JEn Questions Bank EE Machine
JEn Questions Bank EE Machine
JEn Questions Bank EE Machine
QUESTION BANK
1. Two transformers of the same type, using the 7. The efficiency of a 100 KVA transformer is 0.98
same grade of iron and conductor materials, are at full as well as at half load. For this transformer
designed to work at the same flux and current at full load the copper loss.
densities; but the linear dimensions of one are two
(a) is less than core loss.
times those of the other in all respects. The ratio
of KVA of the two transformers closely equals : (b) is equal to core loss.
(a) 16 (b) 8 (c) is more than core loss
(c) 4 (d) 2 (d) none of the above
2. When a transformer winding suffers a short-circuit, 8. The magnetising current in a transformer is rich
the adjoining turns of the same winding experience: in :
(a) an attractive force (b) a repulsive force
(a) 3rd harmonic (b) 5th harmonic
(c) no force (d) none of the above
(c) 7th harmonic (d) 13th harmonic
3. Supply to one terminal of a -Y connected
three-phase core type transformer which is on 9. In a constant voltage transformer (CVT), the
no-load, fails. Assuming magnetic circuit symmetry, output voltage remains constant due to :
voltages on the secondary side will be : (a) capacitor (b) input inductor
(a) 230, 230, 115 (b) 230, 115, 115 (c) saturation (d) tapped windings
(c) 345, 115, 115 (d) 345, 0, 345 10. If an ac voltage wave is corrupted with an
4. Auto-transformer is used in transmission and arbitrary number of harmonics, then the overall
distribution : voltage waveform differs from its fundamental
(a) when operator is not available frequency component in terms of :
(b) when iron losses are to be reduced (a) only the peak values
(c) when efficiency considerations can be ignored (b) only the rms values
(d) when the transformation ratio is small (c) only the average values
5. Keeping in view the requirement of parallel (d) all the three measures (peak, rms and
operation, which of the 3-phase connections given average values)
below are possible?
11. A 220/440 V, 50 KVA single phase transformer
(a) delta-delta to delta-star operates on 220 V, 40 Hz supply with secondary
(b) delta-delta to star-delta winding. Then :
(c) star-star to delta-star (a) the eddy current loss and hysteresis loss of
the transformer decrease
(d) delta-star to star-delta
(b) the eddy current loss and hysteresis loss of
6. The laws of electromagnetic induction (Faraday’s the transformer increase
and Lenz’s law) are summarized in the following
(c) the hysteresis loss of the transformer
equation
increases while eddy current loss remains
di the same
(a) e = iR (b) e L
dt
(d) the hysteresis loss remains the same
d whereas eddy current loss decreases
(c) e (d) none of these
dt
jK.f
(c) - jKfIm (d) - I .
m
Vg 2
27. CRGO laminations in a transformer are used to
minimise :
40 Turns
(a) Eddy - Current loss
Ideal Transformer
(b) Hysteresis loss
(a) 20 (b) 40 (c) Both eddy - current and hysteresis loss
(c) 80 (d) 120 (d) Ohmic loss
22. A transformer has at full-load iron loss 900W and 28. The useful flux of a transformer is 1 Wb. When it
copper loss of 1,600W. At what % of load, is loaded at 0.8 pf lag, then its mutual flux
transformer will have maximum efficiency
(a) May decrease to 0.8 Wb
(a) 133% (b) 125%
(b) May increase to 1.01 Wb
(c) 75% (d) 66.6%
(c) Remains constant
23. A transformer is so designed that primary and
secondary have.......... (d) May decrease to 0.99 Wb
(a) High leakage reactance 29. If, in a transformer Pc = core loss and Psc = full
(b) Large resistance
load ohmic loss, then maximum kVA delivered to
(c) Tight magnetic coupling the load at maximum efficiency is equal to rated
(d) Good electric coupling kVA multiplied by
24. Which of the following 3-phase connections of
transformer create disturbances in communication 2
Pc Pc
systems? (a) (b)
Psc Psc
(a) Star/Delta (b) Delta/Star
(c) Star/Star (d) Delta/Delta
Pc Psc
25. Leakage fluxes of a transformer may be minimized (c) (d)
Psc Pc
by
(a) Keeping the magnetizing current of the 30. The hysteresis and eddy - current losses of 1-phase
minimum transformer working on 200 V, 50 Hz supply are
(b) Reducing the reluctance of the iron core to Ph and Pe respectively. The percentage decrease
the minimum in these losses when operated on a 160 V, 40 Hz
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ENGINEERS ACADEMY
62 | Transformers Junior Engineer
supply would respectively be (b) Increase by a factor of 1.2
(a) 32, 36 (b) 20, 36 (c) remain unaltered
(c) 25, 50 (d) 40, 80 (d) decrease marginally
31. In a transformer if primary leakage impedance is 35. A bank of three identical single phase 250 KVA 11
neglected, then kV / 230 V transformer is used to provide 400V
1. Magnetizing current lags the applied voltage Low tension supply from a 11 kV, 3 phase substation
V1 by 900 the effective KVA rating of the bank will be
2. Core loss current lags V1 by 900 (a) 250 (b) 250 3
0
3. Exciting current lags V1 by 90
(c) 500 (d) 750
4. Core loss current is in phase with V1
36. when a short circuit test is conducted on a single
5. Exciting current lags V1 by about 800
phase transformer, 30% of the rated voltage is
6. Magnetizing current statements are
required to allow full load current. The short circuit
from these the correct statement are power factor is found to be 0.2 . The percentage
(a) 1, 4, 5 (b) 3, 4, 6 regulation of UPF is
(c) 1 , 4 (d) 1 , 2 , 6 (a) 30 (b) 29.5
32. A 50 Hz transformer having equal hysteresis and
(c) 15 (d) 6
eddy current losses at rated excitation is operated
at 45 Hz at 90% of rated voltage compared to rated 37. A auto transformer has V1, I1 as input quantities
operating point, the core loss under this condition and V2, I2 as output quantities with V2 < V1 the VA
(a) reduces by 90 % (b) reduces by 19 % conducted from input to output is
(a) Low % age impedance and high I2R loss to 38. A auto transformer has V1, I1 as input quantities
core loss ratio and V2, I2 as output quantities with V2<V1 the VA
(b) High % age impedance and high I2R loss to transformed from primary to secondary is
core loss ratio
(a) V1 I 2 (b) V2 I1
(c) High % age impedance and low I2R loss to
core loss ratio
(c) V1 I1 V2 I 2 (d) V1 V2 I1
2
(d) Low % age impedance and low I R loss to
core loss ratio 39. A 400 V / 200 V transformer has a full load voltage
34. A transformer designed for operation of 60 Hz regulation of x p.u. at 0.8 pf lagging if this
supply is worked on 50 Hz supply system without transformer is used as on auto transformer with
changing its voltage and current ratings. When voltage rating 400 V / 600 V or 200 V / 600 V ,
compared with full load efficiency at 60 Hz, the then its voltage regulation would be
transformer efficiency on full load at 50 Hz will be
(a) Increase marginally
# 100-102, Ram Nagar, Bambala Puliya
Email : info @ engineersacademy.org
Pratap Nagar, Tonk Road, Jaipur-33
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Ph.: 0141-6540911, +91-8094441777
ENGINEERS ACADEMY
EE : Electrical Machines Transformers | 63
V1 V2 V1 V2
x x 2x x (b) K, K
(a) , (b) , V2 V1
3 3 3 3
V1 V2 V1 V2
(c) K, K
V1 V2
2x 2x x 2x
(c) , (d) ,
3 3 3 3 V1 V2 V1 V2
(d) K, K
V2 V2
40. Which of the following statement are incorrect ?
1. Maximum voltage regulation of transformer 43. Two transformer of identical voltage but of different
occurs at leading power factor capacities are operating in parallel for satisfactory
2. Voltage regulation of a transformer is maximum load sharing.
when load power factor (lagging) angle has (a) Impedance must be equal
the same value as the angle of equivalent (b) per unit impedance must be equal
impedance
(c) per unit impedance and X/R ratio must be equal
3. Voltage regulation of a transformer can be
negative at leading power factor. (d) Impedance and X/R ratio must be equal
4. Voltage regulation of a transformer at zero 44. Two transformer to be operating in parallel have
power factor is always zero. their secondary no load emfs Ea for transformer A
and Eb for transformer B. As Ea is somewhat more
(a) 1 and 3 (b) 2and 3
then Eb, a circulating current IC is established at
(c) 2 and 4 (d) 1 and 4 no load which tends to
41. In an auto transformer of voltage ratio V1/V2 and E a Eb
V1 > V2 , the fraction of power transferred (a) boost both Ea and Eb with Ic =
Z ea Z eb
inductively is
E a Eb
(b) boost Ea and buck Eb with Ic =
V V2 Z ea Z eb
(a) V V (b) V
1 2 1
E a Eb
(c) Buck Ea and boost Eb with Ic =
Z ea Z eb
V1 V2 V1 V2
(c) V V (d) V1 E b Ea
1 2 (d) Buck both Ea and Eb with Ic =
Z ea Z eb
42. A single phase transformers with KVA rating k 45. Short circuit test an a single phase transformer gave
V1 the following date
has voltage rating of V this transformer can be
2 30V at 50 Hz, 20 A Pf = 0.2 lagging
connected as an auto transformer to get two If S.C. test is performed on 30V, 25 Hz, then short
V1 V2 V1 V2 circuit current
possible voltage ratings of V2
and V1
the
(a) decrease at Pf < 0.2
respective KVA ratings as an auto transformer are (b) increases at a P.f. < 0.2
ANSWER SHEET
1. Ans. (a) 3. Ans. (b)
KVA E I, Given the two Transformers are of 1:1
same current density.
115V
I = J A, where J is current density which is given 230V
230V
as constant.
I A (cross sectional area of conductor)
As the linear dimensions are doubled (radius of 115V
conductor becomes double)
Cross sectional area of conductor becomes 4 230V
times and hence the current carrying capability.
115V
I2 A 2 4A1
= 4 115V
I1 A1 A1
E = 4.44 Bm A n f Tph
If the primary applied voltage is 230V then for a
E An core type 3-phase –Y transformer with one primary
[An = net area of core material] supply terminal disconnected, secondary side phase
Voltages are of the form 230V, 115V, 115 V.
E2 A 2 4A1 4. Ans. (d)
= 4
E1 A1 A1
Advantages of auto transformer will be more
with % tapping nearer to 100% (or)
[ A2 = L2 b 2 2L1 2b1
transformation ratio nearer to unity i.e., smaller.
4L1b1 4A1 ] In autotransformer transformation ratio is always
[(L.V)/(H.V)].
KVA 2 E 2 I2
= 5. Ans. (d)
KVA1 E1I1
– Y & Y – transformers both belong to ± 30º
16E1I1 phasor group hence they can be used for parallel
= 16 operation.
E1I1
6. Ans. (c)
2. Ans. (a)
Faraday’s Law: induced emf a rate of change
of flux linkages
dN
e
dt
Lenz’s Law: induced emf opposes the basic
According to Lorentz’s principle, two conductors cause which produces it
carrying current in the same direction will
d N
experience force of attraction and the conductors e =
dt
carrying current in opposite direction will
d
experience force of repulsion. e =
dt
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Ph.: 0141-6540911, +91-8094441777
ENGINEERS ACADEMY
EE : Electrical Machines Transformers | 67
7. Ans. (c) 12. Ans. (c)
V
1.6 Wi
1
= Wh1 Wh1 2Wh1
Wh 1 f
f When frequency = 45 Hz & applied voltage =
V11.6 f 0.6 90% of rated voltage
V1 f
Ph1 340w
1 Ph2 120w
V2 f2
100 50
V1 V2
V2 40
f1 f 2
V2 80V
Ph V 1.6 * f 0.6
55. Ans. (c)
Total Iron loss
for lagging load the primary power factor is less 1.6 0.6 2
than the load power factor. V f V
Pi Ph1 2 2 Pe1 2
V1 f1 V1
56. Ans. (b)
V1 240V . f1 60Hz and V2 200V , f 2 50 Hz 1.6 0.6 2
200 50 200
340 120
V1 V2 220 60 200
4 425W
f1 f 2
59. Ans. (c)
V No emf in core
constant
f
eddy current loss = 0W
Ph f 60. Ans. (a)
Given data SC test - 1 : 16 V, 50 Hz
Ph 2 f 2
cos sc = 0.25 lag
Ph1 f1
Isc =35 A
Ph 2 50 30 25W SC test -2 : 16 V , 75 Hz , cos sc = ?
60
For SC test -1
57. Ans. (a)
Vsc 16
Pi 90w 60 Hz Z 01 0.45
I sc 35
Pi 52w 40 Hz
R 01 cos sc Z 01
Bm = constant 0.25 0.457 0.114
Pi 0.9 50 0.01 50
2 0.114
cos sc 0.17l a g
0.654