Electrical Machine - IES Academy

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IES Academy
Transformer
Chapter-1
Iron loss does not depend upon the load but upon the applied voltage.
IES-4. Ans. (a)
p1
75 3
IES-5. Ans. (a) η= where, η= =
p2
100 4
∴ p1
9
p
=
2
16
IES-6. Ans. (d) Air-core means non-iron core so there will be no hysteresis losses.
IES-7. Ans. (d)
IES-8. Ans. (d) Magnetizing flux φα V f
but core loss depends upon the frequency.
IES-9. Ans. (a) Load for η
max
= full load
Iron loss
copper loss
IES-10. Ans. (b) Hysterisis loss not depend on thickness of lamination and time.
2 2 2
IES-11. Ans. (d) Eddy current loss P
e
=kB
e maxf t
V
2 2
Bmax
α . So, P
e
= k
e
V f
f
where, t = thickness; f = frequency; and B
max
= flux density
So, P
e
will be same or no change in eddy current loss.
IES-12. Ans. (b) Load for η
max
=
iron loss 1
full load = 100
copper loss 2
= 70.7 kVA
ES-13. Ans. (d) Iron losses not change due to load variations copper losses ∝ (loading in Pu) 2 .
2 2 2
IES-14. Ans. (d) P
e
= K
e
B f t where t is thickness.
IES-15. Ans. (c) A part of the winding being common, leakage flux and therefore leakage
reactance is less.
V
IES-16. Ans. (b) Answer would depend upon the voltage ratio a = 1 and type of autotransformer
i.e., step-up or step-down.
V2
For step-up transformer
⎛ 1 ⎞
( KVA ) = ( KVA)
Auto ⎜ ⎟
[where a < 1]
two-wdg
⎝1-a
⎠
For step-down transformer
⎛ a ⎞
( KVA ) = ( KVA)
Auto ⎜ ⎟
[where a > 1]
two-wdg
⎝a-1⎠
1
Let, N1 = N2 (Ideal two winding transformer) for step-up transformer a= 2
then ( KVA ) = 2 × ( KVA )
auto
two-wdg
For step-down transformer = 2, then ( KVA ) = 2 × ( KVA )
IES-17. Ans. (c) Let V
1
(primary) > V 2
(secondary)
auto
two-wdg
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