Transformers Questions and Answers – No Load Operation of Transformer

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This set of Transformers Questions and Answers for Entrance exams focuses on “No Load Operation of Transformer”.

1. What is the no-load current drawn by transformer?
a) 0.2 to 0.5 per cent
b) 2 to 5 per cent
c) 12 to 15 per cent
d) 20 to 30 per cent
View Answer

Answer: b
Explanation: The no load current is about 2-5% of the full load current and it accounts for the losses in a transformer. These no-load losses include core(iron/fixed) losses, which contains eddy current losses & hysteresis losses and the copper(I2*R) losses due to the no Load current.
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2. Purpose of no-load test on a transformer is ___________
a) Copper loss
b) Magnetising current
c) Magnetising current and loss
d) Efficiency of the transformer
View Answer

Answer: c
Explanation: No-load current is little bit greater than actual magnetizing current. Total no-load current supplied from the source has two components, one is magnetizing current which is utilized for magnetizing the core and other component is consumed for compensating the core losses in transformer.

3. No-load current in a transformer ________________
a) Lags behind the voltage by about 75°
b) Leads the voltage by about 75°
c) Lags behind the voltage by about 15°
d) Leads the voltage by about 15°
View Answer

Answer: a
Explanation: No-load current lags behind the voltage by an angle which is near to 900. Thus, angle between no-load current and magnetizing current is very small. No-load current has another component which is in phase with voltage.

4. Which of the following statement is true for no-load current of the transformer?
a) has high magnitude and low power factor
b) has high magnitude and high power factor
c) has small magnitude and high power factor
d) has small magnitude and low power factor
View Answer

Answer: d
Explanation: Since no-load current lags voltage by the angle of nearly 900, power factor being equal to cosine of the angle between current and voltage, it will be equal to value which is near to 0. Thus, power factor will be low.

5. In no-load test we keep secondary terminals __________
a) Shorted
b) Shorted via fixed resistor
c) Open
d) Shorted via variable resistors
View Answer

Answer: c
Explanation: In no-load test, as we don’t require any load, we are not allowed to connect any resistor (fixed/variable) to the transformer secondary. We don’t short the secondary terminals either.
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6. Maximum value of flux established in a transformer on load is equal to _________
a) E1/ (4.44*f*N1)
b) E1/ (4.44*f*N2)
c) E2/ (4.44*f*N1)
d) Cannot define
View Answer

Answer: a
Explanation: E1/ (4.44*f*N1). The emf induced in the primary due to applied voltage to primary winding is equal to change in flux with respect to time multiplied by number of turns in the primary. So, by solving this equation we get, E1= (4.44*f*φ*N1).

7. Induced emf in the primary of transformer is equal to terminal voltage applied at primary.
a) True
b) False
View Answer

Answer: a
Explanation: Induced emf in the primary is approximately equal to the applied voltage. Ideally there lies a very small difference in the values, but it is neglected because winding resistance in the transformer is of very small order.

8. For a linear B-H relationship, which option is correct?
a) The exciting current is equal to core loss current
b) The exciting current is equal to magnetizing current
c) The exciting current is equal to de-magnetizing current
d) The exciting current is equal to cross-magnetizing current
View Answer

Answer: b
Explanation: For a linear B-H relationship it is assumed that, there are no losses present in the core like eddy current losses and hysteresis losses are neglected. Thus, core loss current is equal to 0, which ultimately confirms exciting current is purely magnetizing one.

9.Third harmonic current in transformer at no-load is ______________
a) 3% of exciting current
b) 10% of exciting current
c) 25% of exciting current
d) 35% of exciting current
View Answer

Answer: d
Explanation: The effect of saturation nonlinearity is to create a family of odd-harmonic components in the exciting current, the predominant being the third harmonic; this may constitute as large as 35–40% of the exciting current.
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10. Ii in no-load test is responsible for ______________
a) Production of flux
b) Reactive power drawn from the supply
c) Active power drawn from the supply
d) No significance
View Answer

Answer: c
Explanation: It will be assumed here that the current Io and its magnetizing component Im and its core-loss component Ii are sinusoidal on equivalent rms basis. In other words, Im is the magnetizing current and is responsible for the production of flux, while Ii is the core-loss current responsible for the active power being drawn from the source to provide the hysteresis and eddy-current loss.

11. The parallel circuit model is drawn because _________________
a) Conductance Gi accounts for core-loss current
b) Inductive susceptance Bm accounts for magnetizing current
c) Gi for core – loss current and Bm for magnetizing current
d) Cannot say
View Answer

Answer: c
Explanation: The parallel circuit model of exciting current can be easily imagined wherein conductance Gi accounts for core-loss current Ii and inductive susceptance Bm for magnetizing current Im. Both these currents are drawn at induced emf E1 = V1 for resistance-less, no-leakage primary coil; even otherwise E1 =V1.

Sanfoundry Global Education & Learning Series – Transformers.

To practice all areas of Transformers for Entrance exams, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn