Engineering Jobs   »   Quiz: Electrical Engineering 12 May 2021

Quiz: Electrical Engineering 12 May 2021

Quiz: Electrical Engineering
Topic: Transformer

Each question carries 1 mark.
Negative marking: 1/4 mark
Time: 10 Minute

Q1. Total core loss is also referred as
(a) Eddy current loss
(b) Hysteresis loss
(c) Copper loss
(d) Magnetic loss

Q2. Which of the given test determine the iron loss of the transformer?
(a) Short circuit test
(b) Back-to-back test
(c) Open circuit test
(d) Both option a & b

Q3. A 10 kVA, 400 V/ 200 V, single-phase transformer with a percentage resistance of 3% and percentage reactance of 6% is supplying a current of 50 A to a resistive load. The value of the load voltage is
(a) 192 V
(b) 194 V
(c) 196 V
(d) 390 V

Q4. Which of the following is/are the requirements of a good transformer oil?

  1. Low viscosity
  2. Low volatility
  3. High viscosity
  4. High volatility
    (a) Only 1 and 2
    (b) Only 3 and 4
    (c) Only 1 and 4
    (d) Only 2 and 3

Q5. While conducting short circuit test on a transformer the following side is short circuited:
(a) High voltage side
(b) Low voltage side
(c) Primary side
(d) Secondary side

Q6. The frictional loss in a transformer is
(a) 10 %
(b) 50 %
(c) 0 %
(d) More than 50 %

Q7. The iron loss in a 100 KVA transformer is 1 kW and full load copper losses are 2
kW. The maximum efficiency occurs at a load of
(a) 100 KVA
(b) 70.7 KVA
(c) 141.4 KVA
(d) 50 KVA

Q8. The iron loss per unit frequency in a ferromagnetic core, when plotted against
frequency, is a
(a) Straight line with positive slope
(b) Straight line with negative slope
(c) Parabola
(d) Constant

Q9. The secondary winding of which of the given transformer is always kept closed?
(a) Voltage transformer
(b) Current transformer
(c) Step-up transformer
(d) Power transformer

Q10. A 40 KVA transformer has a core loss of 400 W and full-load copper loss of 800 W. The fraction of rated load at maximum efficiency is
(a) 50 %
(b) 62.3 %
(c) 70.7 %
(d) 100 %

S1. Ans.(d)
Sol. Core loss consisting of eddy current loss and hysteresis loss.
i.e., core loss= eddy current loss+ hysteresis loss.
Core loss is mainly due to magnetic properties. So, core loss is also called magnetic loss.
Copper loss is due heat generated due to resistance of winding.

S2. Ans.(c)
Sol. An open circuit test is also known as a no-load test. It is used to find the iron loss at the rated applied voltage to the primary side. Iron loss of the transformer is not varying with a change in load. The iron loss also helps to determine the voltage regulation and efficiency of the transformer. Open circuit test of a transformer is carried out on the low voltage side by keeping the high voltage side open.

S3. Ans.(b)
Sol. % regulation= (R cos⁡ϕ+X sin⁡ϕ)×100
For resistive load: ?=0⁰
∴% regulation=3×1+6×0=3%=200×3/100=6 V
∴Terminal voltage=200-6=194 V

S4. Ans.(a)
Sol. Transformer Oil is basically an Insulating Oil.
An Insulating Oil must have the following properties:
Excellent dielectric properties resulting in minimum power loss.
High resistivity leading to better insulation values between windings.
High flash point and thermal stability facilitating reduction in evaporation losses.
Low viscosity
Low volatility etc.

S5. Ans.(b)
Sol. During short-circuit test of a transformer, low voltage side is short-circuited and during open circuit test, high voltage side is open circuited.

S6. Ans.(c)
Sol. As we know, the transformer is a static device based on the principle of mutual
induction with no rotating part, so there is no frictional loss.

S7. Ans.(b)
Sol. Load for maximum efficiency.
(S_η)max =S_full load √(P_i/P_cufl )
=100 kVA√(1/2)=70.7 kVA

S8. Ans.(a)
Sol. Iron loss given by
∴P_i/f=a+bf………STRAIGHT line with positive slope

S9. Ans.(b)
Sol. The secondary winding of the current transformer is always kept closed because if it open-circuited, the current transformer would develop an extremely high voltage across the secondary terminals. It may lead to damage to the transformer insulation as well as arcing across the terminal.

S10. Ans.(c)
Sol. η=√(Pi/Pcufl )=√(400/800) ×100=70.7 %

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