Vb N 1

C

B

Vb N 2

b2

V 120 180 c1 2

c2

Slope of V. R curve for lag p.f loads is greater than slope of V. R curve for lead p.f loads. V.R is zero is possible only for leading power factors close to unity. 13. The full load copper loss of a transformer is 1600watts. At half-load, the copper loss will be (APTranso-12) (a) 400watts (b) 80watts (c) 1600 watts (d) 6400watts 13. Ans: (a) Sol: Copper loss at any fraction “x” of full – load = x2  full – load copper losses. given data: - X 

1 2

 Copper – losses at half – full load

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:4: Electrical Machines  1

=    2

2

1600 =

1600 = 400 W 4

14. A 440/110V, 1-phase Transformer has a primary resistance of 0.03Ω and secondary resistance of 0.02Ω. Its iron losses at normal input are 150watts. The secondary current at which the maximum efficiency will occur is (APTranso-12) (a) 63.56A (b) 70.28A (c) 82.80A (d) 46.39A 14. Ans: (c) Sol: Secondary current corresponding to max efficiency, I 2 m 

Wi R 02

Where Wi = iron losses. R02 = Secondary equivalent resistance. Given data : - R1 = 0.03 , R2 = 0.02 , Wi = 150 W R02 = R2 + R1  110  = 0.02  0.03     440  I 2m 

17. In a 50 kVA, 1-phase Transformer, the iron loss is 500W and full load copper loss is 800W. The efficiency of the Transformer at full load at 0.8p.f lagging is (in %) (APTranso-12) (a) 92 (b) 89.56 (c) 96.85 (d) 79.82 17. Ans: (c) E 2 I 2 cos  2  100 Sol: Efficiency,  %  E 2 I 2 cos  2  Pc  p i Where E2 I2 = Name plate VA rating cos 2 = p. f of load, Pc = cupper losses Pi = iron losses Given data: E2 I2 = 50 kVA cos 2 = 0.8 lag Pc = 800 W Pi = 500 W % 

2

= 0.021875

150 = 82.80 A 0.021875

15. Which of the following Transformers should never have the secondary open circuited while the primary is energized? (APTranso-12) (a) Power transformer (b) Potential transformer (c) Current transformer (d) Auto transformer 15. Ans: (c) Sol: If secondary of current transformer is opened, maximum flux is induced in core so that secondary induced voltage is increase and finally damages the insulation on secondary. So that the secondary of C.T can never be open circuited when primary is excited. 16. An Isolation Transformer has primary to secondary turn’s ratio of (APTranso-12) (a) 1:1 (b) 1:2 (c) 2:1 (d) 1:3 16. Ans: (a) Sol: Isolation transformers are provide electrical isolation between primary & secondary ACE Engg. Publications

without change in voltage and currents levels.

50 k  0.8  100 50 k  0.8  500  800

 % = 96.85%. 18. In core type furnace the secondary winding has (APTranso-12) (a) No turns (b) More number of turns (c) Less number of turns (d) Equal to primary turns 18. Ans: (c) Sol: The arc voltage is of the order of 50 V to 100 V i.e, secondary voltage is of this order in order to obtain large powers required for melting metals, the secondary current will be of the order of several hundreds or thousand amperes and it should be made of less No. of turns with thick wire. 19. A 220V/12-0-12V transformer has an emf/turn of 1V. The number of turns on secondary would be (APSPDCL-12) (a) 12 with centre tap (b) 220 with centre tap (c) 24 with no centre tap (d) 24 with centre tap 19. Ans: (d) Sol: 12 V Vs



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:5: APPSC Previous Questions

(a) 33.33A (c) 66.66A 22. Ans: (a) Sol: MVA  3VL I L 220V / 12 V – 0V – 12V Emf/turn = 1 V by the voltage notation of 220 V/ – 12 V – 0 V – 12 V, it is concluded that it’s center taped transformer. On secondary each half should have “12 V”. So EMF/ turn = 1 V for 12 V, we require “12” turns [for each half]. So total “24” turns with center tap. 20. An ideal transformer has N1 = 100 turns N2 = 200 turns with a mutual flux of m(t)=0.05(t22t). The induced emf of secondary in volts is (APSPDCL-12) (a) 5(t1) (b) 10(t1) (c) 5(t21) (d) 20(t1) 20. Ans: (d) Sol: Induced emf of secondary = N 2 = 200



d dt



d  0.05[ t 2  2 t ] dt

= – 20 [t – 1] 21. The following is the apparent disadvantage of auto transformer as compared to twowinding transformer (APSPDCL-12) (a) power rating is greater (b) efficiency is low (c) conductive isolation is not present (d) voltage regulation is low 21. Ans: (c) Sol: Disadvantages of auto-transformers:1. Auto – transformers are not suitable where electrical isolation is required between primary and secondary. 2. Auto – transformers are not suitable for turns – ratio is greater than 3:1 3. If any open circuit at common part occurs it should trip otherwise there will be high voltage at L.V which results in damage of load. 22. A 230/2300V, Y/ 3-phase transformer is rated at 230 kVA. Its rated secondary current/phase is (APSPDCL-12) ACE Engg. Publications

IL 

I ph 

(b) 133.33A (d) 30.33A

230  103 100  3  2300 3

I L 100   33.3A 3 3

23. If a transformer core has air gaps, then (HMWS-12) (a) reluctance of magnetic path is decreased (b) hysteresis loss is decreaed (c) magnetizing current is greatly increased (d) eddy current is increased 23. Ans: (c) Sol: If air gaps are present in the transformer core, the reluctance is increases and magnetizing current is also increases. 24. When the primary of a transformer is connected to a dc supply (HMWS-12) (a) primary draws small current (b) core losses are increased (c) primary leakage reactance is increased (d) primary may burn out 24. Ans: (d) Sol: When transformer is excited with d.c source, the magnetizing component increases and the core getting saturated and induced emf in both primary and secondary windings is zero and the primary winding draws a very high current from source which may burn the primary winding. 26. A 20-turn iron-cored inductor is connected to a 100V, 50Hz source. The maximum flux density in the core is 1Wb/m2. The crosssectional area of the core is (HMWS-12) (a) 0.152m2 (b) 0.345m2 (c) 0.0225m2 (d) 0.56m2 26. Ans: (c) Sol: Induced voltage, E = 4.44 N Bm A f volts Given data: N = 20, E = 100 V, F = 50 Hz, Bmax = 1 Wb/m2 100 = 4.44  20  1  A  50 A = 0.0225 m2 27. The no-load pf of a transformer is small because (HMWS-12)

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:6: Electrical Machines

(a) iron loss component of I0 is large (b) magnetizing component of I0 is large (c) magnetizing component of I0 is small (d) copper losses are high 27. Ans: (b) Sol: Transformer has poor no-load power factor because it’s magnetic component of current is very high when compared to iron loss component of current. 28. A 230/2300V transformer takes no load current of 5A at 0.25 power factor lagging. The core loss is (HMWS-12) (a) 300.2W (b) 192.5W (c) 212.6W (d) 287.5W 28. Ans: (d) Sol: W0 = V1 I0 cos 0= Iron loss = 230  5  0.25 = 287.5 Watt 29. In a 3-phase delta transformer, one of the phases has burnt up. The transformer will then supply (HMWS-12) (a) zero output (b) rated output (c) 66.67% of its rated output (d) 86.6% of its rated output 29. Ans:( * ) Sol: All answers wrong with one transformer in a - bank burnt, the remaining two can supply

57.7%

  

transformer rating transformers.

1  100  of 3 

without

the

3-

overloading

30. We can find ……… of the transformer with open-circuit test (HMWS-12) (a) copper losses (b) total equivalent resistances (c) turns ratio (d) total equivalent leakage reactance 30. Ans: (c) 31. A transformer has full-load copper loss of 400W. The copper loss at half full-load will be (HMWS-12) (a) 50W (c) 400W 31. Ans: (d) ACE Engg. Publications

(b) 200W (d) 100W

Sol: Cupper losses at half full-load  1  2 



2

 400 = 100 W

32. In short circuit test on a transformer, we generally short circuit (HMWS-12) (a) low voltage winding (b) high voltage winding (c) either low or high voltage winding (d) neither low nor high voltage winding 32. Ans: (a) 33. The function of oil in a transformer is (TRANSCO-AE–12) (a) to provide insulation and cooling (b) to provide protection against lightning (c) to provide protection against short circuit (d) to provide lubrication 33. Ans: (a) 34. The low voltage winding of a 400/230V, 1phase, 50 Hz transformer is to be connected to a 25 Hz, the supply voltage should be (TRANSCO-AE–12) (a) 230 V (b) 460 V (c) 115 V (d) 65 V 34. Ans: (c) V Sol: Bmax  1 . As f is reduced Bmax increases. f Hence the magnetizing component of current increases, which may damage the transformer due to increased current. V Hence 1 must be maintained constant. f V11 V12 230   V2   25  115 V f1 f2 50 35. The core flux of a practical transformer with a resistive load (TRANSCO-AE–12) (a) is strictly constant with load changes (b) Increases linearly with load (c) Increases as the square root of the load (d) decreases with increased load 35. Ans: (c) Sol: Consider the approximate equivalent circuit o of a transformer shown in figure.1 I  I 0 +

Im V1 jXm

+ req jxeq + –   I req  jx  eq  o  V  V 0

Resistive load

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–

core losses neglected Fig.1

:7: APPSC Previous Questions

(a) not possible (b) possible at unity power factor load (c) possible at leading power factor load (d) possible at lagging power factor load 37. Ans: (c) Sol: The approximate expression for voltage regulation of a transformer is Let V the load voltage be the rated value. Then V1  V  I (req + j xeq) As load increases, I increases, V1 increases, and Im the magnetizing current which determines the core flux increases. Hence core flux increases. Its mathematical relation with the load current I is obtained from the following phasor diagram. tan

1

x

eq

r eq



V1  I Fig.2

2 2 I req  x eq

V0o

We have 1/2

2 V1   V 2  I 2 (req2  x eq )  2VI req V1 and hence core flux do increase, as I increases, and the increase is nonlinear. Options (a), (b) and (d) are wrong. Hence (c) should be the answer.

36. In transformers, which of the following statements is valid? (TRANSCO-AE–12) (a) In an open circuit test, copper losses are obtained while in short circuit test, core losses are obtained. (b) In an open circuit test, current is drawn at high power factor. (c) In a short circuit test, current is drawn at zero power factor (d) In an open circuit test, current is drawn at low power factor. 36. Ans: (d) Sol: The Power factor under no load condition is  0.2. The Power factor of T/F under S.C Test is  0.5 to 0.6. 37. In a transformer, zero voltage regulation at full load is (TRANSCO-AE–12) ACE Engg. Publications

V.R 

I(req cos   x eq sin ) V

, ‘+’ sign

for lagging power factors and ‘–’ sign for leading power factors. 0 ≤  ≤ 90 and so both cos and sin are always positive. Other terms in the above expression are also positive. So using the ‘+’ sign (lagging loads) voltage regulation can never be zero. But using the ‘–’ sign (leading loads), regulation can become zero if req cos  = xeq sin . 38. A single-phase transformer has a turns ratio of 1:2, and is connected to a purely resistive load as shown in the figure. The magnetizing current drawn is 1 A, and the secondary current is 1A. If core losses and leakage reactances are neglected, the primary current is (TRANSCO-AE–12) 1: 2

1A

(a) 1.41 A (b) 2 A (c) 2.24 A (d) 3 A 38. Ans: (c) Sol: The load current referred to primary is 2A, and the magnetizing current is 1A. These two have a phase difference of 90 w.r.t each other. Hence the primary current is (2 2  12 ) = 2.24 A. 40. The iron loss in a 100 kVA transformer is 1 kW and full load copper losses are 2 kW. The maximum efficiency occurs at a load of (SSC-JE-12) (a) 70.7 kVA (b) 141.4 kVA (c) 50 kVA (d) 100 kVA 40. Ans: (a) Sol: Iron loss in a transformer = 1kW

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:8: Electrical Machines

Full load copper losses =2 kW 1 Max. efficiency occurs at = 100 kvA 2 = 70.7 kVA 41. Following graph shows the loss characteristics of a sheet of ferromagnetic material against varying frequency f. Pi is the iron loss at frequency f, hysteresis and eddy current losses of the sheet at 100 Hz are (SSC-JE-12) Pi/f Slope = 0.001 0.01 f

(a) 10W, 50W (b) 1W,5 W (c) 1W,10W (d) 10 W,100W 41. Ans: (c) Sol: Pi = Af + Bf2 P  i = A + Bf f By using graph, A = 0.01 B = 0.001 Hysteresis Loss at 100 Hz = Af= 1W Eddy current loss at 100 Hz = Bf2 = (0.001)104 = 10 W 42. Hysteresis losses are present in iron core coil when (SSC-JE-12) (a) the current is unsymmetrical alternating only (b) the current in the coil is d.c. only (c) the current in the coil is sinusoidal only (d) the current in the coil is alternating 42. Ans: (d) 43. Eddy current loss in ferromagnetic core is proportional to (SSC-JE-12) (a) square root of frequency (b) frequency (c) reciprocal of frequency (d) square of frequency 43. Ans: (d) 2 2 2 Sol: Eddy current loss w e  KB m f t ACE Engg. Publications

Where, K =

2 ; 6

Bmax = Maximum flux density. f = frequency of eddy current t = Thickness of lamination 44. The magnetic materials that are used to prepare permanent magnets should have (SSC-JE-12) (a) small hysteresis loop (b) high retentivity (c) low coercive force (d) steeply rising magnetization curve 44. Ans: (d) 45. In a 1-phase transformer, the copper loss at full load is 600 Watts. At half of the full load the copper loss will be (SSC-JE-12) (a) 75 Watts (b) 600 Watts (c) 300 Watts (d) 150 Watts 45. Ans: (d) Sol: Copper loss at full load = 600 W AT half full load = x2(full load loss) 1 =  600 4 = 150 W 46. An autotransformer used with a sodium vapour lamp should have high (SSC-JE-12) (a) leakage reactance of windings (b) VA rating (c) transformation ratio (d) winding resistance 46. Ans: (a) 47. In an auto-transformer, the number of turns in primary winding is 210 and in secondary winding is 140. If the input current is 60 A, the currents in output & in common winding are respectively (SSC-JE-12) (a) 40 A, 100 A (b) 90A, 30A (c) 90 A, 150 A (d) 40 A,20A 47. Ans: (b) 60A Sol:

210V

90A 30A 140V

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:9: APPSC Previous Questions

210  6 = 140  I2  I2 = 90 A 48. A 3-phase transformer has its primary connected in delta and secondary in star. Secondary to primary turns ratio per phase is 6. For a primary voltage of 200 V, the secondary voltage would be (SSC-JE-12) (a) 693 V (b) 1200 V (c) 58 V (d) 2078 V 48. Ans: (d) Sol:

50. Two parallel conductors carrying current in opposite directions will exert on each other (SSC-JE-13) (a) an attractive force (b) a repulsive force (c) an axial force (d) no force 50. Ans: (b) Sol: 1 1 Force b/w them is repulsive

Vph  Y  Vph   

6

 Vph(Y) = 6. Vph() = 6  200 = 1200 V V(Y) Line = 1200 3 V = 2078 V 49. The emf induced in a coil is given by e  N

d dt

Where e is the emf induced, N is the number of turns and d is the instantaneous flux linkage with the coil in time dt. The negative sign in the expression is due to (SSC-JE-13) (a) Hans Christian Oersted (b) Andre-Marie Ampere (c) Michael Faraday (d) Emil Lenz 49. Ans: (d) Sol: The direction of statically induced emf is such that the current due to this emf will flow through a closed circuit in such a direction that it which in turn produce some flux according to Electro Magnetic Theory and this flux must opposes the changes in main field flux which is the cause for production of emf as well as current. This is called Lenz law. ACE Engg. Publications

(1+2



2

weak

Strong flux

F

Repulsion

weak

F

If the two parallel conductors carrying current in opposite direction, then there will be a force of repulsion between them.

52. Silicon content in iron lamination is kept within 5% as it (SSC-JE-13) (a) makes the material brittle (b) reduces the curie point (c) increases hysteresis loss (d) increases cost 52. Ans: (a) 53. The high-voltage and low-voltage winding resistances of a distribution transformer of 100 kVA, 1100/220 volts, 50 Hz are 0.1  and 0.004  respectively. The equivalent resistances referred to high-voltage side and low-voltage side are respectively (SSC-JE-13) (a) 2.504  and 0.2  (b) 0.2  and 0.008  (c) 0.10016  and 2.504  (d) 0.008  and 0.10016  53. Ans: (b) Sol: Equivalent resistance refer to high voltage  0.004 side = 0.1  1 / 5  2 = 0.1 + 0.1 = 0.2 

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: 10 : Electrical Machines

Refer to Low voltage side = 0.004 

0.1 52

= 0.008  54. If the frequency of input voltage of a transformer is decreased keeping the magnitude of the voltage unchanged, then (SSC-JE-13) (a) both hysteresis loss and eddy current loss in the core will increase (b) hysteresis loss will increase but eddy current loss will decrease (c) hysteresis loss will increase but eddy current loss will remain unchanged (d) hysteresis loss will decrease but eddy current loss will remain unchanged 54. Ans: (c) Sol: V1 = Constant, frequency is increased Bm 

V f 

Wh  V1.6f0.6 We  V2 By keeping applied voltage constant, if the frequency of operation is increased in a transformer, then there hysteresis loss increases. Eddy current loss will remain unchanged 55. Power factor of a transformer on no load is poor due to (EPDCL-14) (a) Magnetizing reactance of the transformer (b) Open circuited secondary (c) Low primary winding resistance (d) Low no-load current 56. Ans: (a) Sol: At no load magnetizing current is high magnetizing reactance is low So power factor angle is high power factor is low 57. During short circuit test the core losses are negligible. This is because (EPDCL-14) (a) The voltage applied across the high voltage side is a fraction of its rated voltage and so is the mutual flux (b) The current on the low voltage side is very small (c) The power factor is high (d) Iron becomes fully saturated 57. Ans: (a) ACE Engg. Publications

58. The efficiency of a power transformer at relatively light loads is quite low. This is due to (EPDCL-14) (a) Small copper losses (b) Small secondary output (c) High fixed loss in comparison to the output (d) Poor power factor 58. Ans: (c) Sol: At light load, the cu loss less than iron loss. So iron are relatively high for given output 59. A 2 kVA transformer has iron loss of 150 W and full-load copper loss of 250 W. The maximum efficiency of the transformer would when the total loss is (EPDCL-14) (a) 500 W (b) 400 W (c) 300 W (d) 100 W 59. Ans: (c) Sol: max occurs at variable loss = constant loss Cu loss = 150 W  Total loss = 150 + 150 = 300 W 60. In a auto transformer, power is transferred, through (EPDCL-14) (a) Conduction process only (b) Induction process only (c)Both Conduction and Induction processes (d) Mutual coupling 60. Ans: (c) Sol: In two winding transformer power transferred only by induction. In Auto Transformer power is transferred by both Conduction and Induction process 61. The turns ratio of autotransformer is N ac  1.5 a N bc b

V1

V2 c

Considering equal loads and for the same heating in the windings, the ratio of equivalent resistance of auto transformer to that of two wining transformer (rating V1/V2) is (APSPDCL-14) (a)

1 2

(b)

1 3

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: 11 : APPSC Previous Questions 1 6

(c)

1 9

(d)

61. Ans: (d)

R

K  Rauto  K auto    2W 2 R R2W  K auto  2    K 2W    2

Sol:

2

The waveform of secondary induced emf would be (a) Volts 75V 0.1 0.12

90V

1 N se Where, K2-W = 2 N pe 3 K auto   1.5 2

(b)

 1   2 

2

Volts



Volts 75V 0.1 0.12

(c)

1  R2W =  1 k 

225V

(d)

Volts

 

2 1   R2W 3

62. Ans: (c)

N1 N1=200

N2

c d

0.009 twb 0.06

0.1 < t < 0.12 sec,  =  0  0.009     t  0.12   0.12  0.1 

 0.009  75 V  0.06 0.06  t  0.1 sec, E 2  500   0  0 Volts 0  t  0.06sec, E 2  500  

  0.009   0.02 

0.1  t  0.12 sec, E2  500  

N2=500

=225 V



ACE Engg. 0.009 Publications 0

d dt

0< t< 0.06sec,  

62. The core of a two winding transformer is subjected to magnetic flux variation shown below: (APSPDCL-14)

t

0.1 0.12

30

Sol: E2   N 2 1 R 1 R2W  auto  9 R2W 9

+75V 0.06

2



a

t

0.06

2



b

t

0.1 0.12

30



Method: 2



90V 0.06

2  Rauto  K 2W    1 1   R2W  K auto  2  3  2  3 2 9    2

Rauto

t

0.06

0.06

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0.1

0.12

t (sec)

: 12 : Electrical Machines

So option (c) is the correct choice (Check Diagram) 63. A 3.3 kV/400 V, single phase transformer required 330 V to allow full load current at sc  750 . Its voltage regulation at full load, 0.707 lagging is (APSPDCL-14) (a) 4.33% (b) 8.66% (c) 16.2% (d) 14.3% 63. Ans: (b) Sol: Short circuit voltage, VSC = 330 V % Z = % rated voltage required to produced rated short circuit. V % Z  SC 100 E1 330

= 3.3 k  100 =1% SC = 75  SC = tan– 1

%X %R

%X  % R = 0.288 % %Z %X sin SC   % X = 0.9659% %Z cos SC 

% Reg = %R cos   % X sin  = 0.288  0.707 + 0.96 + 0.9659  0.707 = 8.66 % (OR) VSC cos SC   2  %Reg = E2 SC  75o

2 = 45o VSC = 330 V E2 = 400 % Re g  



330 cos 75o  45o 400



330 3 = 8.66%  400 2

64

In a three phase delta transformer, one phase burns up. The transformer will supply (ISRO-14) (a) 57.7% of its rating (b) zero output (c) 63% of its output rating (d) at full output rating 64. Ans: (a)

ACE Engg. Publications

Sol: All answers wrong with one transformer in a - bank burnt, the remaining two can supply

57.7% 





1  100  of 3 

transformer rating without transformers.

the

3-

overloading

65. In an auto transformer, power is transferred through (ISRO-14) (a) conduction process alone (b) induction process alone (c) both conduction and induction processes (d) mutual coupling 65. Ans: (c) Sol: In an auto-transformer power is transferred through both conduction and induction. There is no electrical isolation between primary & secondary. 66. Buchholz relay is a (ISRO-14) (a) voltage sensitive device (b) current sensitive device (c) frequency sensitive device (d) gas actuated device 66. Ans: (d) Sol: Buchholz relay is used to protect the transformer from all internal faults. It is a gas actuated relay. 67. In Scott connection, if the ratio of the main transformer is k, then the teaser transformer has transformation ratio of (ISRO-14) (a) 2k / 3 (b) 3k / 2 (c) K / 3 (d) K/2 67. Ans: (a) 68. With core type transformers, the limbs are stepped so as to (ISRO-14) (a) reduce the iron material and therefore iron loss (b) provide better cooling (c) reduce the conductor material and therefore I2R loss (d) provide more mechanical strength to the core 68. Ans: (c) 69. While conducting short-circuit test on a transformer the following side is short circuited (ISRO-14)

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: 13 : APPSC Previous Questions

(a) High voltage side (b) low voltage side (c) primary side (d) secondary side 69. Ans: (b) Sol: As rated current is less in H.V side, it is convenient to conduct the test in H.v side by short circuiting the L.V winding terminals. B lags behind H. 70. A delta-star transformer has a phase to phase voltage transformation ratio of a : 1 [ delta phase : star phase]. The line to line voltage ratio of star-delta is given by: (SSC–JE– S1-14) (a)

3 a

(c)

3 a

(b) a (d)

3 1

a 1

70. Ans: (c) Sol:  – Y  a : 1 (phase turns ratio) Y: 1 : a (phase turns ratio) 3 : a (line turns ratio) 71. A 10  resistive load is to be impedance matched by a transformer to a source with 6250  of internal resistance. The ratio of primary to secondary turns of transformer should be: (SSC–JE– S1-14) (a) 10 (b) 15 (c) 20 (d) 25 71. Ans: (d) Sol: According to Max. Power transfer theorem 6250 = K2 × 10  K2 = 625  K = 25 72. Hysteresis is the phenomenon in the magnetic circuit by which (SSC–JE– S2-14) (a) H lags behind B (b) B lags behind H (c) B and H are always same (d) setting up a constant flux is done  72. Ans: (b) Sol: The BH curve(b)for magnetic circuit is Saturation (H) ACE Engg. Publications

Hysteresis

V1 , V2 V1 > V2, the fraction of power transferred inductively is proportional to (SSC–JE– S2-14) (a) V1/ (V1 + V2) (b) V2/ V1 (c) (V1 – V2)/ (V1 + V2) (d) (V1 – V2)/ V1 73. Ans: (d) Sol: Power transferred by inductively is = (1 –k) Ptotal  V    1  2  Ptotal V1  

73. In an autotransformer of voltage ratio

 V1  V2   Ptotal V1  

 

74. Stepped core is used in transformer in order to reduce (SSC–JE– S2-14) (a) volume of iron (b) volume of copper (c) iron loss (d) reluctance of core 74. Ans: (b) Sol: The cross section of transformer core is made of steps instead of square shape due to following advantages. 1. Cruciform core reduces the diameter of circum circle. 2. Due to less dia of circum circle, the insulating material required is less. 3. Length per one turn is less so that amount of copper required for winding can be reduced. 4. Due to above reasons size, weight and cost of transformer is less with cruciform core. 5. Cruciform core has more utilization factor.

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: 14 : Electrical Machines

75. The north pole of a magnet is moved away from a metallic ring. The induced current in the ring flows (SSC–JE– S2-14) (a) clockwise (b) anticlockwise (c) first anticlockwise and then clockwise (d) first clockwise and then anticlockwise 75. Ans: (b) Sol: The direction of induced emf or current can also be find by applying Flat palm rule i.e., If four fingers are placed along the conductor such that the flux enters perpendicular palm and thumb indicates direction of motion of conductor, then the four fingers indicates the direction of induced emf or current.  The induced current in the ring flows anticlockwise. 76. Low voltage windings are placed nearer to the core in the case of concentric windings because (SSC–JE– S2-14) (a) it reduces hysteresis loss (b) it reduces eddy current loss (c) it reduces insulation requirement (d) it reduces leakage fluxes 76. Ans: (c) Sol: In concentric winding procedure LV is placed nearer to the core to get following advantages. 1. If LV winding is placed nearer to the core, then amount of insulation required for the transformer can be reduced. 2. The amount of copper required for windings can also be reduced. 3. Due to above two advantages size, weight and cost of transformer can be reduced. 4. By placing L.V winding nearer to the core, leakage flux in the transformer core can be reduced, there by increases power transfer capability of transformer. 77. If K is the phase-to-phase voltage ratio, then the line-to-line voltage ratio in a 3-phaes Y transformer is (SSC–JE– S2-14) (a) K (b) K / 3 ACE Engg. Publications

(c) 3 K (d) 78. Ans: (c) Sol: K phase to phase voltage ratio. Line-line Y- transformer 3V / V 

3/K

3K

79. A transformer has at full load, iron loss of 900 watts and copper loss of 1600 watts. Then the transformer will have a maximum efficiency at a load of (HMWS-15) (1) 66.6% (2) 125% (3) 75% (4) 133% 79. Ans: (c) Sol: given Iron loss Wi = 900W Copper loss Wcu = 1600W Under maximum efficiency condition Wi  100 = 900  100 =75% Load = Wcu 1600 80. A transformer possesses a percentage resistance and a percentage reactance of 1% and 4% respectively. Its voltage regulation at power factor 0.8 lagging and 0.8 leading would be (HMWS-15) (1) 4.8% and – 1.6% (2) 3.2% and –1.6% (3) 3.2% and –3.2% (4) 2.4% and – 0.8% 80. Ans: (2) Sol: given that %R = 1% %X = 4% Voltage regulation = %Rcos%Xsin + for lag  for lead Now at 0.8 lagging p.f = (1)  0.8 + 4  0.6 = 3.2 % Now at 0.8 leading p.f = 1  0.8  4  0.6 =  1.6 % 81. A 100V/10V, 50VA transformer is converted to 100V/110V auto transformer, the rating of the auto transformer is (HMWS-15) (1) 100 VA (2) 500 VA (3) 110 VA (4) 550 VA 81. Ans: (4) Sol: k 

L.V 100  H.V 110

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: 15 : APPSC Previous Questions

Auto transformer rating = winding

1  two 1  k 

1  50 10 transformer= =550 1 11

VA 82. A transformer takes a current of 0.6 A and absorbs 64 W when the primary is connected to its normal supply of 200v, 50Hz, the secondary being on open circuit. The iron loss component of current is (HMWS-15) (1) 0.32 A (2) 0.43A (3) 1 A (4) 0.2 A 82. Ans: (1) Sol: Given that I0 = 0.6 A, P0 = 64 W, V = 200V, P0 = V1I0cos0 P0  V1  I w  (Iw = I0cos0 iron loss component of current) P 64 Iw  0   0.38 V1 200 83. In case of power transformer, the no load current in terms of full load primary current is (HMWS-15) (1) 40 to 50% (2) 15 to 30% (3) 30 to 40% (4) 3 to 5% 83. Ans: (4) Sol: The no load current of a power transformer is 3% to 5% of full load primary current. 84. The core of a transformer is made of (TGenco-15) (a) silicon steel (b) annealed copper (c) seasoned wood (d) aluminum 84. Ans (a) 85. The core of a transformer is assembled with laminated sheets to reduce (TGenco-15) (a) hysteresis loss (b) eddy-current loss (c) magnetic noise (d) magnetizing current Ans(b) 86. The emf induced in the secondary winding of a 50 Hz single-phase transformer having 1000 turns on its secondary is 222 V. The ACE Engg. Publications

maximum flux density in the core is 0.1 Wb/m2. The cross-sectional area of the core is (TGenco-15) (a) 0.1 m2 (b) 0.01 m2 (c) 1 m2 (d) 0.001 m2 86. Ans: (b) Sol: e.m.f induced per turn 222  0.222V 1000 emf / turn  4.44  Bmax  A net  f

=

0.222 = 4.440.1Anet Anet = 0.01m2

volts / turn

50

87. An additional condition for parallel operation of three-phase transformers over singlephase transformers is that (TGenco-15) (a) the transformers should belong to the same vector group (b) ratios of the winding resistance to resistances for the transformers should be equal (c) the transformers should have the same kVA ratings (d) the transformers should not belong to the same vector group 87. Ans: (a) Sol: Conditions for parallel operating of two transformers are given below. (1) Two transformers have same phase sequence (2) Two transformers are connected to same polarities. (3) Two transformers are belong to same vector group. 88. The inrush current of a transformer at no load is maximum if the supply voltage is switched on (TSTransco-15) (a) at peak voltage value (b) at zero voltage value (c) at half voltage value (d) at 0.866 time voltage value 88. Ans (b) 89. A transformer has negative voltage regulation when its load power factor is (TSTransco-15) (a) Zero (b) Unity (c) Leading (d) Lagging 89. Ans: (c)

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: 16 : Electrical Machines

Sol: Negative voltage regulation is possible only with leading power factor load. 90. The below figure shows two coils with coupling coefficient of 0.6, L1= 0.4H and L2 = 2.5H. The mutual inductance M is equal to (TSTransco-15) M

+ v1

L1

+ L2

v2 –

–

(a) 0.6H (b) 2.9 H (c) 2.1 H (d) 1.45 H 90. Ans: (a) Sol: mututal inductance, M = k L1L 2 Coupling co-efficient, k = 0.06 Self-inductance, L1 = 0.4H Self-inductance, L2 = 2.5H M  0.06 2.5  0.4

M = 0.6H 91. Two transformers, each having iron loss of Pi watts and full-load copper loss of P c, are put to back to back test and full-load current is allowed to flow through the secondaries, the total input power will be (TSTransco-15) (a) 2Pi (b) Pc (c) Pi + Pc (d) 2(Pi + Pc) 91. Ans: (d) Sol: back-to back test (or) sumpner’s test is conducted on transformers for determining the temperature rise. It required two identical transformers. Two transformers, each having iron losses = Pi Two transformers, each having full-load cupper loss = PC The total input power, P = 2[Pi +Pc] 92. The desirable properties of transformer core material are (TSTransco-15) (a) low permeability and low hysteresis loss (b) high permeability and high hysteresis loss (c) high permeability and low hysteresis loss (d) low permeability and hysteresis loss 93. Ans: (c) Sol: Desirable properties of transformers core are ACE Engg. Publications

(i) Permeability of transformer core high (ii) Iron losses in transformer core is less (iii) The resistance of transformer winding is zero (iv)Magnetization curve (or) B-H curve is linear for transformer core 94. The phase difference between any two successive third harmonic voltages in 3- transformer is : (TSSPDCL-15) (a) zero radians (b) /2 radians (c) /3 radians (d) 2/3 radians 94. Ans: (a) Sol: Va = Vmsin3t Vb = Vm sin (3t 360) = Vm sin3t VC = Vm sin 3(t + 120) = Vm sin 3t Phase difference  = 0 0  = 0 95. At full-load of a transformer, the iron loss and copper loss are 3000 W and 4000 W respectively. Then total loss at maximum efficiency is: (TSSPDCL-15) (a) 7000 W (b) 6000 W (c) 8000 W (d) 4000 W 95. Ans: (b) Sol: At maximum efficiency Pcu= Piron Total losses = 2Piron = 2  3000 = 6000 W 96. Two 3- transformers cannot be operated in parallel, if their: (TSSPDCL-15) (a) kVAs are different (b) phase sequences are different (c) % impedances are different (d) voltage ratios are different 96. Ans: (b) 97. A transformer operates 24 hours day at fullload. It’s full-load efficiency is: (TSSPDCL-15) (a) equal all-day efficiency (b) more than all –day efficiency (c) less than all-day efficiency (d) equal to maximum efficiency 97. Ans: (a)

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: 17 : APPSC Previous Questions

98. From an open circuit test on a transformer, the no load power factor was determined. Out of the following choices which is the most likely value. (TSNPDCL-15) (a) 0.9 (b) unity (c) 0.8 (d) 0.4 98. Ans: (d) Sol: No load power factor of transformer varies from 0.2lag to 0.4lag . 99. Consider the following statements (1) Magnetising current in a 1-  transformer is sinusoidal (2) Magnetising currents in the 3- phase supply lines of a 3- phase transformer are sinusoidal (3) Magnetising current in a 1-  transformer is non sinusoidal but the induced voltages are sinusoidal (TSNPDCL-15) The correct statements are (a) (1) and (3) (b) (2) only (c) (2) and (3) (d) (1) only 99. Ans: (c) Sol: statement 1) Magnetising current in a 1- transform is non-sinusoidal i.e., peaky wave. statement 2. Magnetising currents in the 3phase supply lines of a 3- phase transformer are sinusoidal, because 3rd harmonic current can not follow from line to line, which is true . Statement 3: Magnetising current in a 1 Transformer is non-sinusoidal (Peaky) but the induced voltages are sinusoidal, which is true So option ‘C’ is correct 100. In a star-delta connected 3 transformer, supplied with 11 kV on star side, the line

ACE Engg. Publications

current is 20 A. per phase turns ratio is 11. The secondary line voltage and line current are : (TSNPDCL-15) (a) 577 V, 381 A (b) 550 V, 220 A (c) 635 V, 381 A (d) 1 kV, 220 A 100. Ans: (a) V1ph N1  11 Sol: given turns ratio = = N2 V2 ph 11  10 3 V2,ph = = 577 3  11 I 2 ph

I1ph

= 11  I2ph = 11 20

I2 line =

3

1120= 381 A

101. In a transformer the core is laminated to reduce: (TSNPDCL-15) (a) copper losses in the core (b) hysterisis losses only (c) hysterisis and eddy current losses (d) eddy current losses only 101. Ans: (d) Sol: In a transformer the core is laminated to reduce, eddy current losses only. 102. A short circuit test on a 1 , 4 kVA, 200/400 V, 50 Hz transformer gave following results HV side : 15 V, 10 A, 80 W. The percentage regulation on full load unity power factor is : (TSNPDCL-15) (a) 2 (b) 4 (c) 1 (d) –2 102. Ans: (a) Sol: The percentage regulation on full load unity power factor =% R from short circuit test % R WSC

80

 100 = 2% = VA rating  100 = 4000

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