LMRC JE Based Quiz Electrical Engineering(Transformer)

Q1. If the percentage impedance of the two transformers working in parallel are different, then—
(a) Parallel operation will not be possible.
(b) Transformer will be overheated.
(c) Power factors of both the transformers will be same.
(d) Parallel operation will still be possible, but the power factors at which the two transformers operate will be different from the power factor of the common load.

Q2. An ideal transformer has an input voltage of 480V. The output current and voltage are 10A and 120V respectively. Determine the value of input current-
(a) 25 A
(b) 2.5 A
(c) 0.25 A
(d) 0.025 A

Q3. The no-load current drawn by transformer is usually what percentage of the full-load current?
(a) 0.2 to 0.5 %
(b) 12 to 15 %
(c) 20 to 30 %
(d) 2 to 5 %

Q4. The secondary winding of which of the following transformers is always kept closed?
(a) Current transformer
(b) Voltage transformer
(c) Power transformer
(d) Step-down transformer

Q5. Power transformers are designed to have maximum efficiency at–
(a) No – load
(b) 50% load
(c) 80% load
(d) Full load

Q6. Which of the following combination of 3 – ϕ transformers can be successfully operated in parallel–
(a) γ – ∆ and ∆ – ∆
(b) ∆ – ∆ and ∆ – γ
(c) ∆ – γ and ∆ – ∆
(d) ∆ – γ and ∆ – γ

Q7. The leakage flux in a transformer depends upon the value of–
(a) Applied voltage
(b) Load current
(c) Frequency
(d) Mutual flux

Q8. Transformer core are laminated in order to–
(a) reduce copper loss
(b) reduce Hysteresis loss
(c) both a and b
(d) minimize eddy current loss

Q9. If a 500 KVA, 200HZ transformer is operated at 50HZ, its KVA(s) rating will be–
(a) 250 KVA
(b) 125 KVA
(c) 1000 KVA
(d) 2000 KVA

Q10. Which of the following will improve the mutual coupling between primary and secondary circuit?
(a) Transformer oil of high break-down voltage
(b) High reluctance magnetic core
(c) Winding material of high resistivity
(d) Low reluctance magnetic core

Solutions

Transformer

S1. Ans.(d)

S2. Ans.(b)
Sol. V₁I₁ = V₂I₂
⇒ I₁ =

S3. Ans.(d)
Sol. Current required to set up magnetic flux in a magnetic material is rather low. Hence no-load primary current is a very small fraction of the rated full load current

S4. Ans.(a)
Sol. Because if CT secondary is not shorted then it can develop a very high voltage across secondary which may damage transformer insulation.

S5. Ans.(d)
Sol. Power transformers are not directly connected to consumer. Therefore, load fluctuation is very less so transformer can operate on full-load.

S6. Ans.(d)
Sol. For parallel operation phase difference should be zero.

S7. Ans.(b)
Sol. leakage flux × load current
In a transformer, core flux is the difference of primary flux and secondary flux, which are opposite to each other in direction.
Due to increased current, both increases but differences remain same. Hence leakage flux increases.

S8. Ans.(d)
Sol. Because eddy current loss is proportional to the square of the diameter of the core.
By lamination net effective diameter of the core reduces and hence core loss minimized.

S9. Ans.(b)
Sol. For constant load;
KVA ie S is directly proportional to induced emf and emf is directly proportional to frequency
⇒ S × F
So,

⇒ S₂ = 125 KVA

S10. Ans.(d)
Sol. Magnetic Reluctance is same as resistance but it stores energy.
Lower the magnetic reluctance lower will be opposition to flux.

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