Question 1(a) [3 marks]

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

Answer:

Term	Definition
Resistance	The property of a material that opposes the flow of electric current, measured in ohms (Ω)
Electrical Energy	The ability to do work by electrical means, measured in joules (J) or kilowatt-hours (kWh)
Electrical Power	The rate at which electrical energy is transferred or converted, measured in watts (W)

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

Question 1(b) [4 marks]

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]

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]

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]

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

Answer:

Term	Definition
Amplitude	Maximum displacement of a waveform from its mean position, measured in volts or amperes
Frequency	Number of complete cycles occurring in one second, measured in hertz (Hz)
Time Period	Time taken to complete one cycle of waveform, measured in seconds (s)

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

Question 2(b) [4 marks]

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:

     +--[10Ω]--[20Ω]--[30Ω]--+
     |                        |
   (100V)                     |
     |                        |
     +------------------------+

Solution:

Parameter	Calculation	Result
Equivalent Resistance	R = 10Ω + 20Ω + 30Ω	60Ω
Circuit Current	I = 100V/60Ω	1.67A
Voltage across 10Ω	V₁ = 1.67A × 10Ω	16.7V
Voltage across 20Ω	V₂ = 1.67A × 20Ω	33.3V
Voltage across 30Ω	V₃ = 1.67A × 30Ω	50.0V
Power in 10Ω	P₁ = 1.67² × 10	27.8W
Power in 20Ω	P₂ = 1.67² × 20	55.6W
Power in 30Ω	P₃ = 1.67² × 30	83.4W

Mnemonic: "REÇVP" - Resistances Equivalent Causes Voltage and Power division

Question 2(c) [7 marks]

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:

    │    ╭─╮   ╭─╮   ╭─╮   ╭─╮
    │   ╱   ╲ ╱   ╲ ╱   ╲ ╱   ╲
    │  ╱     V     V     V     ╲
────┼─╱───────────────────────╲─────
    │╱       ╱╲       ╱╲       ╲
    V       V  ╲     V  ╲       V
    │      ╱    ╲   ╱    ╲
    │     ╱      ╲ ╱      ╲
    │    ╰─╯     ╰─╯      ╰─╯

    --- Voltage waveform
    --- Current waveform (identical phase)

Vector Diagram:

         │
         │
         V (voltage)
         │
         │
─────────┼────────→
         │        I (current)
         │
         │

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

Question 2(a) OR [3 marks]

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

Answer:

Term	Definition
Cycle	One complete repetition of a periodic waveform from start point to same point again
Form Factor	Ratio of RMS value to average value of AC waveform (For sine wave = 1.11)
Peak Factor	Ratio of maximum value to RMS value of AC waveform (For sine wave = 1.414)

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

Question 2(b) OR [4 marks]

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:

         ┌─[20Ω]─┐
         │       │
     +───┼───────┼───+
     │   │       │   │
    (60V) ├─[30Ω]─┤  │
     │   │       │   │
     │   └─[50Ω]─┘   │
     │               │
     +───────────────+

Solution:

Parameter	Calculation	Result
Current in 20Ω	I₁ = 60V/20Ω	3A
Current in 30Ω	I₂ = 60V/30Ω	2A
Current in 50Ω	I₃ = 60V/50Ω	1.2A
Total Current	I = 3A + 2A + 1.2A	6.2A
Equivalent Resistance	1/Req = 1/20 + 1/30 + 1/50	9.68Ω
Power in 20Ω	P₁ = 60V × 3A	180W
Power in 30Ω	P₂ = 60V × 2A	120W
Power in 50Ω	P₃ = 60V × 1.2A	72W

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

Question 2(c) OR [7 marks]

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:

           Current
    │      ╭─╮     ╭─╮     ╭─╮     ╭─╮
    │     ╱   ╲   ╱   ╲   ╱   ╲   ╱   ╲
    │    ╱     ╲ ╱     ╲ ╱     ╲ ╱     ╲
────┼───╱───────V───────V───────V───────╲─
    │  ╱         ╲       ╲       ╲       ╲
    │ ╱           ╲       ╲       ╲       ╲
    │╱             ╲       ╲       ╲       ╲
    V               ╰─╮     ╰─╮     ╰─╮     ╰
    │                 │       │       │
    │                 │       │       │
    │                 │       │       │
    │                 V       V       V     
    │                ╱ ╲     ╱ ╲     ╱ ╲   Voltage
    │               ╱   ╲   ╱   ╲   ╱   ╲ 
    │              ╱     ╲ ╱     ╲ ╱     ╲

Vector Diagram:

         │ I (current)
         │
         │
         │
─────────┼────────→
         │
         │
         │
         V V (voltage)

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

Question 3(a) [3 marks]

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

Answer:

Term	Definition	Formula
RMS Value	Root Mean Square value - equivalent DC value producing the same heating effect	Vrms = 0.707 × Vmax for sine wave
Average Value	Mean value of all instantaneous values over half cycle	Vavg = 0.637 × Vmax for sine wave

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

Question 3(b) [4 marks]

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)

Parameter	Calculation	Result
Maximum value	Imax = 25 A	25 A
RMS value	Irms = Imax/√2 = 25/1.414	17.68 A
Average value	Iavg = 2Imax/π = 2×25/3.14	15.92 A
Angular frequency	ω = 314 rad/s	314 rad/s
Frequency	f = ω/2π = 314/6.28	50 Hz
Time period	T = 1/f = 1/50	0.02 s

Mnemonic: "SMART" - Sine's Maximum divided by root 2 equals RMS Then 2/π for Average

Question 3(c) [7 marks]

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]

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]

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

Parameter	Calculation	Result
Maximum EMF	Emax = 100 V	100 V
Angular Frequency	ω = 2π50 = 314 rad/s	314 rad/s
Frequency	f = 50 Hz (directly from equation)	50 Hz
Time Period	T = 1/f = 1/50	0.02 s

Mnemonic: "FAST" - Frequency And period are reciprocals, Sin's Top value is maximum

Question 3(c) OR [7 marks]

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

Answer:

Delta (Δ) Connection:

graph LR
    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]

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

Answer:

Term	Definition
M.M.F. (Magnetomotive Force)	The force that produces magnetic flux in a magnetic circuit, measured in ampere-turns (AT)
Reluctance	The magnetic equivalent of resistance, opposition to magnetic flux, measured in AT/Wb
Flux	The total magnetic field passing through a surface, measured in webers (Wb)

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

Question 4(b) [4 marks]

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

Answer:

Power Type	Symbol & Unit	Definition
Apparent Power	S (VA)	Vector sum of active and reactive power
Active Power	P (W)	Actual work-producing power consumed by the load
Reactive Power	Q (VAR)	Power that oscillates between source and load

Power Triangle:

          ^ Q (Reactive Power)
          │
          │
          │
          │           S (Apparent Power)
          │         /
          │        /
          │       /
          │      /
          │     /
          │    θ
          │   /
          └──/───────────>
             P (Active Power)

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]

Compare electric and magnetic circuit.

Answer:

Parameter	Electric Circuit	Magnetic Circuit
Force	EMF (V)	MMF (AT)
Opposition	Resistance (Ω)	Reluctance (AT/Wb)
Flow	Current (A)	Flux (Wb)
Ohm's Law	V = IR	MMF = Φ × S
Medium	Conductor	Ferromagnetic material
Energy	Stored in electric field	Stored in magnetic field
Leakage	Negligible	Significant
Path	Conductors	Usually closed loop
Material Property	Conductivity	Permeability
Current Flow	Electron flow	No particle flow

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

Question 4(a) OR [3 marks]

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]

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:

Component	Symbol	Unit	Meaning
Active Power	P	Watt (W)	Real power doing useful work
Reactive Power	Q	VAR	Power oscillating between source and load
Apparent Power	S	VA	Vector sum of P and Q
Power Factor	cos φ	-	Ratio of active to apparent power (P/S)

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]

Differentiate statically and dynamically induced E.M.F.

Answer:

Parameter	Statically Induced EMF	Dynamically Induced EMF
Definition	EMF induced due to change in current in the primary coil	EMF induced due to relative motion between conductor and magnetic field
Mechanism	Change in linkage flux	Cutting of magnetic flux
Movement	No physical movement required	Requires relative motion
Examples	Transformer, inductor	Generator, motor
Faraday's Law	e = -N(dΦ/dt)	e = Blv
Application	Power transfer without motion	Power generation through motion
Energy Conversion	Electrical to magnetic and back	Mechanical to electrical or vice versa

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

Question 5(a) [3 marks]

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

Answer:

Term	Definition
Solar Cell	Basic photovoltaic unit that converts sunlight directly into electricity through semiconductor material
Solar Panel	Collection of solar cells connected in series/parallel in a frame
Solar Array	Multiple solar panels connected together to form a larger electricity-generating unit

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

Question 5(b) [4 marks]

Differentiate HAWT and VAWT.

Answer:

Parameter	Horizontal Axis Wind Turbine (HAWT)	Vertical Axis Wind Turbine (VAWT)
Axis Orientation	Parallel to ground	Perpendicular to ground
Efficiency	Higher (35-45%)	Lower (15-30%)
Wind Direction	Needs to face the wind	Works with wind from any direction
Generator Location	At the top of tower	Can be placed at ground level
Space Required	More	Less
Noise	Higher	Lower
Examples	Propeller-type, widely used commercially	Darrieus, Savonius designs

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

Question 5(c) [7 marks]

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:

Solar Panels: Convert sunlight to DC electricity
Charge Controller: Regulates battery charging, prevents overcharging
Battery Bank: Stores energy for use when sunlight isn't available
Inverter: Converts DC to AC power for household appliances
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]

Explain the need of green energy for our planet.

Answer:

Need for Green Energy:

Sustainability: Renewable sources won't deplete unlike fossil fuels
Pollution Reduction: Minimizes air and water pollution from burning fossil fuels
Climate Change: Reduces greenhouse gas emissions that cause global warming
Energy Security: Decreases dependence on imported fuels
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]

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]

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:

Wind Turbine: Converts wind's kinetic energy to mechanical energy
Generator: Transforms mechanical rotation to electrical energy
Controller: Regulates power output and protects from high winds
Battery: Stores excess energy (for off-grid systems)
Inverter: Converts DC to AC for consumption
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