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
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