Comprehensive Study Notes for National Certificate in Electrical Power Engineering

National Certificate in Electrical Power Engineering Study Notes

LO2 – Describe Electric Circuits

2.1 SI Units of Charge, Force, Work and Power

  • The International System of Units (SI) provides standard units for physical quantities in electrical engineering.

    • Quantity: Current
      Symbol: I
      SI Unit: Ampere (A)
      Formula: I = \frac{Q}{t}

    • Quantity: Charge
      Symbol: Q
      SI Unit: Coulomb (C)
      Formula: Q = I \times t

    • Quantity: Force
      Symbol: F
      SI Unit: Newton (N)
      Formula: F = m \times a

    • Quantity: Work
      Symbol: W
      SI Unit: Joule (J)
      Formula: W = V \times Q

    • Quantity: Power
      Symbol: P
      SI Unit: Watt (W)
      Formulas:

    • P = \frac{W}{t}

    • P = V \times I

    • Quantity: Potential
      Symbol: V
      SI Unit: Volt (V)
      Formula: V = \frac{W}{Q}

    • Quantity: EMF
      Symbol: E
      SI Unit: Volt (V)
      Formula: E = V + Ir

    • Quantity: Resistance
      Symbol: R
      SI Unit: Ohm (Ω)
      Formula: R = \frac{V}{I}

    • Quantity: Conductance
      Symbol: G
      SI Unit: Siemens (S)
      Formula: G = \frac{1}{R}

    • Quantity: Energy
      Symbol: E
      SI Unit: Joule (J)
      Formula: E = P \times t

Charge, Work, and Power Calculations

  • Charge Calculation: Q = I \times t

    • Example: If I = 3 \, A and t = 10 \, s, then Q = 3 \times 10 = 30 \, C

  • Work Calculation: W = V \times Q

    • Example: If V = 12 \, V and Q = 5 \, C, then W = 12 \times 5 = 60 \, J

  • Power Calculation:

    • Formulas:

    • P = V \times I

    • P = I^2 R

    • P = \frac{V^2}{R}

    • Example:

    • Given V = 12 \, V, I = 3 \, A, R = 4 \, Ω,

    • All forms yield same power:

    • P = 12 \times 3 = 36 \, W

    • P = 9 \times 4 = 36 \, W

    • P = \frac{144}{4} = 36 \, W

  • Electrical Energy Calculation: E = P \times t

    • Energy in Joules when P is in Watts and t is in seconds.

    • Example: For a 2 kW heater operating for 3 hours,
      E = 2 \times 3 = 6 \, kWh = 21,600,000 \ J

2.2 Resistance vs Conductance

  • Resistance Calculation:
    R = \frac{V}{I}

  • Current Calculation - Ohm's Law:
    I = \frac{V}{R}

  • Voltage Calculation - Ohm's Law:
    V = I \times R

  • Conductance Calculation:
    G = \frac{1}{R}

  • Resistance vs Conductance:

    • Resistance (R): Opposition to current flow, expressed in Ohms (Ω).

      • Higher R means less current.

    • Conductance (G): Ease of current flow, expressed in Siemens (S).

      • Higher G means higher current flow.

2.3 Electrical Power vs Energy

  • Power: Rate of Energy Use

    • Unit: Watt (W)

    • Formula: P = V \times I

  • Energy: Total Energy Used

    • Unit: Joule (J) or kWh

    • Formula: E = P \times t

Linear and Non-Linear Devices

  • Linear Devices: Obey Ohm's Law; resistance remains constant (e.g., resistors).

  • Non-Linear Devices: Do not obey Ohm's Law; resistance varies with voltage or temperature (e.g., diodes, thermistors, LDRs).

Resistance and Resistivity

  • Resistive Formula: R = \frac{\rho L}{A}

    • Where

    • \rho (rho) is resistivity (Ω·m),

    • L is length (m),

    • A is cross-sectional area (m²).

  • Factors Affecting Resistance:

    • Longer wire: Increases R.

    • Thicker wire (larger A): Decreases R.

    • High resistivity material: Increases R.

    • Higher temperature (metals): Increases R.

  • Example Calculation:

    • For Copper: \rho = 1.7 \times 10^{-8} \ Ω·m, L = 5 \, m, A = 2 \times 10^{-6} \ m²,
      R = \frac{1.7 \times 10^{-8} \times 5}{2 \times 10^{-6}} = 0.0425 \ Ω

Temperature Coefficient of Resistance

  • Formula: R_T = R_0 (1 + \alpha T)

    • Where

    • R_0 = resistance at 0°C,

    • \alpha = temperature coefficient (per °C),

    • T = temperature (°C).

  • Behavior:

    • Metals (\alpha positive): Resistance increases with temperature.

    • Semiconductors (\alpha negative): Resistance decreases with temperature.

    • Example Calculation:

    • For a resistor with R_0 = 50 \, Ω, \alpha = 0.004/°C, T = 75°C,
      R_T = 50(1 + 0.004 \times 75) = 50 \times 1.3 = 65 \, Ω

Resistor Colour Coding

  • 4-band Resistor Color Code:

    • Bands:

    • Band 1 & 2: Digits

    • Band 3: Multiplier

    • Band 4: Tolerance

  • Example:

    • Brown, Black, Red, Gold

    • Values: 1, 0, ×100, ±5% = 1,000 Ω = 1 kΩ ±5%

Multiples and Sub-multiples

  • Prefix Table:

    • Mega (M): \times 10^6

    • Kilo (k): \times 10^3

    • Milli (m): \times 10^{-3}

    • Micro (µ): \times 10^{-6}

    • Nano (n): \times 10^{-9}

    • Pico (p): \times 10^{-12}

  • Conversion Examples:

    • 3 MΩ = 3,000,000 Ω

    • 4.7 kΩ = 4,700 Ω

Main Effects of Electric Current

  • Heating Effect: ext{Heat} = I^2 R t (Joules).

    • Used in: heaters, toasters, fuses.

  • Magnetic Effect: Current produces a magnetic field around a conductor.

    • Basis for: motors, relays, electromagnets.

  • Chemical Effect: Current through an electrolyte causes electrolysis.

    • Used in: electroplating, battery charging, metal refining.

Fuses

  • Fuse Design: A thin wire that melts and breaks the circuit when current exceeds safety level.

  • Calculation for Fuse Size: I = \frac{P}{V}

    • Choose the next standard fuse size above this value.

  • Example:

    • For a 500 W appliance on 240 V:
      I = \frac{500}{240} = 2.08 \, A

    • Choose a 3 A fuse.

LO3 – Batteries and Alternative Sources of Energy

The Simple Cell

  • Construction: Two different metal electrodes in an electrolyte.

    • Chemical reaction drives electrons from the negative terminal (zinc) to the positive terminal (copper).

  • Polarisation Issue: Hydrogen bubbles form on the copper electrode, increasing internal resistance and reducing current.

    • Solution: Use a depolarizer (e.g., manganese dioxide).

  • Local Action Issue: Impurities in zinc cause energy wastage even when no current is drawn.

    • Solution: Amalgamate zinc with mercury.

EMF and Internal Resistance

  • Formula for EMF:
    E = V + Ir

  • Terminal Voltage Formula:
    V = E - Ir

  • Circuit Current Formula: I = \frac{E}{R + r}

    • Where

    • E = EMF (V),

    • V = terminal voltage (V),

    • I = current (A),

    • r = internal resistance (Ω),

    • R = external resistance (Ω).

  • Analogue: EMF represents salary; internal resistance represents tax; terminal voltage represents take-home pay.

  • Example:

    • E = 6 \, V, r = 0.5 \, Ω, R = 2.5 \, Ω

    • Results: I = \frac{6}{3} = 2 \, A and V = 6 - (2 \times 0.5) = 5 \, V (1 V lost internally).

Cell Capacity

  • Formula:
    ext{Capacity (Ah)} = I \times t

  • Duration Calculation:
    t = \frac{ ext{Capacity}}{I}

  • Example:

    • Capacity = 60 Ah, I = 4 \, A

    • Time: t = \frac{60}{4} = 15 \, hours

Primary vs Secondary Cells

  • Features of Primary Cells:

    • Not rechargeable: No

    • Reaction: Irreversible

    • Internal Resistance: Higher

    • Cost: Cheap

    • Examples: Zinc-carbon, alkaline

  • Features of Secondary Cells:

    • Rechargeable: Yes

    • Reaction: Reversible

    • Internal Resistance: Lower

    • Cost: Expensive

    • Examples: Lead-acid, Li-ion, NiCd

  • Applications:

    • Primary: Torches, remotes

    • Secondary: Cars, phones, UPS

Lead-Acid Battery (Secondary Cell)

  • Composition:

    • Positive Plate: Lead dioxide (PbO2)

    • Negative Plate: Spongy lead (Pb)

    • Electrolyte: Dilute sulphuric acid (H₂SO₄)

  • Nominal Voltage per Cell:

    • 2 V → 12 V battery = 6 cells in series

  • Charging and Discharging Process:

    • Discharging: Chemical → Electrical

    • Charging: Electrical → Chemical

    • Electrolyte: Weaker during discharging, stronger during charging

    • Plates: Both convert to lead sulphate during discharge, restored to original form during charge

Safe Disposal of Batteries

  • Disposal Instructions:

    • Never throw in ordinary waste bins due to toxic contents.

    • Dispose at designated recycling points.

    • Avoid burning or crushing due to explosion risks and toxic gas release.

    • Handle leaking batteries with protective gloves.

Renewable Energy Sources

  • Solar (PV): Converts sunlight to electricity via panels.

    • Formula: P = \eta \times A \times G

  • Wind: Wind drives turbines.

    • Formula: P = \frac{1}{2}\rho A v^3

  • Hydropower: Flowing water drives turbines.

    • Formula: P = \rho g Q h

  • Geothermal: Earth's heat produces steam.

  • Biomass: Organic material undergoes combustion/processing.

  • Hydropower Example:

    • Given: Q = 2 \, m³/s, h = 10 \, m

    • Calculation: P = 1000 \times 9.81 \times 2 \times 10 = 196200 \, W \approx 196.2 \, kW

LO4 – Series and Parallel Networks

Series Circuits

  • Rules:

    • Same current throughout the circuit.

    • Voltages add up.

    • Resistances add up.
      R_T = R_1 + R_2 + R_3

    • Current Calculation:
      I = \frac{V_T}{R_T}

    • Voltage Calculation:
      V_1 = I \times R_1,
      V_2 = I \times R_2,
      V_3 = I \times R_3

    • Total Voltage Calculation:
      V_T = V_1 + V_2 + V_3

  • Example:

    • For resistors R_1 = 2 \, Ω, R_2 = 3 \, Ω, R_3 = 5 \, Ω, and V = 20 \, V:

      • R_T = 10 \, Ω

      • I = 2 \, A

      • Calculated Voltages:

      • V_1 = 4 \, V

      • V_2 = 6 \, V

      • V_3 = 10 \, V

      • Verification: 4 + 6 + 10 = 20 \, V ✓

Potential Divider

  • Formula: V_{out} = V_{in} \times \frac{R_2}{(R_1 + R_2)}

    • Explanation: The output voltage is taken across R_2. The larger the R_2 relative to R_1, the greater the voltage received by R_2.

  • Example:

    • Given: V_{in} = 15 \, V, R_1 = 5 \, Ω, R_2 = 10 \, Ω

    • Calculation:
      V_{out} = 15 \times \frac{10}{15} = 10 \, V

Parallel Circuits

  • Rules:

    • Same voltage across each branch.

    • Currents add up.

    • Total resistance is less than the smallest branch resistance.

    • Formula for total resistance:
      \frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}

  • Shortcut for Two Resistors:
    R_T = \frac{R_1 \times R_2}{(R_1 + R_2)}

  • Current Calculations:

    • I_T = \frac{V}{R_T}

    • For branch currents: I_1 = \frac{V}{R_1}, I_2 = \frac{V}{R_2}

    • Verification:
      I_T = I_1 + I_2 + I_3

  • Example:

    • Given: R_1 = 4 \, Ω, R_2 = 6 \, Ω, R_3 = 12 \, Ω, V = 12 \, V

    • Calculation:

    • \frac{1}{R_T} = \frac{3}{12} + \frac{2}{12} + \frac{1}{12} = \frac{6}{12}

    • Thus R_T = 2 \, Ω

    • And I_T = 6 \, A

      • For Branches:

      • I_1 = 3 \, A, I_2 = 2 \, A, I_3 = 1 \, A

  • Verification:

    • Total Currents: 3 + 2 + 1 = 6 \, A ✓

Current Division

  • Formulas:

    • For resistors R_1 and R_2:
      I_1 = I_T \times \frac{R_2}{(R_1 + R_2)}
      I_2 = I_T \times \frac{R_1}{(R_1 + R_2)}

    • Note: Current through smaller resistor increase in comparison to larger resistor.

  • Example:

  • Given: I_T = 12 \, A, R_1 = 8 \, Ω, R_2 = 4 \, Ω:

    • Calculation:

    • I_1 = 12 \times \frac{4}{12} = 4 \, A

    • I_2 = 12 \times \frac{8}{12} = 8 \, A

    • Verification:

    • Total: 4 + 8 = 12 \, A ✓

Relative and Absolute Voltages

  • Definitions:

    • Absolute Voltage: Measured from earth (0 V reference).

    • Relative Voltage: Voltage difference between two points — can be positive or negative.

    • Example:

    • If Point A = 10 V, Point B = 4 V:

      • V_A relative to B = 10 - 4 = +6 \, V

      • V_B relative to A = 4 - 10 = -6 \, V

Wiring Lamps – Series vs Parallel

  • Comparison:

    • Series:

    • Brightness: Dimmer (voltage is shared)

    • One lamp failure: All go out

    • Voltage per lamp: \frac{V_{supply}}{n}

    • Total resistance: Increases with each lamp

    • Applications: Decorative/fairy lights

    • Parallel:

    • Brightness: Full brightness

    • One lamp failure: Others stay on

    • Voltage per lamp: Full supply voltage

    • Total resistance: Decreases with each lamp

    • Applications: Home & commercial lighting.

LO5 – Capacitors and Capacitance

What is a Capacitor?

  • A capacitor is a device that stores electrical charge. It consists of two conducting plates separated by an insulating material known as a dielectric. It allows AC signals to pass through but blocks DC once fully charged.

Capacitance

  • Formula for Capacitance: C = \frac{Q}{V}

    • Rearranged:

    • Q = C \times V

    • V = \frac{Q}{C}

    • Units: Capacitance is measured in Farads (F), charge in Coulombs (C), and voltage in Volts (V).

    • Example Calculation:

    • For a 100 µF capacitor charged to 50 V:
      Q = 100 \times 10^{-6} \times 50 = 0.005 C = 5 \, mC

Electric Field Strength

  • Formula for Electric Field Strength: E = \frac{V}{d}

    • Where V = voltage across plates (V) and d = distance between plates (m).

    • Example Calculation:

    • For V = 500 \, V and d = 0.005 \, m,
      E = \frac{500}{0.005} = 100000 \, V/m = 100 \, kV/m

Electric Flux Density

  • Formula for Electric Flux Density: D = \frac{Q}{A}

    • Where D = flux density (C/m²), Q = charge on plate (C), and A = plate area (m²).

    • Example Calculation:

    • For Q = 0.002 \, C and A = 0.01 \, m²,
      D = \frac{0.002}{0.01} = 0.2 \, C/m²

Permittivity

  • Formula for Permittivity: \epsilon = \epsilon_0 \times \epsilon_r

    • Where \epsilon_0 = 8.85 \times 10^{-12} \, F/m (constant) and \epsilon_r is the relative permittivity of the dielectric (given).

    • Approximate Values for Common Materials:

    • Air/vacuum: \epsilon_r = 1

    • Paper: \epsilon_r = 2 - 2.5

    • Polyester: \epsilon_r = 3 - 4

    • Mica: \epsilon_r = 5 - 7

    • Ceramic: \epsilon_r = 10 - 1000+

Parallel Plate Capacitor

  • Capacitance Formula for Parallel Plates: C = \frac{\epsilon_0 \epsilon_r A}{d}

    • Influence on Capacitance:

    • Increase plate area A → C increases

    • Decrease plate separation d → C increases

    • Use dielectric with higher \epsilon_r → C increases

    • Example Calculation:

    • For \epsilon_r = 4, A = 0.02 \, m², d = 0.002 \, m,
      C = \frac{8.85 \times 10^{-12} \times 4 \times 0.02}{0.002} = 354 \ pF

Capacitors in Parallel and Series

  • Capacitors in Parallel:

    • Total Capacitance:
      C_T = C_1 + C_2 + C_3

    • Same voltage across all capacitors.

    • Example:

    • For C_1 = 4 \, µF, C_2 = 6 \, µF, C_3 = 10 \, µF,
      C_T = 20 \, µF

  • Capacitors in Series:

    • Total Capacitance:
      \frac{1}{C_T} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}

    • Shortcut for Two Capacitors:
      C_T = \frac{C_1 \times C_2}{(C_1 + C_2)}

    • Same charge on each capacitor.

    • Total Voltage Calculation:
      V_T = V_1 + V_2 + V_3

    • Example:

    • For C_1 = 6 \, µF, C_2 = 3 \, µF, C_3 = 2 \, µF, \frac{1}{C_T} = \frac{1}{6} + \frac{1}{3} + \frac{1}{2} = 1

      • Thus, C_T = 1 \, µF

Dielectric Strength

  • Formula for Maximum Voltage: ext{Max V} = ext{Dielectric Strength} \times d

    • Where d is the plate separation (m).

    • Example Calculation:

    • For dielectric strength of 20 kV/mm and d = 0.5 mm,
      ext{Max V} = 20 \times 0.5 = 10 kV

    • Note: Exceeding this voltage permanently damages the capacitor.

Energy Stored in a Capacitor

  • Energy Formula: E = \frac{1}{2} C V^2 = \frac{Q^2}{2C} = \frac{QV}{2}

    • Example Calculation:

    • For C = 200 \, µF and V = 100 \, V:
      E = \frac{1}{2} \times 200 \times 10^{-6} \times 10,000 = 1 \, J

RC Time Constant

  • Formula for Time Constant:
    \tau = R \times C

  • Interpretation: Time constant \tau indicates how fast a capacitor charges/discharges through a resistor.

  • Charging/Discharging Characteristics:

    • 1\tau → 63.2% charge

    • 2\tau → 86.5% charge

    • 3\tau → 95.0% charge

    • 4\tau → 98.2% charge

    • 5\tau → ≈ 100% charge (fully charged)

    • 1\tau → 36.8% remaining

    • 2\tau → 13.5% remaining

    • 3\tau → 5.0% remaining

    • 4\tau → 1.8% remaining

    • 5\tau → ≈ 0% (fully discharged)

  • Example Calculation:

    • For R = 22 \, kΩ and C = 47 \, µF:
      \tau = 22000 \times 47 \times 10^{-6} = 1.034 \, s

    • Fully charged after 5\tau ≈ 5.17 \, s

Practical Types of Capacitors

  • Types and Applications:

    • Electrolytic: Uses aluminum oxide dielectric; suitable for large capacitance in power supplies.

    • Ceramic: Uses ceramic dielectric; suitable for high-frequency circuits.

    • Polyester (film): Uses plastic film dielectric; general-purpose applications.

    • Mica: Uses mica dielectric; suitable for precision and high frequency.

LO6 – Magnetism

6.1 Electromagnetism

Magnetic Field Due to Electric Current

  • Principle: When current flows through a conductor, a circular magnetic field is generated around it.

  • Right-Hand Grip Rule: Grip the conductor with your right hand; the thumb points in the direction of current, and the fingers curl in the direction of the magnetic field.

  • Magnetic Field Strength Formula: H = \frac{NI}{l}

    • Where

    • N = number of turns,

    • I = current (A),

    • l = length of coil (m).

    • Example Calculation:

    • For N = 500, I = 2 \, A, l = 0.1 \, m:
      H = \frac{500 \times 2}{0.1} = 10,000 \, A/m

Magnetic Flux Density

  • Formula: B = \mu_0 \mu_r H

    • Units: Tesla (T)

    • Where:

    • \mu_0 = 4 \pi \times 10^{-7} H/m,

    • \mu_r = relative permeability of core material.

Force on a Current-Carrying Conductor

  • Formula: F = BIL

    • Where:

    • B = magnetic flux density (T)

    • I = current (A)

    • L = length in field (m)

    • Example Calculation:

    • For B = 0.8 \, T, I = 5 \, A, L = 0.3 \, m:
      F = 0.8 \times 5 \times 0.3 = 1.2 \, N

Fleming's Left-Hand Rule (for Motors)

  • Rule Summary:

    • First Finger = Field Direction

    • Second Finger = Current Direction

    • Thumb = Motion/Force Direction

Force on a Moving Charge

  • Formula: F = q v B

    • Where:

    • q = charge (C)

    • v = velocity (m/s)

    • B = flux density (T)

    • Example Calculation:

    • For q = 1.6 \times 10^{-19} \, C, v = 2 \times 10^{6} \, m/s, B = 0.3 \, T:
      F = 1.6 \times 10^{-19} \times 2 \times 10^{6} \times 0.3 = 9.6 \times 10^{-14} \, N

DC Motor – Torque

  • Torque Formula: T = B I N A

    • Where:

    • T = torque (Nm)

    • N = number of coil turns

    • A = area of coil (m²)

    • Example Calculation:

    • For B = 0.5 \, T, I = 2 \, A, N = 50, A = 0.01 \, m²:
      T = 0.5 \times 2 \times 50 \times 0.01 = 0.5 \, Nm

  • Function of Commutator: It reverses current direction every half turn, keeping rotation continuous.

6.2 Electromagnetic Induction

  • Faraday's Law:

  • Formula: E = N \times \frac{\Delta \Phi}{\Delta t}

    • Where:

    • N = number of turns,

    • \Delta \Phi = change in magnetic flux (Webers or Wb),

    • \Delta t = time (s).

  • Lenz's Law: The induced current flows in a direction that opposes the change causing it, hence the negative sign in the full formula.

  • Fleming's Right-Hand Rule (for Generators):

    • First Finger = Field

    • Second Finger = Induced Current

    • Thumb = Motion

  • Example Calculation:

    • For N = 200, \Delta \Phi = 0.05 \, Wb, \Delta t = 0.1 \, s:
      E = 200 \times \frac{0.05}{0.1} = 100 \, V

  • Inductance of a Coil:

    • Formula:
      L = \frac{N^2 \mu A}{l}

    • Where:

    • L = inductance (Henrys, H)

    • N = number of turns

    • \mu = permeability of core material (H/m)

    • A = cross-sectional area (m²)

    • l = length of coil (m)

    • Example Calculation:

    • For N = 100, \mu = 4\pi \times 10^{-7} \ H/m, A = 0.01 \, m², l = 0.2 \, m:
      L \approx 0.63 \, mH

  • Energy Stored in an Inductor:

    • Formula:
      E = \frac{1}{2} L I^2

    • Example Calculation:

    • For L = 2 \, H and I = 3 \, A:
      E = \frac{1}{2} \times 2 \times 3^2 = 9 \, J

  • Mutual Inductance:

  • Formula for induced EMF: E_2 = M \times \frac{\Delta I_1}{\Delta t}

    • Where M = mutual inductance (H). The changing field of coil 1 induces an EMF in coil 2, applied in transformers.

    • Example Calculation:

    • For M = 0.5 \, H, \Delta I = 4 \, A, \Delta t = 0.2 \, s:
      E = 0.5 \times \frac{4}{0.2} = 10 \, V

LO7 – Electrical Measuring Instruments and Measurements

Introduction: Forces in Analogue Instruments

  • Deflecting Force: Causes pointer movement proportional to the quantity measured.

  • Controlling Force: A hairspring returning the pointer to zero when current stops.

  • Damping Force: Prevents pointer oscillation, ensuring steady readings.

Moving-Coil Instrument

  • Mechanism: A permanent magnet provides a fixed field; current through a coil creates a force (F = BIL) that deflects the pointer against the hairspring.

  • Characteristics:

    • Measures DC only

    • Uniform (linear) scale

    • High accuracy and sensitivity

    • Not suitable for AC without a rectifier

Moving-Iron Instrument

  • Mechanism: Current through a coil magnetizes iron pieces inside; they either attract or repel, deflecting the pointer.

  • Characteristics:

    • Measures both AC and DC

    • Non-uniform (cramped lower end) scale

    • Robust and cheap

    • Less accurate than moving-coil

    • Affected by external magnetic fields.

Moving-Coil Rectifier Instrument

  • Mechanism: Can only read DC; adding a rectifier (diode bridge) converts AC to pulsating DC.

  • Characteristics:

    • Measures AC (via rectifier)

    • More accurate than moving-iron for AC

    • Uniform scale

Comparison Table: Measuring Instruments

Feature

Moving-Coil

Moving-Iron

Moving-Coil Rectifier

Measures

DC only

AC and DC

AC (via rectifier)

Accuracy

High

Moderate

Good

Scale

Uniform

Non-uniform

Uniform

Cost

Higher

Lower

Moderate

Sensitivity

High

Lower

High

External field effect

Low

High

Low

Construction

Permanent magnet + coil

Coil + iron pieces

Moving-coil + rectifier

Connection of Meters in Circuits

  • Instrument | Measures | Connected | Internal Resistance

  • Ammeter | Current (A) | In SERIES | Very LOW

  • Voltmeter | Voltage (V) | In PARALLEL | Very HIGH

  • Ohmmeter | Resistance (Ω) | Directly across component | —

  • Wattmeter | Power (W) | Current coil in series; voltage coil in parallel | —

Connection Notes:

  • Ammeter: Should have very low resistance so it does not affect circuit current.

  • Voltmeter: Should have very high resistance to draw negligible current.

Master Formula Reference – All Learning Outcomes

  • Formula Table:
    | Quantity Found | Formula | Learning Outcome |
    |----------------|-------------|------------------|
    | Charge (C) | Q = I \times t | LO2 |
    | Work / Energy (J) | W = V \times Q | LO2 |
    | Power (W) | P = V \times I, P = I^2R, P = \frac{V^2}{R} | LO2 |
    | Electrical energy (J / kWh) | E = P \times t | LO2 |
    | Voltage (V) − Ohm's Law | V = I \times R | LO2 |
    | Current (A) − Ohm's Law | I = \frac{V}{R} | LO2 |
    | Resistance (Ω) − Ohm's Law | R = \frac{V}{I} | LO2 |
    | Conductance (S) | G = \frac{1}{R} | LO2 |
    | Resistance from material properties | R = \frac{\rho L}{A} | LO2 |
    | Resistance at temperature T | R_T = R_0(1 + \alpha T) | LO2 |
    | Fuse current rating | I = \frac{P}{V} | LO2 |
    | Terminal voltage (V) | V = E − Ir | LO3 |
    | Circuit current with internal resistance | I = \frac{E}{(R + r)} | LO3 |
    | Cell capacity (Ah) | Capacity = I \times t | LO3 |
    | Hydropower (W) | P = \rho g Q h | LO3 |
    | Total series resistance | R_T = R_1 + R_2 + R_3 | LO4 |
    | Total parallel resistance | \frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} | LO4 |
    | Parallel resistance (2 resistors) | R_T = \frac{R_1 \times R_2}{(R_1 + R_2)} | LO4 |
    | Potential divider output | V_{out} = V_{in} \times \frac{R_2}{(R_1 + R_2)} | LO4 |
    | Current division (branch 1) | I_1 = I_T \times \frac{R_2}{(R_1 + R_2)} | LO4 |
    | Capacitance (F) | C = \frac{Q}{V} | LO5 |
    | Electric field strength (V/m) | E = \frac{V}{d} | LO5 |
    | Electric flux density (C/m²) | D = \frac{Q}{A} | LO5 |
    | Parallel plate capacitance | C = \frac{\epsilon \epsilon_r A}{d} | LO5 |
    | Total parallel capacitance | C_T = C_1 + C_2 + C_3 | LO5 |
    | Total series capacitance | \frac{1}{C_T} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} | LO5 |
    | Energy in capacitor (J) | E = \frac{1}{2} C V^2 | LO5 |
    | RC time constant (s) | \tau = R \times C | LO5 |
    | Magnetic field strength (A/m) | H = \frac{NI}{l} | LO6 |
    | Magnetic flux density (T) | B = \mu \mu_r H | LO6 |
    | Force on conductor (N) | F = BIL | LO6 |
    | Force on charge (N) | F = qvB | LO6 |
    | Torque on motor coil (Nm) | T = BINA | LO6 |
    | Induced EMF — Faraday's Law | E = N \times \frac{\Delta \Phi}{\Delta t} | LO6 |
    | Inductance of coil (H) | L = \frac{N^2 \mu A}{l} | LO6 |
    | Energy in inductor (J) | E = \frac{1}{2} LI^2 | LO6 |
    | Mutual inductance EMF (V) | E = M \times \frac{\Delta I}{\Delta t} | LO6 |