Ques.11.A three-phase power transformer is provided with star-delta connections. In order to protect against fault, connection for current transformer should be in (SSC 2010)
Star-star
Delta-star
Delta-delta
Star-delta
Answer.2. Delta-star
Explanation
The currents on the two sides of a 3-phase transformer, in general, differ in both magnitude and phase. The CT connection must compensate for these differences, as a percentage relay must compare currents which, under normal conditions, are identical (in magnitude and phase). In star/star or delta/delta transformers, the currents on the two sides have the same phase angle, the three CTs on both sides are, therefore, identically connected (star or delta) and an appropriate choice of CT ratios would be sufficient for magnitude equalization.
In the case of a star/delta transformer, the currents on the two sides also differ in phase by 30°. This is compensated by connecting CTs in Delta on the star side, and in the star on the delta side of the transformer
Ques.12. 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 2011)
11 W, 20 W
30 W, 45 W
16 W, 30 W
22 W, 40 W
Answer.4. 22 W, 40 W
Explanation
Core loss = Eddy current loss + Hysteresis loss
= Kef2 + Kcf
At 50 Hz, Pc = 46 watt
46 = Ke (50)2 + Kh x 50………. 1
At 70 Hz, Pc = 80 watt
70 = Ke (70)2 + Kh x 70…………… 2
From equation (1) and (2),
Ke = 0.0111
Kh = 0.363
Now at 60 Hz,
Pe = Ke x 602
= 0.0111 x 3600 = 39.96 ≅ 40 Watt
Ph = 0.36306 x 60
= 21.78≅ 22 Watt
Ques.13. A 100 kVA single phase transformer exhibits maximum efficiency at 80% of full load and the total loss in the transform under this condition is 1000 W. The ohmic losses at full load will be (SSC 2011)
781.25 watt
1250 watt
1562.5 watt
12500 watt
Answer.1. 781.25 watt
Explanation
At maximum efficiency
Core loss = Copper loss
i.e Pc = Pcu
Total loss = Pc + Pcu = 1000 watt
Pcu = 500 watt
Copper loss = X2 x Copper loss at full load
Where
X stands for loading condition of the transformer
at 80% of the full load then Pcu at full load
Pcu = 500/0.82 = 781.25 watt
Ques.14. 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 2011)
50%
62.3%
70.7%
100%
Answer.3. 70.7%
Explanation
Pi = iron loss(core loss) = 400 Watt,
Pc = 800 Watt(copper loss),
Maximum efficiency occurs at fraction x of full load such that
x = √(Pi ⁄ Pc) = √(400/800)
=0.707 x 100 = 70.7 %
Ques.15. In a 1-phase transformer, the copper loss at full load is 600 watts. Half of the full load the copper loss will be (SSC 2012)
150 watts
75 watts
600 watts
300 watts
Answer.1. 150 watts
Explanation
Copper loss is proportional to the square of load current. At half load, load current becomes half as voltage remains the same, so the copper loss will become (1/2)2 i.e 1/4 times of full load copper loss.
So (Pcu)full load = 600 watts
(Pcu) half load = 600/4 = 150 watts
Ques.16.In an auto-transformer, the number of turns in the primary winding is 210 and in the secondary winding is 140. If the input current is 60 A, the current in output and in common winding are respectively. (SSC 2012)
40A, 20A
40A, 100 A
90A, 30 A
90A, 150A
Answer.3. 90A, 30 A
Explanation
The auto-transformer ratio is given as
I1/I2 = N2/N1
60/I2 = 140/210
I2 = 90A
Current in common Winding Ic
Ic = (N2 – N1)I1/N2
= 60 x (210 – 140)/140
= 30 A
Ques.17. 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 2012)
2078 V
693 V
1200 V
58 V
Answer.1. 2078 V
Explanation
Transformer ratio is given by
N1/N2 = V1/V2
0r
N2/N1 = V2/V1 6/1 = V2/200 V2 = 1200 volt per phase
For star connection line voltage is
VL = √3 x 1200 = 2078.46 V
Ques.18. 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 2012)
100 KVA
70.7 KVA
141.4 KVA
50 KVA
Answer.2. 70.7 KVA
Explanation
Full load KVA = 100 KVA
Iron loss = 1 KW
Copper Loss = 2 KW
The kVA loading corresponding to maximum efficiency
Ques.19. The iron loss per unit frequency in a ferromagnetic core, when plotted against frequency, is a (SSC 2012)
Constant
A straight line with a positive slope
Straight-line with a negative slope
Parabola
Answer.2. A straight line with a positive slope
Explanation
Both hysteresis loss and eddy current loss give rise to heat in a magnetic circuit. The two losses are usually taken together and are called ‘iron loss’.
Pi = PE + PH
Pi = f2B2max + f B1.6max
Iron losses thus vary with both frequency and magnetic flux density. The power transformer is likely to be fed at the constant frequency and at the constant voltage, which means that magnetic flux density is almost constant. Thus iron losses in a transformer are constant from no load to full load.
Ques.20. The 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 2012)
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