Electrical Machine - I (3140913) MCQs

MCQs of Transformers

Showing 41 to 50 out of 155 Questions
41.
When a 400 Hz transformer is operated at 50 Hz, its kVA rating is.
(a) reduced to 1/16.
(b) reduced to 1/4.
(c) reduced to 1/2.
(d) reduced to 1/8.
Answer:

Option (d)

42.
The primary of a 220/110 V, 50 Hz transformer is connected to a 110V, 60 Hz supply. The secondary output voltage is
(a) 5.5
(b) 55
(c) Zero
(d) 220
Answer:

Option (b)

43.
A 1 φ, 50 Hz transformer is having a voltage ratio of 300/3000 volts. If Emf per turn is 10 volts find primary and secondary number of turns.
(a) 300, 3000
(b) 100, 1000
(c) 30, 300
(d) 300, 30
Answer:

Option (c)

44.
In an ideal transformers
(a) Both primary and secondary induced EMFs are self-induced emf
(b) Both primary and secondary induced EMFs are mutually induced emf
(c) Primary induced emf is self-induced and secondary induced emf is mutually induced
(d) Primary induced emf is mutually induced and secondary induced emf is self-induced
Answer:

Option (c)

45.
If a transformer has 50 primary turns and 10 secondary turns. What is the reflective resistance of the secondary load resistance (250 Ω)?
(a) 25
(b) 6250
(c) 250
(d) 62500
Answer:

Option (b)

46.
An ideal transformer is one which has
(a) No winding resistance
(b) No leakage reactance
(c) No losses
(d) All of the above
Answer:

Option (d)

47.
In an ideal transformer, the magnetizing current lags behind applied primary voltage by an angle of
(a) 90
(b) 75
(c) -90
(d) 45
Answer:

Option (a)

48.
No load primary current in a transformer
(a) lags behind applied voltage somewhat less than 90 degree
(b) lags behind applied voltage by exactly 90 degree
(c) leads the applied voltage by exactly 90 degree
(d) leads the applied voltage somewhat less than 90 degree
Answer:

Option (a)

49.
In one 132/33 kV transformer the LV resistance is 0.02 per unit. The resistance referred to the HV side is
(a) 0.32 pu
(b) 0.02 pu
(c) 0.00125 pu
(d) None of the all
Answer:

Option (b)

50.
If a transformer has turns ration K, the primary and secondary currents are I1 and I2 respectively and magnetizing current and core loss component of no load current Iμ and Iw respectively, then
(a) I1=KI2
(b) I1=KI2+Iw
(c) I1=KI2+Iw+Iμ
(d) I1=I2K+Iw+Iμ
Answer:

Option (c)

Showing 41 to 50 out of 155 Questions