Basic Electrical Engineering | PYQs | 2023 | beu

Q.1 Answer the following questions briefly:


(a) State Kirchhoff’s Current Law and Kirchhoff’s Voltage Law.

Answer
Kirchhoff’s Current Law (KCL)

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:
KCL Daigram-  BEE 2024 BEU PYQ Solution
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:
KVL Formula - ✅ BEE 2024 BEU PYQ Solution BEE 2024 BEU PYQ Solution – Basic Electrical Engineering Previous Year Questions with Answers


(b) Define Network and Circuit.

Answer

Network:

  • A combination of various electrical elements like resistors, capacitors, and inductors.

Circuit:

  • A closed network in which current can flow.
  • Every circuit is a network, but every network is not a circuit.


(c) State the Maximum Power Transfer Theorem.

Answer

Maximum power is transferred from source to load when load resistance $R_L$ = source resistance $R_th$.

Used to get the most efficient power output.


(d) State the steps to solve the Superposition Theorem.

Answer
Procedure:
  1. Consider one source at a time (make others 0: voltage source = short, current source = open).
  2. Calculate output (voltage or current) due to that single source.
  3. Repeat for all sources one by one.
  4. Add algebraically all the individual effects to get the final result.


(e) Write two differences between Real Power and Reactive Power.

Answer
Real Power (P)Reactive Power (Q)
It does actual work (watts).It does no actual work, only oscillates energy.
Unit is Watt (W).Unit is VAR (Volt-Amp Reactive).


(f) Define Resonance.

Answer
  • Resonance occurs when the inductive reactance $X_L$ equals capacitive reactance $X_C$ in an AC circuit.
  • At resonance, the impedance is minimum and current is maximum.


(g) Draw the B-H characteristic of ferromagnetic material.

Answer


(h) Write the types of DC Motors and Generators.

Answer

DC Motors:

  1. Shunt Motor
  2. Series Motor
  3. Compound Motor

DC Generators:

  1. Shunt Generator
  2. Series Generator
  3. Compound Generator


(i) Why is Earthing important for an electrical circuit?

Answer
  • To protect human life from electric shock.
  • It provides a safe path for fault current to flow into the ground.
  • Prevents damage to equipment during insulation failure.


(j) What are the basic differences between a Generator and a Motor?

Answer
GeneratorMotor
Converts mechanical to electrical energy.Converts electrical to mechanical energy.
Works on the principle of electromagnetic induction.Works on the principle of Lorentz force.

Q.2 (a) Derive an expression for the current and impedance for a series RL and RC circuit excited by a sinusoidally alternating voltage. Draw the phasor diagrams.

Answer
Series RL Circuit

Given: An AC voltage source is applied across a resistor $R$ and an inductor $L$ in series:

$$ v(t) = V_m \sin(\omega t) $$

Impedance (Z):
  • Inductive reactance: $$ X_L = \omega L $$
  • Total impedance: $$ Z = R + j\omega L $$
  • Magnitude: $$ |Z| = \sqrt{R^2 + (\omega L)^2} $$
Current (I):

Using Ohm’s Law:

$$ I = \frac{V}{Z} = \frac{V_m \angle 0^\circ}{|Z| \angle \theta} $$

Where $$ \theta = \tan^{-1}\left( \frac{\omega L}{R} \right) $$

Current lags voltage by angle $\theta$.

Phasor Diagram (RL):
Series RC Circuit

Given: An AC voltage source is applied across a resistor $R$ and a capacitor $C$ in series:

$$ v(t) = V_m \sin(\omega t) $$

Impedance (Z):
  • Capacitive reactance: $$ X_C = \frac{1}{\omega C} $$
  • Total impedance: $$ Z = R – j\frac{1}{\omega C} $$
  • Magnitude: $$ |Z| = \sqrt{R^2 + \left(\frac{1}{\omega C}\right)^2} $$
Current (I):

Using Ohm’s Law:

$$ I = \frac{V}{Z} = \frac{V_m \angle 0^\circ}{|Z| \angle -\theta} $$

Where $$ \theta = \tan^{-1}\left( \frac{1}{\omega C R} \right) $$

Current leads voltage by angle $\theta$.

Phasor Diagram (RC):


(b) A series circuit consisting of a 10 Ω resistor, a 100µF capacitor, and a 10 mH inductor is driven by a 50 Hz AC voltage source of maximum value 100V. Calculate the equivalent impedance, current in the circuit, and the phase angle.

Answer

Q.3 (a) A load resistance $R_L$ is connected across the source $V_s$ with internal resistance $R_{int}$ in series with the source. Obtain the condition that the power transferred to the load from the source is maximum.

Answer

Given: A voltage source with internal resistance $R_{int}$ is connected to a load resistance $R_L$.

Step 1: Current through Load

The total resistance in the circuit is:

$$ R_{total} = R_{int} + R_L $$

Using Ohm’s Law:

$$ I = \frac{V_s}{R_{int} + R_L} $$

Step 2: Power Delivered to Load

Power across $R_L$ is:

$$ P_L = I^2 R_L = \left( \frac{V_s}{R_{int} + R_L} \right)^2 R_L $$

So,

$$ P_L = \frac{V_s^2 R_L}{(R_{int} + R_L)^2} $$

Step 3: Condition for Maximum Power

Differentiate $P_L$ with respect to $R_L$:

$$ \frac{dP_L}{dR_L} = \frac{V_s^2 \left[(R_{int} + R_L)^2 – 2R_L(R_{int} + R_L)\right]}{(R_{int} + R_L)^4} = 0 $$

Numerator simplifies to:

$$ (R_{int} + R_L)^2 – 2R_L(R_{int} + R_L) = 0 $$

Factor:

$$ (R_{int} + R_L)(R_{int} – R_L) = 0 $$

Therefore,

$$ \boxed{R_L = R_{int}} $$

Conclusion:

Maximum power is transferred when the load resistance equals the internal resistance of the source:

$$ \boxed{R_L = R_{int}} $$

(b) A resistance R is connected in series with a parallel circuit comprising two resistances of 12 Ω and 8 Ω respectively. The total power dissipated in the circuit is 70 W when the applied voltage is 20 V. Calculate R.

Answer


Q.4 (a) With a neat diagram, explain the construction and principle of a single-phase transformer. What are the characteristics of an ideal transformer?

Answer
Construction of Single-Phase Transformer
  • Core: Laminated silicon steel sheets provide a low-reluctance path and reduce eddy current loss.
  • Windings: Primary (input) and secondary (output) windings made of copper wire.
  • Insulation: Used between windings and the core to avoid short circuits.
  • Cooling: Oil or air cooling helps in heat dissipation.
Basic Transformer Construction-  BEE 2024 BEU PYQ Solution
Working Principle

A transformer works on the principle of mutual induction:

“When alternating current flows through the primary winding, it produces a changing magnetic flux in the core. This changing flux links with the secondary winding and induces an EMF in it.”

According to Faraday’s Law of Electromagnetic Induction, the EMF induced is:

$$ e = -N \frac{d\phi}{dt} $$

EMF Equation

The RMS value of the induced EMF is given by:

$$ E = 4.44 f N \phi_m $$

  • Where:
  • $E$ = Induced EMF (RMS)
  • $f$ = Frequency of AC (Hz)
  • $N$ = Number of turns
  • $\phi_m$ = Maximum magnetic flux (Wb)
Characteristics of Ideal Transformer
  • No copper, core, or leakage losses.
  • Perfect magnetic coupling between windings.
  • 100% efficiency.

Efficiency:

$$ \eta = \frac{\text{Output Power}}{\text{Input Power}} = 100\% $$

(b) Derive the EMF equation of a transformer.

Answer
Derivation of the EMF Equation OF Transformer-  BEE 2024 BEU PYQ Solution

Q.5 (a) Draw a neat sketch of a DC generator and label the component parts. Name the material used for each component part.

Answer
Construction of a DC Generator

A DC generator consists of the following main parts:

1. Yoke
  • Acts as the outer frame and supports the magnetic poles.
  • Material: Cast iron or cast steel.
2. Pole Core and Pole Shoe
  • Pole core holds the field winding and pole shoe spreads the magnetic flux uniformly.
  • Material: Laminated steel.
3. Field Winding
  • Produces magnetic flux when current flows through it.
  • Material: Copper wire.
4. Armature Core
  • Mounted on the shaft and rotates inside the magnetic field.
  • Provides a low reluctance path for flux.
  • Material: Laminated soft iron.
5. Armature Winding
  • EMF is induced in this winding due to rotation.
  • Material: Copper wire.
6. Commutator
  • Converts AC induced in armature to DC at the output terminals.
  • Material: Copper segments insulated by mica.
7. Brushes
  • Conducts current from commutator to external circuit.
  • Material: Carbon or graphite.
8. Shaft
  • Transfers mechanical energy from prime mover to the armature.
  • Material: Mild steel.
9. Bearings
  • Reduces friction and supports the shaft.
  • Material: High carbon steel.


(b)
Explain the operating principle of a three-phase induction motor.

Answer

A three-phase induction motor operates based on the principle of electromagnetic induction and the formation of a rotating magnetic field (RMF).

1. Rotating Magnetic Field (RMF)

When a balanced three-phase AC supply is applied to the stator windings, a rotating magnetic field is produced. This field rotates at a constant speed known as synchronous speed, given by the formula: $$N_s = \frac{120 \times f}{P}$$

Where:
$N_s$ = Synchronous speed in RPM
$f$ = Supply frequency in Hz
$P$ = Number of poles

2. Induced EMF in Rotor

The rotating magnetic field cuts the stationary rotor conductors. According to Faraday’s law of electromagnetic induction, an EMF is induced in the rotor. Since the rotor circuit is closed, current starts flowing in the rotor conductors.

3. Torque Production

The rotor current interacts with the rotating magnetic field, producing a force (as per Lorentz force law) and hence a torque. This torque causes the rotor to rotate in the same direction as the rotating magnetic field.

4. Slip

The rotor never reaches synchronous speed. It always rotates at a slightly lower speed than NsN_s, and this difference is called slip, given by: $$s = \frac{N_s – N_r}{N_s}$$

Where:
$s$ = Slip (dimensionless or %)
$N_r$ = Rotor speed in RPM


Q.6 (a) With neat diagrams, explain various types of fuses used in electrical wiring systems.
Remarks: according to new syllabus, this topic is removed.


(b) Write a detailed note on Fuse and Circuit Breaker.

Remarks: according to new syllabus, this topic is removed.

Q.7 (a) State and explain Thevenin’s Theorem for a DC circuit. Write applications, advantages, and limitations of Thevenin’s Theorem.

Answer

Thevenin’s Theorem states that:
Any linear bilateral DC network, no matter how complex, can be replaced by an equivalent circuit consisting of a single voltage source $V_{th}$ in series with a resistance $R_{th}$, connected to the load.

To apply Thevenin’s Theorem:

  1. Remove the load resistor $R_L$ from the circuit.
  2. Find the open-circuit voltage across the terminals where RLR_L was connected. This is the Thevenin voltage VthV_{th}.
  3. Replace all independent voltage sources with short circuits and current sources with open circuits. Then find the equivalent resistance across the open terminals. This is Thevenin resistance $R_{th}$.
  4. Draw the Thevenin equivalent circuit: a voltage source VthV_{th} in series with resistance $R_{th}$, connected to the load $R_L$.

Applications of Thevenin’s Theorem
  • Used to simplify complex DC circuits
  • Helps in analyzing the effect of varying load resistance
  • Useful in power systems for load flow studies
  • Frequently used in electronics for small-signal analysis
Advantages
  • Reduces complex circuits to simple two-element equivalents
  • Saves time in analyzing the circuit for different load conditions
  • Makes power transfer calculations easy
Limitations
  • Applicable only to linear, bilateral circuits
  • Not valid for circuits with nonlinear components like diodes or transistors
  • Cannot be applied directly to time-varying or non-steady-state circuits


(b) State Norton’s Theorem.

Answer

Norton’s Theorem states that:
Any linear bilateral DC network can be replaced by an equivalent circuit consisting of a single current source $I_N$ in parallel with a resistance $R_N$, connected across the load.

Explanation
  • The current source $I_N$​ is the short-circuit current flowing through the load terminals when they are shorted.
  • The resistance $R_N$​ is the internal resistance of the network seen from the open load terminals, calculated by deactivating all independent sources:
    • Voltage sources → short circuit
    • Current sources → open circuit

Thus, any complex DC circuit can be simplified to a current source in parallel with a resistor, which makes it easier to analyze the behavior of the circuit across varying loads.

Q.8 A three single-phase balanced load connected in three-phase three-wire star form. With the help of phasor diagram, obtain the relationship between line and phase quantities of voltage and current.

Answer


Q.9 Write short notes on any two of the following
(a) Batteries and its types and application

Remarks: according to new syllabus, this topic is removed.


(b) Losses in Transformer

Answer

1. Core Losses (Iron Losses)

  • Occur in the transformer core.
  • Constant losses (do not vary with load).
  • Consist of:
    • Hysteresis Loss: Due to repeated magnetization. Reduced by using silicon steel.
    • Eddy Current Loss: Due to circulating currents in the core. Reduced by laminating the core.

2. Copper Losses (I²R Losses)

  • Occur in primary and secondary windings.
  • Vary with square of load current: $P_{cu} = I^2 R$
  • Zero at no-load, maximum at full-load.


(c) Differences between Balanced, Unbalanced and Faulty Electrical Systems

Answer

1. Balanced Electrical System

  • All three-phase voltages and currents are equal in magnitude and 120° apart in phase.
  • Load on all three phases is identical.
  • No current flows through the neutral in a star-connected system.
  • System operates efficiently and stably.

2. Unbalanced Electrical System

  • Phase voltages or currents are unequal in magnitude and/or not 120° apart.
  • Load on the three phases is not the same.
  • Neutral current exists in star-connected systems.
  • Causes voltage drops and unbalanced power distribution.

3. Faulty Electrical System

  • Occurs due to abnormal conditions like short circuit, open circuit, or insulation failure.
  • Causes sudden rise or fall in current/voltage.
  • Leads to equipment damage if not protected.
  • Requires protection systems like fuses, relays, and circuit breakers.


(d) Working and application of Synchronous Generator

Answer

Working of Synchronous Generator

  • A synchronous generator works on the principle of Faraday’s law of electromagnetic induction.
  • The rotor (field winding) is excited with DC current, creating a rotating magnetic field.
  • When the rotor rotates (by a prime mover like turbine or engine), the magnetic field cuts the stator conductors.
  • An EMF is induced in the stator windings, which is AC in nature.
  • The frequency of the generated voltage is proportional to the rotor speed and number of poles.

Formula: $$f = \frac{N \times P}{120}$$

Where:
$f$ = frequency in Hz,
$N$ = rotor speed in RPM,
$P$ = number of poles

Applications of Synchronous Generator

  • Used in power plants to generate electricity.
  • Employed in hydro, thermal, nuclear, and gas power stations.
  • Suitable for large-scale power generation due to constant speed operation.
  • Used in industries requiring constant voltage and frequency.

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