Fundamentals of Electrical Engineering (DI01000101) - Winter 2024 Solution

Solution guide for Fundamentals of Electrical Engineering (DI01000101) Winter 2024 exam

Question 1(a) [3 marks]

Explain ohm's law with its limitation and application.

Answer:

Table: Ohm's Law Summary

AspectDescription
StatementCurrent through conductor is directly proportional to voltage
FormulaV = I × R
UnitsV (Volts), I (Amperes), R (Ohms)

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]

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:

goat

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]

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

Answer:

Table: Kirchhoff's Laws Comparison

LawStatementMathematical FormApplication
KVLSum of voltages in closed loop = 0ΣV = 0Series circuits
KCLSum of currents at node = 0ΣI = 0Parallel circuits

KVL Diagram:

KCL Diagram:

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]

Differentiate statically induced emf and dynamically induced emf

Answer:

Table: Static vs Dynamic EMF

ParameterStatically Induced EMFDynamically Induced EMF
CauseChanging magnetic fieldRelative motion between conductor and field
FieldTime-varying, conductor stationarySteady field, conductor moving
ExamplesTransformer, inductorGenerator, motor
Formulae = -N(dΦ/dt)e = BLv
ApplicationsAC circuits, power suppliesPower generation, motors

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]

Explain various types of losses in transformer.

Answer:

Table: Transformer Losses

Loss TypeCauseLocationCharacteristics
Iron LossHysteresis + Eddy currentsCoreConstant, frequency dependent
Copper LossI²R heatingWindingsVariable with load
Stray LossLeakage fluxOverallMinimal

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]

Explain working principle of transformer.

Answer:

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

Diagram:

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]

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

SymbolParameterUnits
ERMS EMFVolts
fFrequencyHz
NNumber of turns-
ΦₘMaximum fluxWeber
4.44Form factor constant-

Transformation Ratio:

K = E₂/E₁ = N₂/N₁

Mnemonic: "Four-Forty-Four Flux Formula" (4.44 factor)

Question 2(a OR) [3 marks]

Write application of transformer.

Answer:

Table: Transformer Applications

ApplicationPurposeVoltage Level
Power transmissionReduce transmission lossesStep-up (400kV)
DistributionSafe voltage for consumersStep-down (230V)
IsolationElectrical isolation1:1 ratio
Electronic circuitsDC power suppliesStep-down

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]

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

SymbolParameterUnits
EbBack EMFVolts
TTorqueN-m
φFlux per poleWeber
NSpeedRPM
IaArmature currentAmperes
KMotor constant-

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

Question 2(c OR) [7 marks]

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

Answer:

Construction Components:

Table: DC Motor Parts

ComponentFunctionMaterial
StatorProvides magnetic fieldCast iron/steel
Rotor/ArmatureRotating partSilicon steel laminations
CommutatorCurrent direction reversalCopper segments
BrushesCurrent collectionCarbon
Field windingsElectromagnetsCopper wire

Construction Diagram:

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]

Explain construction of transformer.

Answer:

Table: Transformer Construction

ComponentMaterialFunction
CoreSilicon steel laminationsMagnetic flux path
Primary windingCopper/AluminumInput energy
Secondary windingCopper/AluminumOutput energy
InsulationVarnish/PaperElectrical isolation
TankSteelOil containment & cooling

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]

Explain application of DC motor

Answer:

Table: DC Motor Applications

Motor TypeSpeed CharacteristicApplications
ShuntConstant speedFans, pumps, lathes
SeriesVariable speedTraction, cranes
CompoundModerate variationElevators, compressors

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]

Explain different types of DC motor.

Answer:

Table: DC Motor Classification

TypeField ConnectionSpeed-TorqueApplications
ShuntParallel to armatureConstant speed, low starting torqueFans, pumps
SeriesSeries with armatureVariable speed, high starting torqueTraction
CompoundBoth series & shuntModerate characteristicsGeneral purpose

Shunt Motor Diagram:

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]

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

RatioTypeVoltage ChangeApplications
K > 1Step-upIncreasesPower transmission
K < 1Step-downDecreasesDistribution
K = 1IsolationSameSafety isolation

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]

Write application of autotransformer.

Answer:

Table: Autotransformer Applications

ApplicationAdvantageVoltage Range
Motor startingReduced starting current50-80% of rated
Voltage regulationFine voltage adjustment±10% variation
LaboratoryVariable voltage source0-110% of input
Power systemsEconomic transmissionClose voltage ratios

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]

Explain speed control of DC shunt motor

Answer:

Table: Speed Control Methods

MethodRangeEfficiencyApplications
Armature controlBelow rated speedHighPrecise speed control
Field controlAbove rated speedHighConstant power drives
Resistance controlBelow rated speedLowSimple applications

Armature Control Diagram:

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]

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:

Table: Vector Parameters

ParameterSymbolUnitsDescription
MagnitudeEmVoltsMaximum EMF
Angular velocityωrad/sRotation speed
Phase angleφDegreesInitial phase
Frequencyf = ω/2πHzCycles per second

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]

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

Answer:

Table: AC Parameters Definition

TermDefinitionFormulaUnits
RMS ValueEffective value producing same heatingIm/√2Amperes
Average ValueMean value over half cycle2Im/πAmperes
FrequencyNumber of cycles per secondf = 1/THz
Time PeriodTime for one complete cycleT = 1/fSeconds

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]

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

Answer:

Star Connection Diagram:

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

ParameterRelationship
Line VoltageVL = √3 × Vph
Line CurrentIL = Iph
PowerP = √3 × VL × IL × cosφ

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]

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:

goat

Table: Current Vector Elements

ElementSymbolDescription
MagnitudeImPeak current value
PhaseφLeading/lagging angle
Angular velocityωRotation speed
RMS valueI = Im/√2Effective current

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]

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

Answer:

Table: AC Current Parameters

TermDefinitionFormulaTypical Value
Form FactorRMS/Average value ratioIrms/Iavg1.11 (sine wave)
Peak FactorPeak/RMS value ratioIm/Irms1.414 (sine wave)
Angular VelocityRate of phase changeω = 2πf314 rad/s (50Hz)
AmplitudeMaximum instantaneous valueImPeak current

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]

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

Answer:

Delta Connection Diagram:

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

ParameterRelationship
Line VoltageVL = Vph
Line CurrentIL = √3 × Iph
PowerP = √3 × VL × IL × cosφ

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

Question 5(a) [3 marks]

Explain AC through pure resistor with necessary circuit and waveform.

Answer:

Circuit Diagram:

Waveform:

goat

Table: AC through Resistor

ParameterRelationshipPhase
Ohm's LawV = IRSame phase
PowerP = VI = I²RAlways positive
ImpedanceZ = RPurely resistive

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]

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

Answer:

Table: AC Circuit Parameters

TermDefinitionFormulaUnits
ImpedanceTotal opposition to AC currentZ = √(R² + X²)Ohms
Phase AngleAngle between V and Iφ = tan⁻¹(X/R)Degrees
Power FactorCosine of phase anglePF = cosφ = R/Z-
Reactive PowerPower in reactive componentsQ = VI sinφVAR

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]

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

Answer:

Table: Protective Devices

DeviceProtection AgainstApplication
FuseOvercurrentLow/Medium voltage
MCBOverload, Short circuitDomestic/Commercial
ELCBEarth leakageSafety protection
RelayVarious faultsIndustrial systems
Surge arresterOvervoltageTransmission lines

MCB (Miniature Circuit Breaker) - Detailed Explanation:

Construction:

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

ParameterValuePhase Relationship
Current amplitudeIm = Vm/ωLCurrent lags voltage by 90°
Inductive reactanceXL = ωL = 2πfLFrequency dependent
PowerP = 0 (average)No net power consumption

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

Question 5(b OR) [4 marks]

Explain concept of power and power triangle in AC circuit.

Answer:

Types of Power:

Table: AC Power Components

Power TypeSymbolFormulaUnitsDescription
Active PowerPVI cosφWattsUseful power
Reactive PowerQVI sinφVARCirculating power
Apparent PowerSVIVATotal power

Power Triangle:

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:

goat

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:

goat

Table: Switch Positions for Staircase Control

S1 PositionS2 PositionLamp Status
UpUpON
UpDownOFF
DownUpOFF
DownDownON

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)