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• Jatin Aggarwal

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Stream / Branch of Engineering :- Electrical Engineering Subject :- Electrical Machines Topic:- Single Phase Transformer Note:- If not mentioned in the question, question belongs to topic - Single Phase Transformer, otherwise topic is mentioned in the question itself Numbering of questions is corresponding to the selection of some questions from my personal question bank , therefore are NOT from 1 to 55

Transformer (Basic)

Q2. (Done) Sub Topic Principle of Transformer Action

Difficulty Level 2

Source Book (EM PSB)

Time to solve (sec) 90

For the circuit shown as in the figure, the turns ratio from Primary to secondary for maximum power transfer shall be _ _ _ _ :1.

(a) 2 Ans:- (b)

(b) 2.5

(c) 3.8

(d) Can not be determined

Range is 2.4 to 2.6

Solution: For maximum power transfer, source and load impedance should be conjugate to each other. Accordingly, real part resistance of both source and load shall be equal. Therefore, R [20∠ 60 °] =

N1 2 ∗R ¿ ] N2

( )

N1 2 ∗R ¿ ] R [10 + j17.32] = N2

( )

N1 2 ∗1.6 =>10 = N2

( )

N N 1 2 10 = =6.25 => 1 =2.5 N2 1.6 N2

( )

Q5. (Done) Sub Topic Losses and Efficiency

Difficulty Level 2

Source Book (EM PSB)

Time to solve (sec) 120

A single phase transformer has eddy current loss of 100W and hysteresis of 120W. For the same supply voltage, if both lamination thickness and operating frequency are reduced by 10%, new value of core loss shall be..............watt. (Take Steinmetz constant = 2.) (a) 214.33

(b) 187.22

(c) 198

(d) 200.45

Ans:- (a),

Range 214.0 to 214.5

Solution: -

Pc =k h V x f 1−x +k e V 2 t 2 (core loss) =(Hysteresis loss)+(Eddy current loss)

where V and f =>supply voltage and frequency, k h&k e=> constants, x =>Steinmetz constant New Hysteresis loss = 120∗( 0.90 )1−2=133.33New Eddy Current Loss = 100∗( 0.9 )2=81.00 So total core loss (new) = 133.33 + 81 = 214.33 watt

Q6. (Done) Sub Topic Losses and Efficiency

Difficulty Level 3

Source Book (EM PSB)

Time to solve (sec) 150

For a 2000/200V, 20KVA, 1-φ transformer, the load current at which maximum efficiency occurs is 90A. Its equivalent resistance is 0.015pu. Then its half rated load efficiency at 0.8 pf shall be _ _ %. (a) 96.17

(b) 98.05

(c) 86,46

(d) 93.49

Ans:- (a) Range 96.0 to 96.30, Solution: Rated current of HV & LV sides are

20000 20000 =10 A & =100 A respectively. Therefore 90A 2000 200

current belongs to secondary winding. So pu current at ηmax =

90 =0.9 pupu cu loss at 100

ηmax =( I 2 pu ) 2 ( r e 2 pu ) = ( 0.9 )2∗0.015 = 0.01215pu As core loss = cu loss at ηmax , therefore core loss = 0.01215pu

ηh alfload∧ pf =0.8 =

=

( HflKVA∗powerfactor)inpu output ¿ output+losses ( Hfl KVA∗Powerfactor ¿inpu ) + ( culossat h flinpu )+ ( Ironlossinpu )

1 ∗0.8 2

(

1 ∗0.8 + ( 0.52∗0.015 ) + ( 0.01215 ) 2

)

= 96.177%

Q14.(Done) Sub Topic Parallel Operation

Difficulty Level 1

Source Book (EM PSB)

Time to solve (sec) 45

Two number, two windings transformers having KVA ratings as 100KVA, and 50KVA with equal voltage ratios are working in parallel. If leakage impedance of first transformer is 0.02 pu, then the pu leakage impedance of second transformer on its own base so that load is shared by these transformers in proportion to their KVA ratings shall be _ _ _ _ _ pu. (a) 0.04 Ans:- (c) ,

(b) 0.01

(c) 0.02

(d) 2

Range 0.02 to 0.02

Solution: - To meet the required condition of sharing of load in proportion to their KVA rating, pu leakage impedance of transformers on their own base should be equal. Accordingly, pu impedance of 2nd transformer should be equal to 2% on its own base.

Q23. (Done) Sub Topic Principle of Transformer Action

Difficulty Level 3

Source Book (EM PSB)

Time to solve (sec) 90

A, B & C are three windings of a transformer having 200, 100 and 60 turns. When a voltage of 100V is applied across A, then voltage of terminal ‘x’ w.r.t ‘y’ as shown in the figure shall be_ _

(a) 80

(b) -80

(c) -20

(d) 20

Ans:- (d), Range 20 to 20 Solution: -

To find out voltage across xy, it is necessary to find out the instantaneous polarities and value of voltage induced across each winding. Polarity making: -If terminal a is marked +ve in the A winding, then current enters at a, flux developed due to this current shall be Clockwise as per Right hand thumb rule. To counter this flux due to current in the ‘A’ winding, NOTIONAL direction of current in winding ‘B’ shall be from z to x and current in winding ‘C’ shall be from w to y. These NOTIONAL direction of currents shall develop flux in the anti-clockwise. As current in winding ‘B’ leaves at terminal x and in the winding ‘C’ leaves at terminal y, So they are marked +ve w.r.t their other terminals. Amplitude of induced voltage: Across windingB=

100 60 ∗100=50 V , Across windingC= ∗100=30V Voltage Vxy: -Now, if 200 200

we traverse from terminal y to x (ywzx) then voltage Vxy is V xy=−30+50=20 V

Q25. (Done) Sub Topic Losses and Efficiency

Difficulty Level 3

Source Book (EM PSB)

Time to solve (sec) 150

In a test for determination of losses of a 440V, 50Hz transformer, the total iron losses were found to be 2500 watts at normal voltage & frequency. When the applied voltage and frequency were 220V and 25Hz, the iron losses were found to be 850 watts. Then Hysteresis losses at normal voltage and frequency shall be _ _ _ watt. (a) 1000

(b) 800

(c) 900

(d) 850

Ans:- (c), Range 900 to 900 Solution: -

( Pc ) TotalIronLoss=Ph ( HysteresisLoss ) + Pe ( EddyCurrentLoss) -- (At normal voltage& frequency) Pc =k h V x f 1−x +k e V 2 ---------- (1) At normal frequency & voltage, Pc = 2500W, So 2500=P h+ P e ------------ (2) 1 x 1 1−x 1 ∗( ) =Ph∗ 2 2 2

()

Sonew Ph =Ph1=Ph∗

So 850=

1250=

()

x+1−x

=

Ph 1 2 Pe New P e =Pe1 =P e∗ = 2 2 4

()

Ph Pe -------------- (3) + 2 4

Ph Pe -----------(2) + 2 2

=>−400=

−Pe =¿ P e=1600 watt, 4

Putting this value of Pe in (2)

2500=P h+ 1600=¿ Ph=900 watt

Q29. (Done) Sub Topic Voltage Regulation

Difficulty Level 2

Source Book (EM PSB)

Time to solve (sec) 120

A 200 KVA, 1-φ transformer has full load Cu loss of 800W and the leakage reactance is 0.032 pu. Then pu voltage regulation of transformer at 0.7 pu load at power factor 0.6 leading shall be _ _ _ _ _ per unit. (a) -0.0151

(b) -0.0162

Ans:- (b), Range -0.0161 to -0.0164

(c) -0.0171

(d) -0.0181

Solution: - puvoltageregulationatleadingpu=

E2−V 2 I 2 r e2 cos θ2−I 2 x e2 sin θ2 = E2 E2

If I 2=x I fullload =x I fl [x => Percentage of full load], E2 is rated voltage So

¿

puvoltageregulation

x I fl r e2 cos θ2−x I fl x e2 sinθ 2 I r I x ¿ x fl e2 cos θ2− fl e2 sin θ2 ¿ x [ r epu cos θ 2−x e2 sin θ2 ] E2 E2 E2

[( ) ( ) ]

r epu =

r e2 (actual ) I fl r e 2( actual ) I 2fl r e 2 (actual ) FullloadCuloss 800 r epu = =0.004 So putting values = = = 200∗1000 E2 E 2 I fl RatedVA ( E2|I fl )

of x = 0.7, r epu =0.004 , x epu=0.032 cos θ 2=0.6 and sin θ2=0.8

puVR=0.7 [0.004∗0.6−0.032∗0.8]

= -0.01624

Q37. (Done) Sub Topic Equivalent Circuit

Difficulty Level 1

Source Book (EM PSB)

Time to solve (sec) 60

A 100 KVA, 1000/400V, 1-φ transformer, when excited at rated voltage on hv side, draws a no load current of 3.0A at 0.5 lagging pf. If it is excited from the lv side at rated voltage, the no load current, power factor and power input shall be

(a) 1.2 Ampere, 0.2 lagging, 96 watt (b) 1.2 Ampere, 0.5 lagging, 240 watt (c) 7.5 Ampere, 0.2 lagging, 600 watt (d) 7.5 Ampere, 0.5 lagging, 1500 watt Ans:(d), Range 0 to 0 Solution: (i)

I at lv side=

3.0∗Turns at hv side 3∗1000 = =7.5 Amp(ii) At no load, effect of leakage turns at lv side 400

reactance shall be negligible. Therefore power factor shall be same both the sides i.e. 0.5 lag (iii)

power

input

shall

be

same

irrespective

powerinput=VI cos θ ( At hv side )=1000∗3∗0.5=1500 watt

Q38.(Done)

of

hv

or

lv

side.

Therefore

Sub Topic Principle of Transformer Action

Difficulty Level 1

Source Book (EM PSB)

Time to solve (sec) 75

A 50 KVA, 6600/230V, 1-φ transformer has hv and lv winding resistances of 7Ω and 0.008Ω respectively. With lv winding open, a current of 0.3A at a pf of 0.3 lagging is recorded on hv side at rated voltage. Then to get maximum efficiency, load current shall be _ _ _ _ _ _ Ampere at the primary side. (a) 189.7

(b) 43.7

(c) 5.69

(d) 6.61

Ans:- (d), Range 6.58 to 6.65 Solution: During open circuit, power drawn is approximately equal to iron loss (if rated voltage is being applied). So Iron Loss Pc is Pc =6600∗0.3∗0.3=594 watt

(

Equivalent resistance at hv side=7 + ( 0.008 )∗

6600 2 =13.58 ΩAt maximum efficiency, Cu loss = 230

)

Iron Loss So

Culoss=I 2∗13.58=594

=> I =

√

594 =6.61 Ampere 13.58

Q40.(done) Topic :- Three Phase Transformer Sub Topic Type and Construction of Three Phase Transformers

Difficulty Level 2

Source Book (GTEM PSB)

Time to solve (sec) 120

A 50Hz, 3-φ, Core type Y /∆ transformer of rating 11KV/440V has square core section core with a circumscribing circle of 0.4m diameter. Then the number of turns per phase on high voltage winding shall be _ _ _ _ _. Maximum flux density is 1.25T and insulator occupies 10% of the core area. (a) 318

(b) 551

Ans:- (a), Range 317 to 318 Solution: -

(c) 286

(d) 496

In the adjacent figure, a circle of dia (0.4m = b) is shown circumscribing the square of side a. Then from right angled triangle opq a b 0.4 0.4 pq= = cos 45° =¿ a=b cos 45 °= , So area of square ¿ 2 2 √2 √2

2

( )

As 10% area is used in insulation. Therefore available net cross sectional area of core is ¿

0.4 2 ∗0.9=0.072 m2=A i √2

( )

emf per turn per p h ase=√ 2 π Bm A i f =4.44∗1.25∗0.072∗50=20. Per phase voltage in the high voltage winding is

So number turns per phase in the winding is

11000 √3

11000 =317.5 ≅ 318 √ 3∗20

Q41. (Done) Sub Topic Principle of Transformer Action

Difficulty Level 2

Source Amit K Jain

Time to solve (sec) 75

Three cores made up of iron(Fe), aluminium(Al) and wood(Wd) have the same dimensions and are wound with the same number of turns. If these cores are fed from the same voltage at a fixed frequency, then corresponding to the amplitude of flux produced in the cores, mark the right option. (a) ϕ fe >ϕ Al> ϕwd (b)ϕ fe ϕfe > ϕwd (d) ϕ Al=ϕ fe=ϕ wd Ans:(d) Range 0 to 0

Solution: Max induced flux=

Applied Voltage √ 2 πfNA

As applied voltage, source frequency, number of turns in the winding and cross sectional are of the core are same. Therefore induced flux shall be same in all the cores.

Q44. (Done) Sub Topic Losses and Efficiency

Difficulty Level 2

Source Amit K Jain

Time to solve (sec) 90

Three cores made up of iron(Fe), aluminium(Al) and wood(Wd) have the same dimensions and are wound with the same number of turns. If these cores are fed from same voltage at a fixed frequency, then corresponding to amplitude of eddy current loss (P e) in the cores, mark the right option. (a) ( Pe ) Al < ( Pe )Fe < ( Pe )wd (b) ( Pe ) Al >( Pe )Fe >( Pe )wd (c) ( Pe ) Fe ( Pe )wd Ans: (b) Range 0 to 0 Solution: In addition to windings, emf is also induced in the transformer cores. This emf cause development of circulating currents which are called eddy current and loss caused by them (i 2r) is called eddy current loss. For the applied voltage being same across all the transformers, induced emf shall also be almost same in all cores. Therefore eddy current loss (

V2 ) shall be inversely proportional to resistance of the core R

material. For the given core dimensions, resistance of core shall be directly proportional to resistivity of core material. As resistivity of Al, Fe and Wd are in increasing order, so eddy current loss shall be opposite to this order.

Q47.(Done) Sub Topic Principle of Transformer Action

Difficulty Level 3

Source Book (EM PSB)

Time to solve (sec) 150

A 1-ϕ transformer is designed to operate at rated voltage and frequency as 230V & 50Hz respectively. If its primary voltage is increased by 10% at no load, then match the change in various operating components as mentioned below: -

Operating components

Change

(list -1)

(list - 2)

i

Operating maximum flux of the core(ϕ m ¿

p

Increases

ii

q

Reduces

iii

Exciting current (i e) Core Loss

r

Remains Same

iv

Operating power factor

v

Effect of third harmonics in the exciting current

(a) i-p,

ii-q,

iii-q,

iv-r,

v-r

(b) i-p,

ii-p,

iii-p,

iv-q, v-p

(c) i-q,

ii-q,

iii-q,

iv-r, v-p

(d) i-q,

ii-p,

iii-p,

iv-q, v-r

Ans: (b) Range 0 to 0 Solution: ϕ m :−¿ As V ∝ϕ m f ; V being increased and f is constant, therefore ϕ mwill also increase. i e:-As component of shunt branch are same, but applied voltage is being increased, so i eshall increase. Or, in other words, to develop larger flux, more exciting current shall be required. Core loss: AsCore loss=k h V x f 1−x +k e V 2. f is constant but voltage V is being increased. Therefore, from above relation, Core loss shall be increased (As x is approximately 1.65). Power factor: - As transformer draws now more exciting current, therefore its power factor shall reduce. Exciting current carries inductive characteristics. Effect of third harmonics on the exciting current: - Due to increase in exciting current, third harmonics shall be more prominent now.

Q49. (Done) Sub Topic Principle of Transformer Action

Difficulty Level 3

Source Amit K Jain

Time to solve (sec) 150

Primary winding of a 1-φ transformer is energised from fixed sinusoidal voltage with secondary open circuited. If the shaded portion of the iron core is removed, then mark the correct match for the change in below mentioned operating components.

Operating components p

(list - 2) Increases

ii

(i e) Core flux

q

Reduces

iii

No load power

r

Remains Same

i

(list -1) Exciting current

Change

factor

(a) i-p, ii-r, iii-q (b) i-p, ii-p, iii-q (c) i-q, ii-p, iii-p (d) i-q, ii-q, iii-p Ans:-(a) Range 0 to 0 Solution: Core Flux: - As V ∝ϕf ; as applied voltage and frequency are fixed in the original core and during the shaded portion removed condition, therefore flux will remain same. Exciting Current: - After removal of shaded portion, in the cut section, reluctance of the magnetic core shall be increased. As mmf =i e N=ϕ∗Rl (ϕ → constant , Rl →incresed ¿ .Therefore to maintain sameϕ, high mmf shall be required, i.e. i eshall be increased. Power factor: - As exciting current increases, so power factor shall be reduced.

Q55.(Done) Sub Topic Losses & Efficiency

Difficulty Level 2

Source Book (EM PSB)

Time to solve (sec) 120

A 100KVA, 11KV/400V, 3-φ transformer has its maximum efficiency of 98% when it delivers 80KVA at upf and rated voltage. Then efficiency of the transformer for rated KVA output at 0.8 pf lagging shall be _ _ _ _ _%. (a) 97.45

(b) 97.75

(c) 97.95

(d) 97.65

Ans:-(a), Range 97.3 to 97.55 Solution: Total losses=

losses =

( 1η −1)∗output=( 0.981 −1)∗80,000=1632.65 wAs at max η,

core losses = ohmic

1632.65 =816.32 w 2

Rated load ohmic losses = So transfer efficiency η=

1 2 ∗816.32=1275.51 w 0.8

( )

100,000∗0.8 ∗100=97.452 % 80,000+ 1275.5+ 816.32

Q57. (Done) Sub Topic Parallel Operation

Difficulty Level 3

Source Book (EM PSB)

Time to solve (sec) 150

Two 1-φ, ideal transformers T1 and T2 of turns ratio 4: 1 and 3: 1 are connected in parallel across a 120V, 50Hz supply. Their secondaries are connected in series and supplying power to a resistor R of 10Ω. Then impedance seen by the source, current and power supplied by the source shall be.

(a) 29.39Ω, 4.08A, 490watt (b) 4.575Ω, 14.28A, 65.3watt (c) 420.3Ω, 1.165A, 490watt (d) 64.1Ω, 1.02A, 65.3watt Ans:-(a) Range 0 to 0 Solution: Voltage induced across ab=

120 120 =30Voltage induced across cd = =40So voltage across ad is 4 3

V ad =V cd +V ab (Polarities of c and a are same), V ad =30+40=70 V So Power consumed by resister ¿

V 2ad 70 2 = 490watt = R 10

Both the transformers are ideal, so power consumption seen by Primary side shall be equal to the power consumption at Secondary side. So power supplied by source is 490watt. Current supplied by source=

Power supplied 490 = =4.08 Amp Voltage of source 120

Impedance seen by source =

120 =29.38 Ω 4.08

Q61. (Done) Sub Topic Transformer as Magnetically Coupled Device

Difficulty Level 4

Source Book (EM PSB)

Time to solve (sec) 180

A single-phase two winding transformer gave the following test results: (i) H.V. winding (590) turns when energised from 230V, 50Hz supply, takes a no-load current of 0.35A and induced e.m.f across open circuited l.v. winding is 110V. (ii) L.V. winding (295 turns) when energised from 115V, 50Hz supply takes a no-load current of 0.72A and induced e.m.f across open circuited h.v. winding is 226V. Then leakage inductance of H.V. side shall be _ _ _ _ _ _ (mH). (a) 974

(b) 1090

Ans:-(d),

(c) 97

(d) 92

Range 90 to 95

Solution: Self inductance L=

ψ i

E=√ 2 π fN ϕmax =√ 2 πf ψ max Here ψ max are the maximum flux-linkages. ∴ Maximum value of flux-linkage with h.v. winding ψ m 1=

L1=

V1

√ 2 πf

=

ψm1 L1 = √2 I

230 Self-inductance of h.v. winding √ 2 π (50 )

[√

230 ∗ 2 π ( 50 )

] [√

E 1 =2.092 H Max value of mutual flux linkages¿ 2 Mutual 2∗0.35 √2 πf

]

inductance M=

E2

1 = 2I

(√ ) (√ ) (√ 2 πf

∗

110 ∗ 2 π ( 50 )

1 ) [ √ 2∗0.35 ]=1 H .

then primary (HV) leakage inductance = L1 – aM where L1 = 2.092H; M = 1H and a=

N 1 590 = =2 N 2 295

Therefore Primary (HV) leakage inductance shall be = 2.092 – (2 * 1) = 0.092H = 92mH

Q63 (Done) Sub Topic Transformer as Magnetically Coupled Device

Difficulty Level 2

Source Book (EM PSB)

Time to solve (sec) 60

Mutual inductance between Primary and Secondary windings in case of a transformer can be increased by (a) Increasing Primary or Secondary Turns (b) Increasing permeance offered to the mutual flux (c) Decreasing leakage flux (d) By adopting any of the above measures Ans:-(d) Range 0 to 0 Solution: As mutual Inductance M is given by M = N1N2ᴧ where N1 and N2=> Py and Sy turns ᴧ -> Permeance So, by adopting any of the above mentioned measure, M can be increased.

Q65(Done) Sub Topic Parallel operation

Difficulty Level 2

Source Book (EM PSB)

Time to solve (sec) 90

Two single phase transformers rated 1000KVA and 500KVA have per unit leakage impedance of (0.02 +j0.06) and (0.025 + j0.08) respectively. What is the largest KVA load that can be delivered by the parallel combination of these transformers without overloading any of them? (a) 1825.3KVA (b) 1377.44KVA (c) 1523.87KVA (d) 1447.56KVA Ans:-(b) , Range 1375 to 1380 Solution: -

|Z e |=|0.02+ j0.06|=0.06325 1

|Z e |=|0.025+ j0.08|=0.0838These values are based on their own KVA ratings. As Z e < Z e , so 2

1

2

transformer 1 reaches its rated KVA first. Therefore largest KVA load on both the transformers shall = 1377.34 KVA

be( KVA )Total =1000+

0.06325 ∗500 0.0838

Q66.(Done) Sub Topic Parallel operation

Difficulty Level 3

Source Book (EM PSB)

During parallel operation of two 1-φ transformers, if |Z e |=|Z e | but a

b

Time to solve (sec) 150

x ea x eb ≠ then which of the r ea r eb

following statements are true (i) The total KVA output is equal to the sum of KVA ratings of individual transformers (ii) The KVA shared by transformers A and B are equal (iii) Transformer with greater leakage impedance angle operates at a poor power factor as compared to the other with a lower leakage impedance angle. (a) All the statements are true (b) Only i and ii are true (c) Only ii and iii are true (d) Only iii is true Ans:-(c), Range 0 to 0 Solution: In the given condition of |Z e |=|Z e | but a

b

x ea x eb ≠ ; KVA shared by transformer A and B shall be r ea r eb

equal. But total KVA output is less than the sum of KVA ratings of individual transformer. Statement (iii) is also correct

Q67. (Done) Sub Topic Autotransformer

Difficulty Level 3

Source Amit K Jain

Time to solve (sec) 150

Equivalent two winding transformer diagram for a step down Auto-transformer with input side voltage & current V1 and I1 AND output side voltage & current V2 and I2 can be shown by one of the below mentioned one figure. Mark the correct figure.

(a) Fig. A (b) Fig. B (c) Fig. C (d) Fig. D Ans:-(d), Range 0 to 0 Solution: One winding diagram of 1-φ auto-transformer (step-down) is given as shown.

transformer, V2< V1 and therefore I2> I1. Accordingly direction of current in AB & BC are marked as transformer action occur in AB and BC section, So AB can be considered as Primary winding and BC can be considered as Secondary & therefore two winding equivalent diagram can be drawn as in the fig D. AB winding has N 1 turns, V1 – V2 voltage across it and I1 current flowing through it. BC winding has N 2 turns, I2 – I1 current flowing through it.

Q69. (Done) Sub Topic Autotransformer

Difficulty Level 2

Source Book (EM PSB)

Time to solve (sec) 120

A 1-φ two winding transformer has 2.5% loss at rated load. If this transformer is used as an auto-transformer, then at rated load of auto-transformer, percentage losses in case of autotransformer shall be: (a) 2.5% (b) Less than 2.5% (c) Less than 2.5% in case of step-down auto-transformer only (d) More than 2.5% Ans:- (b), RANGE 0 to 0 Solution: Whether it is a step-down or step-up transformer, KVA rating of auto-transformer is more than the two winding transformer from which it is made.

( 1−k1 )∗KVA rating of two winding transformerWhere

KVA rating∈case ofautotransformer=

k=

V 1 V2 ∨ and is always less than 1. V 2 V1

As at full load, voltage across same coils and current through them remains same from two winding to auto-transformer. Therefore, losses in the transformer at a given load shall remain irrespective of that they are used as two winding transformer or as auto-transformer. As loss in watt remains same, but KVA base has increased in the case of auto-transformer,

therefore percentage loss reduces in case of auto-transformer irrespective of it being step-up or step-down.

Q70.(Done) Sub Topic Autotransformer

Difficulty Level 2

Source Book (EM PSB)

Time to solve (sec) 90

A 1-φauto-transformer with tapes 0%, 25%, 75% and 100% positions is connected to 400V primary supply and feeds two secondary load resistors connected as shown in the figure. Then the power supplied by 400V source shall be _ _ _ _ _ watt. Consider ideal core and windings.

(a) 450 Ans(b)

(b) 675

(c) 225

(d) 375

Range 675 to 675

Solution: (i) Voltage across 200Ω resistors = 300V So Power drawn by 200Ω resistor ¿

3002 =450 watt 200

(ii) Voltage across 400Ω resistors = 300V So Power drawn by 400Ω resistor ¿

3002 =225 watt 400

In ideal conditions powersupplied by source = Power consumed by load = 450 + 225 = 675 watt So power supplied by source is 675 watt.

Q 72 (Done) Sub Topic Autotransformer

Difficulty Level 3

Source Book (EM PSB)

Time to solve (sec) 180

A 20KVA, 2300/230V, two winding transformer is to be used as an Auto-transformer with 2300/2530V configuration. If the efficiency of two winding transformer at 0.6pf, full load is

96%. Then efficiency of Auto-transformer at same power factor and at full load shall be _ _ _ _ %. (a) 97.36

(b) 86.42

(c) 99.62

(d) 98.85

Ans (c), Range 99.5 to 99.7 Solution: 20000 20000 ) and 86.95A ¿ 2300 230 respectively. Therefore, winding diagram for 2300/2530 configuration becomes as shown below.

(

Rated current in 2300V and 230V windings are 8.695A (=

)

So KVA rating of Auto-transformer becomes = 2530 X 86.95 = 220KVA (When no loss in considered). So input at full load shall be 220 KVA for Auto-transformer for two winding transformer at full load and at 0.6pf Losses=

{ 1η −1}∗output , so Losses in 2- winding = {0.961 −1} X ( 20000) X ( 0.6)

= 500Watt.

For Auto-transformer As in case of Auto-transformer also, windings are at the sated voltage and current, so loss remains same 500Watt for same power factor and load. Therefore, for Auto-tr

η=1−

500 = 99.623% 220000× 0.6+500

Q 74 (done) Sub Topic Autotransformer

Difficulty Level 2

Source Book (EM PSB)

Time to solve (sec) 60

In the figure of Auto-transformer shown, transformer action occurs between the two windings. Mark the correct two windings between which transformer action occurs.

(a) AC and BC (b) AC and AB (c) AB and BC (d) None of above Ans:- (c), Range 0 to 0 Solution: -

Transformer action occurs between winding section AB and BC. To check the same mmf of both the sections may be calculated, if mmf comes out to be same, then transformer action is established. Mmf of AB = I1 (N1 – N2) = I1N1 – I1N2

(as, I1N1 = I2N2)

= I2N2 – I1N2

= (I2 – I1) N2 = mmf of BC

Therefore transformer action occurs between winding AB and BC.

Q75 (done) Sub Topic Autotransformer

Difficulty Level 1

Source Book (EM PSB)

Time to solve (sec) 45

Whether an Auto-transformer is a constant flux device? (a) Yes (b) No (c) Depends upon the configuration (d) Yes, in case of set-up configuration only Ans:-(a), Range 0 to 0 Solution: Auto-transformer is a re-arrangement of two minding transformer, where flux in the core depends upon the applied voltage. As in auto-transformer also, voltage across the windings are rated one, therefore flux remains same and does not vary with the change in load current.

Q77 (Done) Sub Topic Losses and Efficiency

Difficulty Level 4

Source Amit K Jain

Time to solve (sec) 180

The ohmic, hysteresis and eddy current losses in a transformer at 50Hz and 60Hz are shown below:At 50 Hz: Pcu1 = 1.6%; Ph1 = 0.9%; Pe1 = 0.6% At 60 Hz: Pcu2= 1.6%; Ph2= 0.806%; Pe2= 0.6% If total loses at 60 Hz has to be same as the total loses at 50Hz, then the output at 60Hz shall be _ _ _ _ times to the output at 50Hz. Supply voltage to remain same. (a) 1.028

(b) 1.034

(c) 0.993

(d) 0.989

Ans (a) Range – 1.025 to 1.030 Solution: The core loss depends upon voltage and frequency, therefore core loss shall not vary with variation of load and they will remain as per the given data. For the total losses to remain same, ohmic losses alone can vary. Since Total losses at 50Hz = Total losses at 60Hz

1.6 + 0.9 + 0.6 =New Ohmic losses at 60 Hz +0.6 + 0.806 So permissible value of ohmic loss at 60Hz = 3.1 – 1.406 = 1.694% As ohmic loss are proportional to the square of current, so

{

NewPermissibleCurrent ( OriginalC urrent )

} { 2

=

New Permissible Current = √

1.694 NewO h micLoss = OriginalO h micLoss 1.60

}

1.694 * original current = (1.028) (original current) 1.6

Therefore for the same voltage, output at 60Hz = 1.028 (output at 50Hz)

Q78 (Done) Sub Topic Losses & Efficiency

Difficulty Level 4

Source Book (EM PSB)

Time to solve (sec) 180

An 11/0.4 KV, 25Hz, 1-ɸ transformer has ohmic, hysteresis and eddy current losses of 1.8, 0.8 and 0.3% respectively. What do these losses become, if the transformer is operated from 22KV, 50Hz supply system. The current remains same in both the cases. (a) 1.8%, 1.6%, 1.2% (b) 1.8%, 0.8%, 0.6% (c) 0.9%, 1.6%, 1.2% (d) 0.9%, 0.8%, 0.6% Ans:-(d), Range 0 to 0 Solution: At 50Hz, the voltage is doubled (=

( 2211 )), but current remains same, therefore output P at 50Hz 2

is double the output at 25Hz, i.e. base value for loss is doubled. (a) As current is same, cu loss in watt will remain same, however as base value has been doubled, so loss in (%) will be halved i.e.

1.8 =0.9 % 2

(b) Phα vxf1-x , As v and f both are doubled so Phα (2) x (2)1-x α 2x+1-x α 21 α 2 So Ph in watt is doubled but base being doubled, in percentage, it remains same. So Ph new = 0.8% (c) Pe α v2 α (2)2 α 4

So in watts, Pe becomes 4 times to previous value, but as base has been doubled, so in percentage it comes 2 times of earlier value. So Pe new = 2∗0.3=0.6 % So new values of losses are PCu = 0.9%; Ph = 0.8%; Pe = 0.6%

Q80 (Done) Sub Topic Losses & efficiency

Difficulty Level 1

Source Amit Kumar Jain

Time to solve (sec) 45

Eddy current loss in a transformer Pe , at any given voltage depends upon the frequency of supply(f) as :(a) Pe α f (b) Pe α f2 (c) Pe α

1 f

(d) Pe = constant Ans:-(d), Range 0 to 0 Solution: As Pe = Kef2Bm2

---------------- (1)

And voltage of supply α fBm --------(2) From (1) & (2), Pe α V2. So for given supply voltage, i.e. V is constant, therefore Pe shall remain constant.

Q81 (Done) Sub Topic Principle of Transformer Action

Difficulty Level 2

Source Amit K Jain

Time to solve (sec) 120

In a 1-ɸ transformer, if mutual flux (ɸ) is measured with very high resolution measuring equipment, then it is observed that (a) ɸ remains same for all the loads (b) Its marginally high in case of lagging Pf load vis-à-vis in case of loading Pf load (c) Its marginally high in case of leading Pf load vis-à-vis in case of lagging Pf load (d) Its highest at the upf and then reduces for leading or lagging Pf

Ans:-(c), Range 0 to 0 Solution: Due to primary leakage impedance drop, the emf E 1 (or E2) and therefore, the mutual flux ɸ decreases for lagging power factor loads and may increase for leading power factor leads.

Q83(Done) Topic :- Three Phase Transformer Sub Topic Equivalent Circuit

Difficulty Level 3

Source Book (EM PSB)

Time to solve (sec) 180

A 200KVA, 11000/400V, ∆/ϒ distribution transformer gave the following open circuit test result:Open circuit test: - 400V, 9A, 1.5Kw Then core loss resistance (Rc) and magnetising reactance (Xm) referred to hv side shall be respectively:(a) 242.2KΩ and 60.02 KΩ (b) 80.76 KΩ and 20.01 KΩ (c) 106.8 Ω and 26.47 Ω (d) 5.085 KΩ and 1.26 KΩ Ans:-(a), Range 0 to 0 Solution: Open circuit test. This circuit is performed on the l.v. side, ∴ Per p h ase voltage V 1= Per p h ase core loss P c =

400 =231 V .Per p h ase exciting current I e=9 A . √3

1500 =500 W .Now V 1 I e cos θ0 =Pc 3

∴ Corelosscurrent =I e cos θ0=I c =

P c 500 = =2.165 A . V 1 231

Magnetisingcurrent I m=√ I 2e −I 2c = √9 2−(2.165)2=8.73 A . RcL=

X mL=

V 1 231 = =106.8 Ω, I c 2.165

V 1 231 = =26.47 ΩCore loss resistance referred to h.v. side I m 8.73

RcH =RcL

(

X mH =X cL

Perp h asevoltageon h . v . side 2 11,000 2 RcH =106.8 =242.2 kΩ. Perp h asevolatgeon h . v . side 231

(

)

(

)

11,000 2 ( 11,000 2 = 26.47 ) =60.02 kΩ. 231 231

)

(

)

Q84.(Done) Sub Topic Auto Transformer

Difficulty Level 3

Source Book (EM PSB)

Time to solve (sec) 150

A 120:480V, 10KVA transformer is to be used as an auto-transformer as shown in the figure to supply load at 480V side, then KVA rating of Auto-transformer shall be _ _ _ _ KVA.

(a) 30 Ans (b),

(b) 50

(c) 12.48

(d) 40

Range 50-50

Solution: -

Rated current of 120V coil is

10000 =83.33 A ) and of 480V coil is ( =20.8 A ) ( 10000 120 480

Rated current drawn by 120V winding is 83.33 and input voltage is = 480 + 120 = 600V

Therefore rated KVA shall be 600∗83.33=50 KVA

Q85. (Done) Sub Topic Voltage Regulation

Difficulty Level 2

Source Book (EM PSB)

Time to solve (sec) 120

When a short circuit test is conducted on a single phase transformer, 30% of the rated voltage is required to allow full load current. The short circuit power factor is found to be 0.2. Then percentage regulation at UPF is (a) 30

(b) 29.5

(c) 15

(d) 6

Ans:- (d), Range 6 to 6 Solution: As 30% voltage is required to draw rated current diagram short circuit test. Therefore Ze pu = 0.3 and pf in sc is 0.2, Cosθ e = 0.2

Therefore Impedance Triangle becomes as above Repu = ZepuCosθ e = 0.3∗0.2=0.06 So voltage regulation at upf= repu * 1 = 0.06 = 6%

Q86. (Done) Topic :- Three Phase Transformer Sub Topic Type and Construction of Three Phase Transformers

Difficulty Level

Source

Time to solve (sec)

2

Book (GTEM PSB)

90

Three 1-φ, 11000 / 220V transformer are connected to form a 3-φtransformer bank. The high voltage side is connected in star, and low voltage side is in delta. What is the turns ratio of 3-φ transformer? (a) 50 : 1

(b) 50√ 3 : 1

Ans:- (b), Range 0 to 0 Solution: -

(c) 50/√ 3 : 1

(d) 50√ 2: 1

Turns ration in 3-φ transformers is given as ratio of line voltages. As at star connection side Line voltage = P h ase voltage∗√ 3 = 11000∗√ 3 V At Delta connection side Line voltage = Phase voltage = 220V So T /Ratio=

Line voltage at star connection Line voltage at Deltaconnection

=

11000∗√3 = 50√ 3 : 1 220

Q90.(Done) Sub Topic Open Circuit and Short Circuit Test

Difficulty Level 3

Source Amit K Jain

Time to solve (sec) 120

When short circuit test on a transformer is performed at 25V, 50Hz; then drawn current ' I ' 1 is at a lagging pf-angle of ϕ 1. If the test is performed at 25V, 25Hz and the drawn current ' I ' 2 is at a lagging pf-angle of ϕ 2, then (a) I 2> I 1∧ϕ2< ϕ1 (b) I 2< I 1∧ϕ2< ϕ1 (c) I 2> I 1∧ϕ2> ϕ1 (d) I 2< I 1∧ϕ2> ϕ1 Ans (a) Range 0 to 0 Solution: At the reduced supply frequency, value of leakage impedance (x e = 2 πfl) shall reduce, therefore at the same voltage, current drawn (I2 =

V ) shall increase. ℜ+ jxe

As now circuit becomes less inductive (due to reduction of leakage impedance), therefore power factor improves, i.e. power factor angle φ reduces.

Q91.(Done) Sub Topic Losses and Efficiency

Difficulty Level 2

Source Amit K Jain

Time to solve (sec) 90

The efficiency of a transformer at full load 0.8 pf lag is 90%. Its efficiency at full load 0.8 pf lead will be

(a) somewhat less than 90% (b) somewhat more than 90% (c) 90% (d) 91% Ans (a), Range 0 to 0 Solution: As at leading power factor, for the given supply voltage, induced voltage E, increases marginally and therefore flux in the core φ increases, which marginally increase the core loss and therefore cause nominal reduction in efficiency.

Q95. (done) Sub Topic Open Circuit and Short Circuit Test

Difficulty Level 1

Source Book (EM PSB)

Time to solve (sec) 45

Following statements are made regarding the short-circuit test on a 1-phase transformer, correct statements shall make the correct answer: 1. H.V. side is short circuited 2. It is performed at rated current 3. It gives the core loss 4. L.V. side is short-circuited 5. It is performed at rated voltage 6. It gives ohmic loss on the side instruments are connected 7. It helps in the calculation of voltage regulation From these, the correct answer is (a) 1, 2, 6, 7 (b) 2, 3, 4, 7 (c) 4, 5, 6, 7 (d) 2, 4, 7 Ans:-(d), Range 0 to 0 Solution: Given option satisfied the required condition of short circuit test to run rated current while short circuting the LV side, So that amplitude of short circuit current supplied from HV side is less and easily measurable with the available laboratory equipment.

Q99.(done)

Topic :- Three Phase Transformer Sub Topic Tertiary Winding

Difficulty Level

Source

Time to solve (sec)

2

Book (GTEM PSB)

120

The function of using delta-connected tertiary winding in a star-star transformers is (A) to permit the flow of sufficient fault current for the operation of protective devices (B) to permit the single-phase loading (C) to supress harmonic voltages (D) to provide low-reactance path for zero-sequence currents. From above, the correct answer is (a) A, B, C (b) A, B, D (c) B, C, D (d) all Ans:-(d), Range 0 to 0 Solution: (a) Delta winding permits flow of fault current, triplen harmonic currents and zero sequence currents, therefore causing operation of protective device and keeping the emf almost sine wave, so avoiding adverse effect of harmonic voltages. (b) Example of single phase load in case of a 1:1 phase to phase turns ratio Y /∆ Transformer is shown below, mmf balance has been marked

From above, it is evident that 1-φ load can be supplied by ∆ winding.

Q102 (done) Topic :- Three Phase Transformer Sub Topic Open Delta or V-Connection

Difficulty Level

Source

Time to solve (sec)

2

Book (GTEM PSB)

120

Two 100 kVA single-phase transformers are connected in open delta. The maximum balanced 3phase load that the combination can deliver without overloading either unit is.....................KVA (a) 173.2

(b) 200

(c) 300

(d) 115.5

Ans:-(a), Range 173 to 173.5 Solution: Let’s say rated phase voltage & phase current be Vph and Iph for each secondary winding of transformer. Then in closed Delta: Line voltage VL= Vph, IL= √ 3 Iph So VA rating of bank of 3-transformers in delta√ 3 VL IL = 3 Vph Iph It is given that Vph Iph = 100KVA. So capacity of closed Delta configuration is 300KVA In case of open Delta:Line voltage VL= Vph, IL= Iph So VA rating of open delta configuration is = √ 3 VL IL = √ 3 * 100 = 173.2KVA. So open delta configuration can deliver 173.2 KVA.

Q105(done) Sub Topic Excitation Phenomena

Topic :- Three Phase Transformer Difficulty Level

Source

Time to solve (sec)

2

Book (GTEM PSB)

120

For a sinusoidal magnetizing current in a single-phase transformer, the waveforms of flux and induced emfs in both the windings are respectively (a) peaky, sinusoidal (b) peaky, flat-topped

(c) flat-topped, peaky (d) sinusoidal, peaky Ans:-(c) Solution: In a transformer (a) For a sinusoidal applied voltage, the flux is a sine wave and the magnetising current due to saturation is peaky containing a pronounced third harmonic current. This condition holds good if transformer permits the flow of harmonic magnetising current. (b) However, if harmonic magnetising currents are not permitted to flow in the transformer, then for a sinusoidal magnetising current, the flux wave is flat topped and the induced emfs are in primary and secondary windings are peaky. The condition given in the question matches with the second situation.

Q107 (done) Sub Topic Parallel Operation of Three Phase Transformer

Topic :- Three Phase Transformer Difficulty Level 3

Source Book (GTEM PSB)

Time to solve (sec) 120

Consider the following statements: (A) Dy1 and Yd11 can operate in parallel (B) Dy11 and Yd1 can operate in parallel (C) Dy1 and Yd1 can operate in parallel (D) Yd1 and Yz1 can operate in parallel. From above, the correct statement(s) is / are (a) C, D (b) A, B (c) A, C, D (d) all Ans:-(d), Range 0 to 0 Solution: (i) Configuration of same phasor groups can be operated in parallel. Accordingly, as (Dy1 and Yd1) AND (Yd1 and Yz1) belonging to same phasor group can be put for parallel operation. (ii) In exception to above mentioned general rule, transformers with the below mentioned configurations can also be operated in parallel

(a) Dy1 and Yd11 & (b) Dy11 and Yd1

Q108(done) Topic :- Three Phase Transformer Sub Topic Connection and Phasor Groups

Difficulty Level 2

Source Book (GTEM PSB)

Time to solve (sec) 90

Tick the correct combination of phasor groups from the following: (a) Yd6, Yy1 (b) Dd6, Yd1 (c) Yd11, Dy6 (d) Yy11, Dy0. Ans:-(b) Solution: Phasor groups Yy1, Dy6 and Dy0 are not valid. In the option (b), Dd6 and Yd1 are valid groups. Therefore, option (b) is right answer.

Q115. (done) Three Phase Transformer Sub Topic Connection and Phasor Groups

Difficulty Level 3

Source Book (GTEM PSB)

Time to solve (sec) 120

A Yyo transformer is fed from 3-φ supply leads marked as A2 B2 C2. If the primary leads are cyclically interchanged from A2, B2, C2 to B2, C2, A2 then match the following.

i

List-1 Primary

ii

phasors Secondary line

phasors Correct match is (a) i-q, ii-q (b) i-r, ii-r (c) i-p, ii-r (d) i-p, ii-q

Ans:-(d), Range 0 to 0 Solution: -

List-2 line

p

remain unaffected

q r

rotate through a time phase angle of 120° anticlockwise rotate through a time phase of angle of 120° clockwise

Connection diagram and phasor diagram after cyclic change are shown as below

Since the primary voltage those obtained from the supply, their phase orientation cannot change. Now B2 is connected to A2, therefore lv phasor nb2 occupies the position of na2 and so on. In view of this the lv line voltages are displaced by a time phase angle of 120° anti-clockwise.

Q116 (done) Three Phase Transformer Sub Topic Parallel Operation

Difficulty Level 3

Source Book (GTEM PSB)

Time to solve (sec) 180

Two, 3-φ transformers, rated at 500KVA and 450KVA respectively are connected in parallel to supply a load of 1000 KVA at 0.8 pf lagging. Leakage Impedance (pu) of first transformer is 0.025 + j 0.06 and of second transformer is 0.016 + j 0.07. Then the pf at which the transformer 1 operates_ _ _ __ . (a) 0.843

(b) 0.863

(c) 0.883

(d) 0.823

Ans (a) , Range 0.840 to 0.85 Solution: As KVA ratings of both the transformers are different, so their pu leakage impedances are also on different base. To bring them on common base of (say) 500KVA, new leakage impendences are Ze1 (pu) = 0.025 + j 0.06 (As it is already on 500 KVA base) Ze2 (pu) =

500 (0.016 + j 0.07) = 0.0178 + j 0.0778 = 0.0798 ∟77.11 ° 450

(As pu impedance is proportional to the KVA of base) So Ze1 + Ze2 = 0.0428 + j 0.1398 = 0.1443 ∟72.75° S = 1000 ∟ -cos−1 0.8 = 1000 ∟-36.9° So load shared by transformer 1 is given as ´ ´ ´S1 = S∗( Ze 2 ) =1000 ∟−36.9 °∗0.0798∟ 77.11 ´ 1+ Ze ´2 0.1443∟ 72.75 Ze So angle of ´S1 is – 36.9 + 77.11 – 72.75 = -32.54° So transformer 1 is working at lagging pf (due to –ve sign) and cos 32.54 = 0.843

Q118. (done) Three Phase Transformer Sub Topic Parallel Operation

Difficulty Level 3

Source Book (GTEM PSB)

Time to solve (sec) 120

A 3-phase step down transformer is connected in Y /∆ and has a turn ratio per phase of 30 to 1. Then corresponding ratio of a second transformer for parallel operation connected in ∆ /Y to work on the same line voltages on each side shall be? (a) 30:1

(b) 10:1

(c) 90:1

(d) 30√ 3:1

Ans:-(c), Range 0 to 0 Solution: For working of two transformers in parallel, ratio of line voltages should be same. In case of γ /∆ transformer

So, if V voltage at line is applied to Y transformer, at ∆ side, line voltage shall be

V √3∗30

Let’s say per phase voltage ratio, Py to Sy is K: 1 in case of ∆ /γ transformer then

However, for parallel operation, if i/p voltage to both the transformers are same, then o/p voltages should also be same. So

V √3 V = => K = (√ 3)2 * 30 = 90 K √ 3∗30

So required turn in case of ∆ /γ transformer shall be 90: 1

Q119 (done) Three Phase Transformer Sub Topic Excitation

Difficulty Level 3

Source Book (GTEM PSB)

Time to solve (sec) 180

For magnetic saturation in a transformer (without hysteresis), mark the correct match, if transformer is energised with a sinusoidal voltage and it is working at no load:-

i

List -1 magnetising current ¿ i φ

ii

flux φ

iii

Power loss in the transformer

p q

List -2 peaked in nature symmetrical about its maximum value

r s t u

consists of series of add harmonics consists of series of even harmonics due to fundamental current component is zero due to harmonic current components is zero

(a) i-p, q, r;

ii-q;

iii-t, u

(b) i-p, s;

ii-p, q, r;

iii-t

(c) i-s;

ii-p, q;

iii-u

(d) i-p, r;

ii-s;

iii-t, u

Ans:-(a), Range 0 to 0 Solution: Without hysteresis loss, but during magnetic saturation, if transformer is energised with sinusoidal voltage. Then flux induced in the transformer shall also be sinusoidal. However. Energising) current –im shall be drawn corresponding to B-H or (φ-im) curve and the waveform comes out to be as shown below: -

¿ i φis peaky but symmetrical about the centre and consists of series of odd harmonics. But flux φ is symmetrical to its maximum value. In the absence of hysteresis, power drawn from the source is zero. Power associated with fundamental voltage V1 and odd harmonic of ¿ i φ is always zero.

Q123(done) Three Phase Transformer Sub Topic Excitation

Difficulty Level 3

Source Book (GTEM PSB)

Time to solve (sec) 150

The voltage applied to the primary winding of an unloaded 1-φ transformer is given by V = 400 cos wt +100 cos 3 wt

The primary has 500 turns and frequency of the fundamental component of the applied voltage is 50Hz. Then maximum value of flux in mwb shall be: (a) 8.664 (b) 2.334 (c) 2.758 (d) 7.332 Ans:-(b), Range 2.32 to 2.35 Solution: The applied voltage

v = −N

dφ dt

So φ =

−1 vdt N ∫

But as V =400 cos wt +100 cos 3 wt , φ=

−1 (¿ 400 cos wt +100 cos 3 wt)dt ¿ 500 ∫

= =

100 4 1 [ sin wt + sin3 wt ] 500 w 3w 1000 1 4 sin wt + sin3 wt mwb 5∗2 π∗50 3w

[

]

As maximum value of flux occurs at wt = 90° then maximum value of flux φ m=

2 1 4− = 2.334 mwb π 3

[ ]

Q124. (done) Three Phase Transformer Sub Topic Harmonics in Three Phase Transformer

Difficulty Level 3

Source Book (GTEM PSB)

Time to solve (sec) 150

For a 3-φ transformer having star connection arrangement, mark the correct statement (a) Line voltages are free from triplen harmonics but phase voltages contain them (b) Line voltages contain triplen harmonics but phase voltages are free from them (c) Line and phase voltages both contain triplen harmonics (d) Line and phase voltages both do not contain triplen harmonics Ans:-(a), Range 0 to 0 Solution: Triplen harmonic voltages are represented as below for each phase: -

VA3 = Vm3 sin 3 nwt Vb3 = Vm3 sin 3 n ( wt −120° ) = Vm3 sin ( 3 nwt−360° n ) = Vm3 sin ( 3 nwt ) Vc3 = Vm3 sin 3 n ( wt −240 ° ) = Vm3 sin ( 3 nwt−720° n ) = Vm3 sin ( 3 nwt ) From above it is evident that all the triplen harmonic voltages are available in phase. So, triplen voltages of equal magnitude appear in phase and so their scalar sum shall be carried out rather than vector sum.

For line voltage, trace the path anc, then from a to n, there is voltage drop of E 3n and from n to c, there is voltage gain of E3n, so net gain from a to c is zero, i.e. triplen voltages are absent in the line voltages in star configuration.

Q126 (done) Three Phase Transformer Sub Topic Connection and Phasor Groups

Difficulty Level 2

Source Book (GTEM PSB)

Time to solve (sec) 120

The secondary of a transformer in interstar has an output voltage of 400V. If this secondary is reconnected in star, then the output voltage becomes (a) 400V

(b) 461.87V

Ans:-(b), Range 460 to 462 Solution: -

(c) 692.82V

(d) 450V

An interstar winding is shown as below. Half parts of the winding have been shown, if voltage across every half part is E (say) then live

voltage, viz.Va4b4= 3E Therefore as line voltage in interstar is given as 400, so 3E=400 => E =

400 . 3

But in star connection, two half parts make the phase winding, so phase voltage in star connection shall be

2∗400 . As line voltage in Y is = √ 3 phase voltage. So line voltage in Y shall 3

be 2∗√ 3∗400 =461.87 3

Q129. (done) Three Phase Transformer Sub Topic Connection and Phasor Groups

Difficulty Level 2

A delta-zigzag three-phase transformer can have the symbol (a) Dz0 or Dz1 (b) Dz0 or Dz6 (c) Dz1 or Dz11 (d) Dz0 only Ans:-(b), Range 0 to 0 Solution: -

Source Book (GTEM PSB)

Time to solve (sec) 120

Delta-zigzag can have phase difference of 0° or 180°, therefore possible configuration are Dz0 or Dz6.

Q130 (done) Three Phase Transformer Sub Topic Connection and Phasor Groups

Difficulty Level 4

Source Book (GTEM PSB)

Time to solve (sec) 180

Three single phase transformers are connected to form a 3 - φ transformer bank. The transformers are connected in the following manner. The transformer connection will be

represented by. (a) Yy0

(b) Yy6

(c) Yd1

(d) Yd6

Ans(b), Range 0 to 0 Solution:As in primary winding A, B, and C are inter connected, so in phasors also they have to be shown interconnected, so in phasors also they have to be shown interconnected. After drawing primary phasors as discussed above, secondary phasors have to be drawn in parallel to the primary phasors but making connections as shown in the secondary winding diagram

From above, it is evident that there is phase difference of 180°between line voltages of Vc2B2 and vc1b1. Therefore given configuration isYy6.

Q131 Three Phase Transformer Sub Topic Connection and Phasor Groups

Difficulty Level 4

Source Book (GTEM PSB)

Time to solve (sec) 180

Three single phase transformers are connected to form a 3-φ . Transformer bank. The transformers are connected in the following manner: -

The transformer connection will be represented by (a) Dy1 (b) Dy11 (c) Dy0 (d) Dy6 Ans:-(b) Range 0 to 0 Solution: -

(i) Primary phasors A1A2; B1B2; C1C2 are to be connected as shown in the winding diagram in the question. (ii) Thereafter, secondary phasors are to be drawn in parallel to the primary phasors but connections are to be made as shown in the secondary winding in the question.

(iii) Now line voltage are to be compared on both the sides. In both the primary and secondary windings, output is taken from A2B2C2 and a2b2c2. Therefore let’s compare VC2B2 and Vc2b2 phasors. Then angle between Vc2b2 and VC2B2 is 30° and Vc2b2 is leading by 30° to VC2B2. As primary winding is in ∆ and secondary is inγ ; secondary (lv) line phasors are 30° ahead to primary (Hv) phasors. Therefore given configuration is Dy11.

Q135. (done) Three Phase Transformer Sub Topic Open Delta or V-Connection

Difficulty Level 4

Source Book (GTEM PSB)

Time to solve (sec) 180

A bank of three-single-phase transformers has its primary in star and secondary in delta. The primary is fed from 3-wire supply and the secondary per phase voltage is 230V. If the secondary delta

is

opened,

a

voltage

of

380V

is

measured

across

open

delta.

In case the neutrals of primary and sources are connected together, the voltage across opencircuited delta becomes (a) 126.67V (b) zero (c) 103.33V (d) 302.48V

Ans:-(b), Range 0 to 0 Solution: When the neutral of primary and neutral of source are not connected, then the third harmonic current cannot flow in the star connected winding of Yd transformer. So the magnetising current derived from the supply must be a sine wave with other than triplen harmonics. But a sinusoidal magnetising current requires flat topped φ with triplen harmonics emfs in both primary and secondary winding phases. Therefore when voltage across open delta is measured then it is 3 E3 (if per phase 3rd harmonic voltage is E3 ). So 3 E3 =380=¿ E3 =126.69V . However, when source and Y neutrals are connected, then triplen harmonic currents can flow in theY. Therefore magnetising current becomes peaky and flux and voltage induced becomes sinusoidal (i.e. without triplen harmonics), so voltage measured across open delta now shall be zero, as triplen harmonic voltage shall be absent.

Q141(done) Three Phase Transformer Sub Topic Single Phasing in Three Phase Transformer

Difficulty Level 3

Source Book (GTEM PSB)

Time to solve (sec) 150

A 1-φ load is connected between two phases of secondary side of a ∆ /γ transformer, as shown below: -

Load current is 1 Ampere as shown. Phase to phase turns ratio from Primary to Secondary is 1. Then line current in supply wire ‘A’ of Primary shall be..........Ampere. Ideal transformers considered. (a) 1

(b) 2

Ans:- (b), Range 2 to 2

(c) 3

(d) 0

Solution: -

To calculate the line current in ‘A’ supply, wire, mmf balance of load current has to be carried out. For mmf balance, mmf of a particular phase of Primary and Secondary side should be equal. As both the mmf’s counter each other (of Primary and Secondary phases), therefore current direction are shown reversed in respective phases. As turns ratio is 1, therefore ratio of mmf is proportional to current in the phases. Accordingly current in ‘A’ supply line shall be 2 Ampere.

Q145. done Three Phase Transformer Sub Topic Single Phasing in Three Phase Transformer

Difficulty Level 2

Source Book (GTEM PSB)

Time to solve (sec) 150

A Y /∆, 3-φ transformer is being supplied V voltage at the primary side by a feeder. If one supply line feeding the Y gets disconnected, then line voltages at the secondary side shall be? Primary to Secondary phase turns ratio is 1. (a) 0.5V, 0.5V, 0 (b) 0.5V, 0.5V, 0.5V (c) V, 0.5V, 0.5V (d) V, 0, 0.5V Ans:-(a), Range 0 to 0 Solution:If one supply line feeding the Y gets disconnectd thenY is fed from 1-φ supply of line voltage V. Under such conditions, voltage across two phases shall be V/2 and there shall be no voltage across third phase. As primary to Secondary turn’s ratio is 1, therefore same voltage shall appear across phase winding of secondary. This is shown below: -

From above, it is evident that line voltages at the secondary side shall be

V V , and zero. 2 2