BEE 2024 BEU PYQ Solution – Basic Electrical Engineering Previous Year Questions with Answers

Welcome to the BEU PYQ Solution for Basic Electrical Engineering (BEE) 2024. In this post, we provide step-by-step solutions to important Basic Electrical Engineering Previous Year Questions with Answers (PYQs) from the Bihar Engineering University (BEU) exam. BEU BEE previous year questions

Q.1 Choose the correct option for the following (BEE 2024 beu pyq solution):

(a) Which of the following network theorems is used to simplify a circuit by replacing a network with a single equivalent voltage source and a series resistance?

• (i) Superposition Theorem
• (ii) Norton’s Theorem
• (iii) Thevenin’s Theorem
• (iv) Maximum Power Transfer Theorem

Answer: (iii) Thevenin’s Theorem

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(b) What is the primary function of the squirrel cage rotor in a three-phase induction motor?

• (i) To generate a rotating magnetic field
• (ii) To reduce the slip of the motor
• (iii) To provide mechanical power
• (iv) To enhance the efficiency of the motor

Answer: (iii) To provide mechanical power

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(c) What is the relationship between the voltage and current in a purely inductive AC circuit?

• (i) Voltage leads current by 90 degrees
• (ii) Voltage lags current by 90 degrees
• (iii) Voltage and current are in phase
• (iv) Voltage and current are 180 degrees out of phase

Answer: (i) Voltage leads current by 90 degrees

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(d) What is the power factor in an AC circuit that has a real power of 200W and an apparent power of 250VA?

• (i) 0.8
• (ii) 0.55
• (iii) 0.67
• (iv) 1.05

Answer: (i) 0.8

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(e) In a magnetic circuit, if the magnetomotive force (MMF) increases, what happens to the magnetic flux, assuming the reluctance remains constant?

• (i) The magnetic flux decreases
• (ii) The magnetic flux increases
• (iii) The magnetic flux remains unchanged
• (iv) The reluctance increases

Answer: (ii) The magnetic flux increases

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(f) Which of the following is true for the analysis of a parallel magnetic circuit?

• (i) The total magnetizing force is the sum of individual forces
• (ii) The flux is shared among different branches based on reluctance
• (iii) The total reluctance is the sum of individual reluctances
• (iv) The magnetic flux is the same in all branches

Answer: (ii) The flux is shared among different branches based on reluctance

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(g) What is the EMF equation of a transformer?

• (i) E = 4.44 f N A Bmax
• (ii) E = 4.44 f N Bmax
• (iii) E = 2 f N A Bmax
• (iv) E = f N A Bmax

Answer: (i) E = 4.44 f N A Bmax

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(h) Which of the following losses in a transformer is due to the resistance of the winding?

• (i) Core loss
• (ii) Hysteresis loss
• (iii) Eddy current loss
• (iv) Copper loss

Answer: (iv) Copper loss

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(i) Which of the following components is used to protect an electrical installation from overcurrent?

• (i) MCB (Miniature Circuit Breaker)
• (ii) MCCB (Molded Case Circuit Breaker)
• (iii) ELCB (Earth Leakage Circuit Breaker)
• (iv) Switch Fuse Unit (SFU)

Answer: (i) MCB (Miniature Circuit Breaker)

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(j) Which type of battery is commonly used for backup power in electrical installations due to its high capacity and long life?

• (i) Nickel-cadmium battery
• (ii) Lead-acid battery
• (iii) Lithium-ion battery
• (iv) Alkaline battery

Answer: (ii) Lead-acid battery

BEE 2024 BEU PYQ Solution – Basic Electrical Engineering Previous Year Questions with Answers

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Q2. Circuit Theorems and Applications

(a) Explain Kirchhoff’s Voltage Law (KVL) and Kirchhoff’s Current Law (KCL) in detail. Provide an example of how these laws are used in analyzing a simple DC circuit.

Answer:

Kirchhoff’s Laws

Kirchhoff’s Voltage Law (KVL)

According to Kirchhoff’s Voltage Law, in any closed circuit or mesh, the algebraic sum of all the EMFs and voltage drops will be ze

Mathematical Representation:

Example:

• Battery: V=10V
• Resistors: R₁=4Ω, R₂=6Ω

Step 1: Find Total Resistance

R_total = R₁ + R₂ = 4Ω + 6Ω = 10Ω

Step 2: Find Total Current

I = V R_total = 10V 10Ω = 1A

Step 3: Find Voltage Drops

V₁ = I R₁ = 1A × 4Ω = 4V V₂ = I R₂ = 1A × 6Ω = 6V

Step 4: Apply KVL

V_battery – V₁ – V₂ = 10V – 4V – 6V = 0

Thus, Kirchhoff’s Voltage Law is verified.

Kirchhoff’s Current Law (KCL)

Statement:

According to Kirchhoff’s Current Law, the algebraic sum of all the currents meeting at a point or a junction will be zero.

Mathematical Representation:

Example:

• Current entering: I₁=5A
• Current splitting: I₂=3A, I₃=?

Apply KCL:

I₁ = I₂ + I₃ 5A = 3A + I₃ I₃ = 2A.

Thus, Kirchhoff’s Current Law is satisfied.

Steps to Analyze a DC Circuit Using Kirchhoff’s Laws

1. Apply KCL at junctions to find unknown currents.
2. Apply KVL in loops to find unknown voltages.
3. Use Ohm’s Law (V=IR) to relate voltage, current, and resistance.

(b) Discuss the procedure and steps involved in the application of the Maximum Power Transfer Theorem for analyzing a DC circuit. Provide an example where these theorems are applied.

Answer:

According to the maximum power transfer theorem, the maximum power flow through load resistor R_L can be achieved when the load resistor equals the thevenin’s equivalent resistance of the circuit.

Mathematical Condition:

R_L = R_th

Procedure for Applying the Maximum Power Transfer Theorem:

1. Find the Thevenin equivalent circuit:
   • Determine V_th (Thevenin voltage).
   • Determine R_th (Thevenin resistance) by removing the load and calculating equivalent resistance.
2. Set R_L equal to R_th to maximize power.
3. Calculate the maximum power using the formula:

Example Application:

Consider a DC circuit with a voltage source and a series resistor:

• V_th=24V
• R_th=6Ω

Step 1: Set Load Resistance for Maximum Power

R_L = R_th = 6Ω

Step 2: Calculate Maximum Power

P_max = 24W

BEE 2024 BEU PYQ Solution – Basic Electrical Engineering Previous Year Questions with Answers

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Q3. Sinusoidal Waveforms and Impedance in AC Circuits

(a) Explain the representation of sinusoidal waveforms in AC circuits. Discuss the importance of peak, RMS, and average values, and how they are used in the analysis of AC circuits.

Answer:

Sinusoidal waveforms are represented by their amplitude, frequency, and phase. Peak value indicates the highest voltage or current, RMS value is used for power calculations, and the average value helps in waveform analysis.

Representation of Sinusoidal Waveforms in AC Circuits

Sinusoidal waveforms in AC circuits are represented as: v(t) = Vm​ sin ( ωt + ϕ )

• Vm​ = Peak value (maximum amplitude)
• ω = Angular frequency (2πf)
• ϕ = Phase angle

Importance of Peak, RMS, and Average Values

1. Peak Value (Vm)

• Maximum voltage or current in a cycle.
• Used for transient and peak power calculations.

2. RMS Value (Vrms)

3. Average Value (Vavg)

Usage in AC Circuit Analysis

• Peak values help determine insulation and breakdown voltage.
• RMS values are used for power and heating effect calculations.
• Average values are useful in rectifier circuits and signal processing.

(b) Discuss the significance of impedance in AC circuits. Explain the impedance of series and parallel R-L-C circuits and how resonance is achieved in series R-L-C circuits.

Answer:

Significance of Impedance in AC Circuits

Impedance (Z) is the total opposition to AC current, combining resistance (R) and reactance (X): Z=R+jX

where j is the imaginary unit. Impedance determines voltage-current relationships and power dissipation in AC circuits.

Impedance of Series R-L-C Circuit

For a series circuit with resistance (R), inductance (L), and capacitance (C): Z=R+j(ωL−1/ωC)

where:

• ωL is inductive reactance (XL)
• ωC is capacitive reactance (XC)
• Net reactance: X=XL−XC

Total impedance:

Impedance of Parallel R-L-C Circuit

For a parallel R-L-C circuit, the total admittance (Y) is: Y=1/R+j(1/XL−1/XC)

Impedance is the reciprocal of admittance: Z=1/Y

Parallel circuits are analyzed using conductance and susceptance.

Resonance in Series R-L-C Circuit

Resonance occurs when inductive and capacitive reactances cancel out: XL=XC ωL=1/ωC

Resonant frequency:

At resonance:

• Impedance is purely resistive (Z=R)
• Current is maximum
• Voltage across L and C can be much higher than source voltage (voltage magnification effect)

BEE 2024 BEU PYQ Solution – Basic Electrical Engineering Previous Year Questions with Answers

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Q4. Magnetic Circuits and Their Analysis

(a) Explain the analogy between electric circuits and magnetic circuits. Discuss the concept of magnetomotive force (MMF), reluctance, and magnetic flux in a magnetic circuit.

Answer:

Magnetic circuits have similarities with electric circuits:

Magnetomotive Force (MMF)

MMF is the force that drives magnetic flux through a magnetic circuit. It is given by: MMF=NI

where:

• N = Number of turns of coil
• I= Current in amperes.

Reluctance (𝓡)

Reluctance is the opposition to magnetic flux in a magnetic circuit, similar to resistance in an electric circuit: R=lμA\mathcal{R} = \frac{l}{\mu A}R=μAl​

where:

• l = Length of magnetic path
• μ = Permeability of material
• A = Cross-sectional area

Higher reluctance means lower magnetic flux for the same MMF.

Magnetic Flux (Φ)

Magnetic flux (Φ) is the total magnetic field passing through a given area: Φ=BA

where:

• B = Magnetic flux density
• A = Cross-sectional area

Flux is analogous to electric current and follows the relation: Φ=MMF/R.

(b) Discuss the steps involved in the analysis of series and parallel magnetic circuits. How is the total reluctance determined in each case?

Answer:

Steps to Analyze Series and Parallel Magnetic Circuits

🔹 Series Magnetic Circuit

Identify the Magnetic Path – Determine the core material, length, and cross-sectional area.
➡ Calculate Magnetomotive Force (MMF) – Use the formula: MMF=N×I

➡ Find the Reluctance (R\mathcal{R}) – Use the formula: R=l/μA

➡ Apply Magnetic Ohm’s Law – Calculate the magnetic flux: Φ=MMF/Rtotal

➡ Determine Total Reluctance – Add all reluctances like resistances in series: Rtotal=R1+R2+R3+…

➡ Verify with Kirchhoff’s Flux Law – Ensure that the total flux remains constant in the circuit.

🔹 Parallel Magnetic Circuit

➡ Identify Different Magnetic Paths – Determine all parallel branches and their reluctances.
➡ Calculate MMF for Each Branch – MMF remains the same across parallel branches.
➡ Find Individual Reluctances (R) – Use: R=l/μA

➡ Apply Magnetic Ohm’s Law – Compute the flux for each branch using: Φ=MMF/R

➡ Determine Total Reluctance – Use the parallel formula: 1/Rtotal=1/R1+1/R2+…

➡ Verify with Kirchhoff’s Flux Law – Ensure the sum of fluxes in branches equals the total flux.

BEE 2024 BEU PYQ Solution – Basic Electrical Engineering Previous Year Questions with Answers

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Q5. Transformers and Induction Motors

(a) Describe the working principle of a transformer. Discuss the differences between an ideal transformer and a practical transformer, including losses and efficiency.

Answer:

Working Principle of a Transformer

A transformer works on the principle of mutual induction.

🔹 Mutual Induction:

• When an AC voltage is applied to the primary winding, it creates an alternating magnetic flux (Φ) in the core.
• This flux links with the secondary winding, inducing an EMF (electromotive force) due to Faraday’s law of electromagnetic induction.
• If the secondary circuit is closed, current flows, transferring electrical power without direct electrical contact between the two windings.

Differences Between Ideal and Practical Transformer

Ideal Transformer	Practical Transformer
100% Efficiency (No losses)	Less than 100% efficiency (due to losses)
No core losses	Hysteresis & Eddy current losses occur
No resistance in windings	Some energy lost as heat due to winding resistance
All flux links secondary winding	Some flux leaks, reducing efficiency
Perfect Voltage Regulation (Output voltage = Input voltage × Turns ratio)	Voltage slightly drops under load

Losses in a Practical Transformer

1. Copper Losses – Due to resistance in windings.
2. Iron (Core) Losses – Includes Hysteresis loss and Eddy current loss.
3. Leakage Flux – Some magnetic flux does not link with the secondary winding.
4. Stray Losses – Due to imperfect magnetic coupling.

Efficiency of a Transformer

The efficiency (η) is given by: η=(Output Power/Input Power)×100%

In a well-designed transformer, efficiency can be 95-98%.

(b) Describe the working principle, construction, and efficiency of a three-phase induction motor. Discuss the significance of the torque-slip characteristic and its application in industrial motors.

Answer:

Three-Phase Induction Motor

1. Working Principle

• A three-phase induction motor works on the principle of electromagnetic induction.
• When a 3-phase AC supply is given to the stator winding, it creates a rotating magnetic field (RMF).
• This RMF induces an electromotive force (EMF) in the rotor winding due to Faraday’s law of electromagnetic induction.
• The induced EMF generates a current in the rotor, which interacts with the stator’s magnetic field to produce torque.
• This causes the rotor to rotate in the direction of the RMF, but at a slightly lower speed than synchronous speed (Ns).

2. Construction

A three-phase induction motor consists of several key components that work together to convert electrical energy into mechanical energy. The main parts are:

1. Yoke

• The yoke is the outer frame of the motor, providing mechanical support to internal components.
• Made of cast iron or steel to ensure high magnetic permeability and protect the motor from external damage.
• Provides a low reluctance path for the magnetic field, ensuring efficient operation.

2. Rotor (Armature)

• Contains coils (armature winding) where current flows.
• Rotates within the magnetic field generated by the stator.
• Converts electrical energy into mechanical (rotational) energy.
• Typically made of laminated steel to reduce eddy current losses.

3. Commutator

• Reverses the direction of current flow in the armature winding.
• Ensures unidirectional torque production in the motor.
• Made of copper segments, insulated from each other.
• Wears over time due to brush contact, requiring periodic maintenance.

4. Brushes

• Conduct current between the external circuit and the rotating commutator.
• Usually made of carbon for good conductivity and low wear.
• Require replacement as they wear out over time.
• Minimize friction while maintaining good electrical contact.

5. Field Windings

• Wound around the poles of the stator to create a magnetic field.
• Can be series-wound, shunt-wound, or compound-wound, depending on motor design.
• The strength of the magnetic field determines the torque produced.
• Consumes electrical power to generate the magnetic field.

6. Shaft

• Transmits the mechanical energy generated by the rotor to an external load.
• Made of steel or other durable materials to handle torque.
• Supports rotational movement with minimal deflection.

3. Efficiency of Three-Phase Induction Motor

• The efficiency of an induction motor depends on the power losses in the motor.
• Major losses include:
  • Copper Losses (due to winding resistance).
  • Iron Losses (due to magnetic hysteresis and eddy currents).
  • Friction and Windage Losses (due to mechanical friction in bearings and air resistance).
• Efficiency Formula: Efficiency= (Output Power/Input Power)×100
• Efficiency is generally high (85-95%) for large motors used in industries.

4. Torque-Slip Characteristic and Its Significance

• Slip (S) is the difference between synchronous speed (Ns) and rotor speed (Nr), given by: S=((Ns−Nr)/Ns)×100
• Torque-Slip Relationship:
  • Low Slip (S < 1%) → Small torque, motor runs almost at synchronous speed.
  • Moderate Slip (1-6%) → Good torque, motor runs efficiently.
  • High Slip (>6%) → Torque decreases, motor overheats, and performance drops.
• Application in Industries:
  • Used in pumps, fans, compressors, conveyors, and industrial machines.
  • Self-starting and requires low maintenance, making it ideal for continuous operations.

BEE 2024 BEU PYQ Solution – Basic Electrical Engineering Previous Year Questions with Answers

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Q6. DC Motors and EMF Induction

(a) Explain the construction, working, and speed control methods of a separately excited DC motor. How does the torque-speed characteristic change with varying field current?

Answer:

A separately excited DC motor is a type of DC motor where the field winding is powered by a separate DC source, allowing independent control of speed and torque.

1. Construction

A separately excited DC motor consists of the following main parts:

a) Yoke (Frame)

• The outer casing of the motor, usually made of cast iron or steel.
• Provides mechanical support and protects internal components.

b) Field Winding

• A set of coils wound around the poles that produce a magnetic field.
• Powered by a separate DC power supply, allowing independent control of the motor’s operation.

c) Armature (Rotor)

• The rotating part of the motor where mechanical power is developed.
• Contains armature windings, placed in slots of a laminated iron core to reduce energy losses.

d) Commutator

• A copper ring made of multiple segments, insulated from each other.
• Helps in reversing the direction of current in the armature winding, ensuring continuous rotation.

e) Brushes

• Made of carbon, these conduct electricity between the external circuit and the rotating commutator.
• Wear out over time and require periodic replacement.

f) Shaft

• A steel rod connected to the armature, used to transmit mechanical power to an external load.

2. Working Principle

• When a DC voltage is applied to the armature winding, current flows, creating a magnetic field.
• The interaction between the armature current and the magnetic field from the field winding generates a force on the armature conductors.
• This force produces torque, which makes the armature rotate.
• The commutator and brushes ensure that the current direction in the armature windings is always correct, maintaining continuous rotation.

3. Speed Control Methods

The speed of a separately excited DC motor can be controlled by varying:

a) Armature Voltage Control

• By changing the supply voltage to the armature, the speed can be varied.
• Higher voltage → Higher speed, Lower voltage → Lower speed.
• Used in applications requiring a wide speed range.

b) Field Current Control

• The speed is inversely proportional to the field current.
• If field current increases, the magnetic field strengthens, reducing speed.
• If field current decreases, the speed increases.
• This method is useful for high-speed applications.

c) Armature Resistance Control

• By adding an external resistor in series with the armature, the voltage drop increases, reducing speed.
• This method is less efficient because energy is lost as heat.

4. Torque-Speed Characteristics with Varying Field Current

• The torque of a DC motor is given by:
  T ∝ ɸ Iₐ (where ɸ = magnetic flux, Iₐ = armature current).
• The speed is given by:
  N ∝ (V – IₐRₐ) / ɸ

Effect of Increasing Field Current (ɸ Increases)

• Torque increases (because more flux interacts with the armature).
• Speed decreases (as speed is inversely proportional to flux).

Effect of Decreasing Field Current (ɸ Decreases)

• Torque decreases (due to weaker magnetic field).
• Speed increases, but at very low flux, instability and excessive speed can occur.

(b) Explain the significance of the EMF equation in a transformer. Derive the equation and discuss the factors affecting the EMF induced in the transformer windings.

Answer:

Significance of the EMF Equation in a Transformer

The EMF equation of a transformer helps to determine the voltage induced in the primary and secondary windings. It is important because:

1. It shows the relationship between voltage, frequency, and number of turns in the windings.
2. It helps in designing a transformer for the required voltage and frequency.
3. It ensures efficient energy transfer from the primary to the secondary side.

Derivation of the EMF Equation

Factors Affecting the Induced EMF in a Transformer

1. Number of Turns (N)
   • More turns → Higher EMF
   • Fewer turns → Lower EMF
2. Supply Frequency (f)
   • Higher frequency → Higher EMF
   • Lower frequency → Lower EMF
3. Magnetic Flux (Φm)
   • Higher flux → Higher EMF
   • Lower flux → Lower EMF
4. Core Material
   • A better magnetic material (like soft iron) improves flux linkage, increasing EMF.
5. Leakage Flux
   • If some flux does not link with both windings, EMF reduces.
   • A properly designed core minimizes leakage.

BEE 2024 BEU PYQ Solution – Basic Electrical Engineering Previous Year Questions with Answers

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Q7. Electrical Installations and Safety

(a) Discuss the components of low-tension (LT) switchgear and their functions in electrical installations. Explain the working principle of MCB, MCCB, and ELCB in ensuring safety.

Answer:

Components of Low-Tension (LT) Switchgear and Their Functions

Low-Tension (LT) switchgear is used in electrical installations to control, protect, and isolate electrical circuits operating at low voltage (typically up to 1,000V). The main components of LT switchgear are:

1. Miniature Circuit Breaker (MCB)

• Function: Protects electrical circuits from overload and short circuits.
• Working: It trips (automatically switches off) when the current exceeds the set limit, preventing damage to wires and equipment.

2. Molded Case Circuit Breaker (MCCB)

• Function: Similar to MCB but designed for higher current ratings and added protection.
• Working: Protects against overloads, short circuits, and sometimes earth faults. It has an adjustable trip setting and can be manually reset.

3. Earth Leakage Circuit Breaker (ELCB)

• Function: Detects leakage current to prevent electric shocks and fires.
• Working: If there is any leakage of current to the earth (like when a person accidentally touches a live wire), it trips and disconnects the circuit, ensuring safety.

4. Fuse

• Function: Protects circuits from high current by breaking the connection when current exceeds a safe limit.
• Working: A thin wire inside melts when excessive current flows, stopping the current and preventing damage.

Working Principle of MCB, MCCB, and ELCB for Safety

1. MCB (Miniature Circuit Breaker)
   • Trips during overload or short circuit to protect wires and appliances.
   • Uses a bimetallic strip (for overload) and electromagnetic coil (for short circuit) to detect excessive current and cut off power.

2. MCCB (Molded Case Circuit Breaker)

• Works like an MCB but for higher currents.
• Provides adjustable protection settings for industrial and commercial use.
• Protects against overload, short circuit, and ground faults.

3. ELCB (Earth Leakage Circuit Breaker)

• Detects current leakage to the ground, which can cause electric shock.
• Trips when leakage exceeds a safe level, ensuring human safety.
• Two types: Voltage-based ELCB (older type) and Current-based ELCB (RCCB), which is more commonly used today.

(b) Explain the significance of energy consumption calculations, power factor improvement, and battery backup in electrical installations. Discuss how these aspects contribute to the efficiency and reliability of an electrical system.

Answer:

Importance of Energy Consumption, Power Factor Improvement & Battery Backup

1. Energy Consumption Calculations

• Helps track electricity usage and reduce wastage.
• Prevents overload and manages costs.
• Formula: Energy (kWh) = Power (kW) × Time (hours)

For example, if a 100W (0.1 kW) bulb runs for 10 hours, energy consumed is: 0.1×10=1 kWh

2. Power Factor Improvement

• Power factor is the ratio of useful power to total power. A low PF leads to energy loss.
• Benefits:
  • Lowers electricity bills.
  • Reduces energy losses and improves efficiency.
• Methods: Use capacitors or power factor correction devices.

3. Battery Backup (UPS)

• Provides power during outages, preventing data loss and equipment damage.
• Ensures uninterrupted operation for critical systems.

BEE 2024 BEU PYQ Solution – Basic Electrical Engineering Previous Year Questions with Answers

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Q8. Circuit Analysis Using Thevenin’s Theorem and Power Factor Calculation

(a) Consider a circuit with a 10 V voltage source in series with a resistor R1= 4Ω, and this combination is in parallel with a resistor R2=6Ω. Use Thevenin’s Theorem to find the Thevenin equivalent voltage and Thevenin resistance as seen from the terminals of the parallel resistors.

Answer:

Using Thevenin’s Theorem,

(b) A single-phase AC circuit consists of a resistor R=10Ω and an inductor L=0.1H connected in series across a 230 V, 50 Hz supply. Calculate the power consumed by the circuit, and determine the power factor.

Answer:

BEE 2024 BEU PYQ Solution – Basic Electrical Engineering Previous Year Questions with Answers

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Q9. Short Notes on Any Two of the Following:

(a) Synchronous generators and its application

Answer:

A synchronous generator is a type of AC generator that converts mechanical energy into electrical energy. It operates at a constant speed, synchronized with the frequency of the power system (e.g., 50 Hz or 60 Hz). The rotor rotates at the same speed as the magnetic field of the stator, maintaining synchronization.

Working Principle

• A prime mover (like a steam turbine, diesel engine, or water turbine) rotates the rotor.
• The rotor has electromagnets that create a rotating magnetic field.
• This induces an alternating current (AC) in the stator windings.
• The frequency of the generated AC depends on the speed of rotation and the number of poles in the generator.

Applications of Synchronous Generators

1. Power Stations – Used in thermal, hydro, and nuclear power plants to generate electricity.
2. Industrial Plants – Provides power backup and ensures a steady supply in large industries.
3. Ships & Aircraft – Used in marine and aviation systems to generate reliable AC power.
4. Wind Turbines – Converts wind energy into electrical energy for the power grid.
5. Diesel Generators – Used for emergency backup in hospitals, offices, and commercial buildings.

(b) Significance of Real Power, Reactive Power, and Apparent Power

Answer:

Significance of Real Power, Reactive Power, and Apparent Power

In electrical systems, power is classified into three types: Real Power (P), Reactive Power (Q), and Apparent Power (S). Understanding these is important for efficient power management.

1. Real Power (P) – Measured in Watts (W)

• Also called Active Power, it is the actual power used by electrical devices to perform useful work, such as lighting, heating, and running motors.
• It is responsible for energy consumption and is converted into mechanical work or heat.
• Formula: P=V×I×cos⁡θ

Significance:

• Determines the efficiency of electrical appliances.
• Helps in reducing energy wastage and improving system performance.

2. Reactive Power (Q) – Measured in VAR (Volt-Ampere Reactive)

• This power does not perform useful work but is needed to maintain voltage levels in AC systems.
• It is mainly used by inductive loads like motors and transformers to establish magnetic fields.
• Formula: Q=V×I×sin⁡θ

Significance:

• Helps in the proper functioning of inductive devices.
• Too much reactive power can cause voltage drops and power losses.

3. Apparent Power (S) – Measured in VA (Volt-Amperes)

• It is the total power supplied by the source, combining both real and reactive power.
• It represents the total demand on the power system.
• Formula: S=V×I

Significance:

• Determines the total capacity required for generators, transformers, and cables.
• Helps in designing efficient electrical systems with proper power factor correction.

(c) Time domain analysis of first-order RL Circuit

Answer:

Time Domain Analysis of a First-Order RL Circuit

A first-order RL circuit has a resistor (R) and an inductor (L) in series with a voltage source. The current changes over time when the circuit is switched on or off.

1. Current Response in RL Circuit

• When the switch is closed (Charging Phase)

• At t = 0: i(0)=0i(0) = 0
• At t → ∞: i(∞)=V/R
• When the switch is opened (Discharging Phase) i(t)=I0e^−(R/L)t
  • Current decreases gradually to zero.

2. Time Constant (τ)

τ=L/R

• After τ seconds: Current reaches 63% of its final value.
• After 5τ: Nearly 99% of the transition is complete.

3. Importance

Helps in designing circuits with controlled current changes.

Used in filters, motor drives, and transient analysis.

(d) Star and Delta Conversion for DC Circuit

Answer:

BEE 2024 BEU PYQ Solution – Basic Electrical Engineering Previous Year Questions with Answers

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Conclusion:

In this post, we provided detailed solutions for BEE 2024 BEU (Basic Electrical Engineering) previous year questions. Understanding these solutions will help you strengthen your core concepts and improve your exam performance. Keep practicing and exploring related topics to enhance your knowledge.

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