Fundamentals of Electrical Engineering (DI01000101) - Winter 2024 Solution

Question 1(a) [3 marks]

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Explain ohm’s law with its limitation and application.

Answer:

Table: Ohm’s Law Summary

Limitations:

• Temperature dependency: Resistance changes with temperature
• Non-linear materials: Does not apply to semiconductors, diodes
• AC circuits: Modified form needed for reactive components

Applications:

• Circuit analysis: Calculate unknown voltage, current, or resistance
• Power calculations: P = V²/R, P = I²R

Mnemonic: “Voltage Is Really Important” (V = I × R)

Question 1(b) [4 marks]

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Explain faraday’s law of electromagnetic induction with necessary figure.

Answer:

Faraday’s Laws:

• First Law: EMF is induced when magnetic flux changes through conductor
• Second Law: Magnitude of EMF equals rate of flux change

Mathematical Expression:

e = -N × (dΦ/dt)

Diagram:

Applications:

• Transformers: Mutual induction principle
• Generators: Mechanical to electrical energy conversion
• Inductors: Self-induced EMF opposes current changes

Mnemonic: “Flux Change Generates EMF” (dΦ/dt = EMF)

Question 1(c) [7 marks]

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Explain kirchhoff’s voltage law and kirchhoff’s current law with necessary diagram.

Answer:

Table: Kirchhoff’s Laws Comparison

KVL Diagram:

graph LR
    A[+] --> B[V1]
    B --> C[R1]
    C --> D[V2]
    D --> E[R2]
    E --> A
    
    style A fill:#f9f,stroke:#333,stroke-width:2px
    style C fill:#bbf,stroke:#333,stroke-width:2px
    style E fill:#bbf,stroke:#333,stroke-width:2px

KCL Diagram:

graph TD
    A[I1] --> B((Node))
    C[I2] --> B
    B --> D[I3]
    B --> E[I4]
    
    style B fill:#f96,stroke:#333,stroke-width:4px

Key Points:

• KVL: Algebraic sum considers voltage polarities
• KCL: Considers current directions (incoming vs outgoing)
• Applications: Circuit analysis, finding unknown values

Mnemonic: “Voltage Loops, Current Nodes” (KVL for loops, KCL for nodes)

Question 1(c OR) [7 marks]

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Differentiate statically induced emf and dynamically induced emf

Answer:

Table: Static vs Dynamic EMF

Static EMF Types:

• Self-induced: Same coil creates and experiences flux change
• Mutually induced: One coil affects another coil

Dynamic EMF Factors:

• Magnetic field strength (B): Tesla
• Conductor length (L): Meters
• Velocity (v): m/s

Mnemonic: “Static Stays, Dynamic Dances” (Static = stationary, Dynamic = motion)

Question 2(a) [3 marks]

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Explain various types of losses in transformer.

Answer:

Table: Transformer Losses

Iron Losses:

• Hysteresis loss: Magnetic domain reversal energy
• Eddy current loss: Circulating currents in core

Copper Losses:

• Primary winding: I₁²R₁
• Secondary winding: I₂²R₂

Mnemonic: “Iron Core, Copper Coil” (Location of main losses)

Question 2(b) [4 marks]

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Explain working principle of transformer.

Answer:

Working Principle: Mutual electromagnetic induction between primary and secondary windings through common magnetic core.

Diagram:

graph LR
    AC[AC Supply] --> P[Primary Coil N1]
    P --> C[Iron Core]
    C --> S[Secondary Coil N2]
    S --> L[Load]
    
    style C fill:#f96,stroke:#333,stroke-width:3px
    style P fill:#bbf,stroke:#333,stroke-width:2px
    style S fill:#bfb,stroke:#333,stroke-width:2px

Operation Steps:

• Step 1: AC current in primary creates alternating flux
• Step 2: Flux links secondary through core
• Step 3: Changing flux induces EMF in secondary
• Step 4: Secondary EMF drives current through load

Key Relations:

• Voltage ratio: V₂/V₁ = N₂/N₁
• Current ratio: I₁/I₂ = N₂/N₁

Mnemonic: “Primary Produces, Secondary Supplies” (Energy transfer direction)

Question 2(c) [7 marks]

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Derive emf equation of transformer.

Answer:

Given Parameters:

• N₁: Primary turns, N₂: Secondary turns
• Φₘ: Maximum flux, f: Frequency

EMF Derivation:

Step 1: Flux Variation

Φ = Φₘ sin(2πft)

Step 2: Rate of Flux Change

dΦ/dt = 2πfΦₘ cos(2πft)

Step 3: Maximum Rate

(dΦ/dt)ₘₐₓ = 2πfΦₘ

Step 4: RMS EMF Formula

E₁ = 4.44 × f × N₁ × Φₘ
E₂ = 4.44 × f × N₂ × Φₘ

Table: EMF Equation Components

Transformation Ratio:

K = E₂/E₁ = N₂/N₁

Mnemonic: “Four-Forty-Four Flux Formula” (4.44 factor)

Question 2(a OR) [3 marks]

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Write application of transformer.

Answer:

Table: Transformer Applications

Industrial Applications:

• Welding transformers: High current, low voltage
• Instrument transformers: Measurement and protection
• Audio transformers: Impedance matching

Mnemonic: “Power Distribution Isolation Electronics” (Main application areas)

Question 2(b OR) [4 marks]

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Write equation for back emf and torque of D.C motor.

Answer:

Back EMF Equation:

Eb = (φ × Z × N × P) / (60 × A)

Simplified Form:

Eb = K × φ × N

Torque Equation:

T = (φ × Z × Ia × P) / (2π × A)

Simplified Form:

T = K × φ × Ia

Table: Symbol Definitions

Mnemonic: “Back EMF opposes, Torque proposes” (EMF opposes supply, torque drives rotation)

Question 2(c OR) [7 marks]

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Explain construction and working of D.C. motor with necessary figure

Answer:

Construction Components:

Table: DC Motor Parts

Construction Diagram:

graph TB
    subgraph "DC Motor"
        N[N Pole] --> A[Armature]
        A --> S[S Pole]
        C[Commutator] --> B[Brushes]
        F[Field Winding] --> N
        F --> S
    end
    
    style A fill:#f96,stroke:#333,stroke-width:3px
    style C fill:#bbf,stroke:#333,stroke-width:2px

Working Principle:

• Step 1: Current flows through armature conductors
• Step 2: Magnetic field interacts with current
• Step 3: Force generated by Fleming’s left-hand rule
• Step 4: Commutator reverses current direction
• Step 5: Continuous rotation maintained

Force Equation:

F = B × I × L

Mnemonic: “Current Creates Circular motion” (Current interaction produces rotation)

Question 3(a) [3 marks]

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Explain construction of transformer.

Answer:

Table: Transformer Construction

Core Types:

• Shell type: Windings surrounded by core
• Core type: Core surrounded by windings

Cooling Methods:

• Air cooling: Small transformers
• Oil cooling: Large transformers with radiators

Mnemonic: “Core Carries Current Carefully” (Core design importance)

Question 3(b) [4 marks]

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Explain application of DC motor

Answer:

Table: DC Motor Applications

Industrial Applications:

• Shunt motors: Machine tools requiring constant speed
• Series motors: Electric vehicles, starting heavy loads
• Compound motors: Rolling mills, punch presses

Advantages:

• Easy speed control: Voltage/field control
• High starting torque: Series motors
• Reversible operation: Change field/armature polarity

Mnemonic: “Shunt Stays, Series Speeds” (Speed characteristics)

Question 3(c) [7 marks]

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Explain different types of DC motor.

Answer:

Table: DC Motor Classification

Shunt Motor Diagram:

graph LR
    V[DC Supply] --> A[Armature]
    V --> F[Field Winding]
    A --> V
    F --> V
    
    style A fill:#f96,stroke:#333,stroke-width:2px
    style F fill:#bbf,stroke:#333,stroke-width:2px

Characteristics:

• Shunt: Speed ∝ (V - IaRa)/φ
• Series: High starting torque, speed varies with load
• Compound: Combines advantages of both types

Speed Control Methods:

• Armature control: Vary armature voltage
• Field control: Vary field current
• Resistance control: Add external resistance

Mnemonic: “Shunt Steady, Series Strong, Compound Combined” (Key characteristics)

Question 3(a OR) [3 marks]

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Explain transformation ratio of transformer.

Answer:

Definition: Transformation ratio (K) is the ratio of secondary to primary voltage or turns.

Mathematical Expression:

K = N₂/N₁ = E₂/E₁ = V₂/V₁

Table: Transformation Ratio Types

Current Relationship:

I₁/I₂ = N₂/N₁ = K

Power Relationship:

P₁ = P₂ (Ideal transformer)

Mnemonic: “Turns Tell Transformation” (Turns ratio determines voltage ratio)

Question 3(b OR) [4 marks]

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Write application of autotransformer.

Answer:

Table: Autotransformer Applications

Advantages:

• Economy: Less copper and iron required
• Efficiency: Higher than two-winding transformer
• Size: Compact design
• Regulation: Better voltage regulation

Limitations:

• No isolation: Common electrical connection
• Safety: Higher fault current

Mnemonic: “Auto Adjusts Advantageously” (Automatic voltage adjustment benefit)

Question 3(c OR) [7 marks]

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Explain speed control of DC shunt motor

Answer:

Table: Speed Control Methods

Armature Control Diagram:

graph LR
    V[Variable DC] --> R[Rheostat]
    R --> A[Armature]
    A --> V
    V --> F[Field Winding]
    F --> V
    
    style R fill:#f96,stroke:#333,stroke-width:2px
    style A fill:#bbf,stroke:#333,stroke-width:2px

Speed Equations:

• Armature control: N ∝ (V - IaRa)/φ
• Field control: N ∝ V/φ
• Resistance control: N ∝ (V - Ia(Ra + Rext))/φ

Modern Methods:

• Chopper control: PWM voltage control
• Ward-Leonard system: Motor-generator set
• Electronic control: Thyristor/IGBT drives

Characteristics:

• Smooth control: Stepless speed variation
• Efficiency: Armature control most efficient
• Cost: Field control economical

Mnemonic: “Armature Accurate, Field Fast, Resistance Rough” (Control characteristics)

Question 4(a) [3 marks]

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Explain vector representation of alternating EMF.

Answer:

Vector Representation: Alternating EMF can be represented as a rotating vector (phasor) with constant magnitude and angular velocity.

Mathematical Form:

e = Em sin(ωt + φ)

Diagram:

graph LR
    subgraph "Phasor Diagram"
        O((O)) --> E[Em]
        O --> A[ωt]
    end
    
    style E fill:#f96,stroke:#333,stroke-width:3px
    style A fill:#bbf,stroke:#333,stroke-width:2px

Table: Vector Parameters

Advantages:

• Visual representation: Easy to understand phase relationships
• Mathematical simplification: Complex calculations made easier

Mnemonic: “Vectors Visualize Voltage Variation” (Phasor representation benefits)

Question 4(b) [4 marks]

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Define following terms w.r.t Alternating current: RMS value, Average value, Frequency, Time period

Answer:

Table: AC Parameters Definition

Mathematical Relations:

• Form Factor: RMS/Average = π/2√2 = 1.11
• Peak Factor: Peak/RMS = √2 = 1.414
• Angular frequency: ω = 2πf

Practical Values:

• RMS current: Used for power calculations
• Average current: Used for DC equivalent
• Frequency: 50 Hz (India), 60 Hz (USA)

Mnemonic: “Really Mean Square, Average Frequency Time” (Key AC parameters)

Question 4(c) [7 marks]

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Derive equation for relation between line and phase voltage and current in star connection

Answer:

Star Connection Diagram:

graph TD
    R[R Phase] --> N[Neutral N]
    Y[Y Phase] --> N
    B[B Phase] --> N
    
    R --> LR[Line R]
    Y --> LY[Line Y]  
    B --> LB[Line B]
    
    style N fill:#f96,stroke:#333,stroke-width:3px

Voltage Relations:

Phase Voltages: VR, VY, VB (with respect to neutral) Line Voltages: VRY, VYB, VBR (between lines)

Phasor Analysis:

VRY = VR - VY

For balanced system:

• Phase voltages are equal in magnitude: VR = VY = VB = Vph
• Phase difference = 120°

Vector Addition: Using phasor diagram and cosine rule:

VL = √(Vph² + Vph² - 2Vph·Vph·cos(120°))
VL = √(2Vph² + Vph²) = √3 × Vph

Final Relations:

Table: Star Connection Relations

Current Relations: In star connection, line current equals phase current:

IL = Iph

Mnemonic: “Star Scales Voltage, Same current” (√3 factor for voltage, current unchanged)

Question 4(a OR) [3 marks]

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Explain vector representation of alternating current.

Answer:

Vector Representation: AC current represented as rotating phasor with magnitude and phase angle.

Mathematical Expression:

i = Im sin(ωt + φ)

Phasor Diagram:

Table: Current Vector Elements

Applications:

• Circuit analysis: Phase relationships between voltage and current
• Power calculations: Real and reactive power components

Mnemonic: “Current Circles Continuously” (Rotating phasor concept)

Question 4(b OR) [4 marks]

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Define following terms w.r.t Alternating current: Form factor, Peak factor, Angular velocity, Amplitude

Answer:

Table: AC Current Parameters

Mathematical Relations:

• Form factor: Indicates waveform shape
• Peak factor: Shows crest factor
• Angular velocity: Links frequency and phase
• Amplitude: Determines RMS and average values

Practical Significance:

• Design considerations: Peak factors for insulation
• Waveform analysis: Form factors for distortion
• Synchronization: Angular velocity for timing

Mnemonic: “Form Peak Angular Amplitude” (Four key factors)

Question 4(c OR) [7 marks]

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Derive equation for relation between line and phase voltage and current in delta connection

Answer:

Delta Connection Diagram:

graph LR
    subgraph "Delta Connection"
        A[A] --> B[B]
        B --> C[C]
        C --> A
    end
    
    A --> IA[IA]
    B --> IB[IB]
    C --> IC[IC]
    
    style A fill:#f96,stroke:#333,stroke-width:2px
    style B fill:#bbf,stroke:#333,stroke-width:2px
    style C fill:#bfb,stroke:#333,stroke-width:2px

Voltage Relations: In delta connection, line voltage equals phase voltage:

VL = Vph

Current Analysis: Each line current is vector sum of two phase currents.

For Line Current IA:

IA = IAB - ICA

Phasor Analysis: For balanced system with phase currents equal in magnitude:

• IAB = ICA = ICB = Iph
• Phase difference between currents = 120°

Vector Subtraction:

IA = IAB - ICA = IAB - (-ICA)

Using phasor diagram:

IL = √(Iph² + Iph² - 2Iph·Iph·cos(60°))
IL = √(2Iph² - Iph²) = √3 × Iph

Final Relations:

Table: Delta Connection Relations

Mnemonic: “Delta Doubles current, Same voltage” (√3 factor for current, voltage unchanged)

Question 5(a) [3 marks]

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Explain AC through pure resistor with necessary circuit and waveform.

Answer:

Circuit Diagram:

graph LR
    AC[AC Source] --> R[Resistor R]
    R --> AC
    
    style AC fill:#f96,stroke:#333,stroke-width:2px
    style R fill:#bbf,stroke:#333,stroke-width:2px

Waveform:

Table: AC through Resistor

Characteristics:

• Current and voltage in phase: No phase difference
• Power consumption: Continuous power dissipation
• Resistance unchanged: Same as DC value

Mnemonic: “Resistor Refuses phase Shift” (No phase difference)

Question 5(b) [4 marks]

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Define following terms w.r.t Alternating current: Impedance, Phase angle, Power factor, Reactive power

Answer:

Table: AC Circuit Parameters

Power Relations:

• Active Power: P = VI cosφ (Watts)
• Reactive Power: Q = VI sinφ (VAR)
• Apparent Power: S = VI (VA)

Power Triangle:

S² = P² + Q²

Practical Significance:

• High power factor: Efficient power utilization
• Low power factor: Higher current for same power
• Reactive power: No net energy transfer

Mnemonic: “Impedance Phase Power Quadrature” (Four key AC parameters)

Question 5(c) [7 marks]

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Enlist different protective device and explain construction and working of any one protective device.

Answer:

Table: Protective Devices

MCB (Miniature Circuit Breaker) - Detailed Explanation:

Construction:

graph TD
    subgraph "MCB Construction"
        C[Contacts] --> A[Arc Chamber]
        A --> B[Bimetallic Strip]
        B --> M[Magnetic Coil]
        M --> T[Trip Mechanism]
    end
    
    style C fill:#f96,stroke:#333,stroke-width:2px
    style B fill:#bbf,stroke:#333,stroke-width:2px
    style M fill:#bfb,stroke:#333,stroke-width:2px

Components:

• Fixed and moving contacts: Current carrying parts
• Bimetallic strip: Thermal protection
• Electromagnetic coil: Magnetic protection
• Arc quenching chamber: Arc extinction
• Operating mechanism: Manual/automatic operation

Working Principle:

Overload Protection:

• Current heats bimetallic strip
• Strip bends and trips mechanism
• Time-delay characteristic protects against temporary overloads

Short Circuit Protection:

• High fault current creates strong magnetic field
• Electromagnetic force operates trip mechanism
• Instantaneous operation for safety

Advantages:

• Reusable: Reset after fault clearance
• Reliable operation: Dual protection mechanism
• Easy maintenance: Accessible contacts

Mnemonic: “MCB Magnetically Controls Both” (Thermal and magnetic protection)

Question 5(a OR) [3 marks]

#

Derive equation of AC current passing through pure inductor

Answer:

Given: Pure inductor with inductance L, applied voltage v = Vm sin(ωt)

Voltage-Current Relationship:

v = L × (di/dt)

Substituting applied voltage:

Vm sin(ωt) = L × (di/dt)

Integration:

di = (Vm/L) sin(ωt) dt
i = -(Vm/ωL) cos(ωt) + C

At steady state, C = 0:

i = -(Vm/ωL) cos(ωt)
i = (Vm/ωL) sin(ωt - 90°)

Table: Pure Inductor Characteristics

Mnemonic: “Inductor Impedes, Current lags” (XL opposes current, 90° lag)

Question 5(b OR) [4 marks]

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Explain concept of power and power triangle in AC circuit.

Answer:

Types of Power:

Table: AC Power Components

Power Triangle:

graph LR
    O((O)) --> P[P = VI cosφ]
    O --> Q[Q = VI sinφ]
    P --> S[S = VI]
    
    style P fill:#f96,stroke:#333,stroke-width:2px
    style Q fill:#bbf,stroke:#333,stroke-width:2px
    style S fill:#bfb,stroke:#333,stroke-width:2px

Mathematical Relations:

S² = P² + Q²
Power Factor = P/S = cosφ

Significance:

• Active power: Does useful work (heating, mechanical)
• Reactive power: Maintains magnetic/electric fields
• Power factor: Efficiency indicator

Mnemonic: “Power Triangle: Please Qualify Students” (P, Q, S components)

Question 5(c OR) [7 marks]

#

Explain wiring of lamp control from one place and staircase type.

Answer:

1. Lamp Control from One Place:

Circuit Diagram:

Components:

• SPST Switch: Single pole, single throw
• Live wire control: Switch in live wire for safety
• Simple on/off: Basic control mechanism

2. Staircase Wiring (Two-Way Control):

Circuit Diagram:

Table: Switch Positions for Staircase Control

Working Principle:

• Two-way switches: SPDT (Single Pole Double Throw)
• Common terminal: Connected to live and lamp
• Strappers: Link switches together
• Toggle action: Either switch can control lamp

Applications:

• Staircase lighting: Control from top and bottom
• Long corridors: Control from both ends
• Bedroom lighting: Control from bed and door

Advantages:

• Convenience: Control from multiple locations
• Energy saving: Easy switching reduces wastage
• Safety: No need to walk in dark

Installation Points:

• Proper earthing: All metal parts earthed
• Cable rating: Adequate current capacity
• Switch height: Standard 4 feet from floor

Mnemonic: “Two-way Toggles, Two places” (Two switches, two locations)