Electrical Machine - IES Academy
Electrical Machine - IES Academy Electrical Machine - IES Academy
India’s No. 1 IES Academy Transformer Chapter-1 So, V 1 + V 2 = 2640 ----- (i) V 1 – V 2 = 2160 ----- (ii) From (i) & (ii) V 1 = 2400V & V 2 = 240V ∴ V 1 : V 2 = 10 : 1 IES-18. Ans. (d) V 1 = 2000 – 200 = 1800 V due to opposite polarity connection 1800 × 50 ( kVA ) auto = 1800 × 50 ⇒ I 2 = = 45A 2000 IES-19. Ans. (b) IES-20. Ans. (c) ⎛ V ⎞ L ⎛ 161 ⎞ IES-21. Ans. (d) S trans =Sin ⎜1- ⎟×100% = Sin 1 100% 30% of Sin ⎝ V ⎜ − × = H ⎠ 230 ⎟ ⎝ ⎠ IES-22. Ans. (b) Power transferred conductively = 3 × 0.8 = 2.4 kW. IES-23. Ans. (c) V = 0.08 X 1 pu ∴ V actual = 0.08 x 2000 = 160 V ( 400) 2 IES-24. Ans. (b) R actual = 0.02 × = 0.8Ω 4000 5000 IES-25. Ans. (b) Core-loss component = =22.73A 220 IES-26. Ans. (d) Core losses will be lesser in case of using high frequency ferrite core. IES-27. Ans. (d) P c = 10W and P m = 10 W as no-load components of power are equally divided. 160 V OC( HV) =160 V ⇒ V OC(LV) = =80 V 2 2 2 V V As P= c andP m= R ωL C ⎛ 80 ⎞ ∴ P c (at40Hz)= ⎜ × 10 100 ⎟ = 6.4 W ⎝ ⎠ 2 ⎛ 80 ⎞ 50 and P m ( at40Hz ) = ⎜ × 10 × =8W 100 ⎟ ⎝ ⎠ 40 ∴ W 1 = P c +P m = 6.4 + 8 = 14.4 W 10 Similarly I SC( LV) = 10A ⇒ I SC( HV) = = 5A 2 2 as = I R P cu 2 ⎛5 ⎞ ∴ P cu ( at 40 Hz ) = ⎜ ×25 =25W 5 ⎟ ⎝ ⎠ ∴ W 2 = 25 W IES. 28. Ans. (a) IES-29. Ans. (c) IES-30. Ans. (a) To ensure co-phasor primary and secondary windings. IES-31. Ans. (d) IES-32. Ans. (c) V bd = V bc + V cd = 230∠ 0 ° +115 ∠ -120 ° =115 3 V m 2 www.iesacademy.com E-mail: iesacademy@yahoo.com Page-52 25, 1 st Floor, Jia Sarai, Near IIT. New Delhi-16 Ph: 011-26537570, 9810958290
India’s No. 1 IES Academy Transformer Chapter-1 IES-33. Ans. (a) A delta-connected tertiary reduces the impedance offered to the zero sequence currents thereby allowing a larger earth-fault current to flow for proper operation of protective equipment. Further it limits voltage imbalance when the load is unbalanced. It also permits the third harmonic current to flow thereby reducing third-harmonic voltages. When used for this purpose, the tertiary winding is called a stabilizing winding. IES-34. Ans. (b) V 1 = 11000 × 13V (star-connection) V 2 = 220 V (delta-connection) N1 V1 ∴ V 1 : V 2 = 19052 : 220 V and = = 50 3 N2 V2 IES-35. Ans. (a) IES-36. Ans. (a) 800 IES-37. Ans. (a) S1 = = 484.5kVA ⎛ 1 1 ⎞ 0.0198× ⎜ + 0.0198 0.0304 ⎟ ⎝ ⎠ S = 800 = 800 −S ⎛ 1 1 ⎞ 0.0304× ⎜ + 0.0198 0.0304 ⎟ ⎝ ⎠ 2 1 = 315.5 kVA IES-38. Ans. (a) The currents carried by two transformers (also their kVA loadings) are proportional to their ratings if their ohmic impedances (or their pu impedances on a common base) are 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. IES-39. Ans. (d) IES-40. Ans. (a) 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. IES-41. Ans. (b) Due to transformer action, the circulating current will flow both in the primaries and secondaries. It is usual in practice to keep the circulating current less than 10% of the rated current, consequently the transformer turns ratios must be as nearly equal as is possible. IES-42. Ans. (c) Load sharing not depend on η and magnetizing current. Equivalent impedance also have resistance so option B. Leakage reactance is also not correct. 1 IES-43. Ans. (a) φ = B.A. i.e. A (area of the core) ∝ for constant φ . B www.iesacademy.com E-mail: iesacademy@yahoo.com Page-53 25, 1 st Floor, Jia Sarai, Near IIT. New Delhi-16 Ph: 011-26537570, 9810958290
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India’s No. 1<br />
<strong>IES</strong> <strong>Academy</strong><br />
Transformer<br />
Chapter-1<br />
<strong>IES</strong>-33. Ans. (a) A delta-connected tertiary reduces the impedance offered to the zero<br />
sequence currents thereby allowing a larger earth-fault current to flow for proper<br />
operation of protective equipment. Further it limits voltage imbalance when the<br />
load is unbalanced. It also permits the third harmonic current to flow thereby<br />
reducing third-harmonic voltages. When used for this purpose, the tertiary<br />
winding is called a stabilizing winding.<br />
<strong>IES</strong>-34. Ans. (b) V<br />
1<br />
= 11000 × 13V (star-connection)<br />
V<br />
2<br />
= 220 V (delta-connection)<br />
N1 V1<br />
∴ V<br />
1<br />
: V<br />
2<br />
= 19052 : 220 V and = = 50 3<br />
N2 V2<br />
<strong>IES</strong>-35. Ans. (a)<br />
<strong>IES</strong>-36. Ans. (a)<br />
800<br />
<strong>IES</strong>-37. Ans. (a) S1<br />
= = 484.5kVA<br />
⎛ 1 1 ⎞<br />
0.0198× ⎜ +<br />
0.0198 0.0304<br />
⎟<br />
⎝<br />
⎠<br />
S = 800<br />
= 800 −S<br />
⎛ 1 1 ⎞<br />
0.0304× ⎜ +<br />
0.0198 0.0304<br />
⎟<br />
⎝<br />
⎠<br />
2 1<br />
= 315.5 kVA<br />
<strong>IES</strong>-38. Ans. (a) The currents carried by two transformers (also their kVA loadings) are<br />
proportional to their ratings if their ohmic impedances (or their pu impedances on<br />
a common base) are inversely proportional to their ratings or their per unit<br />
impedances on their own ratings are equal. The ratio of equivalent leakage<br />
reactance to equivalent resistance should be the same for all the transformers. A<br />
difference in this ratio results in a divergence of the phase angle of the two<br />
currents, so that one transformer will be operating with a higher, and the other<br />
with a lower power factor than that of the total output; as a result, the given<br />
active load is not proportionally shared by them.<br />
<strong>IES</strong>-39. Ans. (d)<br />
<strong>IES</strong>-40. Ans. (a) The ratio of equivalent leakage reactance to equivalent resistance should be<br />
the same for all the transformers. A difference in this ratio results in a divergence<br />
of the phase angle of the two currents, so that one transformer will be operating<br />
with a higher and the other with a lower power factor than that of the total<br />
output; as a result the given active load is not proportionally shared by them.<br />
<strong>IES</strong>-41. Ans. (b) Due to transformer action, the circulating current will flow both in the<br />
primaries and secondaries. It is usual in practice to keep the circulating current<br />
less than 10% of the rated current, consequently the transformer turns ratios<br />
must be as nearly equal as is possible.<br />
<strong>IES</strong>-42. Ans. (c) Load sharing not depend on η and magnetizing current. Equivalent<br />
impedance also have resistance so option B. Leakage reactance is also not correct.<br />
1<br />
<strong>IES</strong>-43. Ans. (a) φ = B.A. i.e. A (area of the core) ∝ for constant φ .<br />
B<br />
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