Important Multiple choice Question Of Transformer 2022

Answer 4. All of the above

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.2. Leading

Explanation:

Voltage regulation is the change in secondary terminal voltage from no load to full load at a specific power factor of load and the change is expressed in percentage.

E2 = no-load secondary voltage

V2 = full load secondary voltage

Voltage regulation for the transformer is given by the ratio of change in secondary terminal voltage from no load to full load to no load secondary voltage.

Voltage regulation = $\frac{{{E_2} – {V_2}}}{{{E_2}}}$

It can also be expressed as,

Regulation = $\frac{{{I_2}{R_{02}}\cos {\phi _2} \pm {I_2}{X_{02}}\sin {\phi _2}}}{{{E_2}}}$

Regulation depends on the leakage impedance of the transformer and on the power factor of the load.

• At leading power factors, regulation is usually negative; that is, the voltage at the secondary terminals of a transformer is larger at full load than it is when the load is disconnected.
• In such cases, the equipment connected to a transformer’s secondary may be subjected to higher than rated voltages.
• This may occur when the power-factor-correction capacitor banks remain on the network while the plant operates at a reduced load.
• Also, note that for the leading power factor, if the magnitude of the phase angle φ is high, the magnitude of ${{I_2}{X_{02}}\sin \phi }$  may become greater than that of ${{I_2}{R_{02}}\cos {\phi _2}}$. The regulation then becomes negative.

Answer.1. Copper Loss

Explanation:

Short circuit or Impedance test

Short circuit test or Impedance test is performed to determine
⇒ Copper loss at full load
⇒ Equivalent impedance (Z_o1 or Z_o2)
⇒ Leakage reactance (X_o1 or X_o2)

• The figure shows the circuit diagram for conducting the short-circuit test on a transformer. One of the windings of the transformer is short-circuited through an ammeter, while a low voltage is applied to the other winding.
• The applied voltage is slowly increased until full load current flows in this winding. The full load current will then flow in the other winding also.
• Normally the applied voltage is hardly 5 to 7 percent of the rated voltage of this winding. As such the flux established in the core will be quite small and so the iron losses occurring under this condition is negligible.
• Thus, the reading indicated by the wattmeter connected in the circuit gives the full load copper losses of the transformer.

Answer.2. High voltage side

Explanation:

• A short circuit test is conducted to find the copper loss.
• It is calculated under the assumption that core loss is neglected.
• When the SC test is conducted on the LV side it would require a larger voltage to get the rated current.
• Hence core loss cannot be neglected in this case and the wattmeter doesn’t give the copper loss alone.
  Therefore to get accurate results that test is done on the HV side.
• The advantage of conducting the short-circuit test on the HV side is that the rated current value on this side is lesser. This will permit the use of ammeter and wattmeter with a lesser current range.

Answer.2. >140°

Explanation:

Flash Point of Transformer Oil is defined as the temperature at which light hydrocarbon present in the transformer oil starts evaporating causing a flash on the introduction of the source under specified conditions.

• Transformer oil or insulating oil is an oil that is stable at high temperatures and has excellent electrical insulating properties.
• In any case, this temperature should not be allowed to be less than 130 degrees Celsius for transformer oil in service.
• We know that the maximum continuous operating temperature of transformer oil is around 105 degrees Celsius.
• However, during fault conditions, transformer oil temperature may go as high as 130 degrees Celsius.
• Therefore, it is very much required that the flashpoint of oil should be maintained above 130 degrees Celsius to prevent fire during fault conditions.
• As per standard rules, the flash point of transformer oil should be greater than or equal to 140 degrees.
• Due to the relatively high flash point and boiling point of transformer oil, transformer fires can be extinguished quickly by properly designed water spray systems.

Answer.4. Friction loss

Explanation: 

The transformer is a static device that is used to transfer electric power from one circuit to another without changing its frequency. The main function of a transformer is to raise as lower the voltage in a circuit with a corresponding decrease or increase in current at the same frequency. It works on the principle of Faraday’s law of Electromagnetic induction. Transformers have no moving parts, rugged and durable in construction.

Since operation does not involve rotation of an armature, field system, or commutator, rotational, friction loss. and windage losses do not occur and its efficiency is thus high.

For electrical ‘power’ purposes, i.e. transformers operating at 50 or 60 Hz, iron cores are essential and iron losses will occur. Winding copper losses are also present when current is supplied, nonetheless, the transformer is the most efficient of electrical machines and has a full-load efficiency of 95.5% for units of 5 kVA and 97.5% for units up to 1 MVA may be achieved.

Answer.2. Infinite

Explanation:

In a transformer, the primary and secondary windings are not electrically connected therefore the resistance between them is ideally infinite but an autotransformer does the same using a single coil as primary with one or more tap for secondary in different parts of the coil. In this case, the resistance will ideally be zero or a short circuit.

Answer.4. Back to Back test (Sumpner Test)

Explanation:

To determine the maximum temperature rise of a transformer Sumpner’s test is performed. This test can also be performed to find out the efficiency of a transformer. With the help of Sumpner’s test both the open-circuit and short-circuit tests can be performed simultaneously.

• Sumpner’s test is essentially a load test. It requires two identical transformers whose primaries are connected in parallel.
• The two secondaries are connected in series with their polarities in phase opposition. The primary windings are supplied at rated voltage and frequency. A voltmeter, ammeter, and wattmeter are connected to the input.
• As the two secondaries are connected in phase opposition, the two secondary EMFs oppose each other and no current can flow in the secondary circuit.
• A regulating transformer excited by an ac mains supply is used to inject voltage into the secondary winding. The injected voltage is adjusted till the ammeter A, reads full load secondary current. The secondary current causes full load current to flow through the primary windings.
• The wattmeter W₁, indicates total core losses, W₂ indicates total copper losses, and ammeter A₁ indicates the total no-load current of the two transformers.
• Thus by this method, we can load the transformer to full load but the supplying energy is only equal to that required for the losses only. This test can be continued for a long time to determine the maximum temperature rise of a transformer.
• The transformers are kept in this condition for 48 hours and the temperature is noted on an hourly basis and a curve is plot which should become constant after some time also it should be within limit.

Efficiency at full load is

η = (Full Load output)/(Full Load output + Core loss + copper loss)

η = W_full/(W_full + W₁ + W₂)

Answer.3. KVA

Explanation:

The rating of any electrical machine shows its ability to carry the mechanical load without showing any signs of overheating.

There are two types of losses in a transformer

⇒ Iron Losses
⇒ Copper loss

• Copper losses ( I²R)depends on the Current which passes through transformer winding.
• Iron Losses (Core Losses or Insulation Losses) depends on Voltage.
• This shows that the iron and copper losses depend only on the supply voltage and the current flowing through the winding respectively.
• Since the power factor is dependent on the load which is not stable.
• Since these losses do not depend on the phase angle between the supply voltage and the current, the transformer rating expressed as a product of voltage and current is called the VA rating of the transformer.
• KVA tells you the current rating of the transformer, regardless of the power factor of the load.
• Hence, total transformer loss depends on volt-ampere (VA) and not on phase angle between voltage and current

Therefore, the transformer rating is expressed as kVA and not as kW.

Answer.3. High Voltage Winding

Explanation:

• The number of turns on the primary and secondary windings depends upon their respective voltages.
• A high-voltage winding has far more turns than a low-voltage winding.
• On the other hand, the current in a high-voltage winding is much smaller, enabling us to use a smaller size conductor.
• The result is that the amount of copper in the primary and secondary windings is about the same.
• If a transformer has more turns in the secondary winding than in the primary winding, the secondary voltage is higher than the primary voltage and it is a step-up transformer.
• If a transformer has more turns in the primary winding than in the secondary winding, the primary voltage is higher than the secondary voltage. This means that the voltage has been decreased, or “stepped down.”