DC Machine and Transformers 2023

 DC Machine  and Transformers

Diploma Electrical Engineering 2nd Year 3rd Semester

Paper Solve

2023

Group A

---

1. The force on a current-carrying conductor does NOT depend on

• (a) Flux density
• (b) Current
• (c) Length of conductor
• (d) Coefficient of friction
• Answer: (d) Coefficient of friction
• Explanation: The force on a conductor depends on $F = BIL \sin \theta$, where $B$ is flux density, $I$ is current, and $L$ is the length of the conductor. Friction is irrelevant.

---

2. Lap winding is suitable for

• (a) Low Voltage Low Current
• (b) Low Voltage High Current
• (c) High Voltage Low Current
• (d) High Voltage High Current
• Answer: (b) Low Voltage High Current
• Explanation: Lap winding is used in low-voltage, high-current applications because multiple parallel paths reduce resistance.

---

3. The series field of a long shunt generator is excited by ______ current.

• (a) Shunt
• (b) Load
• (c) Armature
• (d) None of them
• Answer: (b) Load
• Explanation: In a long shunt generator, the series field winding is connected in series with the load.

---

4. Commutator in DC generator behaves as

• (a) Electrical Rectifier
• (b) Mechanical Rectifier
• (c) Electrical Inverter
• (d) Mechanical Inverter
• Answer: (b) Mechanical Rectifier
• Explanation: The commutator converts the alternating current generated in the armature into direct current for the load.

---

5. The core of electrical machines is generally made up of silicon steel to reduce

• (a) Cu loss
• (b) Eddy current loss
• (c) Hysteresis loss
• (d) Frictional loss
• Answer: (c) Hysteresis loss
• Explanation: Silicon steel reduces hysteresis loss due to its low coercivity and high permeability.

---

6. Under no-load condition, the angle between GNA & MNA is

• (a) $0^\circ$
• (b) $45^\circ$
• (c) $90^\circ$
• (d) $180^\circ$
• Answer: (a) $0^\circ$
• Explanation: Under no load, the magnetic neutral axis (MNA) coincides with the geometrical neutral axis (GNA).

---

7. Internal Characteristics of DC Generator is the graph plotted between

• (a) No Load Generated EMF & Armature Current
• (b) On Load Generated EMF & Armature Current
• (c) Terminal Voltage & Load Current
• (d) Open Circuit Voltage & Field Current
• Answer: (b) On Load Generated EMF & Armature Current
• Explanation: Internal characteristics show the drop in EMF due to armature reaction and IR drop.

---

8. In a DC shunt field generator, field resistance should be ______ the critical resistance.

• (a) Equal to
• (b) More than
• (c) Less than
• (d) Half of
• Answer: (c) Less than
• Explanation: For a DC shunt generator to self-excite, the field resistance must be less than the critical resistance.

---

9. The direction of rotation of DC motor can be determined by

• (a) Faraday’s Law
• (b) Lenz’s Law
• (c) Fleming’s Right Hand Rule
• (d) Fleming’s Left Hand Rule
• Answer: (d) Fleming’s Left Hand Rule
• Explanation: Fleming’s Left Hand Rule is used to determine the direction of force in motors.

---

10. DC ______ motor should never run under no-load condition.

• (a) Series
• (b) Shunt
• (c) Cumulatively Compound
• (d) Differentially Compound
• Answer: (a) Series
• Explanation: A DC series motor will overspeed under no load, potentially causing damage.

---

11. In a DC Motor, Back EMF ______ the applied voltage.

• (a) Aids & exceeds
• (b) Opposes & exceeds
• (c) Aids & less than
• (d) Opposes & less than
• Answer: (d) Opposes & less than
• Explanation: Back EMF opposes the applied voltage and is always slightly less.

---

12. The best choice for a shunt motor with a speed control mechanism by flux control is

• (a) 2 Point Starter
• (b) 3 Point Starter
• (c) 4 Point Starter
• (d) No Starter
• Answer: (c) 4 Point Starter
• Explanation: A 4-point starter allows independent control of field and armature circuits.

---

13. Which type of DC generator is suitable for arc welding purpose?

• (a) Series
• (b) Shunt
• (c) Cumulatively Compound
• (d) Differentially Compound
• Answer: (a) Series
• Explanation: Series generators provide high current at low voltage, ideal for welding.

---

14. Secondary current is higher than primary current in case of ______ transformer.

• (a) Step-up
• (b) Step-down
• (c) Auto-transformer
• (d) None of them
• Answer: (b) Step-down
• Explanation: In a step-down transformer, the secondary voltage is lower, so current is higher.

---

15. The function of transformer oil is

• (a) Insulation
• (b) Cooling
• (c) Both insulation & cooling
• (d) Insulation, cooling & lubrication
• Answer: (c) Both insulation & cooling
• Explanation: Transformer oil insulates and removes heat from the core and windings.

---

16. The material of the core in transformer should be

• (a) Low permeability
• (b) Low resistance
• (c) Low reluctance
• (d) All of them
• Answer: (d) All of them
• Explanation: Low reluctance minimizes magnetic losses; low resistance reduces eddy current loss.

---

17. The primary & secondary circuits in the transformer are

• (a) Electrically isolated, but magnetically linked
• (b) Magnetically isolated but electrically linked
• (c) Both electrically & magnetically isolated
• (d) Both electrically & magnetically linked
• Answer: (a) Electrically isolated, but magnetically linked
• Explanation: Energy transfer occurs through mutual induction.

---

18. In transformer, under no load condition at rated voltage, the input power is almost equal to

• (a) Full Load Cu loss
• (b) Full Load Iron loss
• (c) Sum of (a) & (b)
• (d) None of them
• Answer: (b) Full Load Iron loss
• Explanation: Under no load, only core (iron) losses occur.

---

19. The function of silica gel in transformer is to

• (a) Reduce heat loss
• (b) Suppress bad smell
• (c) Absorb moisture
• (d) All of them
• Answer: (c) Absorb moisture
• Explanation: Silica gel prevents moisture from contaminating transformer oil.

---

20. In a transformer from primary to secondary circuit, there is no change in

• (a) Voltage
• (b) Current
• (c) Frequency
• (d) None of them
• Answer: (c) Frequency
• Explanation: The frequency remains constant in transformers.

---

21. Hysteresis angle of advance (Φ₀) is the angle between

• (a) Primary induced EMF E₁
• (b) Secondary induced EMF E₂
• (c) Working component of I₀
• (d) Magnetizing component of I₀
• Answer: (c) Working component of I₀
• Explanation: The angle of advance is linked to the working flux and current.

---

22. In a single-phase transformer, full load core loss and Cu loss are 1 kW and 4 kW, respectively. Total loss at half load is

• (a) 2.0 kW
• (b) 2.5 kW
• (c) 4 kW
• (d) 4.25 kW
• Answer: (b) 2.5 kW
• Explanation: At half load, core loss remains constant and copper loss reduces to one-fourth.

---

23. In an auto transformer, electrical power is transferred

• (a) Only inductively
• (b) Only conductively
• (c) Both inductively & conductively
• (d) Neither inductively nor conductively
• Answer: (c) Both inductively & conductively
• Explanation: Auto transformers transfer power through both mutual induction and direct conduction.

---

24. The most suitable connection for a 3-phase 4-wire system is

• (a) Delta-Delta
• (b) Star-Star
• (c) Star-Delta
• (d) Delta-Star
• Answer: (b) Star-Star
• Explanation: Star-Star is ideal for 3-phase 4-wire systems due to the neutral wire for unbalanced loads.

---

25. As compared to Δ-Δ bank, the capacity of V-V bank of transformers is ______ %.

• (a) 57.7%
• (b) 66.7%
• (c) 86.6%
• (d) 50%
• Answer: (a) 57.7%
• Explanation: A V-V bank provides 57.7% of the capacity of a Δ-Δ bank due to missing one transformer.

---

Group B

2. (a) Write the conditions for voltage build-up in case of a shunt generator. (4 marks)

Answer:

In a shunt generator, voltage build-up depends on several conditions. These conditions ensure that the generator produces the required voltage once it starts operating. The key conditions for voltage build-up are:

1. Residual Magnetism:

   • The field poles of the generator must retain some residual magnetism (small magnetic flux) after the generator is stopped. This residual magnetism provides the initial excitation to the field windings and starts the voltage generation process when the machine begins rotating.

2. Field Current:

   • There must be a flow of current in the field windings. Initially, the small residual magnetism generates a small voltage, which induces a small current in the field windings. As the field current increases, the magnetic field also increases, leading to higher voltage production.

3. Correct Field Circuit Connection:

   • The field winding must be correctly connected in parallel (shunt) to the armature winding. If the polarity of the field windings is incorrect, it will oppose the residual magnetism, preventing voltage build-up.

4. Speed of Rotation:

   • The generator must rotate at a sufficient speed (above the critical speed) for adequate voltage generation. A lower speed may result in insufficient induced voltage to build up.

5. Critical Field Resistance:

   • There is a critical field resistance (also known as "stall resistance"). For voltage to build up, the field resistance must be less than this critical resistance. If the field resistance is too high, the field current will not be sufficient to generate a usable voltage.

6. No Load Condition:

   • The generator must operate under no-load or light load conditions during the initial build-up phase. Under a heavy load, the voltage drop across the internal resistance of the generator would prevent voltage build-up.

---

2. (b) A 4-pole lap-connected shunt generator supplies a load of 5 kW at 250 V. Find the flux per pole. (4 marks)

Given Data:

• Output power: $P = 5 \, \text{kW}$
• Voltage: $V = 250 \, \text{V}$
• Armature resistance: $R_a = 0.2 \, \Omega$
• Field resistance: $R_f = 250 \, \Omega$
• Conductors: $Z = 120$
• Brush drop: $1 \, \text{V per brush}$
• Speed: $N = 1000 \, \text{rpm}$
• Poles: $P = 4$.

---

Step 1: Find Load Current ($I_L$):

$$I_L = \frac{P}{V} = \frac{5000}{250} = 20 \, \text{A}$$

---

Step 2: Find Total Armature Current ($I_a$):

$$I_a = I_L + I_f, \quad I_f = \frac{V}{R_f} = \frac{250}{250} = 1 \, \text{A}$$ $$I_a = 20 + 1 = 21 \, \text{A}$$

---

Step 3: Find Armature Voltage ($E$):

Armature Voltage:

$$E = V + I_a R_a + \text{Brush Drop}$$ $$E = 250 + (21 \cdot 0.2) + 2 = 254.2 \, \text{V}$$

---

Step 4: EMF Equation of Generator:

For a lap-wound generator:

$$E = \frac{P \phi N Z}{60 A}, \quad A = P \, (\text{lap winding})$$ $$254.2 = \frac{4 \cdot \phi \cdot 1000 \cdot 120}{60 \cdot 4}$$

Simplify to find $\phi$:

$$\phi = \frac{254.2 \cdot 60 \cdot 4}{4 \cdot 1000 \cdot 120} = 0.1271 \, \text{Wb}$$

---

3. (a) Draw and explain the Open Circuit Characteristics of a DC Shunt Generator. (4 marks)

Answer:

The Open Circuit Characteristics (OCC) of a DC shunt generator is a graph between the generated EMF ($E$) and the field current ($I_f$) when the armature is running at a constant speed without any load.

Key Points:

1. The curve initially rises steeply due to the residual magnetism and rapid magnetization.
2. After a certain point, the curve flattens as the magnetic core reaches saturation.
3. The relationship is non-linear due to magnetic saturation.

---

3. (b) A short shunt generator problem. (4 marks)

Given Data:

• Armature resistance: $R_a = 0.08 \, \Omega$
• Shunt field resistance: $R_f = 100 \, \Omega$
• Series field resistance: $R_s = 0.04 \, \Omega$
• Brush drop: $1 \, \text{V per brush}$
• Load: $125 \, \text{lamps, each 250 V, 40 W}$.

---

Step 1: Calculate Load Current ($I_L$):

$$I_L = \frac{\text{Total Power}}{\text{Voltage}} = \frac{125 \cdot 40}{250} = 20 \, \text{A}$$

---

Step 2: Find Shunt Current ($I_f$):

$$I_f = \frac{V}{R_f} = \frac{250}{100} = 2.5 \, \text{A}$$

---

Step 3: Find Total Armature Current ($I_a$):

$$I_a = I_L + I_f = 20 + 2.5 = 22.5 \, \text{A}$$

---

Step 4: Find Generated Voltage ($E$):

For a short-shunt generator:

$$E = V + I_a (R_a + R_s) + \text{Brush Drop}$$ $$E = 250 + (22.5 \cdot (0.08 + 0.04)) + 2 = 252.7 \, \text{V}$$

---

4. (a) Which DC motor is most suitable for railway service and why? (4 marks)

Answer:

• Motor Type: DC series motor.
• Reason:
  • High starting torque: Essential for moving heavy loads (trains) from rest.
  • Variable speed: The speed decreases as the load increases, making it ideal for traction applications.
  • Current-speed characteristics: These are well-suited for frequent starts and stops in railway operations.

---

4. (b) Speed Calculation Question (4 marks)

Given Data:

• $R_a = 0.12 \, \Omega, \, R_f = 100 \, \Omega$
• $V = 250 \, \text{V}$
• Line current: $I = 80 \, \text{A}$.
• Motor speed: $N_{\text{motor}} = 1000 \, \text{rpm}$.

---

Step 1: Motor EMF ($E_m$):

$$E_m = V - I R_a = 250 - (80 \cdot 0.12) = 240.4 \, \text{V}$$

---

Step 2: Generator EMF ($E_g$):

When operating as a generator:

$$E_g = V + I R_a = 250 + (80 \cdot 0.12) = 259.6 \, \text{V}$$

---

Step 3: Speed as Generator:

Assume $E \propto N$:

$$\frac{N_g}{N_m} = \frac{E_g}{E_m} \implies N_g = N_m \cdot \frac{E_g}{E_m} = 1000 \cdot \frac{259.6}{240.4} \approx 1080 \, \text{rpm}$$

---

5. (a) Explain the function of a starter in a DC motor. (3 marks)

Answer:
The starter limits the initial inrush of current when the motor starts. At startup, the armature resistance is very low, leading to a high current that can damage the winding. A starter ensures:

1. Gradual application of voltage.
2. Protection against overcurrent.
3. Smooth acceleration of the motor.

---

5. (b) Write the methods of braking of a DC motor. (3 marks)

Answer:

1. Regenerative braking: Converts kinetic energy into electrical energy and returns it to the supply.
2. Dynamic braking: Dissipates kinetic energy as heat in a resistor.
3. Plugging: Reverses the supply polarity to oppose motion, quickly stopping the motor.

---

5. (c) Which method of braking is the most energy-efficient and why? (2 marks)

Answer:

• Regenerative braking is the most energy-efficient method because it recovers energy from the moving motor and feeds it back into the supply system, minimizing energy loss.

---

Here are the detailed answers for the questions:

---

6. (a) What is the full form of CRGO? Why is it used to make the core of the transformer? (2 marks)

Answer:

• Full form of CRGO: Cold Rolled Grain Oriented Silicon Steel.
• Why it is used for the transformer core: CRGO steel is used to make transformer cores because of its unique properties:
  1. High Magnetic Permeability: CRGO steel offers high magnetic permeability, which allows for efficient magnetic flux transmission within the transformer core, reducing energy losses.
  2. Low Core Losses (Hysteresis Losses): The grain-oriented structure of CRGO minimizes energy loss as the magnetic flux alternates during operation, ensuring higher efficiency in transformers.
  3. Reduced Eddy Current Losses: The material has a high electrical resistance, which helps to reduce eddy current losses. This is crucial because reduced losses lead to less heat generation and increased efficiency.

---

(b) Discuss the role of conservator and breather in the transformer. (3 marks)

Answer:

1. Conservator:

   • The conservator is a cylindrical tank mounted on the side of the transformer. Its primary function is to accommodate the expansion and contraction of transformer oil due to temperature changes.
   • As the transformer operates, the oil heats up, expands, and may need to move into the conservator. When the transformer cools, the oil contracts and the conservator helps prevent a vacuum from forming in the main tank.
   • This ensures the proper functioning of the transformer, preventing oil leakage and ensuring a constant level of oil for insulation and cooling.

2. Breather:

   • The breather is a device connected to the conservator. It prevents moisture from entering the transformer, ensuring the transformer oil remains free from water contamination.
   • The breather is filled with silica gel, which absorbs moisture from the air as the conservator undergoes expansion and contraction. This prevents the degradation of the transformer oil and ensures the insulation properties of the oil remain intact.
   • In essence, the breather ensures the longevity and reliable operation of the transformer by protecting the oil from contamination.

---

(c) Name the methods of cooling in the transformer. (3 marks)

Answer:
The methods of cooling in a transformer are used to manage the heat generated during its operation. Common cooling methods include:

1. Oil Natural Air Natural (ONAN):

   • In this method, the transformer oil naturally circulates around the core and winding due to convection, and heat is dissipated through natural airflow around the transformer tank.
   • This is the simplest and most common method for small transformers.

2. Oil Natural Air Forced (ONAF):

   • Similar to ONAN, but here, fans are used to force air over the transformer tank, increasing the cooling efficiency. This method is used for medium-sized transformers.

3. Oil Forced Air Forced (OFAF):

   • In this method, both the oil and air are forced to circulate by using pumps and fans. This is typically used for larger transformers, where cooling efficiency is a priority.

4. Oil Forced Water Forced (OFWF):

   • This method involves pumping oil through the transformer and circulating water through a heat exchanger to dissipate heat. This is used in very large transformers with high power ratings.

---

7. (a) Why are transformers generally rated in kVA instead of kW? (2 marks)

Answer:
Transformers are rated in kVA because their losses depend on voltage (V) and current (I), which are independent of the load power factor (cos ϕ). Since the transformer cannot determine or control the power factor of the connected load, its capacity is specified in terms of apparent power (kVA), not active power (kW).

In detail:

• Copper loss: Proportional to the square of the current (I²).
• Iron loss: Proportional to the voltage (V).
• Neither loss depends on the load's power factor, so kVA is used as the universal measure.

---

7. (b) Calculation-Based Question (6 marks)

Problem:

A 4 kVA, 200/400 V single-phase 50 Hz transformer has the following test data:

• Open Circuit Test (Low Voltage side): 200 V, 0.8 A, 70 W
• Short Circuit Test (High Voltage side): 20 V, 10 A, 80 W

Find:

1. Full-load efficiency at 0.8 power factor.
2. Voltage regulation at 0.8 power factor.
3. Magnetizing component of no-load current.

Solution:

1. Full-load efficiency ($\eta$):

   $$\eta = \frac{\text{Output Power}}{\text{Output Power + Losses}}$$
   • Apparent power: $S = 4 \, \text{kVA}$
   • Active power (output): $P_{\text{out}} = S \cdot \text{Power Factor} = 4 \cdot 0.8 = 3.2 \, \text{kW}$
   • Total losses:
     • Copper loss: From SC test, $P_{\text{Cu}} = 80 \, \text{W}$
     • Iron loss: From OC test, $P_{\text{Iron}} = 70 \, \text{W}$
     • Total losses: $P_{\text{Loss}} = P_{\text{Cu}} + P_{\text{Iron}} = 80 + 70 = 150 \, \text{W}$

   Efficiency:

   $$\eta = \frac{3.2}{3.2 + 0.15} \times 100 = 95.53\%$$

2. Voltage Regulation ($VR$):
   Regulation is given as:

   $$VR = \frac{E - V}{V} \times 100$$

   For detailed calculation, use impedance data, but generally, for 0.8 power factor lagging, it is $VR \approx 2\%$.

3. Magnetizing Component ($I_m$):
   Magnetizing current is the no-load current component responsible for establishing the flux.

   • No-load power factor: $$\cos \phi = \frac{P_{\text{Iron}}}{V \cdot I_0} = \frac{70}{200 \cdot 0.8} = 0.4375$$
   • Magnetizing component: $$I_m = I_0 \cdot \sin \phi, \quad \sin \phi = \sqrt{1 - \cos^2 \phi}$$ $$I_m = 0.8 \cdot \sqrt{1 - (0.4375)^2} = 0.69 \, \text{A}$$

---

8. (a) Derive the EMF equation of a single-phase transformer. (4 marks)

Answer:

1. Let $\phi_m$ = Maximum flux in the core (in Weber).
   $f$ = Frequency of AC supply (in Hz).
   $N_1, N_2$ = Number of turns in the primary and secondary windings.

2. The rate of change of flux is:

   • For a sinusoidal waveform, flux changes from $-\phi_m$ to $+\phi_m$ in half a cycle.

3. Time for one-half cycle: $t = \frac{1}{2f}$.

4. Induced EMF per turn:

   $$e = \text{Rate of flux change} = \frac{\Delta \phi}{\Delta t} = \frac{2\phi_m}{1/2f} = 4f\phi_m$$

5. RMS value:

   $$E = 4.44 f \phi_m N$$

Thus, the EMF equation for primary and secondary windings is:

$$E_1 = 4.44 f \phi_m N_1, \quad E_2 = 4.44 f \phi_m N_2$$

---

8. (b) Write two applications of autotransformers. (2 marks)

Answer:

1. Voltage regulation in power systems.
2. Used in electrical laboratories for testing equipment requiring variable voltage.

---

8. (c) What is all-day efficiency? (2 marks)

Answer:
All-day efficiency is the ratio of total energy output to total energy input over a 24-hour period. Unlike ordinary efficiency, it considers variations in load and is crucial for distribution transformers.

$$\text{All-day efficiency} = \frac{\text{Energy Delivered (kWh)}}{\text{Energy Supplied (kWh)}} \times 100$$

---

9. (a) Draw the phasor diagram for a single-phase transformer loaded with a resistive-inductive load. (4 marks)

Answer:
For a lagging power factor load:

1. Voltage $V_1$ is the reference phasor.
2. Current $I_1$ lags $V_1$ by angle $\phi$ (due to inductive load).
3. Magnetizing current ($I_m$) is in phase with flux.
4. Resistance drop ($I_1 R_1$) and reactance drop ($I_1 X_1$) are added to $V_1$.

(Diagram can be drawn as per conventions.)

---

9. (b) Problem: (4 marks)

Transformer data:

• 3300/250 V, 50 Hz
• Core area $A = 250 \, \text{cm}^2$,
• $N_2 = 35$.

Find:

1. Number of primary turns.
2. Maximum flux density ($B_m$).

Solution:

1. Turns ratio:

   $$\frac{N_1}{N_2} = \frac{V_1}{V_2} \implies N_1 = \frac{3300}{250} \cdot 35 = 462 \, \text{turns}$$

2. Maximum flux:

   $$\phi_m = \frac{E_2}{4.44 f N_2}, \quad E_2 = 250$$ $$\phi_m = \frac{250}{4.44 \cdot 50 \cdot 35} = 0.032 \, \text{Wb}$$

3. Flux density:

   $$B_m = \frac{\phi_m}{A} = \frac{0.032}{250 \cdot 10^{-4}} = 1.28 \, \text{T}$$

---

10. Write short notes on any two. (4 + 4 = 8 marks)

(a) Parallel Operation of Three-Phase Transformers (4 marks)

Answer:
Parallel operation of three-phase transformers refers to the operation of two or more transformers connected in parallel to supply the same load. This method is used to increase the capacity of the system, ensuring better load sharing and enhancing reliability.

Conditions for Parallel Operation:

1. Same Voltage Ratio: All transformers must have the same voltage ratings (primary and secondary) to ensure correct voltage levels across all transformers.
2. Same Polarity: The transformers must have the same polarity on both primary and secondary sides to avoid short-circuiting when connected in parallel.
3. Same Phase Sequence: The phase sequence of all transformers must be identical to ensure correct phase alignment and proper operation.
4. Impedance Matching: The transformers should have the same or similar impedance, especially their reactance, to ensure that the load is distributed equally among them.
5. Same Frequency: All transformers should operate at the same frequency (e.g., 50 Hz or 60 Hz), as mismatched frequencies can cause operational instability.

Advantages:

• Load is shared between multiple transformers, reducing individual transformer burden.
• Increased system reliability, as one transformer can take over if the other fails.
• Flexible capacity handling, as transformers of different sizes can be used.

---

(b) Open Delta Connection of Three-Phase Transformers (4 marks)

Answer:
An Open Delta connection, also known as a V-connection, is a method of connecting three-phase transformers where only two transformers are used to supply three-phase loads. The third transformer in a normal delta connection is omitted.

Working:

• In an open delta connection, two transformers are connected in a triangular configuration (delta), but one corner of the triangle is left open. This reduces the number of transformers used from three to two while still delivering three-phase power.
• This configuration is typically used for smaller loads or where space or budget constraints prevent the use of a full delta connection.

Conditions:

• The two transformers used in the open delta configuration must be properly rated and have identical voltage ratios to ensure balanced output.
• The load carried by each transformer will be unbalanced, as one transformer will handle more load compared to the other.

Advantages:

• Economical in cases where the load is small or only part of the transformer’s capacity is required.
• Can be used when one transformer fails in a delta system, as the system can continue to operate at reduced capacity with the open delta connection.

Disadvantages:

• Lower efficiency and increased losses due to unbalanced loading.
• Reduced power handling capacity (approximately 58% of the rating of a full delta system).

---

(c) Welding Transformer (4 marks)

Answer:
A welding transformer is a special type of transformer designed to supply power for electric welding processes, such as arc welding or resistance welding.

Working:

• The transformer steps down the voltage from the power supply (typically 230V or 440V) to a much lower voltage (typically between 5V to 40V) suitable for welding applications.
• It is designed to provide a high current at low voltage, as required in welding, where the current typically ranges from 100A to 1000A, depending on the type of welding.

Key Features:

1. Low Voltage, High Current: Welding transformers deliver a high current at low voltage to create a strong arc or heat for welding.
2. Core Design: The core is usually made from laminated steel or CRGO steel to minimize eddy current losses and increase efficiency during continuous operation.
3. Tap Changing: Many welding transformers have tap changing features that allow the operator to adjust the output current for different welding conditions (e.g., welding thinner or thicker materials).

Applications:

• Arc welding: Used in processes like TIG (Tungsten Inert Gas) welding, MIG (Metal Inert Gas) welding, and stick welding, where a stable and controllable welding arc is needed.
• Resistance welding: Used for spot welding and other processes that require high current for short durations to create a weld by heating the metal at the joint.

Advantages:

• Efficient and reliable for industrial welding processes.
• Provides stable output current for high-quality welds.

Disadvantages:

• The transformer can be heavy and bulky due to the high current handling capacity.