Transformers Questions and Answers

Answer: C

A step up transformer increases Voltage.

The secondary has more turns than the primary, so the secondary voltage is higher in the ratio of the turns, \(V_{2}/V_{1} = N_{2}/N_{1}\).

Power cannot increase: output power equals input power minus losses, so the current falls as the voltage rises. The power factor is set by the load, and the frequency is the same on both sides, because both windings link the same alternating flux.

Answer: A

In a step down transformer, Secondary turns are less than Primary turns.

The voltage ratio equals the turns ratio, so a lower secondary voltage requires fewer secondary turns. This is the defining feature of a step down transformer.

Secondary power being less than primary power is true of every transformer because of losses, not a feature of stepping down. The phase shift between primary and secondary depends on how the terminals and windings are connected and is not always 180°. Secondary current is more than primary current only on load; at no load the secondary current is zero, so 'always' makes the last option false.

Answer: C

In high frequency transformers Ferrite cores are used.

Eddy current loss rises with the square of frequency, so ordinary laminated steel becomes very lossy at high frequencies. Ferrites are ceramic magnetic materials with very high resistivity, so eddy currents in them are negligible, while their permeability is still high enough to guide the flux.

Carbon and wood are not magnetic materials, and aluminium is a non-magnetic conductor that would itself carry large eddy currents.

Answer: A

The leakage flux depends upon Load current.

Leakage flux is the flux of each winding that does not link the other. It is produced by the winding's own current and passes largely through air, so it is proportional to that current: zero at no load and largest at full load. That is why leakage reactance drop grows with load.

The mutual flux in the core is the part set by voltage and frequency. Leakage flux follows the load current, not the voltage, frequency or power factor.

Answer: A

Transformer oil is used To provide insulation and cooling.

Mineral oil has high dielectric strength, so it insulates the windings from each other and from the tank, and it fills gaps that would otherwise hold air. It also carries heat from the core and windings to the tank walls and radiators by convection.

Lightning protection comes from arresters, and short circuit protection from relays and circuit breakers. A transformer has no moving parts, so the oil lubricates nothing.

Answer: B

The Buchholz relay is used on Oil cooled transformers.

It is a gas-actuated relay fitted in the pipe between the main tank and the conservator. An internal fault decomposes the oil and releases gas, which collects in the relay and operates an alarm; a severe fault drives a surge of oil that trips the breaker.

It works only with an oil-filled tank and a conservator, so it cannot be used on air cooled (dry type) units. Welding and furnace transformers are named for their application, not their cooling, and the relay is not specific to them.

Answer: C

In a distribution transformer, Core losses are less than copper losses.

A distribution transformer is energised round the clock, so its core loss is incurred for 24 hours a day, while it runs at full load only for part of the day. It is therefore designed with a low core loss, so that maximum efficiency (where core loss equals copper loss) falls at a fraction of full load. At full load the copper loss is then larger than the core loss.

Equal losses at full load is the design aim for a power transformer, which runs near full load most of the time. More core loss than copper loss would waste energy all day for little benefit, and 'half' is not a design rule.

Answer: A

A Buchholz relay operates when there is a Large internal fault.

The relay responds to gas generated by faults inside the tank, such as insulation breakdown, winding short circuits or core faults, and to the oil surge that a large internal fault produces.

Saturation of the magnetic circuit and overload produce no gas or oil surge, so they do not operate it; overload is handled by overcurrent and thermal protection. Therefore 'Any of the above' is also wrong.

Answer: C

Zero voltage regulation is possible at a Leading power factor.

Approximately, regulation is \(\frac{I(R\cos\phi \pm X\sin\phi)}{V}\), with the plus sign for lagging and the minus sign for leading loads. With a leading power factor the reactive term subtracts, and at the angle where \(R\cos\phi = X\sin\phi\), that is \(\tan\phi = R/X\), the regulation is zero.

At lagging and unity power factor both terms are positive, so the regulation is positive. At zero power factor leading it would be strongly negative, not zero.

Answer: D

An autotransformer is used When the transformation ratio is small.

In an autotransformer part of the winding is common to primary and secondary, and only part of the power is transformed inductively. The saving in copper compared with a two-winding transformer is a fraction \(k\) of it, where \(k\) is the voltage ratio (\(V_{2}/V_{1} \lt 1\)), so the saving is greatest when the two voltages are close, for example between 220 kV and 132 kV.

Lower copper loss and higher efficiency are consequences of that saving and are only significant when the ratio is near one, so they are not the condition for choosing it. The availability of an operator has nothing to do with it.