Important Multiple choice Question Of Transformer 2022

Answer.1. Silica gel

Explanation

Transformer oil should not be exposed directly to the atmosphere because it may absorb moisture and dust from the environment and may lose its electrical properties in a very short time. To avoid this problem a breather is provided on the top of the conservator. It mainly consists of a silica gel. The silica gel absorbs the moisture content of air so that oil contamination can be prevented. The silica gel which is blue in color turns pink when it absorbs moisture fully. It is replaced periodically as routine maintenance.

Answer.2. More than 30 kV

Explanation

Dielectric strength of transformer oil

Dielectric breakdown test of oil is the most reliable method of testing the condition of oil. Transformer oil greatly reduces in its breakdown strength by the presence of moisture content. The presence of moisture content could be very minutely tested by this test after which if needed oil can be filtered and heated for removal of moisture.

The transformer oil must comply with the Bureau of Indian Standard Specifications IS:335-1983 in respect of its tendency towards sludging, acidity, flashpoint, dielectric strength, etc. Description of all the tests performed to ascertain the suitability of the oil is beyond the scope of this experiment and only dielectric strength verification will be dealt with. The electric strength of transformer oil as per IS:335-1983 should be as given below:

(a) New unfiltered oil 30 kV (RMS)

(b) After filtration of 60 kV (RMS)

If the new unfiltered oil does not withstand 30 kV (RMS), the oil should be filtered and re-tested.

The minimum breakdown voltage of 11kV/433 V transformer oil or dielectric strength of transformer oil at which this oil can safely be used in transformer, is considered to be more than 30 kV.

A typical value of BDV Test result for 220 / 6.6 kV Transformer oil is 65 kV and moisture content should be less than 10 ppm

The electric strength of the oil is tested by finding the breakdown voltage at which there is visible arcıng through the oil across two electrodes.

Answer.2. 4.6%

Explanation

In transformer voltage regulation is given by:

Voltage regulation = x (Rpu cosθ ± Xpu sinθ)

Where x = fraction of load

Rpu = resistance in pu (we can use % resistive drop in fraction)

Xpu = reactance in pu (we can use % reactive drop in fraction)

+ for lagging load

– for leading load

Calculation:

Given that, the transformer is working at 0.8 lagging power factor at full load

⇒ cos θ = 0.8, sin θ = 0.6

x = 1

Rpu = 2%

Xpu = 5%

%V.R. = 1 × (2 × 0.8 + 5 × 0.6)

%V.R = 4.6%

Answer.3. The load current

Explanation

The current transformer is used in conjunction with the current measuring device (such as an ammeter, wattmeter, energy meter etc.). The current transformer is used with low-range ammeters to measure currents in high voltage alternating current circuit. The primary current of the transformer is dictated by the load current.

When the current in a circuit is too high to apply to measuring instruments directly, a current transformer produces a reduced current accurately proportional to the load current in the course, conveniently connected to measuring and recording devices. A current transformer also isolates the measuring instruments from a very high voltage in the monitored circuit.

Answer.4. The thin coating of varnish

Explanation

The purpose of providing the coating of varnish in windings are

• To keep moisture out of the windings to preserve insulation.
• To protect the windings from humming or vibrating when magnetized.
• To increase the electrical insulation and dielectric strength.
• Provide heat dissipation from the coil.

Answer.1. 60 A

Explanation

The ratio between the primary and the secondary currents is 100 A/5 A, or 20:1. In other words, the primary current is 20 times greater than the secondary current.

Note that the number of turns and the current in the primary and secondary windings are related by an inverse proportion. i.e I₁ / I₂ = N₂ / N_1.

A current transformer (CT) has large load currents connected to the primary winding of the transformer and the ammeter connected to the seconds winding.

If the ammeter reads 3A that means the current in the secondary winding is 3A. If the turn ratio is kept constant then-new primary current is

Now I₁ / I₂ = I’₁ / I’₂

100/5 = I’₁ / 3

I’₁ = 60 A

Answer.1. +30°

Explanation

Dy11(Δ-Y) means that

• The high voltage winding of the transformer is in delta connection.
• The low voltage winding of the transformer is star connected.
• Dy11 means 30° degrees lagging or 30° leading ( LV leads HV with 30 degrees).

Advantages of DY11

1. On the high voltage side of the transformer, insulation is stressed only to the extent of 57.7% of line voltage.
2. In this case, the neutral point is stable and will not “float” when the load is unbalanced.
3. There is no distortion of flux because the primary is delta connected which allows a path for the 3rd harmonic currents.

Answer.4. 97%

Explanation

Power transformers are used in power supply circuits because of the efficiency and ease with which they transfer energy. The power transformer is capable of receiving a voltage at one level and delivering it at the same level, some higher level, or some lower level.

The efficiency of a transformer

The efficiency is always less than 100% for a practical transformer. With careful design, the efficiency of a transformer is usually greater than 90%.

No energy is lost in an ideal transformer, which does not exist. For such an ideal transformer, the output power is equal to the input power by the principle of conservation of energy.  In fact, due to energy loss in the transformer, the output power is always less than the input power.

Efficiency = Output power/Input Power

Power transformers used for bulk power transmission are operated near about full load at all times and are therefore designed to have maximum efficiency at full load. On the other hand, the distribution transformer’s supply load varies over the day through a wide range. Such transformers are, therefore, designed to have maximum efficiency at about three-fourths the full load.

Answer.4. Paralleling is done with incorrect polarities

Explanation:-

Parallel operation of transformers

There are a number of requirements that must be satisfied before two or more single-phase transformers can be ‘paralleled -i.e. before they can be connected in parallel with each other, in order to supply the same load. These requirements are

1. Same voltage ratio (turns ratio)
2. Similar percentage impedance
3. Similar kVA rating
4. Same polarity

Same voltage ratio:- If two single-phase transformers with different voltage ratios (or turns ratios) are connected in parallel across a common primary voltage, then their secondary voltages will obviously be different. Under ‘no-load’ conditions (i.e. with no load connected), this will result in a circulating current between the loop formed by the two secondary windings. As the impedance of a transformer’s windings is low, this circulating current can be quite high, resulting in unnecessarily high I²R losses.

Similar percentage impedance:- A transformer’s percentage impedance can be determined by shortcircuiting the secondary winding with an ammeter, and gradually increasing the primary voltage until rated current flows in the secondary. The percentage impedance is then simply the ratio of that particular primary voltage to the rated primary voltage, expressed as a percentage.

So, for example, if a particular transformer has a percentage impedance of, say, ‘5%’, then it would take just 5% of the rated primary voltage to cause 100% of the rated secondary current to flow through the short-circuited secondary winding.

For unequal ratings, the numerical (ohmic) values of their impedances should be in inverse proportion to their ratings to have current in them in line with their ratings. A difference in the ratio of the reactance value to the resistance value of the per-unit impedance results in a different phase angle of the currents carried by the two paralleled transformers; one transformer will be working with a higher power factor and the other with a lower power factor than that of the combined output. Hence, the real power will not be proportionally shared by the transformers.

Similar kVA rating:- Transformers with different kVA ratings will share the load more-or-less in proportion to those ratings (i.e. with each transformer carrying roughly its own share of the load), providing their voltage ratios are identical and their percentage impedances are close. However, it is generally recommended that the kVA-rating of any two transformers should never differ by more than a ratio of 2:1.

Same Polarity:- The ‘polarity’ of a transformer describes the instantaneous direction of the potential difference induced across the secondary terminals of that transformer, relative to that across the primary terminals.

The transformers should have the same polarity: The transformers should be properly connected with regard to their polarity. If they are connected within correct polarities then the two EMFs, induced in the secondary windings that are in parallel, will act together in the local secondary circuit and produce a  dead short circuit.

Answer.1. 160 A

Explanation:-

In a transformer, the current or the voltage steps up and down. But the power transferred is always equal.

The rated power of transformer can be written as:

P (MVA) = √3 V_L × I_L

Given-

Power = 3 MVA. (Full load rated power)

V1 = 11 kV

V2 = 433 V

We can write:

3 × 10⁶ = √3 × 11 × 10³ × I1

I₁ = 157.45 A

I₁ = 160 (Approx)