Transformers — Part 1 MCQs with Explanations

The purpose of providing an iron core in a transformer is to decrease the reluctance of the magnetic path . The iron core is used to improve the efficiency of the transformer by providing a low reluctance path for the magnetic flux. This allows the transformer to transfer electrical energy more effectively between its windings. Key Points: Iron Core: Reduces the reluctance (magnetic resistance) of the magnetic path, enhancing the transformer's efficiency. Reluctance: The resistance to the flow of magnetic flux. By decreasing reluctance, the core helps in better flux linkage between the primary and secondary windings.

Copper loss in a transformer is due to the current flowing through the windings' resistance. It's also known as I 2 R loss (where I is current and R is resistance). Importantly, copper loss varies with the square of the load current . Let's denote the full-load current as I FL . The copper loss at full load (P cu_FL ) is proportional to (I FL ) 2 . P cu_FL = k * (I FL ) 2 = 1600 W (where k is a constant related to the winding resistance). At half-load, the current is I FL / 2. The copper loss at half-load (P cu_HL ) is proportional to (I FL / 2) 2 . P cu_HL = k * (I FL / 2) 2 = k * (I FL ) 2 / 4. Since k * (I FL ) 2 = 1600 W, then P cu_HL = 1600 W / 4 = 400 W. Therefore, the copper loss at half-load is 400 W.

Transformers are static electrical devices, meaning they have no moving parts . Because of this, there are no mechanical losses such as friction or windage losses, which occur only in rotating machines like motors or generators. Core losses (which include hysteresis and eddy current losses) occur whenever the transformer core is energized — that is, even under no-load conditions. These losses depend on the supply voltage and frequency, not on the load. Copper losses occur only when the transformer carries load current and vary with the square of the load current. Since a transformer has no moving parts , the friction loss is zero at all times — even at full load. Hence, the correct answer is Friction loss.

The value of flux involved in the e.m.f. equation of a transformer is Given by: E = 4.44 * f * N * Φmax where: E is the induced e.m.f. f is the frequency of the AC supply N is the number of turns in the coil Φmax is the maximum flux The e.m.f. equation of a transformer involves the maximum value of the magnetic flux (Φmax) because the equation is derived based on the peak value of the sinusoidal flux waveform. Therefore, the correct answer is Option C: maximum value.

Eddy current loss in transformers is proportional to the square of the frequency of the applied voltage. As the frequency of the applied voltage increases, eddy current loss increases . This is because higher frequencies induce greater eddy currents in the core material, leading to increased power loss. Therefore, with an increase in the frequency of the applied voltage, Option B correctly describes that eddy current loss will increase.

In India, the highest voltage used for transmitting electrical power is 765 kV . This ultra-high voltage is employed to efficiently transmit electricity over extremely long distances while minimizing losses. Higher voltage levels result in lower current flow through the transmission lines, which in turn reduces resistive losses. This is a critical factor in power transmission, allowing for efficient and cost-effective distribution of electricity across the country. Options A, B, C, and D have lower voltage ratings and are generally used for sub-transmission and distribution purposes.

A transformer is an electrical device that primarily transforms voltage from one level to another. Here's how it works: Voltage transformation: The main function of a transformer is to change the voltage level of an alternating current (AC). By using electromagnetic induction, it increases (steps up) or decreases (steps down) the voltage. Current and power: While a transformer changes the voltage, it indirectly affects the current. The power in a transformer is conserved (neglecting losses), meaning the power on the primary side is approximately equal to the power on the secondary side, assuming ideal conditions. However, the transformer does not transform power directly but maintains it, adjusted for the changes in voltage and current. Frequency: The frequency of the electrical signal remains constant through the transformer. It is not affected by the transformation process. In summary, a transformer primarily transforms voltage while maintaining the same frequency and indirectly affecting current and power.

The core of a transformer is laminated primarily to reduce eddy current losses . Eddy currents are loops of electrical current induced within conductors by a changing magnetic field in the conductor, which can cause energy losses in the form of heat. Key Points: Reduce eddy current losses : Laminating the core introduces thin insulating layers between the sheets of metal, which restricts the flow of eddy currents and thus minimizes energy losses due to these currents. Hysteresis loss : While laminations help with eddy current losses, hysteresis loss is reduced by using materials with low hysteresis coefficients. Copper losses : These are related to the resistance in the winding of the transformer and are not affected by the core lamination. All above losses : Laminations specifically target eddy current losses; they do not directly reduce hysteresis or copper losses.

A transformer is like a special device that helps to change the voltage of electricity. It has two main coils of wire, called "windings." One winding is where electricity comes in (primary), and the other is where it goes out (secondary). The most important thing to remember for this question is: Power in a transformer stays almost the same! Think of electrical power as the "work" being done. Power (P) is calculated by multiplying Voltage (V) by Current (I) (P = V x I). 1. If the transformer steps down the voltage (makes it lower), then to keep the power the same, the current must go up (become higher). 2. If the transformer steps up the voltage (makes it higher), then to keep the power the same, the current must go down (become lower). Now, imagine a water pipe. If you want more water to flow through it (higher current), you need a wider pipe (larger cross-sectional area), right? It's the same for electricity! * Wires that carry a high current need to be thicker (have a larger cross-sectional area) to avoid getting too hot and to allow the electricity to flow easily. * Wires that carry a low current can be thinner (have a smaller cross-sectional area). So, if the low voltage winding has a low voltage, it means it must be carrying a higher current to keep the power constant. And because it carries higher current, its wires need to be thicker , meaning they have a larger cross-sectional area . The high voltage winding will have a lower current, so its wires can be thinner. Options C and D (Primary and Secondary) are not always the answer because a primary winding could be high voltage or low voltage, and the same goes for the secondary. The crucial factor is the voltage level (and thus the current level) it handles. Therefore, the winding with the more cross-sectional area is the Low voltage winding because it carries the higher current. Correct Answer: Option A: Low voltage winding

Here's an explanation to help you understand how to solve this transformer question: The question involves the concept of the turns ratio of a transformer and how it relates to the impedance transformation . Turns Ratio (N1:N2): In this problem, the turns ratio is given as 1:6. This means for every 1 turn in the primary coil, there are 6 turns in the secondary coil. We can represent this as N1/N2 = 1/6. Impedance Transformation: A transformer changes the apparent impedance (resistance in this case) seen by the source (primary side). The relationship is: Z1 = (N1/N2)^2 * Z2 Where: Z1 is the impedance seen by the source (primary side). Z2 is the impedance of the load connected to the secondary side. N1/N2 is the turns ratio. In our problem: N1/N2 = 1/6 Z2 = 470 Ω Now, let's calculate Z1: Z1 = (1/6)^2 * 470 Z1 = (1/36) * 470 Z1 ≈ 13.05 Ω Therefore, the load resistance as seen by the source is approximately 13 Ω. So the correct answer is Option D: 13 Ω

The correct answer is Option D: high voltage side . Here's why: When we put taps on the high voltage (HV) side of a transformer, it means that part of the winding has higher voltage but lower current for the same amount of power. Because the current is lower on the HV side: 1. The wires needed for the tapping connections and the tap-changing switch (which moves between different tap points) can be thinner . 2. The switch itself (called a tap changer) can be much smaller, lighter, and less complex . 3. This makes the whole tap-changing mechanism significantly cheaper to build and maintain. Now, let's look at why the other options are generally not preferred: If taps were placed on the low voltage (LV) side (Option C), that side of the transformer carries a very high current . 1. Handling such high currents would require very thick wires for the tapping connections. 2. The tap-changing switch would need to be very robust, large, and heavy-duty to safely break and make connections carrying high current. 3. This would make the tap changer extremely expensive and bulky . While Options A (primary side) and B (secondary side) can sometimes be the HV or LV side depending on whether the transformer is step-up or step-down, the general rule is to place taps where the current is lower to save cost and reduce complexity. This is almost always the high voltage side . So, to keep the tap changer economical and practical, it's always placed on the side with lower current, which is the high voltage side .

Under no-load conditions , an induction motor primarily draws a small active power and a large magnetizing current. The magnetizing current is highly reactive in nature, causing the motor to operate at a very low power factor . Since the magnetizing current lags the applied voltage, the power factor is lagging and typically very low, around 0.2 lagging . The power factor cannot be leading because an induction motor does not produce reactive power but rather consumes it. The power factor is also far from unity under no-load conditions due to the dominance of magnetizing current. Hence, the correct answer is 0.2 lagging (Option A).

output voltage fluctuation from no load to full load is least is the correct answer to “A good voltage regulation of a transformer means”. This question is standard in competitive examination syllabi for Transformers.

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 is the correct answer to “If the percentage impedances of the two transformers working in parallel are different, then”. This question is standard in competitive examination syllabi for Transformers.

copper loss = iron loss is the correct answer to “An ideal transformer will have maximum efficiency at a load such that”. This question is standard in competitive examination syllabi for Transformers.

oil cooled transformers is the correct answer to “A Buchholz relay can be installed on”. This question is standard in competitive examination syllabi for Transformers.

(A) and (B) both is the correct answer to “The size of a transformer core will depend on”. This question is standard in competitive examination syllabi for Transformers.

Faraday's laws of electromagnetic induction are not valid since the rate of change of flux is zero is the correct answer to “A transformer cannot raise or lower the voltage of a D.C. supply because”. This question is standard in competitive examination syllabi for Transformers.

winding insulation is the correct answer to “Part of the transformer which is most subject to damage from overheating is”. This question is standard in competitive examination syllabi for Transformers.

2 to 5 per cent is the correct answer to “The no-load current drawn by transformer is usually what per cent of the full-load current”. This question is standard in competitive examination syllabi for Transformers.

absorbing moisture is the correct answer to “The chemical used in breather for transformer should have the quality of”. This question is standard in competitive examination syllabi for Transformers.

to take care of the expansion and contraction of transformer oil due to variation of temper-ature of sur-roundings is the correct answer to “The function of conservator in a transformer is”. This question is standard in competitive examination syllabi for Transformers.

to arrest flow of moisture when outside air enters the transformer is the correct answer to “The function of breather in a transformer is”. This question is standard in competitive examination syllabi for Transformers.

loading of the transformers not in proportion to their kVA ratings is the correct answer to “Two transformers are connected in parallel. These transformers do not have equal percentage impedance. This is likely to result in”. This question is standard in competitive examination syllabi for Transformers.

hysteresis and eddy current losses is the correct answer to “In a given transformer for given applied voltage, losses which remain constant irrespective of load changes are”. This question is standard in competitive examination syllabi for Transformers.

Conservator is the correct answer to “The changes in volume of transformer cooling oil due to variation of atmospheric tem-perature during day and night is taken care of by which part of transformer”. This question is standard in competitive examination syllabi for Transformers.

load is balanced only is the correct answer to “Star/star transformers work satisfactorily when”. This question is standard in competitive examination syllabi for Transformers.

on balanced as well as unbalanced loads is the correct answer to “Delta/star transformer works satisfactorily when”. This question is standard in competitive examination syllabi for Transformers.

Infinity is the correct answer to “In a transformer the resistance between its primary and secondary should be”. This question is standard in competitive examination syllabi for Transformers.

150°C is the correct answer to “Gas is usually not liberated due to dissociation of transformer oil unless the oil temperature exceeds”. This question is standard in competitive examination syllabi for Transformers.

core flux remains practically constant is the correct answer to “In an actual transformer the iron loss remains practically constant from noload to fullload because”. This question is standard in competitive examination syllabi for Transformers.

silica gel is the correct answer to “The chemical used in breather is”. This question is standard in competitive examination syllabi for Transformers.

high inductance arid low resistance is the correct answer to “In case of a double cage induction motor, the inner cage has”. This question is standard in competitive examination syllabi for Transformers.

increase is the correct answer to “If the supply frequency to the transformer is increased, the iron loss will”. This question is standard in competitive examination syllabi for Transformers.

leading is the correct answer to “Minimum voltage regulation occurs when the power factor of the load is”. This question is standard in competitive examination syllabi for Transformers.

Current transformer is the correct answer to “The secondary winding of which of the following transformers is always kept closed”. This question is standard in competitive examination syllabi for Transformers.

its power factor will deteriorate is the correct answer to “If a transformer is switched on to a voltage more than the rated voltage”. This question is standard in competitive examination syllabi for Transformers.

low,low is the correct answer to “The transformer oil should have _____ volatility and _____ viscosity.”. This question is standard in competitive examination syllabi for Transformers.

all of the above is the correct answer to “The degree of mechanical vibrations produced by the laminations of a transformer depends on”. This question is standard in competitive examination syllabi for Transformers.

0.4 mm to 0.5 mm is the correct answer to “The thickness of laminations used in a transformer is usually”. This question is standard in competitive examination syllabi for Transformers.