Fundamentals of Electrical Engineering (4311101) - Summer 2024 Solution

Question 1(a) [3 marks]

#

Define EMF, electric current and power. Also write their units.

Answer:

Mnemonic: “EVA” - EMF in Volts, Current in Amperes, Power in Watts

Question 1(b) [4 marks]

#

Three resistors having resistances of 1000 Ω, 2000 Ω and 3000 Ω respectively are connected in series. Find the equivalent resistance of this series connection. Now these three resistors are connected in parallel. Find the equivalent resistance of this parallel connection.

Answer:

For Series Connection:

Req = R1 + R2 + R3
Req = 1000 Ω + 2000 Ω + 3000 Ω
Req = 6000 Ω

For Parallel Connection:

1/Req = 1/R1 + 1/R2 + 1/R3
1/Req = 1/1000 + 1/2000 + 1/3000
1/Req = 0.001 + 0.0005 + 0.00033
1/Req = 0.00183
Req = 545.45 Ω

Diagram:

graph LR
    A[Input] --- B[1000 Ω]
    B --- C[2000 Ω]
    C --- D[3000 Ω]
    D --- E[Output]

    F[Input] --- G[1000 Ω] --- H[Output]
    F --- I[2000 Ω] --- H
    F --- J[3000 Ω] --- H

Mnemonic: “Series Sum, Parallel Product/Sum” - In series add directly, in parallel take reciprocal sum

Question 1(c) [7 marks]

#

Write the definition of Resistor, Capacitor and Inductor. Draw their symbols and write their units. Also write the use of each device in electrical circuit.

Answer:

Diagram:

Mnemonic: “RCI” - Resistor Controls current, Capacitor stores charge, Inductor stores magnetic energy

Question 1(c OR) [7 marks]

#

State Ohm’s law and write the equation of Ohm’s law with circuit diagram. Write applications of Ohm’s law. Also write the limitation of Ohm’s law.

Answer:

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

Equation: V = I × R

Circuit Diagram:

graph LR
    A[Voltage Source V] --- B[Resistor R]
    B --- C[Current I]
    C --- A

Applications of Ohm’s Law:

• Calculating current, voltage, or resistance in circuits
• Designing electrical and electronic circuits
• Power calculations (P = V × I = I² × R = V²/R)
• Circuit analysis using voltage divider and current divider

Limitations of Ohm’s Law:

• Not applicable for non-linear devices (diodes, transistors)
• Not valid for high-frequency AC circuits
• Not valid for non-metallic conductors
• Does not apply during transient conditions

Mnemonic: “VIR” - Voltage equals current times resistance

Question 2(a) [3 marks]

#

Explain the generation of alternating EMF with the help of necessary diagram and equation.

Answer:

Alternating EMF is generated when a conductor rotates in a magnetic field.

Equation: e = E₀ sin(ωt) = E₀ sin(2πft)

Where:

• e = instantaneous EMF
• E₀ = maximum EMF
• ω = angular velocity (2πf)
• f = frequency
• t = time

Diagram:

graph TD
    A[Magnetic Field] --- B[Rotating Coil]
    B --- C[Slip Rings]
    C --- D[Brushes]
    D --- E[AC Output]

Mnemonic: “RCBS” - Rotation of Coil in magnetic field produces sinusoidal EMF

Question 2(b) [4 marks]

#

Explain the behavior of pure capacitor with AC supply with necessary circuit diagram and equation.

Answer:

Behavior of Pure Capacitor with AC:

• Current leads voltage by 90° in a pure capacitor
• Capacitive reactance (Xc) = 1/(2πfC)
• As frequency increases, reactance decreases
• Stores energy in electric field during charging

Circuit and Waveform:

Equation: I = C × dV/dt

Mnemonic: “CIVIC” - Capacitor’s current Is ahead of Voltage by 90° In Circuit

Question 2(c) [7 marks]

#

An AC voltage is expressed as 300 Sin (628t) V. Find (i) Amplitude (ii) Frequency (iii) Time period (iv) Average value (v) RMS Value (vi) Form Factor and (vii) Peak Factor for this AC voltage.

Answer:

Given: v = 300 Sin(628t) V

Mnemonic: “FART FAFP” - Frequency is Angular frequency divided by 2π, RMS is peak divided by root 2, Time period is 1/f, Form factor is 1.11, Average is 2Vₘ/π, Peak factor is 1.414

Question 2(a OR) [3 marks]

#

Explain the generation of 3-phase alternating EMF.

Answer:

3-phase alternating EMF is generated using three separate coils placed 120° apart in a magnetic field.

Key Points:

• Three identical coils are placed 120° apart
• Each coil produces sinusoidal EMF
• Phases are labeled as R, Y, and B (or U, V, W)
• Phase difference between any two phases is 120°

Diagram:

graph LR
    A[Rotating Magnet] --- B[Three Coils 120° Apart]
    B --- C[Three-Phase Output]
    D[Time] --- E[Three Phase Waveforms]

Mnemonic: “THREE” - Three coils Have 120° Rotating EMF Each

Question 2(b OR) [4 marks]

#

Explain the behavior of pure inductor with AC supply with necessary circuit diagram and equation.

Answer:

Behavior of Pure Inductor with AC:

• Current lags voltage by 90° in a pure inductor
• Inductive reactance (XL) = 2πfL
• As frequency increases, reactance increases
• Stores energy in magnetic field

Circuit and Waveform:

Equation: V = L × dI/dt

Mnemonic: “VLIC” - Voltage Leads current by 90° In inductor Circuit

Question 2(c OR) [7 marks]

#

Define phase voltage, line voltage, phase current and line current for 3-phase AC. (i) Calculate the line voltage for star (Y) connection if the phase voltage is 100V. Also find the line current for star (Y) connection if the phase current is 5A (ii) Calculate the line voltage for delta (Δ) connection if the phase voltage is 100V. Also find the line current for delta (Δ) connection if the phase current is 5A.

Answer:

Star (Y) Connection:

• Line voltage = √3 × Phase voltage
• Line current = Phase current

Calculations:

• Line voltage = √3 × 100 = 173.2 V
• Line current = 5 A

Delta (Δ) Connection:

• Line voltage = Phase voltage
• Line current = √3 × Phase current

Calculations:

• Line voltage = 100 V
• Line current = √3 × 5 = 8.66 A

Diagram:

graph TD
    subgraph Star Connection
    A1((R)) --- B1((Y))
    B1 --- C1((B))
    C1 --- A1
    D1((N)) --- A1
    D1 --- B1
    D1 --- C1
    end

    subgraph Delta Connection
    A2((R)) --- B2((Y))
    B2 --- C2((B))
    C2 --- A2
    end

Mnemonic: “SLIP” - Star: Line voltage is √3 times Phase voltage, In Delta: Phase voltage equals Line voltage

Question 3(a) [3 marks]

#

State and explain Faraday’s laws of electromagnetic induction with necessary diagram and equations.

Answer:

Faraday’s Laws:

1. First Law: When a conductor cuts magnetic flux, EMF is induced
2. Second Law: The magnitude of induced EMF is proportional to the rate of change of magnetic flux

Equation: e = -N × (dΦ/dt) Where: e = induced EMF, N = number of turns, dΦ/dt = rate of change of flux

Diagram:

graph LR
    A[Moving Magnet] --- B[Coil]
    B --- C[Galvanometer]
    D[Changing Magnetic Field] --- E[Induced EMF]

Mnemonic: “FIRE” - Flux change Induces Rapid EMF

Question 3(b) [4 marks]

#

Define amplitude, frequency, time duration and RMS value for alternating quantity.

Answer:

Diagram:

Mnemonic: “AFTR” - Amplitude is peak, Frequency is cycles per second, Time period is 1/f, RMS is 0.707 times peak

Question 3(c) [7 marks]

#

Explain self inductance and mutual inductance. (i) Find the self induction of the coil if total magnetic flux linked with the coil is 5 μWb-turns (micro Wb-turns) for 2 A current given to the coil (ii) Find the self induction of the coil, if the parameters of the coils are as follows: number of turns is 10, relative permeability of the material used for coil is 3, length of the coil is 5 cm and cross sectional area of coil is 2 cm².

Answer:

Self Inductance: Property of a coil to oppose change in current through it by inducing EMF in itself.

Mutual Inductance: Property of one coil to induce EMF in another coil due to change in current.

Part (i):

Self inductance (L) = Flux linkage / Current
L = 5 μWb-turns / 2 A
L = 2.5 μH

Part (ii):

L = (μₒ × μᵣ × N² × A) / l
L = (4π × 10⁻⁷ × 3 × 10² × 2 × 10⁻⁴) / (5 × 10⁻²)
L = (4π × 3 × 100 × 2 × 10⁻⁷) / (5 × 10⁻²)
L = (24π × 10⁻⁵) / (5 × 10⁻²)
L = 24π × 10⁻³ / 5
L = 4.8π × 10⁻³
L = 15.07 μH

Diagram:

graph TD
    subgraph Self Inductance
    A[Current in Coil] --> B[Magnetic Field]
    B --> C[EMF in Same Coil]
    end

    subgraph Mutual Inductance
    D[Current in Coil 1] --> E[Magnetic Field]
    E --> F[EMF in Coil 2]
    end

Mnemonic: “SLIM” - Self inductance Linked with own flux, Induction Mutual between two coils

Question 3(a OR) [3 marks]

#

Define dynamically induced EMF. Explain it with the help of necessary diagram and equation.

Answer:

Dynamically Induced EMF: EMF induced in a conductor due to relative motion between the conductor and magnetic field.

Equation: e = Blv Where: e = induced EMF, B = magnetic flux density, l = length of conductor, v = velocity of conductor

Diagram:

graph LR
    A[Magnetic Field B] --- B[Moving Conductor]
    B --- C[Induced EMF e]
    D[Motion with velocity v] --- B

Mnemonic: “MOVE” - Motion Of conductor in magnetic field produces Voltage Effect

Question 3(b OR) [4 marks]

#

Define cycle, Form Factor and Peak Factor for alternating quantity. Write the value of Form Factor and Peak Factor for sinusoidal alternating quantity.

Answer:

Diagram:

Mnemonic: “CFP” - Cycle is one oscillation, Form factor is 1.11, Peak factor is 1.414

Question 3(c OR) [7 marks]

#

State and explain Lenz’s law. State and explain Fleming’s right hand rule for generator. Find the energy stored in inductor having self inductance of 4 μH, if 3 A of current is flowing through the inductor.

Answer:

Lenz’s Law: The direction of induced EMF is such that it opposes the change in magnetic flux that produces it.

Fleming’s Right Hand Rule:

• Thumb: Direction of motion of conductor
• Index finger: Direction of magnetic field
• Middle finger: Direction of induced current

Energy Calculation:

Energy stored in inductor (W) = (1/2) × L × I²
W = (1/2) × 4 × 10⁻⁶ × 3²
W = (1/2) × 4 × 10⁻⁶ × 9
W = 18 × 10⁻⁶ / 2
W = 9 × 10⁻⁶ Joules
W = 9 μJ

Diagram:

Mnemonic: “LOF” - Lenz’s law Opposes Flux change, Fleming’s rule - thumb Motion, index Field, middle Current

Question 4(a) [3 marks]

#

Define PV cell. Explain the function of PV cell.

Answer:

PV Cell: Photovoltaic cell is a semiconductor device that converts light energy directly into electrical energy.

Function:

• Absorbs photons from sunlight
• Creates electron-hole pairs in semiconductor
• Generates potential difference at p-n junction
• Converts solar energy to electrical energy

Diagram:

graph TD
    A[Sunlight] --> B[PV Cell]
    B --> C[DC Electricity]
    D[P-type Silicon] --- E[N-type Silicon]

Mnemonic: “PASE” - PV cell Absorbs Sunlight to generate Electricity

Question 4(b) [4 marks]

#

Explain the classification of green energy.

Answer:

Diagram:

graph TD
    A[Green Energy] --> B[Solar]
    A --> C[Wind]
    A --> D[Hydro]
    A --> E[Biomass]
    A --> F[Geothermal]

Mnemonic: “SWHBG” - Sun, Wind, Hydro, Biomass, Geothermal energy sources

Question 4(c) [7 marks]

#

Draw and explain the block diagram of solar power system.

Answer:

Solar Power System Components:

Block Diagram:

flowchart LR
    A[Solar Panels] --> B[Charge Controller]
    B --> C[Battery Bank]
    C --> D[Inverter]
    D --> E[Distribution Panel]
    E --> F[Home Appliances]
    E -.-> G[Grid Connection]

Mnemonic: “SCBIDG” - Solar panels, Charge controller, Batteries, Inverter, Distribution, Grid

Question 4(a OR) [3 marks]

#

Define green energy, conventional energy and renewable energy.

Answer:

Diagram:

graph TD
    A[Energy Sources] --> B[Green/Renewable]
    A --> C[Conventional/Non-renewable]

    B --> D[Solar, Wind, Hydro, etc.]
    C --> E[Coal, Oil, Natural Gas]

Mnemonic: “GCR” - Green is Clean, Conventional is Carbon-emitting, Renewable is Replenished

Question 4(b OR) [4 marks]

#

Explain the need of green energy.

Answer:

Need for Green Energy:

Diagram:

graph TD
    A[Need for Green Energy] --> B[Environmental Protection]
    A --> C[Resource Conservation]
    A --> D[Energy Security]
    A --> E[Economic Benefits]
    A --> F[Sustainable Development]

Mnemonic: “ERESS” - Environment, Resources, Energy security, Savings, Sustainability

Question 4(c OR) [7 marks]

#

Draw and explain the block diagram of wind power system with types of turbines.

Answer:

Wind Power System Components:

Types of Wind Turbines:

1. Horizontal Axis Wind Turbine (HAWT) - Blades rotate around horizontal axis
2. Vertical Axis Wind Turbine (VAWT) - Blades rotate around vertical axis

Block Diagram:

flowchart LR
    A[Wind] --> B[Wind Turbine]
    B --> C[Gearbox]
    C --> D[Generator]
    D --> E[Controller]
    E --> F[Transformer]
    F --> G[Grid]

    subgraph "Types of Turbines"
    H[Horizontal Axis]
    I[Vertical Axis]
    end

Mnemonic: “WGGTC” - Wind turns turbine, Gearbox speeds up, Generator produces electricity, Transformer steps up, Controller manages

Question 5(a) [3 marks]

#

Explain the factors affecting the value of resistance of a resistor.

Answer:

Factors Affecting Resistance:

Equation: R = ρ × (l/A)

Where:

• R = Resistance
• ρ = Resistivity
• l = Length
• A = Cross-sectional area

Mnemonic: “TLAM” - Temperature, Length, Area, Material affect resistance

Question 5(b) [4 marks]

#

Define active power, reactive power, apparent power and power factor with the help of power triangle. Write their units.

Answer:

Power Triangle:

Mnemonic: “ARSP” - Active is Real power in Watts, Reactive is Stored power in VAR, S is total VA, PF is cosφ

Question 5(c) [7 marks]

#

State and explain Kirchhoff’s Voltage Law (KVL) and Kirchhoff’s Current Law (KCL) with the help of circuit diagram.

Answer:

Kirchhoff’s Voltage Law (KVL): The algebraic sum of all voltages around any closed loop in a circuit is zero.

Kirchhoff’s Current Law (KCL): The algebraic sum of all currents entering and leaving a node is zero.

Circuit Diagrams:

graph TD
    subgraph KVL
    A1((+)) --- B1[R1]
    B1 --- C1[R2]
    C1 --- D1[R3]
    D1 --- A1
    end

    subgraph KCL
    A2((Node)) --- B2[I1]
    A2 --- C2[I2]
    A2 --- D2[I3]
    A2 --- E2[I4]
    end

KVL Example: V₁ + V₂ + V₃ = 0

KCL Example: I₁ + I₂ = I₃ + I₄

Mnemonic: “VCL” - Voltage around Closed Loop is zero, Currents at a point sum to zero

Question 5(a OR) [3 marks]

#

Write the difference between EMF and potential difference. Also write the difference between cell and battery.

Answer:

Diagram:

Mnemonic: “ESOP” - EMF is Source energy, Open circuit too; Potential difference is Operating energy

Question 5(b OR) [4 marks]

#

Write the relation between AC voltage and AC current for pure resistor, pure capacitor and pure inductor. Draw the vector diagram of AC voltage and AC current for pure resistor, pure capacitor and pure inductor. Also write the value of power factor for pure resistor, pure capacitor and pure inductor.

Answer:

Vector Diagrams:

Mnemonic: “RCI” - Resistor Current In phase, Capacitor current leads, Inductor current lags

Question 5(c OR) [7 marks]

#

Define temperature coefficient of material and write its unit. Explain the effect of temperature on resistance of conductor with the help of temperature coefficient of conductor.

Answer:

Temperature Coefficient: The fractional change in resistance per degree change in temperature.

Unit: per degree Celsius (°C⁻¹) or per Kelvin (K⁻¹)

Effect of Temperature on Resistance:

Equation: R₂ = R₁[1 + α(T₂ - T₁)]

Where:

• R₁ = Resistance at temperature T₁
• R₂ = Resistance at temperature T₂
• α = Temperature coefficient
• T₁, T₂ = Initial and final temperatures

For Conductors (Metals):

• Resistance increases with temperature (positive α)
• Resistance decreases when temperature decreases

For Semiconductors:

• Resistance decreases with temperature (negative α)

Table:

Diagram:

graph LR
    A[Temperature Increase] --> B[Increased Atomic Vibrations]
    B --> C[More Electron Collisions]
    C --> D[Increased Resistance in Metals]

Mnemonic: “TRIP” - Temperature Raises resistance In Proportion to coefficient