# Electrical Machines Test - 1 - PDF Flipbook

Electrical Machines Test - 1

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Electrical
Machines

Test-01Solutions

ELECTRICAL MACHINES

1. What is the maximum efficiency (in %) at unity pf load?

a) 95.1

b) 96.2

c) 96.4

d) 98.1

Solution:

% of F.L at which ηmax occurs = � = 0.696

% = 0.696×300×103×1 × 100
0.696×300×103×1+4122+(0.484)8508

= 96.2%

2. The excitation voltage and load angle will respectively be

a) 0.8 pu and 36.86 degrees’ lag

b) 0.8 pu and 36.86 degrees’ lead

c) 1.17 pu and 30.96 degrees’ lead

d) 1.17 pu and 30.96 degrees’ lag

Solution:

V = 10 p.u, Ia = 0.6 p.u at UPF, XS = 1.0 P.U
Ra = 0
Excitation voltage E = V – jIaXs

= 1.0 – j 0.6 × 1
= 1.1662∠-30.960

1

3. Conventional power frequency equivalent circuit of a two-coiled
transformer is obtained on the assumption of
a) Equal primary and secondary ampere turns
b) Negligible mutual reactance
c) Uniform voltage drop along the windings
d) Presence of leakage current between the windings
Solution:
Leakage current between winding (capacitive reactance) is
neglected for power frequency transformer. For high frequency
transformer, leakage current between winding is taken into
account.

4. Two transformers, with equal voltage ratio and negligible
excitation current, connected in parallel, share load in the ratio
of their kVA rating only if their p.u. impedances (based on their
own kVA) are
a) equal
b) in the inverse ratio of their ratings
c) in the direct ratio of their ratings
d) purely reactive
Solution:
The currents carried by two transformers (also their kVA
impedances (or their pu impedances on a common base) are

2

inversely proportional to their ratings or their per unit
impedances on their own ratings are equal. The ratio of
equivalent leakage reactance to equivalent resistance should be
the same for all the transformers.
A difference in this ratio results in a divergence of the phase
angle of the two currents, so that one transformer will be
operating with a higher, and the other with a lower power factor
than that of the total output; as a result, the given active load is
not proportionally shared by them.
5. Armature torque of a d.c motor is a function of which of the
following factors?
1. Speed
2. Field Flux
3. Armature Current
4. Residual Magnetism
Select the correct answer using the codes given below:
a) 2 and 3
b) 1 and 4
c) 3 and 4
d) 1 and 2
Solution:
In dc motor ∝ . = .

3

6. If the wave form of the voltage impressed on the primary of a

Y-Δ bank contains 5th harmonics. What are the wave forms of

the resultant voltages of the primary and the secondary?

Primary Secondary

a) Peaked Peaked

b) Peaked Flat-topped

c) Flat-topped Peaked

d) Flat-topped Flat-topped

7. A separately excited dc machine is coupled to a 50 Hz, three-

phase, 4-pole induction machine as shown in the figure. The dc

machine is energized first and the machines rotate at 1600 rpm.

Subsequently the induction machine is also connected to a 50

Hz, three-phase source, the phase sequence being consistent

with the direction of rotation. In steady state

a) both machine act as generators
b) the dc machine acts as a generator, and the induction machine

acts as a motor
c) the dc machine acts as a motor, and the induction machine

acts as a generator
d) both machines act as motors

4

Solution:

The synchronous speed in induction machine is

Ns = 120×50
4

= 1500 rpm

But the d.c motor is already driving the induction machine at

above rated speed. Hence the induction machine acts as a

generator and the dc machine acts as a motor driving induction

machine.

8. If P1 and P2 are the iron and copper losses of a transformer at
full load, and the maximum efficiency of the transformer is at

75% of the full load, then what is the ratio of P1 and P2?
a) 9/16

b) 10/16

c) 3/4

d) 3/16

Solution:

= � 12

ℎ , = 75 = 3
100 4

∴ 1 = 9
2 16

5

9. A normal Δ - Δ connected 3-phase transformer is obtained by
connecting three single-phase transformers I, II and III as shown
figure.

If transformer III is removed, the system, would not operate in
"open-delta" mode. Under this condition, if voltage Vab is
designed by V∠600, the voltage Vca is
a) V∠600
b) V∠1800
c) V∠-1800
d) V∠-600
Solution:

� ⃗ = � ⃗ + � ⃗
� ⃗ = � ⃗ + � ⃗

6

� ⃗ = −( � ⃗ + � ⃗ )
= −( cos 600 + cos(−600))

�∴ � � ⃗ � = � � ⃗ � = �
� ⃗ = −

� ⃗ = ∠−1800
10. When are eddy-current losses in a transformer reduced?

a) If laminations are thick
b) If the number of turns in primary winding is reduced
c) If the number of turns in secondary winding is reduced
d) If laminations are thin
Solution:

Pe = KeB2f2t2 where t is thickness
11. A d.c series motor is accidentally connected to single-phase a.c

supply. The torque produced will be
a) of zero average value
b) oscillating
d) pulsating and unidirectional
Solution:
In dc series motor, the ac currents through the field and armature
windings will always be in the same direction. So torque will be
unidirectional but pulsating due to ac.

7

12. When one transformer is removed from a Δ-Δ bank of 30 kVA

transformer, the capacity of the resulting 3-phase trans-former

corresponding input line voltages?

a) 11.5 kVA

b) 17.3 kVA

c) 20 kVA

d) 25.9 kVA

Solution:

Capacity of V-V connection

= 1 × (Rating of Δ-Δ connected transformer)
√3

= 1 × 30 = 17.3
√3

Hence, option (b) is correct.

13. The figure given below shows a two winding core-type

transformer. The instantaneous directions of the primary current

I1, the mutual flux ϕm and the primary leakage flux ϕ1, are as

indicated in the figure; the corresponding directions, at the same

instant, of the secondary current I2, and the secondary leakage

flux Q, would be

8

a) I2, a to b; and ϕ2, c to d
b) I2, b to a; and ϕ2, c to d
c) I2, a to b; and ϕ2, d to c
d) I2, b to a; and ϕ2, d to c
Solution:
• Direction of current I2 will be such that it produces a flux that

opposes the flux ϕm. So I2 flows from b to a.
• Leakage flux will only link with secondary but not with

primary. Leakage flux is from c to d.
14. It facilities supply of single-phase loads. Consider the

following statements concerning the utility of mesh-connected
tertiary windings in star-star transformers
1. It is used to suppress harmonic voltages.
2. It is used to allow flow of earth fault current for operation of

protective devices.
3. It facilities supply of single-phase loads.
4. It provides low-reactances paths for zero sequence currents.
Which of these statements are correct?
a) 1, 2, 3 and 4
b) 1, 2 and 3 only
c) 1, 2 and 4 only
d) 3 and 4 only

9

15. In a d.c compound generator, "flat-compound" characteristic,
required for certain applications, may be obtained by connecting
a variable resistance:
a) Across the series field
b) In series with the series field
c) In parallel with the shunt field
d) In series with the shunt field
Solution:
Variable resistance in parallel with the series field winding is
called diverter.

16. A 3-phase synchronous generator is to be connected to the
infinite bus. The lamps are connected as shown in figure for the
synchronization. The phase sequence of bus voltage is R-Y-B
and that of incoming generator voltage is R'-Y'-B'.

It was found that the lamps are becoming dark in the sequence
La - Lb - Lc. It means that the phase sequence of incoming
generator is

10

a) opposite to infinite bus and its frequency is more than infinite
bus

b) opposite to infinite bus but its frequency is less than infinite
bus.

c) same as infinite bus and its frequency is more than infinite
bus.

d) same as infinite bus and its frequency is less than infinite bus.
Solution:
(i) The circuit used for synchronization is shown in fig.1.

Various voltages are shown in this figure with their
reference polarities. This information is essential for a
proper understanding of lamp behavior.

11

(ii) About rotation of the voltage phasors: From the theory of

representation of sinusoidal quantities by phasors all these

phasors are rotating phasors, rotating in acw direction with

peed depending on frequency.
The machine side phase voltage phasors � , � and �
rotate in the acw direction at 2πf rad/s. Since they rotate in

the same direction at the same speed, they are stationary wrt

each other. If we are considering only these phasors, we can

ignore their rotation.
The phasors � ′ , � ′ and � 1 represent voltages at fr Hz;
and so they rotate in acw direction at 2πfr r/s. Unless f = fr;
they rotate at a different speed with respect to � , � and
� phasors. It is this rotation at different speeds which
causes the observed lamp behavior.

(iii) Lamp voltages: By KVL (in fig.1),
� = � − � ′
� = � − � ′
� = � − � ′

(iv) fr > f: Machine set of phasors rotate faster. It is assumed

that supply set of phasors remain stationary while machines

set of phasors rotates anti clockwise with speed (fr – f).

(v) Machine phase-sequence R’Y’B’, fr > f: In figs.2(a), 2(b),

and 2(c), all the voltage phasors are shown at different
times. The symbol � , common to all the voltages, is omitted

for convenience.

12

(vi) Machine phase-sequence R’B’Y’, fr > f: The phasors are
again shown, at different instants of time, in fig.3.

13

This shows that lamps L1, L2 and L3 are becoming dark in

sequence. Hence phase sequence of incoming generator is

opposite to that of the infinite bus and frequency is more than

that of the incoming bus.

17. A separately excited DC generator has an armature resistance

of 0.1Ω and negligible armature inductance. At rated field

current and rated rotor speed, its open circuit voltage is 200 V.

When this generator is operated at half the rated speed, with half

the rated field current, an uncharged 1000μF capacitor is

suddenly connected across the armature terminals. Assume that

the speed remains unchanged during the transient.

At what time (in microsecond) after the capacitor is connected

will the voltage across it reach 25 V?

a) 62.25

b) 69.3

c) 73.25

d) 77.3

Solution:

At half the rated field current and half the rated speed, induced

emf in the armature = 200 = 50
(2×2)

14

−50 + (0.1) + = 0

(0.1) + = 50

=

(0.1) + = 50

1000 × 10−6 � � 0.1 + = 50

10−4 + = 50

+ 104 = 50 × 104

Apply Laplace Transform on both sides

( ) + 104 ( ) = 50×104

( ) = 50×104
( +104)

1 = +
( +104) +104

= 1 , = 1
104 −104

( ) = 50×104 �1 + ( +1104)�
( +104)

( ) = 50 − 50 −104

= 25 , =?
25 = 50 − 50 −104

50 −104 = 25

−104 = ln 0.5

= −(ln 0.5)10−4 = 6.93 × 10−5 = 69.3

15

18. For obtaining very quick braking of a 3-ϕ, wound-rotor
a) a large external resistance has to be inserted in the rotor-
circuit.
b) a large external resistance has to be inserted in the stator-
circuit.
c) interchange any two terminals of the stator supply.
d) interchange any two terminals of the rotor to the slip-rings.
Solution:
For quick braking interchange any two terminals of the stator
supply.

19. A 25 kVA, 2300/230 V 50 Hz, single-phase transformer
supplies rated load at 230 V. The ratio of its leakage reactance
to resistance is √3. The terminal voltage on reducing the load to
zero is observed to be the same as at rated load. What is the
power factor of the rated load?
a) Unity
b) 0.866 lagging
Solution:
Voltage regulation = 0

16

. = ∈ cos ∅ +∈ sin ∅

=0

Given, ∈ = √3

∈ cos ∅ = −∈ sin ∅

− ∈ = cos ∅
sin ∅

tan ∅ = − ∈

= − 1
√3

⇒ ∅ = −300

cos ∅ = 0.866 ( )

20. Match List-I (Name of test) with List-II (Result) and select

List-I

A. Open circuit and short circuit tests

B. Open circuit and zero power factor tests

C. Slip test

D. Maximum lagging current test

List-II

1. Leakage reactance.

2. Direct axis synchronous reactance.

4. Ratio of direct axis synchronous reactance to quadrature axis

synchronous reactance.

17

Codes:

A BCD

a) 1 2 4 3

b) 1 2 3 4

c) 2 1 4 3

d) 2 1 3 4

21. Match List-I (DC machine quantity) with List-II (Relation)

and select the correct answer using the code given below the

lists:

List-I List-II
A. Developed power 1. ∝

B. Torque 2. ∝
C. Generated e.m.f
D. Speed 3. ∝

4. ∝

Codes:

ABCD

a) 2 4 1 3

b) 3 1 4 2

c) 2 1 4 3

d) 3 4 1 2

18

Solution:

Power developed = ×

Torque ‘T’ = ∅

Back EMF ‘ ’ = ∅ hence speed ∝

22. Two transformers of different kVA ratings working in parallel

share the load in proportion to their ratings when their

a) per unit leakage impedances on the same kVA base are the

same.

b) per unit leakage impedances on their respective ratings are

equal.

c) ohmic values of the leakage impedances are inversely

proportional to their ratings.

d) ohmic values of the leakage magnetizing reactances are the

same.

Solution:

Let transformers T1 and T2 have the same voltage ratings (V/E)

but different VA ratings S1 and S2 respectively. Let their leakage
impedances (ref sec) be 1̅ Ω and 2̅ Ω respectively. When

operating in parallel, the secondary circuit will be as shown.

19

� − 1̅ 1̅ = � − 2̅ 2̅ = � , from which 1̅ = � 2 and 1 = 21. If
2̅ � 1 2

(S1/S2) = (Z2/Z1) and they share a load in proportion to their

ratings.

Thus the condition for the transformers to share a load in

proportion to their VA ratings is for the ohmic values of their

impedances to be inversely proportional to their ratings. (c) is

true.

Again, consider the equation I1Z1 = I2Z2 where currents are in

amp and impedances in ohms. Let Z1 pu and Z2 pu be the pu

values based on their respective ratings.

1( ℎ ) = 1 2
1

2( ℎ ) = 2 2
2

∴ 1 1 2 = 2 2 2
1 2

If 1 = 2 ; 1 = 12, and they share the load in proportion to
2

their ratings. (b) is true.

23. Statement (I): The direct on-line (DOL) starter is used to start

a small dc motor.

Statement (II): DOL starter limits initial current drawn by the

armature circuit.

a) Both Statement (I) and Statement (II) are individually true

and Statement (II) is the correct explanation of Statement (I).

b) Both Statement(I) and Statement (II) are individually true but

Statement (II) is not the correct explanation of Statement (I).

20

c) Statement (I) is true but Statement (II) is false.

d) Statement (I) is false but Statement (II) is true.

24. Assertion (A): The short circuit ratio (SCR) of a three-phase

alternator should be high.

Reason (R): A high value of SCR will decrease the value of

voltage regulation and will increase the maximum power output.

a) Both A and B are true and R is the correct explanation of A

b) Both A and R are true but R is NOT the correct explanation

of A

c) A is true but R is false

d) A is false but R is true

Solution:

∵ ∝ 1 . = − × 100%

= ;

For high value of SCR - should be low then V.R. would be

low and would be high.

25. When core material of resistivity 0.1×10-6Ω-m is used in a

transformer, eddy current loss is 100 W. What is the eddy

current loss in the same transformer, if the core material

resistivity is 0.4×10-6Ω-m?

a) 1600 W

b) 400 W

21

c) 25 W

d) 2.5 W

Solution:

Eddy current loss, ∝ 2 2 3ℎ

So, ∝ 1

1 = 2 = 0.4×10−6
2 1 0.1×10−6

100 = 4 ⇒ 2 = 25
2

26. Match List-I (Equivalent circuit parameter) with List-II

(Values) for a 50 MVA three phase alternator and select the

List-I List-II

A. Armature resistance 1. 1 p.u

B. Synchronous reactance 2. 0.1 p.u

C. Leakage reactance 3. 0.01 p.u

Codes:

AB C

a) 1 2 3

b) 3 1 2

c) 3 2 1

d) 1 3 2

22

27. Assertion (A): In a certain case, a d.c shunt generator failed to

build up the voltage.

Reason (R): Shunt field resistance of a d.c shunt generator

should be more than its critical field resistance value to generate

voltage.

a) Both A and R are true and R is the correct explanation of A.

b) Both A and R are true but R is NOT the correct explanation

of A.

c) A is true but R is false.

d) A is false but R is true.

Solution:

If shunt field resistance is more than the critical field resistance,

there will be no voltage build up.

28. The efficiency of a 100 kVA transformer is 0.98 at full as well

as at half load. For this transformer at full load the copper loss

a) is less than core loss.

b) is equal to core loss.

c) is more than core loss

d) none of the above

Solution:

Efficiency = × .
× . + +

Let P.f = unity,

23

= 100 ×1 … ( )
100 ×1+ + … ( )

ℎ = 50 ×1
50 ×1+ +41

Solving (1) & (2) = 2

29. The figure shows the speed torque characteristics of the

following dc motors.

a) Shunt motor
b) Series motor with weak shunt field
c) Cumulatively compound motor
d) Differentially compound motor
The no-load speed of all the motor is same.
Match the type of motors with corresponding curve using the
codes given below:
Codes:

ABC D
a) 1 4 3 2
b) 2 3 4 1
c) 1 3 4 2
d) 2 4 3 1

24

Solution:
4 → series motor
3 → cumulatively compound
2 → shunt motor
1 → differentially compound as ϕsh and ϕse are in opposite
direction
30. The internal characteristic of a dc generator is plotted between
the
a) armature current and voltage generated after armature

reaction.
b) field current and voltage generated at no load.
c) field current and voltage generated on load.
d) armature current and voltage generated at the output

terminals.