Fundamentals of Electrical Engineering (4311101) - Summer 2023 Solution

Question 1(a) [3 marks]

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Define the following term (1) Resistance (2) Electrical energy (3) Electrical Power

Answer:

Mnemonic: “RIP” - Resistance Impedes Path, Energy Is Potential, Power Is Performance

Question 1(b) [4 marks]

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State and Explain Ohm’s Law & write limitations of it.

Answer:

Ohm’s Law: The current flowing through a conductor is directly proportional to the voltage across the conductor and inversely proportional to its resistance.

Mathematically: V = IR, where:

• V = Voltage (volts)
• I = Current (amperes)
• R = Resistance (ohms)

flowchart LR
    V[Voltage] --> I[Current]
    R[Resistance] --Limits--> I

Limitations of Ohm’s Law:

• Not applicable to non-linear devices (semiconductors, gas discharge tubes)
• Doesn’t hold at high temperatures
• Not valid for unilateral elements (diodes)
• Fails for time-varying currents

Mnemonic: “VIRO” - Voltage Is Resistance times Output current

Question 1(c) [7 marks]

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Explain series and parallel connection of batteries.

Answer:

Series Connection of Batteries:

flowchart LR
    B1[Battery 1] --> B2[Battery 2] --> B3[Battery 3] --> L[Load]
    L --> B1

Characteristics of Series Connection:

• Total Voltage = Sum of individual voltages (V = V₁ + V₂ + … + Vₙ)
• Current = Same through all batteries
• Applications: Higher voltage requirements
• Internal Resistance: Increases (Rₛ = r₁ + r₂ + … + rₙ)

Parallel Connection of Batteries:

flowchart LR
    B1[Battery 1] --> L[Load]
    B2[Battery 2] --> L
    B3[Battery 3] --> L
    L --> B1
    L --> B2
    L --> B3

Characteristics of Parallel Connection:

• Voltage = Same as individual battery (if identical)
• Total Current = Sum of individual currents (I = I₁ + I₂ + … + Iₙ)
• Applications: Higher current capacity required
• Internal Resistance: Decreases (1/Rₚ = 1/r₁ + 1/r₂ + … + 1/rₙ)

Mnemonic: “VSCP” - Voltage Sums in Series, Current Parallels

Question 1(c) OR [7 marks]

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Explain series and parallel connection of Resistors.

Answer:

Series Connection of Resistors:

flowchart LR
    S[Source] --> R1[R1] --> R2[R2] --> R3[R3] --> S

Characteristics of Series Connection:

• Equivalent Resistance = Sum of individual resistances (Rₛ = R₁ + R₂ + … + Rₙ)
• Current = Same through all resistors
• Voltage = Divided across resistors proportional to resistance values
• Power divided as per voltage distribution

Parallel Connection of Resistors:

flowchart LR
    S[Source] --> R1[R1]
    S --> R2[R2]
    S --> R3[R3]
    R1 --> S
    R2 --> S
    R3 --> S

Characteristics of Parallel Connection:

• Equivalent Resistance: 1/Rₚ = 1/R₁ + 1/R₂ + … + 1/Rₙ
• Voltage = Same across all resistors
• Current = Divided inverse-proportionally to resistance values
• Total Current = Sum of individual currents

Mnemonic: “RISE-VICE” - Resistance Increases in Series, Voltage Is Constant in Every parallel

Question 2(a) [3 marks]

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Define: (1) Amplitude (2) Frequency (3) Time period

Answer:

Mnemonic: “AFT” - Amplitude is the Full height, Time period is the Total cycle

Question 2(b) [4 marks]

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10Ω, 20Ω and 30Ω resistors are connected in series and 100V supply is given to them. Find (1) Equivalent resistance (2) Circuit current (3) Voltage drop across each Resistor (4) Power loss in each resistor.

Answer:

Diagram:

Solution:

Mnemonic: “REÇVP” - Resistances Equivalent Causes Voltage and Power division

Question 2(c) [7 marks]

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Explain A.C Through pure Resistor with wave form & vector diagram.

Answer:

In a pure resistive circuit with AC supply:

Key Characteristics:

• Current and voltage are in phase with each other
• Circuit follows Ohm’s Law: V = IR
• Power is always positive (P = VI)
• No reactive power consumed
• Power factor = 1 (cos φ = 1)

Waveform:

Vector Diagram:

Mnemonic: “PARVIP” - Pure AC Resistor has Voltage In Phase with current

Question 2(a) OR [3 marks]

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Define: (1) cycle (2) Form factor (3) Peak factor

Answer:

Mnemonic: “CFP” - Cycle Finishes a Pattern, Form Factor = Vrms/Vavg, Peak Factor = Vmax/Vrms

Question 2(b) OR [4 marks]

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20Ω, 30Ω and 50Ω resistors are connected in parallel and 60V supply is given to them. Find (1) Current in each Resistor. (2) Total current (3) Equivalent resistance (4) Power loss in each resistor.

Answer:

Diagram:

Solution:

Mnemonic: “VICTIM” - Voltage Is Constant, Total current Is the Measure (in parallel)

Question 2(c) OR [7 marks]

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Explain A.C Through pure capacitor with wave form & vector diagram.

Answer:

In a pure capacitive circuit with AC supply:

Key Characteristics:

• Current leads voltage by 90°
• Capacitive reactance Xc = 1/(2πfC)
• Only reactive power (no active power)
• Power factor = 0 (lagging)
• Average power over complete cycle = 0

Waveform:

Vector Diagram:

Mnemonic: “CLEAR-90” - Capacitive Load has Electrical Angle Reaching 90° (current leads voltage)

Question 3(a) [3 marks]

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Define RMS value and average value related to alternating waveform write formula of it.

Answer:

Mnemonic: “RAM” - RMS Averages the Mean square (RMS = 0.707×Vmax, AVG = 0.637×Vmax)

Question 3(b) [4 marks]

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If A.C. current is represented by equation i=25 sin(314t). Calculate (1) R.m.s. value (2) Average value (3) Frequency (4) Time period

Answer:

Given equation: i = 25 sin(314t)

Mnemonic: “SMART” - Sine’s Maximum divided by root 2 equals RMS Then 2/π for Average

Question 3(c) [7 marks]

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Explain star connection of resistors and Derive equation shows relationship between voltage and current in star connection.

Answer:

Star (Y) Connection:

graph TD
    N((N)) --- R1[R1] --- L1((L1))
    N --- R2[R2] --- L2((L2))
    N --- R3[R3] --- L3((L3))
    N((Neutral))

Characteristics of Star Connection:

• Three resistors connected at common point (neutral)
• Line voltage (VL) = √3 × Phase voltage (Vph)
• Line current (IL) = Phase current (Iph)
• For balanced load: IL = Iph
• Total power = 3 × Phase power

Mathematical Relationship:

• Phase voltage: Vph = VL/√3
• Phase current: Iph = IL
• For balanced resistive load: Iph = Vph/R
• Therefore: IL = VL/(√3×R)

Mnemonic: “SLIP-3” - Star Line current Is Phase current, Line voltage is Phase voltage times root-3

Question 3(a) OR [3 marks]

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Explain generation of alternating E.M.F.

Answer:

Generation of Alternating EMF:

graph LR
    subgraph "Rotating Coil in Magnetic Field"
    N[N] --- M((Magnet)) --- S[S]
    end
    M --- R[Rotating Coil]
    R --- EMF[EMF Output]

Process:

• Coil rotates in uniform magnetic field
• Flux linkage changes with angle of rotation
• Rate of change of flux induces EMF
• EMF follows sinusoidal pattern: e = Emax sin(ωt)
• Frequency depends on rotation speed

Mnemonic: “FRAME” - Flux Rotation Alternates Magnetic EMF

Question 3(b) OR [4 marks]

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An alternating EMF is expressed by e= 100 sin2π50t. Find out (1) Max value of EMF (2) Frequency (3) Time period (4) Angular Frequency

Answer:

Given equation: e = 100 sin2π50t

Mnemonic: “FAST” - Frequency And period are reciprocals, Sin’s Top value is maximum

Question 3(c) OR [7 marks]

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Explain star connection and Derive equation shows relationship between voltage and current in delta connection.

Answer:

Delta (Δ) Connection:

graph TB
    L1((L1)) --- R1[R1] --- L2((L2))
    L2 --- R2[R2] --- L3((L3))
    L3 --- R3[R3] --- L1

Characteristics of Delta Connection:

• Three resistors connected in closed loop
• Line voltage (VL) = Phase voltage (Vph)
• Line current (IL) = √3 × Phase current (Iph)
• For balanced load: Vph = VL
• Total power = 3 × Phase power

Mathematical Relationship:

• Phase voltage: Vph = VL
• Phase current: Iph = Vph/R
• Line current: IL = √3 × Iph
• Therefore: IL = √3 × VL/R

Mnemonic: “DELVIr3” - Delta Equal Line Voltage, Its line current equals phase current times root-3

Question 4(a) [3 marks]

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Define (1) M.M.F. (2) Reluctance (3) flux

Answer:

Mnemonic: “MFR” - MMF Flows against Reluctance like current flows against resistance

Question 4(b) [4 marks]

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Explain Apparent, Active and Reactive power in A.C circuits.

Answer:

Power Triangle:

Relationships:

• S = √(P² + Q²)
• P = S × cos θ
• Q = S × sin θ
• Power factor = cos θ = P/S

Mnemonic: “SPARQ” - S is Power Apparent, Real is P, Q is reactive

Question 4(c) [7 marks]

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Compare electric and magnetic circuit.

Answer:

Mnemonic: “VIRO-MSΦS” - Voltage Is to Resistance as MMF is to Reluctance, Our φ flows Similar

Question 4(a) OR [3 marks]

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State and explain Fleming’s left hand rule.

Answer:

Fleming’s Left Hand Rule: Used to find the direction of the force experienced by a current-carrying conductor placed in a magnetic field.

graph TD
    subgraph "Fleming's Left Hand Rule"
    T[Thumb: Force] --- F[Forefinger: Field] --- M[Middle finger: Current]
    end

Application:

• Thumb → Direction of Force (F)
• Forefinger → Direction of magnetic Field (B)
• Middle finger → Direction of Current (I)
• Only works when fingers are perpendicular to each other

Mnemonic: “FBI-Left” - Force, B-field, and I-current directions are shown by the Left hand

Question 4(b) OR [4 marks]

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Draw power triangle and explain each component of it.

Answer:

Power Triangle:

graph LR
    O --- P[Active Power P]
    O --- S[Hypotenuse: Apparent Power S]
    P --- Q[Reactive Power Q]
    P -.- A[Power Factor Angle φ]

Components:

Relationships:

• S² = P² + Q²
• P = S × cos φ
• Q = S × sin φ

Mnemonic: “SPQR” - S is Pythagoras of P and Q, Ratio of P/S is power factor

Question 4(c) OR [7 marks]

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Differentiate statically and dynamically induced E.M.F.

Answer:

Mnemonic: “STIM-DMOV” - STatically Induced needs Magnetic flux change, Dynamically needs MOVement

Question 5(a) [3 marks]

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Define (1) solar cell (2) solar panel (3) solar array

Answer:

Mnemonic: “CPA” - Cell Produces electricity, Panel Arrays cells, Array is collection of panels

Question 5(b) [4 marks]

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Differentiate HAWT and VAWT.

Answer:

Mnemonic: “HAVE” - Horizontal Aligns with wind, Vertical Enjoys omnidirectional wind

Question 5(c) [7 marks]

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Draw and explain the Block diagram of solar power system.

Answer:

Solar Power System Block Diagram:

flowchart LR
    S[Solar Panel] --> C[Charge Controller]
    C --> B[Battery Bank]
    B --> I[Inverter]
    I --> L[AC Load]
    B --> D[DC Load]

Components:

1. Solar Panels: Convert sunlight to DC electricity
2. Charge Controller: Regulates battery charging, prevents overcharging
3. Battery Bank: Stores energy for use when sunlight isn’t available
4. Inverter: Converts DC to AC power for household appliances
5. Loads: AC loads (appliances) and DC loads (LED lights, etc.)

Optional Components:

• Monitoring System: Tracks power generation/consumption
• Grid Connection: Allows selling excess electricity

Mnemonic: “SCBIL” - Solar Collects, Battery Inverts for Loads

Question 5(a) OR [3 marks]

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Explain the need of green energy for our planet.

Answer:

Need for Green Energy:

1. Sustainability: Renewable sources won’t deplete unlike fossil fuels
2. Pollution Reduction: Minimizes air and water pollution from burning fossil fuels
3. Climate Change: Reduces greenhouse gas emissions that cause global warming
4. Energy Security: Decreases dependence on imported fuels
5. Economic Benefits: Creates jobs and reduces health costs related to pollution

Mnemonic: “SPECS” - Sustainable, Pollution-free, Economic, Climate-friendly, Secure

Question 5(b) OR [4 marks]

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Classify green energy and explain any one in detail.

Answer:

Classification of Green Energy Sources:

mindmap
  root((Green Energy))
    Solar
    Wind
    Hydro
    Biomass
    Geothermal
    Tidal

Solar Energy in Detail:

• Working Principle: Photovoltaic effect converts sunlight to electricity
• Components: Solar cells, panels, inverters, batteries
• Applications: Residential power, industrial use, transportation
• Advantages: No pollution, abundant source, low maintenance
• Limitations: Weather dependent, requires storage, initial cost

Mnemonic: “SWHBGT” - Sun Wind Hydro Biomass Geothermal Tidal are green energy types

Question 5(c) OR [7 marks]

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Explain block diagram of wind power system and explain the operation of wind power system.

Answer:

Wind Power System Block Diagram:

flowchart LR
    W[Wind Turbine] --> G[Generator]
    G --> C[Controller]
    C --> B[Battery Storage]
    C --> I[Inverter]
    I --> L[Load]
    C --> GR[Grid Connection]

Operation:

1. Wind Turbine: Converts wind’s kinetic energy to mechanical energy
2. Generator: Transforms mechanical rotation to electrical energy
3. Controller: Regulates power output and protects from high winds
4. Battery: Stores excess energy (for off-grid systems)
5. Inverter: Converts DC to AC for consumption
6. Grid Connection: Feeds excess power to grid or draws when needed

Types of Wind Turbines:

• Horizontal Axis (HAWT): Main commercial type
• Vertical Axis (VAWT): Better for urban settings

Wind Speed Requirements:

• Cut-in speed: 3-5 m/s
• Rated output: 12-15 m/s
• Cut-out speed: 25 m/s (for safety)

Mnemonic: “WGCBIL” - Wind Generates, Controller Balances, Inverter Loads