Previous Year Questions- Transformers - 4 | Electrical Machines - Electrical Engineering (EE) PDF Download

Q56: The resistance and reactance of a 100 kVA, 11000/400 V, △-Y distribution transformer are 0.02 and 0.07 pu respectively. The phase impedance of the transformer referred to the primary is       (2004)
(a) (0.02 + j0.07)Ω
(b) (0.55 + j1.925)Ω
(c) (15.125 + j52.94)Ω
(d) (72.6 + j254.1)Ω
Ans: (d)
Sol: Rated kVA = 100kVA
Rated primary line to line coltage
V_l−l = 11000V
Rated primary phase voltage
V_p = V_l−l = 11000V
Rated primary side line current
Rated primary phase current,
Base phase impedance,
Phase impedance = Base phase impedance × (0.02 + j0.07)  

Q57: A 50 kVA, 3300/230 V single-phase transformer is connected as an autotransformer shown in figure. The nominal rating of the auto- transformer will be      (2004)
(a) 50.0 kVA
(b) 53.5 kVA
(c) 717.4 kVA
(d) 767.4 kVA
Ans: (d)
Sol: Consider two winding transformer
When this transformer is connected as step-up auto transformer
KVA rating of the auto transformer

Q58: A 500 kVA, 3-phase transformer has iron losses of 300 W and full load copper losses of 600 W. The percentage load at which the transformer is expected to have maximum efficiency is      (2004)
(a) 50.00%
(b) 70.70%
(c) 141.40%
(d) 200.00%
Ans: (b)
Sol: Copper loass at any load
where, x = fraction of laod
Maximum occurs when
Copper loss-Iron loss

Q59: Figure shows an ideal three-winding transformer. The three windings 1, 2, 3 of the transformer are wound on the same core as shown. The turns ratio N₁ : N₂ : N₃ is 4 : 2 : 1. A resistor of 10 Ω is connected across winding-2. A capacitor of reactance 2.5 Ω is connected across winding-3. Winding-1 is connected across a 400 V, ac supply. If the supply voltage phasor V₁ = 400∠0°, the supply current phasor I₁ is given by        (2003)
(a) (-10 + j10) A
(b) (-10 - j10) A
(c) (10 + j10) A
(d) (10 - j10) A
Ans: (c)
Sol: Turn ratio, N₁ : N₂ : N₃ is 4 : 2 : 1
Induced emf in winding 1 = E₁ ≈ V₁ = 400∠0°V
As we know,
Current in secondary winding
Current in tertiary winding
I₂ referred to primary side
I₃ reffered to primary side
The suuply currrent

Q60: Figure shows a Δ-Y connected, 3-phase distribution transformer used to step down the voltage from 11000 V to 415 V line-to-line. It has two switches S₁ and S₂. Under normal conditions S₁ is closed and S₂ is open. Under certain special conditions S₁ is open and S₂ is closed. In such a case the magnitude of the voltage across the LV terminals a and c is       (2003)
(a) 240V
(b) 480V
(c) 415V
(d) 0V
Ans: (b)
Sol: As the transformer is Δ − Y connected, it means line-to-line voltage on primary side induces line to neutral for phase voltage on secondary side due to transformer.
When S₂ is closed, B and C are at same potential.
So, V_BC = V_B − V_C = 0
Applying KVL in, Δ connection
Voltage across terminals a and c,

Q61: Figure shows an ideal single-phase transformer. The primary and secondary coils are wound on the core as shown. Turns ratio (N₁/N₂) = 2. The correct phasors of voltages E₁, E₂, currents I₁, I₂ and core flux  ϕ are as shown in        (2003)
(a) (b) (c) (d) Ans: (a)
Sol: As per lenz's law,
The positive direction of this emf opposes the positive current direction.]
But,
Because of resistive load, I₂ is in phase with E₂.
But, I_m is very small. On the basis of above analysis, phasor diagrma is  

Q62: A sinple phase transformer has a maximum efficiency of 90% at full load and unity power factor. Efficiency at half load at the same power factor is       (2003)
(a) 86.70%
(b) 88.26%
(c) 88.90%
(d) 87.80%
Ans: (d)
Sol: P_i = 0.055 S
Full load copper loss $=12𝑃𝑐𝑢=𝑃𝑖$= 1²P_cu = P_i
P_cu = 0.055S
Efficiency of halh load, upf

Q63: A 400V/200V/200V, 50 Hz three winding transformer is connected as shown in figure. The reading of the voltmeter, V, will be      (2002)
(a) 0 V
(b) 400 V
(c) 600 V
(d) 800 V
Ans: (a)
Sol: The two 200 turns winding are connected in additive polarity. Hence the output voltage will be 400V. The difference of this 400 V and the voltage induced in first winding (i.e. 400 turn) is same.

Q64: A 1 kVA, 230V/100V, single phase, 50 Hz transformer having negligible winding resistance and leakage inductance is operating under saturation, while 250 V, 50 Hz sinusoidal supply is connected to the high voltage winding. A resistive load is connected to the low voltage winding which draws rated current. Which one of the following quantities will not be sinusoidal?       (2002)
(a) Voltage induced across the low voltage winding
(b) Core flux
(c) Load current
(d) Current drawn from the source
Ans: (d)

Q65: A 3-phase transformer has rating of 20 MVA, 220 kV (star) - 33 kV (delta) with leakage reactance of 12%. The transformer reactance (in ohms) referred to each phase of the L.V. delta-connected side is      (2001)
(a) 23.5
(b) 19.6
(c) 18.5
(d) 8.7
Ans: (b)
Sol: 
Q66: In the protection of transformers, harmonic restraint is used to guard aainst    (2001)
(a) magnetizing inrush current
(b) unbalanced operation
(c) lightning
(d) switching over-voltages
Ans: (a)

Q67: The core flux of a practical transformer with a resistive load      (2001)
(a) is strictly constant with load changes
(b) increases linearly with load
(c) increases as the square root of the load
(d) decreases with increased load
Ans: (d)

Q68: A single-phase transformer is to be switched to the supply to have minimum inrush current. The switch should be closed at       (2001)
(a) maximum supply voltage
(b) zero supply voltage
(c) 1/√2 maximum supply voltage
(d) 1/2 maximum supply voltage
Ans: (a)
Sol: When the value of input voltage is maximum, rate of change of core flux is minimum, as both are 90° out of phase in case of sinusoidal input.