study notes and tips for BS Electrical Engineering students at Quaid-e-Awam University of Engineering, Science & Technology (QUEST) to excel in their program.tudying Electrical Engineering at Quaid-e-Awam University of Engineering, Science & Technology (QUEST) can be a rewarding experience if you approach it with dedication, hard work, and the right study notes. By following the tips outlined in this article, you can enhance your learning experience and excel in your academic journey. Good luck

Study Notes BS Electrical Engineering At Quaid-e-Awam University of Engineering, Science & Technology (QUEST.

366 Applied Mechanics

Comprehensive Study Notes

Course Overview

Applied Mechanics is a foundational engineering course that applies the principles of mechanics to solve real-world engineering and technological problems. It involves the study and analysis of forces, moments, and their effects on bodies and systems in motion or at rest .

The course typically covers statics (equilibrium of rigid bodies, force systems, friction) and kinematics/dynamics (motion of particles and rigid bodies) .

Typical Assessment Structure:

• Mid-Term Exam: 30%

• Final Exam: 50%

• Quizzes, Assignments & Attendance: 20%

Module 1: Basic Concepts and Force Systems

1.1 Fundamental Concepts

Mechanics: The branch of physics concerned with the behavior of physical bodies when subjected to forces or displacements.

Basic Quantities:

• Mass: Measure of matter in a body

• Force: A push or pull that changes or tends to change a body’s state

• Time: Duration of events

• Space: Geometric region of position

• Scalar vs. Vector: Scalars have magnitude only; vectors have magnitude and direction

1.2 Force Systems

Concurrent Forces: Forces whose lines of action intersect at a single point.

Non-Concurrent Forces: Forces whose lines of action do not intersect at a single point.

Force Resolution: Breaking a force into its rectangular components.

• Fx = F cos θ (horizontal component)

• Fy = F sin θ (vertical component)

Resultant Force:

R = √[(ΣFx)² + (ΣFy)²]

Where ΣFx = sum of horizontal components, ΣFy = sum of vertical components

Methods of Vector Addition:

• Graphical: Triangle law, Parallelogram law, Polygon method

• Analytical: Resolution into components

1.3 Moment of a Force

Moment (Torque): The tendency of a force to rotate a body about a point or axis.

Moment = Force × Perpendicular Distance
M = F × d

Principle of Moments (Varignon’s Theorem): The moment of a resultant force about a point is equal to the sum of the moments of its components about that point .

Couple: Two equal, opposite, and parallel forces with different lines of action. A couple produces pure rotation with no translation .

Module 2: Equilibrium

2.1 Conditions of Equilibrium

For a body to be in equilibrium, both the resultant force and resultant moment must be zero :

Two-Dimensional Equilibrium:

ΣFx = 0  (Sum of horizontal forces)
ΣFy = 0  (Sum of vertical forces)
ΣM = 0   (Sum of moments)

Three-Dimensional Equilibrium:

ΣFx = 0,  ΣFy = 0,  ΣFz = 0
ΣMx = 0,  ΣMy = 0,  ΣMz = 0

2.2 Free Body Diagrams (FBD)

A Free Body Diagram is a sketch showing all external forces acting on a body, with the body isolated from its surroundings .

Steps to Draw FBD:

1. Identify the body of interest

2. Draw the body isolated from all supports

3. Show all applied loads (forces, moments)

4. Show reaction forces at supports

5. Label all forces and dimensions

2.3 Types of Supports and Reactions

Module 3: Truss Analysis

3.1 Introduction to Trusses

A truss is a structure composed of bars united at their extremities by joints, which offer no resistance to rotation .

Assumptions in Truss Analysis:

• Joints are smooth pins

• Loads are applied only at joints

• Members are two-force members (either tension or compression)

Tension vs. Compression:

• Tie (Tension): Member is pulled (arrows point away from joint)

• Strut (Compression): Member is pushed (arrows point toward joint)

3.2 Methods of Truss Analysis

Method of Joints:

• Analyze each joint individually using equilibrium equations (ΣFx = 0, ΣFy = 0)

• Start with a joint having at most two unknown forces

Method of Sections:

• Cut through the truss and analyze a section

• Use equilibrium equations to find internal forces

Example from Textbook:
At joint A with reaction R₁ = 5.5 tonnes (upward):

f_AH sin 45° = 5.5
f_AH = 5.5√2 = 7.78 tonnes (tension) [citation:3]

Module 4: Stress and Strain

4.1 Basic Stress Concepts

Stress (σ): Internal resistance per unit area.

σ = F / A

Where F = applied force, A = cross-sectional area

Types of Stress:

• Tensile Stress: Pulling force (elongation)

• Compressive Stress: Pushing force (shortening)

• Shear Stress: Sliding force parallel to surface

4.2 Strain

Strain (ε): Deformation per unit length.

ε = ΔL / L

Where ΔL = change in length, L = original length

4.3 Hooke’s Law

Hooke’s Law: Stress is directly proportional to strain within elastic limit.

σ = E × ε

Where E = Young’s Modulus (Modulus of Elasticity)

Young’s Modulus (E): Measure of material stiffness.

Bulk Modulus: Ratio of volumetric stress to volumetric strain .

4.4 Material Properties

• Elastic Limit: Maximum stress with no permanent deformation

• Yield Point: Stress at which material begins to deform plastically

• Ultimate Strength: Maximum stress a material can withstand

• Fracture Point: Stress at which material breaks

Module 5: Friction

5.1 Types of Friction

5.2 Laws of Solid Friction

1. Friction force is proportional to normal reaction (F = μN)

2. Friction is independent of contact area

3. Kinetic friction is less than limiting static friction

Coefficient of Friction (μ):

μ = F / N

Angle of Friction (φ):

tan φ = μ

5.3 Inclined Plane Problems

For a body on an inclined plane with angle θ:

• Component of weight parallel to plane: W sin θ

• Component of weight perpendicular to plane: W cos θ

• Force needed to move body up: F = W sin θ + μW cos θ

• Force needed to move body down: F = W sin θ − μW cos θ

Module 6: Kinematics

6.1 Motion Concepts

Displacement: Change in position (vector quantity).

Speed vs. Velocity:

• Speed = distance/time (scalar)

• Velocity = displacement/time (vector)

Acceleration: Rate of change of velocity.

6.2 Equations of Motion (Uniform Acceleration)

v = u + at
s = ut + ½at²
v² = u² + 2as

Where:

• u = initial velocity

• v = final velocity

• a = acceleration

• t = time

• s = displacement

6.3 Curvilinear Motion

Normal and Tangential Acceleration:

• Tangential acceleration: changes speed

• Normal (centripetal) acceleration: changes direction

6.4 Projectile Motion

• Horizontal motion: constant velocity

• Vertical motion: constant acceleration (gravity)

Module 7: Work, Energy, and Power

7.1 Work

Work: Product of force and displacement in direction of force.

W = F × d × cos θ

Where θ = angle between force and displacement

Units: Joule (J) = Newton-meter (N·m)

7.2 Energy

Kinetic Energy (KE):

KE = ½mv²

Potential Energy (PE):

PE = mgh

Work-Energy Theorem: Work done = change in kinetic energy.

W = ΔKE = ½mv² − ½mu² [citation:5]

7.3 Power

Power: Rate of doing work.

P = W / t = F × v

Units: Watt (W) = Joule/second

Types:

• I.H.P. (Indicated Horsepower): Power developed in engine cylinder

• B.H.P. (Brake Horsepower): Power available at output shaft

Module 8: Moment of Inertia

8.1 Definition

Moment of Inertia (I): Measure of resistance to rotational acceleration.

I = Σmr²

Where m = mass, r = distance from axis

8.2 Theorems

Parallel Axis Theorem:

I = I_cm + Ad²

Where I_cm = moment about centroidal axis, A = area, d = distance between axes

Perpendicular Axis Theorem:

I_z = I_x + I_y

(For planar areas)

8.3 Moment of Inertia of Common Shapes

8.4 Flywheel

Flywheel: Stores rotational energy to smooth out power fluctuations.

Energy Stored:

E = ½Iω²

Where I = moment of inertia, ω = angular velocity

Module 9: Simple Machines

9.1 Mechanical Advantage

Theoretical Mechanical Advantage (TMA):

TMA = Load / Effort

Actual Mechanical Advantage (AMA):

AMA = Load / Actual Effort

Efficiency:

η = (AMA / TMA) × 100% [citation:5]

9.2 Common Machines

Summary: Key Formulas

285 Electrical Workshop Practice: Comprehensive Study Notes

COURSE OVERVIEW

Electrical Workshop Practice (EE1104) is a foundational 1-credit practical course (0+1) designed to provide students with hands-on experience in electrical workshop practices, safety procedures, measurement techniques, wiring, and basic fabrication skills . The course emphasizes practical application of electrical engineering principles in a workshop setting, preparing students for advanced laboratory and project work.

Why This Course Matters:

• Safety Foundation: Establishes essential safety habits for all future electrical work

• Practical Skills: Develops competency with tools, instruments, and fabrication techniques

• Industry Readiness: Covers competencies required in professional electrical work

• Project Preparation: Provides basis for senior design projects and PCB fabrication

PART ONE: WORKSHOP SAFETY

1.1 Importance of Safety in Electrical Workshops

Electrical workshops present unique hazards that can cause serious injury or death. The primary objective of safety training is to enable students to demonstrate proper use of safety equipment and adherence to safety regulations in workshop practices, including handling tools and safety gear .

Common Electrical Hazards:

Causes of Electrical Accidents:

• Neglecting to disconnect power during inspection and maintenance

• Using damaged or uninsulated tools

• Ignoring the condition of electrical wiring

• Touching switches or electrical parts with wet hands

• Undersized cables for increased load, causing overheating and fire risk

• Not wearing rubber gloves and shoes when inspecting distribution points

1.2 Personal Protective Equipment (PPE)

The following PPE is essential for electrical workshop activities :

1.3 Electric Shock Treatment and First Aid

Immediate Response to Electric Shock:

1. Do not touch the victim while they are still in contact with the electrical source

2. Disconnect the power source immediately (switch off main breaker or use insulated material to separate victim)

3. Prevent crowding around the victim to avoid disturbance

4. Do not allow stimulants like tea or coffee if bleeding is present

5. Cut clothing rather than pulling it off if it causes pain

6. Provide adequate ventilation; loosen belts, buttons, and tilt head if fainting occurs

7. Perform artificial respiration for extended periods even if apparent death has occurred

8. Transport carefully using a stretcher; call for medical help immediately

9. For burns: Apply Vaseline, cooking oil, or toothpaste to isolate from air; cover with gauze without pressure

1.4 Fire Safety in Workshops

Fire Sources:

• Oil-soaked rags catching sparks

• Lit cigarettes falling on oils

• Electrical equipment safety devices burning

Fire Prevention:

• Store flammable materials in designated areas

• No smoking in the workshop

• Dispose of oil-soaked rags in special containers

• Disconnect all electrical motors when fire occurs

• Ensure fire extinguishers are present and inspected regularly

• Keep escape routes and emergency doors clear

PART TWO: MEASUREMENTS AND PRECISION TOOLS

2.1 Vernier Calipers

Purpose: Precise measurement of internal and external dimensions .

Reading the Vernier Scale:

1. Read the main scale value at the zero of the vernier scale

2. Find the vernier division that aligns with a main scale division

3. Add the vernier reading to the main scale reading

4. Least count = smallest main scale division / number of vernier divisions

2.2 Micrometer

Purpose: Precision measurement of small dimensions with higher accuracy than vernier calipers .

Reading the Micrometer:

1. Read the sleeve (main scale) value

2. Read the thimble (rotating scale) value

3. Least count typically 0.01 mm

2.3 Thread Pitch Gauge

Purpose: Identifying thread pitch of screws and bolts .

Usage: Match the gauge teeth with the thread; the gauge that fits perfectly indicates the thread pitch.

2.4 Error Analysis in Measurements

When using measuring instruments, students should understand error concepts :

Absolute Error: |True Value − Measured Value|

Relative Error: Absolute Error / True Value

Percentage Error: Relative Error × 100%

Example from Resistor Color Code Exercise :

PART THREE: ELECTRICAL CABLES AND SWITCHING DEVICES

3.1 Types of Cables

3.2 Electrical Accessories

3.3 Electrical Wiring Symbols

Standard symbols are used in electrical wiring schematics to represent components :

PART FOUR: WIRING CIRCUITS AND EARTHING

4.1 Wiring Regulations

All electrical installation work must comply with wiring regulations to ensure safety and reliability . Key considerations include:

• Conductor Sizing: Cables must be selected according to current rating and voltage drop

• Protection: Fuses and circuit breakers must be correctly rated

• Earthing: Proper grounding is essential for safety

4.2 Parallel Wiring Circuit Construction

Students learn to assemble parallel electric wiring circuits according to wiring regulations .

Parallel Circuit Characteristics:

• Same voltage across all branches

• Current divides among branches

• If one branch fails, others continue to operate

Basic Parallel Wiring Steps:

1. Prepare a detailed wiring diagram

2. Calculate loads, currents, and cable sizes

3. Select protective devices

4. Mount components (switches, sockets, distribution board)

5. Route cables according to diagram

6. Make connections securely

7. Test continuity, polarity, and insulation

4.3 Earthing Concepts

Purpose of Earthing:

• Provide a low-resistance path for fault currents

• Protect users from electric shock

• Prevent fire hazards from fault currents

Types of Earthing:

• Plate Earthing: Copper or GI plate buried in ground

• Pipe Earthing: GI pipe buried vertically

• Rod Earthing: Copper or GI rod driven into ground

Earth Resistance Test Procedure :

1. Place 3 electrodes in earth, 10m apart

2. Connect terminal E to first electrode

3. Connect terminal P to second electrode

4. Connect terminal C to third electrode

5. Measure resistance after applying voltage

6. Acceptable value: Usually ≤ 5 ohms (varies by application)

4.4 Testing of Electrical Installations

The following tests are performed to verify installation safety and correctness :

PART FIVE: ELECTRIC ARC WELDING

5.1 Introduction to Arc Welding

Electric arc welding is a fabrication process used to join metal work pieces. The process uses an electric arc to generate heat that melts the base metals and filler material, creating a joint upon cooling .

5.2 Types of Joints

Students learn to prepare the following joint types :

5.3 Safety in Welding

Hazards:

• Arc Eye (Flash Burn): UV radiation from arc causes painful eye inflammation

• Burns: Hot metal and sparks

• Fumes: Toxic metal fumes and gases

• Electric Shock: Welding circuit is live

Safety Precautions:

• Wear welding helmet with appropriate shade lens

• Use leather gloves, apron, and protective clothing

• Ensure adequate ventilation

• Check equipment for damage before use

• Keep flammable materials away from work area

PART SIX: LATHE MACHINE FAMILIARIZATION

6.1 Introduction to Lathe Machine

The lathe is a machine tool used for shaping metal and other materials by rotating the work piece against a cutting tool .

6.2 Main Parts of a Lathe

6.3 Lathe Operations

PART SEVEN: CIRCUIT DESIGN AND PCB FABRICATION

7.1 Introduction to Proteus

Proteus is circuit design and simulation software used to :

• Design circuits schematically

• Simulate circuit behavior before fabrication

• Verify functionality

• Generate PCB layouts

7.2 PCB Fabrication Process

Printed Circuit Board (PCB) fabrication involves :

1. Design: Create circuit schematic in Proteus

2. Simulation: Verify circuit operation

3. PCB Layout: Arrange components and route tracks

4. Printing: Transfer layout to copper-clad board

5. Etching: Remove unwanted copper using chemical etchant

6. Drilling: Drill holes for component leads

7. Assembly: Solder components to board

8. Testing: Verify functionality

7.3 Soldering Techniques

Good Soldering Practice:

• Clean surfaces before soldering

• Use appropriate solder and flux

• Heat joint, then apply solder

• Make shiny, smooth joints

• Avoid cold solder joints (dull, irregular)

PART EIGHT: COMPUTER SYSTEM ASSEMBLY

8.1 Assembling and Disassembling a Computer System

Students learn to assemble and disassemble a modern computer system along with its accessories . Key components include:

Safety Precautions:

• Disconnect power before working

• Use anti-static wrist strap

• Handle components by edges

• Keep work area clean and organized

SOLVED EXAM QUESTIONS

Question 1: First Aid for Electric Shock

Problem: A worker in an electrical workshop receives an electric shock and is unconscious. Describe the first aid procedure you would follow.

Solution:

Step 1: Ensure Safety

• Do not touch the victim while they are in contact with the electrical source

• Disconnect the power source immediately (switch off main breaker)

Step 2: Assess and Manage

• Check for breathing and pulse

• If not breathing, begin artificial respiration immediately

• Continue for extended periods even if apparent death has occurred

Step 3: Prevent Complications

• Prevent crowding around the victim

• Loosen belts, buttons, and clothing

• Tilt head if fainting occurs

Step 4: Manage Burns if Present

• Apply Vaseline, cooking oil, or toothpaste to burn to isolate from air

• Cover with gauze without applying pressure

Step 5: Seek Medical Help

• Call for medical assistance immediately

• Transport victim carefully using a stretcher

Question 2: Insulation Resistance Test

Problem: Describe the procedure for performing an insulation resistance test on a domestic electrical circuit. What is the acceptance criterion?

Solution:

Procedure :

1. Switch off the main breaker to isolate the circuit

2. Short circuit all lighting connections (connect live and neutral together)

3. Connect the megger between line and earth

4. Rotate the megger handle to generate test voltage (typically 500V DC)

5. Read the megger dial after the reading stabilizes

Acceptance Criterion:

• The insulation resistance should not be less than 50 MΩ of any single circuit

Interpretation:

• Reading ≥ 50 MΩ: Circuit insulation is satisfactory

• Reading < 50 MΩ: Insulation is degraded; investigate and repair

Question 3: Resistor Color Code and Error Calculation

Problem: A resistor has color bands: Violet, Green, Blue, with a Gold tolerance band. Calculate the coded value, and if the measured value is 76 Ω, determine the absolute and relative error.

Solution:

Step 1: Decode the Color Code

Coded Value: 75 × 10⁰ = 75 Ω (with Gold tolerance ±5%)

Step 2: Calculate Errors

Absolute Error = |True Value − Measured Value| = |75 − 76| = 1 Ω

Relative Error = Absolute Error / True Value = 1/75 = 0.0133

Percentage Error = Relative Error × 100% = 1.33%

Question 4: Continuity/Loop Test

Problem: Describe the continuity/loop test procedure for an electrical installation and state the acceptance criterion.

Solution:

Procedure :

1. Connect all metal-clad switches, metal parts, and conduits with earth

2. Open the main switch (isolate supply)

3. Turn on all switches

4. Connect earth continuity tester with conduit and independent earth

5. Measure the resistance value

Acceptance Criterion:

• The resistance value must not be more than 1 ohm

Purpose: Ensures that all exposed metal parts are effectively earthed, providing protection against electric shock in case of fault.

Question 5: Electrical Fire Prevention

Problem: List the main fire hazards in an electrical workshop and describe prevention measures.

Solution:

Fire Hazards :

Prevention Measures :

1. Store flammable materials in designated areas

2. No smoking in the workshop

3. Dispose of oil-soaked rags in special containers

4. Disconnect all electrical motors when fire occurs

5. Call fire fighting team

6. Train on fire extinguisher use

7. Ensure fire extinguishers are present and inspected regularly

8. Keep calm and open doors for orderly evacuation

9. Keep escape routes and emergency doors clear of obstacles

10. Use correct cable sizes for load requirements

REVIEW QUESTIONS

Theoretical Questions

1. List the essential Personal Protective Equipment (PPE) required for electrical workshop activities and explain the purpose of each .

2. Describe the causes of electrical accidents in workshops and the prevention measures .

3. Explain the procedure for performing an insulation resistance test. What is the acceptance criterion?

4. Describe the types of electric arc welding joints and their applications .

5. What are the main parts of a lathe machine? Describe three common lathe operations .

6. List five materials and accessories used in house wiring and explain their functions .

7. Explain the purpose of earthing in electrical installations. Describe the earth resistance test procedure .

8. Describe the steps involved in PCB fabrication using Proteus software .

9. What are the safety precautions for assembling and disassembling a computer system?

10. Explain the difference between open circuit, short circuit, and continuity tests for electrical installations .

Application Exercises

Exercise 1: A resistor has color bands: Red, Violet, Orange, Gold. Decode the value and calculate the percentage error if measured value is 27,500 Ω.

Exercise 2: You receive an electric shock while working on a live circuit. Describe the immediate first aid steps you would take after being rescued.

Exercise 3: Prepare a wiring diagram for a simple parallel circuit with two switches controlling two lamps from a single power source.

Exercise 4: Describe the procedure to test the earth resistance of an installation. What readings would be acceptable?

Exercise 5: List the steps to fabricate a simple PCB for an LED flasher circuit using Proteus.

GLOSSARY OF KEY TERMS

RECOMMENDED RESOURCES

Primary Textbooks

• Umesh Rathore and Naresh Kumar Sharma. A Textbook of Electrical Workshop Practices. S.K. Kataria & Sons, 1st Ed., 2019

• S. K. Choudhury. Elements of Workshop Technology, Vol. 1. Media Promoters & Publishers

• Chapman. Workshop Technology, Part-I, II, III. CBS 

374 Theory of Electromagnetic Field — Study Guide

Part 1: Key Concepts

1. Introduction and Mathematical Foundations

Course Overview
This course introduces the basic principles of electromagnetic field theory, electromagnetic waves, their radiation, propagation, and detection . It requires knowledge of vector calculus, differential equations, and fundamentals of electrical engineering .

Vector Analysis Tools

• Gradient, Divergence, and Curl: Essential operations for describing field behavior

• Gauss’s Divergence Theorem: Relates volume integrals of divergence to surface integrals

• Stokes’ Theorem: Relates line integrals of curl to surface integrals

• Laplacian Operator: Appears in wave and Laplace equations

2. Electrostatics

Coulomb’s Law
The force between two point charges:

F=14πε0q1q2r2r^F=4πε0​1​r2q1​q2​​r^

Electric Field and Potential

• Electric Field: $\mathbf{E}=-\nabla \Phi$

• Electric Potential: Φ=14πε0qrΦ=4πε0​1​rq​ for a point charge

Gauss’s Law (Electrostatics)

$$\nabla \cdot \mathbf{D}=\rho$$

In integral form:

∮D⋅dA=Qenclosed∮D⋅dA=Qenclosed​

Key Concepts:

• Conductors in electric fields: charge resides on surface, field inside is zero

• Dielectrics: polarization, bound charge, electric displacement field $\mathbf{D}$

• Boundary conditions: E1t=E2tE1t​=E2t​ and D1n−D2n=σD1n​−D2n​=σ

• Capacitance calculation

• Poisson’s and Laplace’s equations: ∇2Φ=−ρ/ε∇2Φ=−ρ/ε and ∇2Φ=0∇2Φ=0

• Method of images for boundary value problems

3. Magnetostatics

Biot-Savart Law

dB=μ0I4πdl×r^r2dB=4πμ0​I​r2dl×r^​

Ampere’s Law

$$\nabla \times \mathbf{H}=\mathbf{J}$$

In integral form:

∮H⋅dl=Ienclosed∮H⋅dl=Ienclosed​

Magnetic Vector Potential

$$\mathbf{B}=\nabla \times \mathbf{A}$$

Key Concepts:

• Magnetic dipoles and dipole fields

• Magnetization, bound current densities

• Magnetic field strength $\mathbf{H}$ and flux density $\mathbf{B}$

• Boundary conditions: H1t=H2tH1t​=H2t​ and B1n=B2nB1n​=B2n​

• Inductance calculation (self and mutual)

• Magnetic energy and energy density

• Ferromagnetic hysteresis

4. Maxwell’s Equations

Differential Form:

∇⋅D=ρ(Gauss’s Law)∇⋅B=0(Gauss’s Law for Magnetism)∇×E=−∂B∂t(Faraday’s Law)∇×H=J+∂D∂t(Ampere-Maxwell Law)∇⋅D∇⋅B∇×E∇×H​=ρ(Gauss’s Law)=0(Gauss’s Law for Magnetism)=−∂t∂B​(Faraday’s Law)=J+∂t∂D​(Ampere-Maxwell Law)​

Integral Form:

∮D⋅dA=Qenc∮B⋅dA=0∮E⋅dl=−dΦBdt∮H⋅dl=Ienc+dΦDdt∮D⋅dA∮B⋅dA∮E⋅dl∮H⋅dl​=Qenc​=0=−dtdΦB​​=Ienc​+dtdΦD​​​

Constitutive Relations:

$$\mathbf{D}=\epsilon \mathbf{E},\mathbf{B}=\mu \mathbf{H},\mathbf{J}=\sigma \mathbf{E}$$

5. Electromagnetic Waves

Wave Equation
From Maxwell’s equations in source-free media:

∇2E=με∂2E∂t2∇2E=με∂t2∂2E​

The same equation holds for $\mathbf{H}$. The wave propagates with velocity:

v=1με=cμrεrv=με​1​=μr​εr​​c​

Plane Waves in Vacuum

• $\mathbf{E}$ and $\mathbf{H}$ are perpendicular to each other and to the direction of propagation

• $\mathbf{E}$ and $\mathbf{H}$ are in phase

• Intrinsic impedance: η0=μ0/ε0≈377 Ωη0​=μ0​/ε0​​≈377Ω

• Relation: $E=cB$

Poynting Vector

$$\mathbf{S}=\mathbf{E}\times \mathbf{H}$$

Represents the directional energy flux (power per unit area). Time-averaged Poynting vector:

⟨S⟩=12Re(E×H∗)⟨S⟩=21​Re(E×H∗)

Polarization, Reflection, and Refraction

• Linear, Circular, and Elliptical Polarization

• Fresnel Equations: Determine reflection and transmission coefficients at interfaces

• Brewster Angle: Angle at which reflected light is completely polarized

• Total Internal Reflection: Occurs when light travels from denser to rarer medium beyond critical angle

6. Potentials and Gauge Transformations

Vector and Scalar Potentials

B=∇×A,E=−∇Φ−∂A∂tB=∇×A,E=−∇Φ−∂t∂A​

Gauge Transformations

• Coulomb Gauge: $\nabla \cdot \mathbf{A}=0$

• Lorentz Gauge: ∇⋅A+με∂Φ∂t=0∇⋅A+με∂t∂Φ​=0

Retarded Potentials

Φ(r,t)=14πε0∫ρ(r′,t−∣r−r′∣/c)∣r−r′∣d3r′Φ(r,t)=4πε0​1​∫∣r−r′∣ρ(r′,t−∣r−r′∣/c)​d3r′

7. Radiation and Antennas

Hertz Dipole Radiation
The Hertz dipole is a short oscillating current element. Far-field radiation pattern:

• Electric field: Eθ∝sin⁡θrEθ​∝rsinθ​

• Magnetic field: Hϕ∝sin⁡θrHϕ​∝rsinθ​

• Power radiated: P∝ω4P∝ω4

Antenna Parameters

• Radiation resistance

• Directivity and gain

• Effective aperture

8. Electromagnetic Fields in Materials

Conductors

• Skin Effect: Current concentrates near the surface at high frequencies

• Skin Depth: δ=2ωμσδ=ωμσ2​​

• Complex Permittivity: εc=ε−jσωεc​=ε−jωσ​

Dielectrics

• Polarization mechanisms

• Frequency-dependent permittivity

• Dielectric loss

Magnetic Materials

• Diamagnetic, paramagnetic, and ferromagnetic responses

• Hysteresis and saturation

Part 2: Solved Examination Questions

Question 1: Coulomb’s Law and Electric Field

Two point charges q1=2×10−6 Cq1​=2×10−6C and q2=−3×10−6 Cq2​=−3×10−6C are separated by 0.5 m in vacuum. Calculate the magnitude and direction of the force on q2q2​.

Solution:
Using Coulomb’s Law:

F=14πε0∣q1q2∣r2F=4πε0​1​r2∣q1​q2​∣​

where 14πε0=9×109 N⋅m2/C24πε0​1​=9×109N⋅m2/C2.

F=9×109×(2×10−6)(3×10−6)(0.5)2F=9×109×(0.5)2(2×10−6)(3×10−6)​F=9×109×6×10−120.25F=9×109×0.256×10−12​F=9×109×2.4×10−11=0.216 NF=9×109×2.4×10−11=0.216N

Direction: Attractive (opposite charges), directed from q2q2​ toward q1q1​.

Answer: $F=0.216\text{N}$, attractive.

Question 2: Gauss’s Law Application

A point charge $Q=5\mu \text{C}$ is at the center of a spherical Gaussian surface of radius $r=10\text{cm}$. Calculate the electric flux through the surface.

Solution:
Gauss’s Law:

ΦE=∮E⋅dA=Qencε0ΦE​=∮E⋅dA=ε0​Qenc​​ΦE=5×10−68.85×10−12=5.65×105 N⋅m2/CΦE​=8.85×10−125×10−6​=5.65×105N⋅m2/C

Note: The flux is independent of the radius of the Gaussian surface.

Answer: ΦE=5.65×105 N⋅m2/CΦE​=5.65×105N⋅m2/C.

Question 3: Ampere’s Law

An infinitely long straight wire carries a current $I=10\text{A}$. Calculate the magnetic field at a distance $r=5\text{cm}$ from the wire.

Solution:
Using Ampere’s Law for an infinite wire:

B=μ0I2πrB=2πrμ0​I​B=(4π×10−7)(10)2π(0.05)B=2π(0.05)(4π×10−7)(10)​B=4×10−60.1=4×10−5 TB=0.14×10−6​=4×10−5T

Answer: B=4×10−5 T=40 μTB=4×10−5T=40μT.

Question 4: Electromagnetic Plane Wave

The magnetic field of a plane wave in vacuum is given by:

B=B0sin⁡(kx−ωt)y^B=B0​sin(kx−ωt)y^​

where y^y^​ is a unit vector pointing in the positive y-direction.

(a) Give the wavelength $\lambda$ of this wave.
(b) Write an expression for the electric field $\mathbf{E}$.
(c) Find the direction and magnitude of the Poynting vector $\mathbf{S}$.

Solution:

(a) Wavelength:

λ=2πkλ=k2π​

(b) Electric Field:
For an electromagnetic wave in vacuum, $\mathbf{E}$ and $\mathbf{B}$ are related by:

E=cB×k^E=cB×k^

where k^k^ is the propagation direction.

The wave propagates in the $+x$ direction (since $kx-\omega t$ implies propagation in $+x$). Given $\mathbf{B}$ is in y^y^​, then $\mathbf{E}$ must be in z^z^:

E=cB0sin⁡(kx−ωt)z^E=cB0​sin(kx−ωt)z^

(c) Poynting Vector:

S=1μ0E×BS=μ0​1​E×BS=1μ0[cB0sin⁡(kx−ωt)z^]×[B0sin⁡(kx−ωt)y^]S=μ0​1​[cB0​sin(kx−ωt)z^]×[B0​sin(kx−ωt)y^​]

Since z^×y^=−x^z^×y^​=−x^:

S=−cB02μ0sin⁡2(kx−ωt)x^S=−μ0​cB02​​sin2(kx−ωt)x^

The magnitude is S=cB02μ0sin⁡2(kx−ωt)S=μ0​cB02​​sin2(kx−ωt), directed in the −x^−x^ direction.

Answer: (a) $\lambda =2\pi /k$; (b) E=cB0sin⁡(kx−ωt)z^E=cB0​sin(kx−ωt)z^; (c) S=−cB02μ0sin⁡2(kx−ωt)x^S=−μ0​cB02​​sin2(kx−ωt)x^ .

Question 5: Maxwell’s Equations in a Conductor

Starting from Maxwell’s equations, derive the wave equation for the magnetic field $\mathbf{H}$ in a conducting medium with conductivity $\sigma$, permittivity $\epsilon$, and permeability $\mu$.

Solution:

Start with Ampere-Maxwell Law:

∇×H=σE+ε∂E∂t∇×H=σE+ε∂t∂E​

Take the curl of both sides:

∇×(∇×H)=σ(∇×E)+ε∂∂t(∇×E)∇×(∇×H)=σ(∇×E)+ε∂t∂​(∇×E)

Use the vector identity:

∇×(∇×H)=∇(∇⋅H)−∇2H∇×(∇×H)=∇(∇⋅H)−∇2H

Since $\nabla \cdot \mathbf{H}=0$ (from $\nabla \cdot \mathbf{B}=0$ and $\mathbf{B}=\mu \mathbf{H}$ for constant $\mu$), we have:

−∇2H=σ(∇×E)+ε∂∂t(∇×E)−∇2H=σ(∇×E)+ε∂t∂​(∇×E)

From Faraday’s Law: ∇×E=−μ∂H∂t∇×E=−μ∂t∂H​.

Substituting:

−∇2H=σ(−μ∂H∂t)+ε∂∂t(−μ∂H∂t)−∇2H=σ(−μ∂t∂H​)+ε∂t∂​(−μ∂t∂H​)−∇2H=−σμ∂H∂t−εμ∂2H∂t2−∇2H=−σμ∂t∂H​−εμ∂t2∂2H​

Rearranging:

∇2H=σμ∂H∂t+εμ∂2H∂t2∇2H=σμ∂t∂H​+εμ∂t2∂2H​​

Answer: The wave equation for $\mathbf{H}$ in a conductor is ∇2H=σμ∂H∂t+εμ∂2H∂t2∇2H=σμ∂t∂H​+εμ∂t2∂2H​ .

EL1 Electrical Network Analysis: Study Notes & Solved Problems

⚠️ Course Identification Notice

The search results confirm that EL1 is a course code used at Imperial College London for “Analysis of Circuits” (also called Circuit Analysis) . The course covers fundamental circuit analysis techniques including nodal analysis, Thévenin/Norton equivalents, and AC phasor analysis.

At NUST (Pakistan), the equivalent course is EE-211 Electrical Network Analysis, which covers phasor analysis, AC power, Laplace transforms, frequency response, and two-port networks . At PEC (Pakistan Engineering Council), the standard Electrical Engineering curriculum includes Network Analysis covering matrices, two-port networks, and filter circuits .

The content below synthesizes these curricula, which match the standard scope of an introductory-to-intermediate electrical network analysis course.

Note: If your specific EL1 course covers different material, please verify your course outline.

Part 1: Core Course Framework

Electrical Network Analysis introduces the fundamental techniques for analyzing linear circuits containing resistors, capacitors, inductors, and sources. The course progresses from DC circuit analysis (Kirchhoff’s laws, nodal/mesh analysis, network theorems) through AC steady-state analysis (phasors, impedance, power) to transient analysis and frequency response .

Core Learning Objectives:
By the end of EL1, you should be able to:

1. Apply Kirchhoff’s laws and Ohm’s law to analyze DC circuits

2. Use nodal and mesh analysis to solve complex networks

3. Apply Thévenin’s and Norton’s theorems for circuit simplification

4. Analyze AC circuits using phasor techniques

5. Compute AC power (real, reactive, apparent) and power factor

6. Analyze transient response of RC and RL circuits

7. Understand frequency response and resonance

Part 2: Key Concepts by Topic

Topic 1: Fundamental Circuit Laws and Concepts

Basic Quantities:

• Charge (q): Measured in Coulombs (C)

• Current (i): Rate of charge flow, measured in Amperes (A). i=dqdti=dtdq​

• Voltage (v): Energy per unit charge, measured in Volts (V)

• Power (p): Rate of energy transfer, measured in Watts (W). $p=vi$

• Energy (w): $w=\int pdt$

Ohm’s Law: $V=IR$, where $R$ is resistance in Ohms (Ω). Conductance $G=1/R$, measured in Siemens (S) .

Passive Sign Convention: When current enters the positive terminal of a component, the power absorbed is $P=VI$. If current exits the positive terminal, the power is $P=-VI$ .

Kirchhoff’s Laws:

• Kirchhoff’s Current Law (KCL): The algebraic sum of currents entering a node is zero. ∑Iin=∑Iout∑Iin​=∑Iout​

• Kirchhoff’s Voltage Law (KVL): The algebraic sum of voltages around any closed loop is zero. ∑Vloop=0∑Vloop​=0

Series and Parallel Combinations:

• Series Resistors: Req=R1+R2+⋯+RnReq​=R1​+R2​+⋯+Rn​

• Parallel Resistors: 1Req=1R1+1R2+⋯+1RnReq​1​=R1​1​+R2​1​+⋯+Rn​1​. For two resistors: Req=R1R2R1+R2Req​=R1​+R2​R1​R2​​ .

Voltage and Current Dividers:

• Voltage Divider: Vx=Vtotal×RxRtotalVx​=Vtotal​×Rtotal​Rx​​

• Current Divider: Ix=Itotal×RotherRx+RotherIx​=Itotal​×Rx​+Rother​Rother​​

Topic 2: Nodal and Mesh Analysis

Nodal Analysis: Uses KCL at each non-reference node to find node voltages .

Procedure:

1. Select a reference node (ground)

2. Label remaining node voltages V1,V2,…,VnV1​,V2​,…,Vn​

3. Write KCL at each non-reference node

4. Solve the resulting simultaneous equations

Example: For a node VXVX​ with a voltage source VSVS​ through resistor R1R1​, and resistors R2,R3R2​,R3​ to ground:

VX−VSR1+VXR2+VXR3=0R1​VX​−VS​​+R2​VX​​+R3​VX​​=0

Mesh Analysis: Uses KVL around each independent loop to find mesh currents.

Supernode: When a voltage source is between two non-reference nodes, combine them into a supernode and write KCL for the combined node .

Supermesh: When a current source is shared between two meshes, combine them and write KVL for the combined loop.

Topic 3: Network Theorems

Linearity and Superposition: For linear circuits with multiple independent sources, the response is the sum of responses to each source acting alone. To apply superposition: set voltage sources to 0 V (short circuit) and current sources to 0 A (open circuit) when removing a source .

Thévenin’s Theorem: Any linear two-terminal network can be replaced by an equivalent circuit consisting of a voltage source VThVTh​ in series with a resistance RThRTh​ .

Finding Thévenin Equivalent:

1. Remove the load

2. Find open-circuit voltage Voc=VThVoc​=VTh​

3. Find RThRTh​ by turning off all independent sources and calculating equivalent resistance, OR by RTh=Voc/IscRTh​=Voc​/Isc​

Norton’s Theorem: Any linear two-terminal network can be replaced by a current source ININ​ in parallel with a resistance RNRN​ .

Relationship: VTh=IN×RNVTh​=IN​×RN​ and RN=RThRN​=RTh​

Maximum Power Transfer: Maximum power is delivered to the load when RL=RThRL​=RTh​. Maximum power Pmax=VTh24RThPmax​=4RTh​VTh2​​ .

Star-Delta (Y-Δ) Transformation: Converts between wye and delta configurations using:

RΔ=R1R2+R2R3+R3R1R1RΔ​=R1​R1​R2​+R2​R3​+R3​R1​​

for conversion from wye to delta .

Topic 4: AC Fundamentals and Phasor Analysis

Sinusoidal Signals: v(t)=Vmcos⁡(ωt+ϕ)v(t)=Vm​cos(ωt+ϕ), where VmVm​ is amplitude, $\omega =2\pi f$ is angular frequency, and $\varphi$ is phase angle.

RMS (Effective) Value: Vrms=Vm2Vrms​=2​Vm​​ for sinusoidal signals .

Phasor Representation: A sinusoidal quantity is represented as a complex number:

Vmcos⁡(ωt+ϕ)↔Vm∠ϕVm​cos(ωt+ϕ)↔Vm​∠ϕ

Impedance:

• Resistor: ZR=RZR​=R

• Inductor: ZL=jωL=jXLZL​=jωL=jXL​, where XL=ωLXL​=ωL

• Capacitor: ZC=1jωC=−jXCZC​=jωC1​=−jXC​, where XC=1ωCXC​=ωC1​

Series RLC Impedance: Z=R+j(XL−XC)Z=R+j(XL​−XC​)

Admittance: $Y=1/Z$, measured in Siemens.

Topic 5: AC Power Analysis

Instantaneous Power: $p(t)=v(t)i(t)$

Average (Real) Power: P=VrmsIrmscos⁡(θ)P=Vrms​Irms​cos(θ), where $\theta$ is the phase difference between voltage and current .

Reactive Power: Q=VrmsIrmssin⁡(θ)Q=Vrms​Irms​sin(θ), measured in VAR (volt-amperes reactive).

Apparent Power: S=VrmsIrmsS=Vrms​Irms​, measured in VA (volt-amperes). $S=P+jQ$

Power Factor: $PF=\cos (\theta )$, where θ=θv−θiθ=θv​−θi​. Unity power factor means $\theta =0$ .

Power Factor Correction: Adding a capacitor in parallel with an inductive load reduces the phase angle, improving power factor .

Topic 6: Transient Analysis (RC and RL Circuits)

RC Circuit Time Constant: $\tau =RC$

Capacitor Charging: vC(t)=VS(1−e−t/τ)vC​(t)=VS​(1−e−t/τ)
Capacitor Discharging: vC(t)=V0e−t/τvC​(t)=V0​e−t/τ

RL Circuit Time Constant: $\tau =L/R$

Inductor Current Rise: iL(t)=IS(1−e−t/τ)iL​(t)=IS​(1−e−t/τ)
Inductor Current Decay: iL(t)=I0e−t/τiL​(t)=I0​e−t/τ

General Solution: The complete response of a first-order circuit is:

x(t)=x(∞)+[x(0+)−x(∞)]e−t/τx(t)=x(∞)+[x(0+)−x(∞)]e−t/τ

where x(0+)x(0+) is the initial value and $x(\infty )$ is the steady-state final value .

Topic 7: Frequency Response and Resonance

Transfer Function: H(jω)=Vout(jω)Vin(jω)H(jω)=Vin​(jω)Vout​(jω)​

Series Resonance: Occurs when XL=XCXL​=XC​, so ω0=1LCω0​=LC​1​. At resonance, impedance is minimum ($Z=R$) and current is maximum.

Parallel Resonance: Occurs when XL=XCXL​=XC​. At resonance, impedance is maximum and current is minimum.

Quality Factor (Q): Q=ω0LR=1ω0RC=1RLCQ=Rω0​L​=ω0​RC1​=R1​CL​​. Higher Q means sharper resonance .

Bandwidth: BW=ω0Q=RLBW=Qω0​​=LR​

Part 3: Solved Problems

Problem 1: Nodal Analysis

Question: Use nodal analysis to find VXVX​ in the circuit where a 14V source is connected through a 1Ω resistor to node VXVX​, and node VXVX​ is connected to ground through 2Ω and 4Ω resistors in parallel.

Solution:

Step 1: Write KCL at node VXVX​.

VX−141+VX2+VX4=01VX​−14​+2VX​​+4VX​​=0

Step 2: Simplify.

VX−14+0.5VX+0.25VX=0VX​−14+0.5VX​+0.25VX​=01.75VX=141.75VX​=14VX=8 VVX​=8 V

Step 3: Alternative method using parallel resistors.
The 2Ω and 4Ω in parallel give 2×42+4=432+42×4​=34​ Ω.

Using voltage divider: VX=14×4/31+4/3=14×4/37/3=14×47=8VX​=14×1+4/34/3​=14×7/34/3​=14×74​=8 V.

Answer: VX=8VX​=8 V .

Problem 2: Superposition

Question: Use superposition to find VXVX​ in a circuit with a 6V voltage source through 4Ω, a 1Ω resistor to ground, and a 4A current source at node VXVX​.

Solution:

Case 1: Voltage source active, current source set to 0 (open).
The 4Ω and 1Ω form a voltage divider:

VX(1)=6×14+1=1.2 VVX(1)​=6×4+11​=1.2 V

Case 2: Current source active, voltage source set to 0 (short).
The 4Ω and 1Ω are in parallel: Req=4×14+1=0.8Req​=4+14×1​=0.8 Ω.