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Irish Electrical Apprentice Mathematics Hub
Welcome to the Irish Electrical Apprentice Mathematics Hub. This resource is designed to help Irish electrical apprentices master the mathematical principles embedded in the off-the-job training phases (Phase 2, Phase 4, and Phase 6).
The electrical trade is highly technical. A solid command of mathematics is essential to passing your Electrical Science, Power Distribution, and Electronics examinations. This hub compiles the core formulas, provides revision guides, and offers realistic, worked exam-style questions conforming to SOLAS standards and the National Rules for Electrical Installations (I.S. 10101:2020).
Study Guides & Question Banks
Select a phase to access its dedicated formulas, explanation guides, and worked mock exam questions:
| Phase | Training Location | Primary Math Focus Areas | Resource Link |
|---|---|---|---|
| Phase 2 | ETB Training Centre | Ohm's Law, Power, Resistivity, Temp Coefficient, Resistors (Series/Parallel), EMF & Internal Resistance, Electromagnetism. | Phase 2 Maths Guide |
| Phase 4 | Technical University / IoT | Kirchhoff's Laws, Single-Phase AC Theory (Peak/RMS/Reactance), R-L-C Series Circuits (Impedance, Power Factor), Star/Delta Basics, Single-Phase Transformers. | Phase 4 Maths Guide |
| Phase 6 | Technical University / IoT | Parallel AC Circuits, Power Factor Correction (kVAR & Capacitance), Advanced 3-Phase Power (Star/Delta), Induction Motors, Cable Sizing (I.S. 10101), Earth Loop Impedance. | Phase 6 Maths Guide |
Quick References & Tools
- Unified Formula Cheat Sheet: A single document consolidating all formulas across all phases, showing definitions, unit symbols, and transposition patterns.
- Curated External Resources: Recommended YouTube channels, online practice exams, revision courses, and key textbooks specific to the Irish curriculum.
- Useful Conversions & Metric Prefixes: A reference guide for metric prefixes, Horsepower-to-kW scaling, loop resistance math, and angular unit switches.
- 40-Question Comprehensive Revision Test: A mock examination containing 40 random mathematical questions covering all phases of your training, with worked answers provided on the last page only.
Top Tips for Success in Apprentice Maths
- Master formula transposition early: You are rarely asked to solve for the default subject of a formula. Learn the rules of algebra to rearrange equations (e.g., changing $V = I \times R$ to $R = \frac{V}{I}$).
- Draw the circuit/vector diagram: In Phase 4 and 6, drawing a phasor diagram or circuit layout takes 30 seconds but prevents 90% of basic calculation errors.
- Understand your units: Keep track of prefixes like milli ($m = \times 10^{-3}$), micro ($\mu = \times 10^{-6}$), kilo ($k = \times 10^{3}$), and mega ($M = \times 10^{6}$). Forgetting to convert milli-amps to amps or micro-farads to farads will lead to incorrect answers.
- Practice daily: Mathematical confidence is muscle memory. Solving just 2 questions a day is more effective than cramming for 12 hours the weekend before your exams.
Unified Electrical Formula Cheat Sheet
This formula sheet compiles all mathematical equations required across Phase 2, Phase 4, and Phase 6 of the Irish Electrical Apprenticeship.
Phase 2: DC Science & Basic Principles
1. Ohm's Law & Power
Used to determine relationships between voltage, current, resistance, and electrical power.
| Base Formula | Transpositions | Variable Meanings & Units |
|---|---|---|
| $V = I \times R$ | $I = \frac{V}{R}$ $R = \frac{V}{I}$ |
$V$: Voltage / Potential Difference (Volts, $\text{V}$) $I$: Current (Amperes, $\text{A}$) $R$: Resistance (Ohms, $\Omega$) |
| $P = V \times I$ | $I = \frac{P}{V}$ $V = \frac{P}{I}$ |
$P$: Power (Watts, $\text{W}$) |
| $P = I^2 \times R$ | $R = \frac{P}{I^2}$ $I = \sqrt{\frac{P}{R}}$ |
$P$: Power (Watts, $\text{W}$) |
| $P = \frac{V^2}{R}$ | $R = \frac{V^2}{P}$ $V = \sqrt{P \times R}$ |
$P$: Power (Watts, $\text{W}$) |
2. Resistivity & Temperature Coefficient
For calculating the resistance of conductors based on material characteristics, dimensions, and temperature.
| Base Formula | Transpositions | Variable Meanings & Units |
|---|---|---|
| $R = \rho \frac{l}{A}$ | $l = \frac{R \times A}{\rho}$ $A = \frac{\rho \times l}{R}$ $\rho = \frac{R \times A}{l}$ |
$\rho$: Resistivity (Ohm-meters, $\Omega\cdot\text{m}$ or $\mu\Omega\cdot\text{mm}$) $l$: Length of conductor (Meters, $\text{m}$) $A$: Cross-sectional area (Square meters, $\text{m}^2$ or $\text{mm}^2$) |
| $A = \frac{\pi d^2}{4}$ | $d = \sqrt{\frac{4A}{\pi}}$ | $d$: Conductor diameter (Meters or Millimeters) |
| $R_{\theta} = R_0(1 + \alpha \theta)$ | $R_0 = \frac{R_{\theta}}{1 + \alpha \theta}$ $\theta = \frac{R_{\theta} - R_0}{R_0 \times \alpha}$ |
$R_0$: Resistance at $0^\circ\text{C}$ ($\Omega$) $R_{\theta}$: Resistance at temperature $\theta$ ($\Omega$) $\alpha$: Temperature coefficient of resistance ($/{}^\circ\text{C}$) $\theta$: Temperature change / value ($^\circ\text{C}$) |
3. Resistor Networks & Internal Resistance
| Base Formula | Transpositions | Variable Meanings & Units |
|---|---|---|
| Series Resistance: $R_T = R_1 + R_2 + R_3 + \dots$ |
None (additive) | $R_T$: Total resistance in series ($\Omega$) |
| Parallel (Two Resistors): $R_T = \frac{R_1 \times R_2}{R_1 + R_2}$ |
$R_2 = \frac{R_T \times R_1}{R_1 - R_T}$ | Product-over-sum method (only for exactly two parallel resistors) |
| Parallel (General): $\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots$ |
$R_T = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots}$ | Reciprocal method (for any number of parallel resistors) |
| EMF & Internal Resistance: $E = I(R + r) = V_{term} + I \cdot r$ |
$I = \frac{E}{R + r}$ $r = \frac{E - V_{term}}{I}$ |
$E$: Electromotive Force / Open-circuit voltage ($\text{V}$) $V_{term}$: Terminal voltage under load ($\text{V}$) $r$: Internal resistance of cell ($\Omega$) $R$: Load resistance ($\Omega$) |
4. Basic Electromagnetism
| Base Formula | Transpositions | Variable Meanings & Units |
|---|---|---|
| Magnetic Flux Density: $B = \frac{\Phi}{A}$ |
$\Phi = B \times A$ $A = \frac{\Phi}{B}$ |
$B$: Flux density (Teslas, $\text{T}$) $\Phi$: Magnetic flux (Webers, $\text{Wb}$) $A$: Pole area ($\text{m}^2$) |
| Force on a Conductor: $F = B \times I \times l \times \sin\theta$ |
$I = \frac{F}{B \cdot l \cdot \sin\theta}$ $B = \frac{F}{I \cdot l \cdot \sin\theta}$ |
$F$: Force (Newtons, $\text{N}$) $l$: Length of conductor in field ($\text{m}$) $\theta$: Angle between conductor and flux lines (usually $90^\circ$, so $\sin 90^\circ = 1$) |
5. Conduit & Trunking Capacity
| Base Formula | Transpositions | Variable Meanings & Units |
|---|---|---|
| Conduit Sizing (Factor Method): $\sum (\text{Qty} \times \text{Cable Factor}) \le \text{Conduit Factor}$ |
None | Cable Factor: Tabulated factor for cable diameter Conduit Factor: Tabulated capacity factor for straight runs or runs with bends |
Phase 4: Single-Phase AC Theory & Transformers
1. Waveform Values & Reactance
| Base Formula | Transpositions | Variable Meanings & Units |
|---|---|---|
| Frequency / Period: $f = \frac{1}{T}$ |
$T = \frac{1}{f}$ | $f$: Frequency (Hertz, $\text{Hz}$) $T$: Period / Time for 1 cycle (Seconds, $\text{s}$) |
| RMS Voltage: $V_{RMS} = V_{peak} \times 0.707$ |
$V_{peak} = \frac{V_{RMS}}{0.707}$ | For pure sine waves ($0.707 = \frac{1}{\sqrt{2}}$) |
| Average Voltage: $V_{avg} = V_{peak} \times 0.637$ |
$V_{peak} = \frac{V_{avg}}{0.637}$ | For pure sine waves ($0.637 = \frac{2}{\pi}$) |
| Inductive Reactance: $X_L = 2\pi f L$ |
$L = \frac{X_L}{2\pi f}$ $f = \frac{X_L}{2\pi L}$ |
$X_L$: Inductive Reactance (Ohms, $\Omega$) $L$: Inductance (Henrys, $\text{H}$) |
| Capacitive Reactance: $X_C = \frac{1}{2\pi f C}$ |
$C = \frac{1}{2\pi f X_C}$ $f = \frac{1}{2\pi C X_C}$ |
$X_C$: Capacitive Reactance (Ohms, $\Omega$) $C$: Capacitance (Farads, $\text{F}$) |
2. Series AC Circuits & Power Triangle
| Base Formula | Transpositions | Variable Meanings & Units |
|---|---|---|
| Series Impedance: $Z = \sqrt{R^2 + X^2}$ (where $X = X_L - X_C$ or $X_C - X_L$) |
$R = \sqrt{Z^2 - X^2}$ $X = \sqrt{Z^2 - R^2}$ |
$Z$: Impedance ($\Omega$) $R$: Resistance ($\Omega$) $X$: Net Reactance ($\Omega$) |
| AC Ohm's Law: $V = I \times Z$ |
$I = \frac{V}{Z}$ $Z = \frac{V}{I}$ |
$V$: Applied Voltage ($\text{V}$) $I$: Total current ($\text{A}$) |
| Power Factor: $\cos\phi = \frac{R}{Z} = \frac{P}{S}$ |
$\phi = \arccos\left(\frac{R}{Z}\right)$ | $\cos\phi$: Power Factor (dimensionless ratio, 0 to 1) |
| Active Power: $P = V \times I \times \cos\phi$ |
$I = \frac{P}{V \cos\phi}$ | $P$: Active/True Power (Watts, $\text{W}$ or $\text{kW}$) |
| Apparent Power: $S = V \times I$ |
$I = \frac{S}{V}$ | $S$: Apparent Power (Volt-Amperes, $\text{VA}$ or $\text{kVA}$) |
| Reactive Power: $Q = V \times I \times \sin\phi$ |
$Q = \sqrt{S^2 - P^2}$ | $Q$: Reactive Power (Volt-Amperes Reactive, $\text{VAr}$ or $\text{kVAr}$) |
| Power Triangle Relation: $S = \sqrt{P^2 + Q^2}$ |
$P = \sqrt{S^2 - Q^2}$ | Derived from Pythagoras's Theorem |
3. Transformers & 3-Phase Basics
| Base Formula | Transpositions | Variable Meanings & Units |
|---|---|---|
| Transformer Turns Ratio: $\frac{N_p}{N_s} = \frac{V_p}{V_s} = \frac{I_s}{I_p}$ |
$V_s = V_p \times \frac{N_s}{N_p}$ $I_s = I_p \times \frac{N_p}{N_s}$ |
$N$: Turns, $V$: Voltage, $I$: Current $p$: Primary winding, $s$: Secondary winding |
| Three-Phase Transformer (Delta-Star): $\frac{N_p}{N_s} = \sqrt{3} \frac{V_{p\_line}}{V_{s\_line}}$ |
None | $\frac{N_p}{N_s}$: Phase turns ratio $V_{p\_line}$: Line voltage primary (Delta) $V_{s\_line}$: Line voltage secondary (Star) |
| Transformer Efficiency: $\eta\% = \frac{P_{out}}{P_{in}} \times 100$ |
$\eta\% = \frac{P_{out}}{P_{out} + P_{cu} + P_{fe}} \times 100$ | $P_{cu}$: Copper Losses (variable, $I^2R$) $P_{fe}$: Iron/Core Losses (constant, $\text{W}$) |
| Star (Wye) Connections: $V_L = \sqrt{3} \times V_P$ $I_L = I_P$ |
$V_P = \frac{V_L}{\sqrt{3}}$ | $V_L$: Line Voltage (typically $400\text{V}$ in IE) $V_P$: Phase Voltage (typically $230\text{V}$ in IE) $I_L, I_P$: Line and Phase current |
| Delta Connection: $V_L = V_P$ $I_L = \sqrt{3} \times I_P$ |
$I_P = \frac{I_L}{\sqrt{3}}$ | Standard motor connection for high torque |
Phase 6: Advanced AC, Motors, & Regulations
1. Parallel AC Currents & Power Factor Correction
| Base Formula | Transpositions | Variable Meanings & Units |
|---|---|---|
| Parallel Branch Currents: $I_T = \sqrt{I_R^2 + (I_C - I_L)^2}$ |
$I_R = \sqrt{I_T^2 - (I_C - I_L)^2}$ | $I_R = V/R$ (resistive/in-phase current) $I_L = V/X_L$ (inductive current) $I_C = V/X_C$ (capacitive current) |
| Power Factor Correction (kVAR): $Q_c = P(\tan\phi_1 - \tan\phi_2)$ |
None | $Q_c$: Capacitor reactive power ($\text{VAr}$ or $\text{kVAr}$) $P$: Total installation Active Power ($\text{W}$) $\phi_1$: Original phase angle ($\arccos \text{PF}_1$) $\phi_2$: Targeted phase angle ($\arccos \text{PF}_2$) |
| Required Capacitance: $C = \frac{Q_c}{2\pi f V^2}$ |
$Q_c = 2\pi f V^2 C$ | $C$: Capacitance in Farads ($\text{F}$) (Multiply by $10^6$ to convert to microfarads $\mu\text{F}$) |
2. Balanced 3-Phase Power & Two-Wattmeter Method
| Base Formula | Transpositions | Variable Meanings & Units |
|---|---|---|
| Three-Phase Active Power: $P = \sqrt{3} \times V_L \times I_L \times \cos\phi$ |
$I_L = \frac{P}{\sqrt{3} \cdot V_L \cdot \cos\phi}$ | $P$: Total 3-Phase Active Power ($\text{W}$) $V_L$: Line-to-line voltage ($\text{V}$) $I_L$: Line current ($\text{A}$) |
| Three-Phase Apparent Power: $S = \sqrt{3} \times V_L \times I_L$ |
$S = \sqrt{P^2 + Q^2}$ | $S$: Total 3-Phase Apparent Power ($\text{VA}$) |
| Three-Phase Reactive Power: $Q = \sqrt{3} \times V_L \times I_L \times \sin\phi$ |
None | $Q$: Total 3-Phase Reactive Power ($\text{VAr}$) |
| Two-Wattmeter Power: $P_T = W_1 + W_2$ |
None | $W_1, W_2$: Wattmeter readings ($\text{W}$) (Note: One reading may be negative if $\text{PF} < 0.5$) |
| Two-Wattmeter Phase Angle: $\tan\phi = \sqrt{3} \frac{W_1 - W_2}{W_1 + W_2}$ |
$\phi = \arctan\left(\sqrt{3} \frac{W_1 - W_2}{W_1 + W_2}\right)$ | To calculate power factor, find $\phi$, then compute $\cos\phi$ |
| Unbalanced Star Neutral Current: $I_{Nx} = I_{L1} - 0.5(I_{L2} + I_{L3})$ $I_{Ny} = 0.866(I_{L3} - I_{L2})$ $I_N = \sqrt{I_{Nx}^2 + I_{Ny}^2}$ |
None | $\vec{I}_N$: Neutral current vector sum $I_{L1}, I_{L2}, I_{L3}$: Line currents (resistive) $I_{Nx}, I_{Ny}$: Rectangular components of neutral current |
3. Induction Motors
| Base Formula | Transpositions | Variable Meanings & Units |
|---|---|---|
| Synchronous Speed: $N_s = \frac{120 \times f}{p}$ |
$p = \frac{120 \times f}{N_s}$ $f = \frac{N_s \times p}{120}$ |
$N_s$: Synchronous magnetic speed (RPM) $f$: Supply frequency (typically $50\text{Hz}$) $p$: Number of stator poles (must be an even number) |
| Percentage Slip: $s\% = \frac{N_s - N_r}{N_s} \times 100$ |
$N_r = N_s(1 - s)$ | $N_r$: Rotor speed (actual shaft RPM) $s$: Fractional slip (decimal, e.g., 0.04 for 4%) |
| Rotor Frequency: $f_r = s \times f$ |
$s = \frac{f_r}{f}$ | $f_r$: Frequency of currents induced in rotor ($\text{Hz}$) |
4. Cable Sizing & Regulations (I.S. 10101:2020)
| Base Formula | Transpositions | Variable Meanings & Units |
|---|---|---|
| Coordination Rule: $I_b \le I_n \le I_z$ |
None | $I_b$: Design Current ($\text{A}$) $I_n$: Protective Device Nominal Current ($\text{A}$) $I_z$: Cable Current Carrying Capacity under installation conditions |
| Adjusted Current Capacity: $I_t \ge \frac{I_n}{C_a \cdot C_g \cdot C_i \cdot C_f}$ |
None | $I_t$: Tabulated current carrying capacity of cable ($\text{A}$) $C_a$: Ambient temperature factor $C_g$: Grouping factor $C_i$: Thermal insulation factor $C_f$: Semi-enclosed fuse factor (0.725 for BS 3036 fuses; 1.0 for MCBs) |
| Voltage Drop (Single Phase): $V_d = \frac{I_b \times L \times (mV/A/m)}{1000}$ |
$(mV/A/m) = \frac{1000 \cdot V_d}{I_b \cdot L}$ | $V_d$: Volts dropped in cable ($\text{V}$) $L$: Route length (Meters) $(mV/A/m)$: Tabulated cable millivolt-drop factor |
| Voltage Drop (Three Phase): $V_d = \frac{\sqrt{3} \times I_b \times L \times (mV/A/m)}{1000}$ |
None | Used for balanced three-phase loads |
| Earth Loop Impedance: $Z_s = Z_e + \frac{(R_1 + R_2) \times L \times C_T}{1000}$ |
None | $Z_s$: Total earth fault loop impedance ($\Omega$) $Z_e$: External earth loop impedance (from supply transformer) $R_1 + R_2$: Conductor loop resistance per meter ($\text{m}\Omega/\text{m}$) $C_T$: Temperature correction factor (typically 1.2 for conductors running at full temp) |
Phase 2 Electrical Mathematics: Revision Guide & Question Bank
This document serves as your guide to mastering the mathematical applications in Phase 2 of the Irish Electrical Apprenticeship.
Key Revision Concepts
1. Ohm's Law & Power
At the heart of electrical science is Ohm's Law, which describes the relationship between Voltage ($V$), Current ($I$), and Resistance ($R$). Power ($P$) represents the rate at which electrical energy is converted to another form (like heat or light).
| Concept / Parameter | Formula | Description & Units |
|---|---|---|
| Ohm's Law | $V = I \times R$ | $V$ = Voltage (Volts, $\text{V}$), $I$ = Current (Amperes, $\text{A}$), $R$ = Resistance (Ohms, $\Omega$) |
| Electrical Power | $P = V \times I$ | Base power formula (Watts, $\text{W}$) |
| Power (Losses) | $P = I^2 \times R$ | Used extensively for calculating cable heat/power losses |
| Power (Voltage/Resistance) | $P = \frac{V^2}{R}$ | Power in terms of voltage and resistance |
WARNING: Exam Trap: Always verify your units before putting them into an equation. Current must be in Amperes ($\text{A}$), not milliamperes ($\text{mA}$). Resistance must be in Ohms ($\Omega$), not kilohms ($\text{k}\Omega$). * $150\text{ mA} = 150 \times 10^{-3}\text{ A} = 0.15\text{ A}$ * $4.7\text{ k}\Omega = 4.7 \times 10^3\ \Omega = 4700\ \Omega$
2. Series & Parallel Resistor Networks
In series circuits, current remains the same through all components and resistance is additive. In parallel circuits, voltage is identical across all branches and current splits.
| Network Type | Formula | Method / Usage |
|---|---|---|
| Series Resistance | $R_T = R_1 + R_2 + R_3 + \dots$ | Additive resistance rule |
| Parallel (Two Resistors) | $R_T = \frac{R_1 \times R_2}{R_1 + R_2}$ | Product-over-Sum rule (exactly two resistors) |
| Parallel (General) | $\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots$ | Reciprocal Method (for any number of resistors) |
TIP: Quick Check: In a parallel network, the total resistance ($R_T$) must always be smaller than the smallest individual resistor in that network. If your calculated $R_T$ is larger than any of the branches, re-run your calculation.
3. Resistivity ($R = \rho \frac{l}{A}$)
The resistance of any electrical conductor depends on three physical factors: 1. Material: Accounted for by its resistivity ($\rho$, Greek letter "rho"). Copper has a very low resistivity ($\approx 1.72 \times 10^{-8}\ \Omega\cdot\text{m}$), making it an excellent conductor. 2. Length ($l$): The longer the wire, the higher the resistance (direct relationship). 3. Cross-Sectional Area ($A$): The thicker the wire, the lower the resistance (inverse relationship).
| Parameter | Formula | Description & Units |
|---|---|---|
| Conductor Resistance | $R = \rho \frac{l}{A}$ | $R$ = Resistance ($\Omega$), $\rho$ = Resistivity ($\Omega\cdot\text{m}$), $l$ = Length ($\text{m}$), $A$ = Area ($\text{m}^2$) |
| Cross-Sectional Area | $A = \frac{\pi d^2}{4} \approx 0.7854 \times d^2$ | Area of a solid circular wire from its diameter $d$ |
4. Temperature Coefficient of Resistance
As a conductor heats up, its atoms vibrate more rapidly, making it harder for electrons to flow, thereby increasing resistance. The rate of change is called the temperature coefficient ($\alpha$).
| Calculation Type | Formula | Description & Variables |
|---|---|---|
| Resistance at Temp $\theta$ | $R_{\theta} = R_0(1 + \alpha \theta)$ | $R_0$ = Resistance at $0^\circ\text{C}$ ($\Omega$), $\alpha$ = Temp coefficient, $\theta$ = Temp ($^\circ\text{C}$) |
| Resistance at Temp $\theta_2$ | $R_2 = R_1 \frac{1 + \alpha_0 \theta_2}{1 + \alpha_0 \theta_1}$ | Finding new resistance when starting from a known temperature $\theta_1$ |
| (Where $\alpha_0$ is the temp coefficient at $0^\circ\text{C}$) |
5. EMF & Internal Resistance
A battery or cell is not a perfect voltage source; it has its own internal resistance ($r$) which causes a voltage drop when supplying current.
| Parameter | Formula | Description & Variables |
|---|---|---|
| Battery EMF | $E = I(R + r)$ | $E$ = Electromotive Force / open-circuit voltage ($\text{V}$) |
| Terminal Voltage | $V_{term} = E - I \cdot r$ | $V_{term}$ = Voltage across battery terminals under load ($\text{V}$) |
| EMF Relation | $E = V_{term} + I \cdot r$ | Total EMF is terminal voltage plus internal volt drop |
| (Where $I \cdot r$ is the internal voltage drop within the battery itself) |
6. Magnetic Force on a Conductor
When a current-carrying conductor is placed inside a magnetic field, it experiences a mechanical force (the motor principle).
| Parameter | Formula | Description & Variables |
|---|---|---|
| Magnetic Force | $F = B \times I \times l \times \sin\theta$ | $F$ = Force ($\text{N}$), $B$ = Flux density ($\text{T}$), $I$ = Current ($\text{A}$), $l$ = Active length ($\text{m}$), $\theta$ = Angle |
| Perpendicular Force | $F = B \times I \times l$ | When conductor is perpendicular to field lines ($\sin 90^\circ = 1$) |
7. Conduit & Trunking Capacity (Space Factor)
When pulling cables through containment like conduits or trunking, it is vital not to overfill them to prevent mechanical damage and overheating.
| Method / Rule | Formula | Description & Suitability |
|---|---|---|
| 45% Space Factor Rule | $\text{Space Factor} = \frac{\text{Total Cross-Sectional Area of Cables}}{\text{Internal Cross-Sectional Area of Containment}} \times 100\% \le 45\%$ | Total cross-sectional area of all cables (with insulation) must not exceed 45% of containment area |
| Unit/Factor Method | $\sum (\text{Quantity} \times \text{Cable Factor}) \le \text{Conduit Capacity Factor}$ | Sum of cable factors must not exceed the containment capacity factor |
Reference Factors (I.S. 10101:2020)
| Cable Size | Stranded Cable Factor | Conduit Diameter | Conduit Capacity Factor (with Bends) | |
|---|---|---|---|---|
| $1.5\text{ mm}^2$ | 22 | $20\text{ mm}$ | 286 | |
| $2.5\text{ mm}^2$ | 30 | $25\text{ mm}$ | 543 | |
| $4.0\text{ mm}^2$ | 43 | $32\text{ mm}$ | 865 |
Worked Example Questions
Let's work through 6 realistic exam-style questions step-by-step.
Question 1: Resistor Networks
Scenario: A circuit consists of a $12\ \Omega$ resistor connected in series with a parallel group of two resistors, rated at $15\ \Omega$ and $30\ \Omega$. The entire combination is connected across a $240\text{V DC}$ supply. 1. Calculate the total equivalent resistance ($R_T$) of the circuit. 2. Calculate the total current ($I_T$) drawn from the supply. 3. Calculate the power ($P_{12}$) dissipated by the $12\ \Omega$ resistor.
Step-by-Step Solution:
Step 1: Calculate the parallel group resistance ($R_p$) Use the product-over-sum rule for the $15\ \Omega$ and $30\ \Omega$ parallel resistors: $$R_p = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{15 \times 30}{15 + 30} = \frac{450}{45} = 10\ \Omega$$
Step 2: Calculate the total resistance ($R_T$) The $12\ \Omega$ resistor is in series with the parallel combination ($R_p = 10\ \Omega$): $$R_T = R_{series} + R_p = 12\ \Omega + 10\ \Omega = 22\ \Omega$$ * Answer 1: Total resistance is $22\ \Omega$
Step 3: Calculate the total current ($I_T$) Using Ohm's Law: $$I_T = \frac{V}{R_T} = \frac{240\text{ V}}{22\ \Omega} \approx 10.91\text{ A}$$ * Answer 2: Total current is $10.91\text{ A}$
Step 4: Calculate the power dissipated by the $12\ \Omega$ resistor Since the $12\ \Omega$ resistor is in series with the main supply line, the full total current ($I_T$) flows through it. Use the power formula: $$P_{12} = I_T^2 \times R = (10.91\text{ A})^2 \times 12\ \Omega \approx 119.03 \times 12 \approx 1428.3\text{ W}\ (1.43\text{ kW})$$ * Answer 3: Power dissipated is $1428.3\text{ W}$
Question 2: Conductor Resistance & Resistivity
Scenario: A single-core copper conductor has a length of $120\text{ meters}$ and a cross-sectional area of $2.5\text{ mm}^2$. The resistivity of copper is $1.72 \times 10^{-8}\ \Omega\cdot\text{m}$. 1. Calculate the electrical resistance ($R$) of the conductor. 2. If the length of this conductor is doubled, what is the new resistance?
Step-by-Step Solution:
Step 1: Convert units to standard SI format * Length ($l$) = $120\text{ m}$ (already in standard units) * Area ($A$) = $2.5\text{ mm}^2$. We must convert square millimeters to square meters: $$1\text{ mm}^2 = 1 \times 10^{-6}\text{ m}^2 \implies 2.5\text{ mm}^2 = 2.5 \times 10^{-6}\text{ m}^2$$ * Resistivity ($\rho$) = $1.72 \times 10^{-8}\ \Omega\cdot\text{m}$
Step 2: Apply the resistivity formula $$R = \rho \frac{l}{A} = (1.72 \times 10^{-8}) \times \frac{120}{2.5 \times 10^{-6}}$$ Simplify the math: $$R = 1.72 \times 10^{-8} \times 48 \times 10^{6} = 1.72 \times 48 \times 10^{-2} = 82.56 \times 10^{-2} = 0.826\ \Omega$$ * Answer 1: The resistance is $0.826\ \Omega$
Step 3: Analyze the effect of doubling length Since resistance is directly proportional to length ($R \propto l$), doubling the length will double the resistance: $$R_{new} = 2 \times R = 2 \times 0.8256 = 1.651\ \Omega$$ * Answer 2: The new resistance is $1.651\ \Omega$
Question 3: Temperature Coefficient
Scenario: The field windings of a DC shunt motor have a resistance of $55\ \Omega$ at a room temperature of $20^\circ\text{C}$. After running under load for several hours, the temperature of the winding increases to $70^\circ\text{C}$. Calculate the resistance of the winding at this operating temperature. (Take the temperature coefficient of copper at $20^\circ\text{C}$ to be $\alpha_{20} = 0.00393 /{}^\circ\text{C}$).
Step-by-Step Solution:
Step 1: Identify the variables * Initial resistance ($R_{20}$) = $55\ \Omega$ * Initial temperature ($\theta_1$) = $20^\circ\text{C}$ * Final temperature ($\theta_2$) = $70^\circ\text{C}$ * Temperature change ($\Delta\theta$) = $\theta_2 - \theta_1 = 70 - 20 = 50^\circ\text{C}$ * Temperature coefficient ($\alpha_{20}$) = $0.00393 /{}^\circ\text{C}$
Step 2: Apply the temperature coefficient formula Since the coefficient is given at the starting temperature ($20^\circ\text{C}$), we can use the simplified change formula: $$R_{\theta} = R_{start} (1 + \alpha_{start} \cdot \Delta\theta)$$ Substitute the values: $$R_{70} = 55 \times [1 + (0.00393 \times 50)]$$ $$R_{70} = 55 \times [1 + 0.1965]$$ $$R_{70} = 55 \times 1.1965 \approx 65.81\ \Omega$$ * Answer: The resistance of the winding at $70^\circ\text{C}$ is $65.81\ \Omega$
Question 4: Battery Internal Resistance
Scenario: A lead-acid battery has a measured open-circuit voltage (EMF) of $12.8\text{V}$. When a starter motor drawing a current of $80\text{ A}$ is connected to the battery, the terminal voltage drops to $11.2\text{V}$ under load. 1. Calculate the internal resistance ($r$) of the battery. 2. Calculate the power wasted as heat inside the battery.
Step-by-Step Solution:
Step 1: Calculate the internal voltage drop ($V_{drop}$) The difference between the open-circuit voltage (EMF, $E$) and the closed-circuit terminal voltage ($V_{term}$) is the drop across the internal resistance: $$V_{drop} = E - V_{term} = 12.8\text{ V} - 11.2\text{ V} = 1.6\text{ V}$$
Step 2: Transpose Ohm's Law to find internal resistance ($r$) Since the load current ($I = 80\text{ A}$) must flow through the battery's internal resistance: $$V_{drop} = I \times r \implies r = \frac{V_{drop}}{I}$$ $$r = \frac{1.6\text{ V}}{80\text{ A}} = 0.02\ \Omega\ (20\text{ m}\Omega)$$ * Answer 1: The internal resistance of the battery is $0.02\ \Omega$
Step 3: Calculate the internal power dissipation ($P_{int}$) The power dissipated as heat inside the battery is: $$P_{int} = I^2 \times r = (80)^2 \times 0.02 = 6400 \times 0.02 = 128\text{ W}$$ * Answer 2: $128\text{ Watts}$ of power is wasted as heat inside the battery.
Question 5: Magnetic Force on a Conductor
Scenario: A conductor carrying a current of $25\text{ A}$ is placed in a uniform magnetic field between pole faces that are $15\text{ cm}$ wide. The magnetic flux density is $0.65\text{ Tesla}$. The conductor is arranged at right angles to the magnetic field. 1. Calculate the mechanical force acting on the conductor. 2. If the conductor is rotated so it lies at an angle of $45^\circ$ to the flux lines, calculate the new force.
Step-by-Step Solution:
Step 1: Identify and convert variables * Current ($I$) = $25\text{ A}$ * Flux density ($B$) = $0.65\text{ T}$ * Active length ($l$) = $15\text{ cm} = 0.15\text{ m}$ * Angle ($\theta$) = $90^\circ$ (perpendicular)
Step 2: Calculate the perpendicular force $$F = B \times I \times l \times \sin(90^\circ)$$ $$F = 0.65 \times 25 \times 0.15 \times 1$$ $$F = 2.4375\text{ Newtons}$$ * Answer 1: The mechanical force is $2.44\text{ N}$
Step 3: Calculate the force at $45^\circ$ $$F_{new} = B \times I \times l \times \sin(45^\circ)$$ Recall that $\sin(45^\circ) \approx 0.7071$: $$F_{new} = 0.65 \times 25 \times 0.15 \times 0.7071 \approx 2.4375 \times 0.7071 \approx 1.724\text{ Newtons}$$ * Answer 2: The force drops to $1.72\text{ N}$
Question 6: Basic Volt Drop Calculation
Scenario: A $230\text{V}$ single-phase sub-circuit feeds a domestic electric oven rated at $4.6\text{ kW}$. The run length of the twin-and-earth cable is $35\text{ meters}$. The cable tables state that the voltage drop factor for the selected cable size is $4.4\text{ mV/A/m}$. 1. Calculate the load current ($I$) drawn by the oven. 2. Calculate the actual voltage drop ($V_d$) in volts. 3. Check if this installation complies with the maximum allowable voltage drop of $4\%$ of the supply voltage for power circuits.
Step-by-Step Solution:
Step 1: Calculate the load current ($I$) Using the single-phase power formula: $$P = V \times I \implies I = \frac{P}{V}$$ $$P = 4.6\text{ kW} = 4600\text{ W}$$ $$I = \frac{4600\text{ W}}{230\text{ V}} = 20\text{ A}$$ * Answer 1: The load current is $20\text{ A}$
Step 2: Calculate the voltage drop ($V_d$) Using the standard millivolt-drop formula: $$V_d = \frac{mV/A/m \times I \times L}{1000}$$ Substitute the parameters: $$V_d = \frac{4.4 \times 20 \times 35}{1000} = \frac{3080}{1000} = 3.08\text{ Volts}$$ * Answer 2: The actual voltage drop is $3.08\text{ V}$
Step 3: Verify compliance * Maximum allowable voltage drop = $4\%$ of $230\text{V}$ $$V_{d\_max} = 230 \times 0.04 = 9.2\text{ V}$$ * Our calculated drop is $3.08\text{ V}$, which is well below $9.2\text{ V}$. * Answer 3: The circuit complies. The drop is only $1.34\%$ ($3.08 / 230 \times 100$), which is less than the $4\%$ limit.
Question 7: Conduit Sizing (Factor Method)
Scenario: A circuit installation requires drawing six $1.5\text{ mm}^2$ stranded copper conductors and four $2.5\text{ mm}^2$ stranded copper conductors through a $6\text{ meter}$ run of conduit that contains two $90^\circ$ bends. 1. Calculate the total cable factor. 2. Determine if a standard $20\text{mm}$ conduit is suitable for this installation. 3. If the installer decides to add two more $2.5\text{ mm}^2$ conductors to this run in the future, determine if the $20\text{mm}$ conduit would still comply, or if they would need to upgrade to a $25\text{mm}$ conduit.
Step-by-Step Solution:
Step 1: Calculate the total cable factor Assign factors to each cable type based on the stranded cable table: * $1.5\text{ mm}^2$ cable factor = $22$ * $2.5\text{ mm}^2$ cable factor = $30$
Calculate the sum of all cable factors: $$\text{Total Cable Factor} = (6 \times 22) + (4 \times 30)$$ $$\text{Total Cable Factor} = 132 + 120 = 252$$ * Answer 1: The total cable factor is $252$
Step 2: Determine suitability of the $20\text{mm}$ conduit For a run with bends, the capacity factor for a $20\text{mm}$ conduit is: * $20\text{mm}$ conduit capacity factor = $286$
Compare the total cable factor against the conduit capacity factor: $$\text{Total Cable Factor } (252) \le 20\text{mm Conduit Capacity } (286)$$ Since $252 \le 286$, the cables will easily fit without exceeding the space factor. * Answer 2: Yes, a $20\text{mm}$ conduit is suitable.
Step 3: Analyze the addition of two more $2.5\text{ mm}^2$ conductors Adding two $2.5\text{ mm}^2$ cables increases the total factor: $$\text{New Cable Factor} = 252 + (2 \times 30) = 252 + 60 = 312$$ Compare this new total with the conduit capacity factors: * $20\text{mm}$ conduit capacity = $286$ (Since $312 > 286$, a $20\text{mm}$ conduit is no longer compliant). * $25\text{mm}$ conduit capacity = $543$ (Since $312 \le 543$, a $25\text{mm}$ conduit is compliant).
- Answer 3: The $20\text{mm}$ conduit would not comply with the additional cables. An upgrade to a $25\text{mm}$ conduit would be required.
Phase 4 Electrical Mathematics: Revision Guide & Question Bank
This document serves as your guide to mastering the mathematical applications in Phase 4 of the Irish Electrical Apprenticeship.
Key Revision Concepts
1. DC Network Analysis: Kirchhoff's Laws
For complex DC networks with multiple loops or sources where Ohm's Law alone cannot solve the circuit, we apply Kirchhoff's Laws:
| Law | Formula | Description |
|---|---|---|
| Kirchhoff’s Current Law (KCL) | $\sum I_{in} = \sum I_{out}$ | The sum of currents entering a junction equals the sum of currents leaving that junction |
| Kirchhoff’s Voltage Law (KVL) | $\sum V_{sources} = \sum (I \times R)$ | Around any closed loop in a circuit, the sum of voltage sources equals the sum of $I \times R$ volt drops |
2. Single-Phase AC Waveforms
AC electricity alternates direction, producing a sinusoidal waveform.
| Parameter | Formula | Description / Value |
|---|---|---|
| Frequency / Period | $f = \frac{1}{T} \iff T = \frac{1}{f}$ | $f$ = Frequency ($\text{Hz}$), $T$ = Period (Seconds, $\text{s}$) |
| RMS Value | $V_{RMS} = V_{peak} \times 0.707 = \frac{V_{peak}}{\sqrt{2}}$ | Effective heating value of AC voltage/current |
| Average Value | $V_{avg} = V_{peak} \times 0.637 = \frac{2 \cdot V_{peak}}{\pi}$ | Average of a half-cycle of sine wave |
| Form Factor | $\text{Form Factor} = \frac{\text{RMS}}{\text{Average}} \approx 1.11$ | Constant ratio for a pure sine wave |
| Instantaneous Value | $v(t) = V_{peak} \sin(2\pi f t) = V_{peak} \sin(\omega t)$ | Voltage at any specific instant in time $t$ ($\omega$ = angular frequency) |
3. AC Reactance ($X_L, X_C$)
In AC circuits, inductors and capacitors oppose current. Unlike pure resistance, this opposition (called reactance) depends on the supply frequency.
| Reactance Type | Formula | Properties |
|---|---|---|
| Inductive Reactance | $X_L = 2\pi f L \quad (\Omega)$ | Opposition from inductors (increases with frequency, current lags by $90^\circ$) |
| Capacitive Reactance | $X_C = \frac{1}{2\pi f C} \quad (\Omega)$ | Opposition from capacitors (decreases with frequency, current leads by $90^\circ$) |
4. Series R-L-C Impedance & Power Factor
When resistance ($R$), inductance ($L$), and capacitance ($C$) are connected in series:
| Parameter | Formula | Description & Units |
|---|---|---|
| Net Reactance | $X = X_L - X_C$ | Total reactive opposition ($\Omega$) |
| Impedance | $Z = \sqrt{R^2 + (X_L - X_C)^2}$ | Vector combination of resistance and net reactance ($\Omega$) |
| Power Factor | $\cos\phi = \frac{R}{Z}$ | Efficiency of power usage (lagging or leading) |
| Phase Angle | $\phi = \arccos\left(\frac{R}{Z}\right)$ | Phase displacement angle between voltage and current |
5. The AC Power Triangle
AC power is divided into three components, represented by a right-angled triangle:
| Power Type | Formula | Units & Meaning |
|---|---|---|
| Active/True Power ($P$) | $P = V \times I \times \cos\phi$ | Watts ($\text{W}$ / $\text{kW}$) - actual work-producing power |
| Apparent Power ($S$) | $S = V \times I$ | Volt-Amperes ($\text{VA}$ / $\text{kVA}$) - total supplied power |
| Reactive Power ($Q$) | $Q = V \times I \times \sin\phi$ | Volt-Amperes Reactive ($\text{VAr}$ / $\text{kVAr}$) - stored/field power |
| Power Triangle Relation | $S = \sqrt{P^2 + Q^2}$ | Pythagorean vector relationship |
6. Single-Phase Transformers
Transformers change voltage and current levels using magnetic coupling.
| Parameter | Formula | Description / Details |
|---|---|---|
| Turns Ratio | $\frac{N_p}{N_s} = \frac{V_p}{V_s} = \frac{I_s}{I_p}$ | $p$ = primary, $s$ = secondary, $N$ = turns, $V$ = voltage, $I$ = current |
| Iron Losses ($P_{fe}$) | Constant core losses | Hysteresis and eddy current losses in steel core |
| Copper Losses ($P_{cu}$) | Variable winding losses ($I^2R$) | Varies with the square of the load current |
| Efficiency ($\eta\%$) | $\eta\% = \frac{P_{out}}{P_{out} + P_{fe} + P_{cu}} \times 100$ | Ratio of output power to total input power |
7. Three-Phase Transformer Winding Ratios
In three-phase systems, transformer windings can be connected in Star (Y) or Delta ($\Delta$).
| Connection Type | Voltage Relation | Current Relation |
|---|---|---|
| Star (Y) Connection | $V_L = \sqrt{3} \times V_P \implies V_P = \frac{V_L}{\sqrt{3}}$ | $I_L = I_P$ |
| Delta ($\Delta$) Connection | $V_L = V_P$ | $I_L = \sqrt{3} \times I_P \implies I_P = \frac{I_L}{\sqrt{3}}$ |
Delta-Star ($\Delta$-Y) Step-Down Transformer Relations
| Parameter | Formula / Relation | Description |
|---|---|---|
| Phase Turns Ratio ($a$) | $a = \frac{N_p}{N_s} = \frac{V_{p\_phase}}{V_{s\_phase}}$ | Winding turns and phase voltage ratio |
| Line Turns Relation | $\frac{N_p}{N_s} = \sqrt{3} \frac{V_{p\_line}}{V_{s\_line}}$ | For Delta-primary and Star-secondary |
| Total Apparent Power ($S$) | $S_{total} = \sqrt{3} \times V_L \times I_L$ | $S_{total} = \sqrt{3} \cdot V_{p\_line} \cdot I_{p\_line} = \sqrt{3} \cdot V_{s\_line} \cdot I_{s\_line}$ |
| Current / Voltage Ratio | $\frac{I_{s\_line}}{I_{p\_line}} = \frac{V_{p\_line}}{V_{s\_line}}$ | Ideal transformer power conservation ($S_{in} = S_{out}$) |
Worked Example Questions
Question 1: DC Circuits (Kirchhoff's Laws)
Scenario: A dual-battery charging circuit can be modeled by two loops. A $12\text{V}$ battery with an internal resistance of $1\ \Omega$ and a $10\text{V}$ battery with an internal resistance of $2\ \Omega$ are connected in parallel to charge a common load resistor of $8\ \Omega$. Calculate the currents $I_1$ (flowing from the $12\text{V}$ battery), $I_2$ (flowing from the $10\text{V}$ battery), and $I_3$ (flowing through the $8\ \Omega$ load resistor).
Step-by-Step Solution:
Step 1: Set up Kirchhoff's Current Law (KCL) at the node Let the current from the $12\text{V}$ battery be $I_1$ and from the $10\text{V}$ battery be $I_2$. They join to flow through the $8\ \Omega$ resistor ($I_3$): $$I_3 = I_1 + I_2 \quad \text{--- (Equation 1)}$$
Step 2: Set up Kirchhoff's Voltage Law (KVL) equations for both loops * Loop 1 (containing the 12V battery and the 8 $\Omega$ resistor): $$12 = I_1(1\ \Omega) + I_3(8\ \Omega)$$ Substitute $I_3 = I_1 + I_2$: $$12 = I_1 + 8(I_1 + I_2)$$ $$12 = 9I_1 + 8I_2 \quad \text{--- (Equation 2)}$$
- Loop 2 (containing the 10V battery and the 8 $\Omega$ resistor): $$10 = I_2(2\ \Omega) + I_3(8\ \Omega)$$ Substitute $I_3 = I_1 + I_2$: $$10 = 2I_2 + 8(I_1 + I_2)$$ $$10 = 8I_1 + 10I_2 \quad \text{--- (Equation 3)}$$
Step 3: Solve the simultaneous equations Multiply Equation 2 by 10, and Equation 3 by 8 to eliminate $I_2$: * $120 = 90I_1 + 80I_2$ * $80 = 64I_1 + 80I_2$
Subtract the second equation from the first: $$120 - 80 = (90I_1 - 64I_1) + (80I_2 - 80I_2)$$ $$40 = 26I_1 \implies I_1 = \frac{40}{26} \approx 1.538\text{ A}$$
Step 4: Solve for $I_2$ and $I_3$ Substitute $I_1 = 1.538\text{ A}$ back into Equation 2: $$12 = 9(1.538) + 8I_2$$ $$12 = 13.842 + 8I_2 \implies 8I_2 = 12 - 13.842 = -1.842$$ $$I_2 = \frac{-1.842}{8} \approx -0.230\text{ A}$$ (The negative sign means current $I_2$ is actually flowing backwards into the 10V battery, charging it).
Now find $I_3$: $$I_3 = I_1 + I_2 = 1.538\text{ A} + (-0.230\text{ A}) = 1.308\text{ A}$$ * Answers: * $I_1 = 1.54\text{ A}$ (discharging from 12V battery) * $I_2 = -0.23\text{ A}$ (charging into 10V battery) * $I_3 = 1.31\text{ A}$ (flowing down the load resistor)
Question 2: AC Waveform Calculations
Scenario: An alternating voltage is represented by the equation: $v = 325 \sin(314.16 \cdot t)$ volts. Calculate: 1. The peak voltage ($V_{peak}$). 2. The supply frequency ($f$). 3. The RMS voltage ($V_{RMS}$). 4. The instantaneous voltage ($v$) when time $t = 5\text{ milliseconds}$ ($0.005\text{ s}$).
Step-by-Step Solution:
Step 1: Identify the standard equation components The standard equation is $v = V_{peak} \sin(2\pi f t)$. Comparing $v = 325 \sin(314.16 \cdot t)$: * Answer 1: $V_{peak} = 325\text{ V}$
Step 2: Calculate the frequency ($f$) The angular frequency $\omega = 2\pi f = 314.16\text{ rad/s}$: $$2\pi f = 314.16 \implies f = \frac{314.16}{2\pi} = \frac{314.16}{6.2832} \approx 50\text{ Hz}$$ * Answer 2: The frequency is $50\text{ Hz}$ (standard Irish grid frequency)
Step 3: Calculate the RMS voltage ($V_{RMS}$) $$V_{RMS} = V_{peak} \times 0.707 = 325 \times 0.707 = 229.8\text{ V}\ (\approx 230\text{ V})$$ * Answer 3: The RMS voltage is $230\text{ V}$
Step 4: Calculate the instantaneous voltage at $t = 0.005\text{ s}$ $$v = 325 \sin(314.16 \times 0.005)$$ $$v = 325 \sin(1.5708\text{ radians})$$
IMPORTANT: Exam Trap: Your scientific calculator must be set to Radians mode when calculating instantaneous voltage using this equation. If you calculate in Degree mode, your answer will be completely incorrect.
- $1.5708\text{ rad} = \frac{\pi}{2}\text{ rad}$ (which is equal to $90^\circ$).
- $\sin(1.5708\text{ rad}) = 1$. $$v = 325 \times 1 = 325\text{ V}$$
- Answer 4: The voltage at $5\text{ ms}$ is $325\text{ V}$ (which is its positive peak value).
Question 3: Series R-L-C Circuit
Scenario: A coil having a resistance of $40\ \Omega$ and an inductance of $0.18\text{ H}$ is connected in series with a $47\ \mu\text{F}$ capacitor across a $230\text{V}, 50\text{Hz}$ AC supply. Calculate: 1. The inductive reactance ($X_L$) and capacitive reactance ($X_C$). 2. The total circuit impedance ($Z$). 3. The current flowing in the circuit ($I$). 4. The circuit power factor ($\cos\phi$) and state whether it is lagging or leading.
Step-by-Step Solution:
Step 1: Calculate Reactances * Inductive Reactance ($X_L$): $$X_L = 2\pi f L = 2 \times \pi \times 50 \times 0.18 = 314.16 \times 0.18 \approx 56.55\ \Omega$$ * Capacitive Reactance ($X_C$): Convert $47\ \mu\text{F}$ to Farads: $47 \times 10^{-6}\text{ F}$. $$X_C = \frac{1}{2\pi f C} = \frac{1}{2 \times \pi \times 50 \times (47 \times 10^{-6})} = \frac{1}{314.16 \times 0.000047} = \frac{1}{0.014765} \approx 67.73\ \Omega$$
Step 2: Calculate Impedance ($Z$) First find the net reactance ($X$): $$X = X_L - X_C = 56.55 - 67.73 = -11.18\ \Omega$$ (The negative reactance tells us capacitive reactance dominates, meaning current will lead voltage).
Now apply Pythagoras: $$Z = \sqrt{R^2 + X^2} = \sqrt{40^2 + (-11.18)^2} = \sqrt{1600 + 125} = \sqrt{1725} \approx 41.53\ \Omega$$ * Answer 2: Circuit impedance is $41.53\ \Omega$
Step 3: Calculate Circuit Current ($I$) Using AC Ohm's Law: $$I = \frac{V}{Z} = \frac{230\text{ V}}{41.53\ \Omega} \approx 5.54\text{ A}$$ * Answer 3: Circuit current is $5.54\text{ A}$
Step 4: Calculate Power Factor ($\cos\phi$) $$\cos\phi = \frac{R}{Z} = \frac{40\ \Omega}{41.53\ \Omega} \approx 0.963$$ Since $X_C > X_L$, the circuit is capacitive, so the current leads the voltage. * Answer 4: Power factor is $0.963$ (leading)
Question 4: AC Power Triangle
Scenario: A single-phase electric motor connected to a $230\text{V}, 50\text{Hz}$ supply draws a current of $12\text{ A}$ at a lagging power factor of $0.72$. Calculate: 1. The Apparent Power ($S$) in VA. 2. The Active Power ($P$) in Watts. 3. The Reactive Power ($Q$) in VAr.
Step-by-Step Solution:
Step 1: Calculate Apparent Power ($S$) $$S = V \times I = 230\text{ V} \times 12\text{ A} = 2760\text{ VA}\ (2.76\text{ kVA})$$ * Answer 1: Apparent power is $2760\text{ VA}$
Step 2: Calculate Active Power ($P$) $$P = V \times I \times \cos\phi = S \times \text{PF}$$ $$P = 2760 \times 0.72 = 1987.2\text{ Watts}\ (1.99\text{ kW})$$ * Answer 2: Active power is $1987.2\text{ W}$
Step 3: Calculate Reactive Power ($Q$) First find the phase angle $\phi$: $$\phi = \arccos(0.72) \approx 43.95^\circ$$ Now find $\sin\phi$: $$\sin(43.95^\circ) \approx 0.694$$ Calculate Reactive Power ($Q$): $$Q = V \times I \times \sin\phi = 2760 \times 0.694 \approx 1915.4\text{ VAr}\ (1.92\text{ kVAr})$$ Alternative method using Pythagoras: $$Q = \sqrt{S^2 - P^2} = \sqrt{2760^2 - 1987.2^2} = \sqrt{7,617,600 - 3,948,964} = \sqrt{3,668,636} \approx 1915.4\text{ VAr}$$ * Answer 3: Reactive power is $1915.4\text{ VAr}$
Question 5: Star & Delta Connections
Scenario: Three identical resistors, each having a resistance of $24\ \Omega$, are connected in Star to a three-phase, $400\text{V}, 50\text{Hz}$ supply. 1. Calculate the phase voltage ($V_p$). 2. Calculate the phase current ($I_p$) and line current ($I_L$). 3. If the three resistors are reconnected in Delta across the same supply, calculate the new line current ($I_{L\_delta}$).
Step-by-Step Solution:
Step 1: Calculate phase voltage ($V_p$) in Star In a Star connection, the line voltage ($V_L = 400\text{V}$) is distributed across phase windings: $$V_p = \frac{V_L}{\sqrt{3}} = \frac{400}{\sqrt{3}} \approx 230.94\text{ V}$$ * Answer 1: Phase voltage is $230.9\text{ V}$
Step 2: Calculate phase and line current in Star Using Ohm's Law for the phase resistor ($R_p = 24\ \Omega$): $$I_p = \frac{V_p}{R_p} = \frac{230.94\text{ V}}{24\ \Omega} \approx 9.62\text{ A}$$ In Star, the line current equals the phase current: $$I_L = I_p = 9.62\text{ A}$$ * Answer 2: Phase current is $9.62\text{ A}$ and Line current is $9.62\text{ A}$
Step 3: Calculate the line current ($I_{L\_delta}$) in Delta In a Delta connection, the phase voltage equals the line voltage ($V_p = V_L = 400\text{V}$): $$I_{p\_delta} = \frac{V_p}{R_p} = \frac{400\text{ V}}{24\ \Omega} \approx 16.67\text{ A}$$ In Delta, the line current is $\sqrt{3}$ times the phase current: $$I_{L\_delta} = \sqrt{3} \times I_{p\_delta} = \sqrt{3} \times 16.67 \approx 28.87\text{ A}$$ * Answer 3: The new line current in Delta is $28.87\text{ A}$ (exactly 3 times the Star line current).
Question 6: Single-Phase Transformer Calculations
Scenario: A single-phase transformer is rated at $5\text{ kVA}, 230\text{V} / 115\text{V}, 50\text{Hz}$. 1. Calculate the full-load primary and secondary currents ($I_p$ and $I_s$). 2. If the secondary has $160\text{ turns}$, calculate the number of primary turns ($N_p$). 3. If the transformer has constant iron losses of $60\text{ W}$ and primary/secondary copper losses of $120\text{ W}$ at full load, calculate the efficiency ($\eta\%$) of the transformer when delivering full load at unity power factor ($\cos\phi = 1.0$).
Step-by-Step Solution:
Step 1: Calculate full-load currents The rating $S = 5\text{ kVA} = 5000\text{ VA}$. * Primary Full-Load Current ($I_p$): $$I_p = \frac{S}{V_p} = \frac{5000\text{ VA}}{230\text{ V}} \approx 21.74\text{ A}$$ * Secondary Full-Load Current ($I_s$): $$I_s = \frac{S}{V_s} = \frac{5000\text{ VA}}{115\text{ V}} \approx 43.48\text{ A}$$ * Answer 1: Primary current is $21.74\text{ A}$; Secondary current is $43.48\text{ A}$
Step 2: Calculate primary turns ($N_p$) Using the turns ratio: $$\frac{N_p}{N_s} = \frac{V_p}{V_s} \implies N_p = N_s \times \frac{V_p}{V_s}$$ $$N_p = 160 \times \frac{230}{115} = 160 \times 2 = 320\text{ turns}$$ * Answer 2: The primary winding has $320\text{ turns}$
Step 3: Calculate transformer efficiency ($\eta\%$) * Output Power ($P_{out}$) at unity power factor: $$P_{out} = S \times \cos\phi = 5000\text{ VA} \times 1.0 = 5000\text{ W}$$ * Total Losses ($P_{losses}$) = Iron losses ($P_{fe}$) + Copper losses ($P_{cu}$): $$P_{losses} = 60\text{ W} + 120\text{ W} = 180\text{ W}$$ * Efficiency ($\eta\%$): $$\eta\% = \frac{P_{out}}{P_{out} + P_{losses}} \times 100$$ $$\eta\% = \frac{5000}{5000 + 180} \times 100 = \frac{5000}{5180} \times 100 \approx 96.53\%$$ * Answer 3: The full-load efficiency is $96.53\%$
Question 7: Three-Phase Transformer Calculations
Scenario: A $150\text{ kVA}$, $10,000\text{V} / 400\text{V}$, $50\text{Hz}$ three-phase delta-star ($\Delta$-Y) step-down transformer is operating at full load. 1. Calculate the full-load primary and secondary line currents ($I_{p\_line}$ and $I_{s\_line}$). 2. Calculate the phase voltage of the primary and secondary windings ($V_{p\_phase}$ and $V_{s\_phase}$). 3. Calculate the transformer's phase turns ratio ($\frac{N_p}{N_s}$). 4. Calculate the primary phase current ($I_{p\_phase}$) at full load.
Step-by-Step Solution:
Step 1: Calculate full-load line currents Apparent power $S = 150\text{ kVA} = 150,000\text{ VA}$ * Primary Line Current ($I_{p\_line}$): $$S = \sqrt{3} \times V_{p\_line} \times I_{p\_line} \implies I_{p\_line} = \frac{S}{\sqrt{3} \times V_{p\_line}}$$ $$I_{p\_line} = \frac{150,000}{\sqrt{3} \times 10,000} = \frac{150,000}{17,320.5} \approx 8.66\text{ A}$$ * Answer 1a: Full-load primary line current is $8.66\text{ A}$
- Secondary Line Current ($I_{s\_line}$): $$I_{s\_line} = \frac{S}{\sqrt{3} \times V_{s\_line}}$$ $$I_{s\_line} = \frac{150,000}{\sqrt{3} \times 400} = \frac{150,000}{692.82} \approx 216.51\text{ A}$$
- Answer 1b: Full-load secondary line current is $216.51\text{ A}$
Step 2: Calculate the phase voltage of the windings * Primary (Delta-connected): In Delta, phase voltage equals line voltage: $$V_{p\_phase} = V_{p\_line} = 10,000\text{ V}$$ * Answer 2a: Primary phase voltage is $10,000\text{ V}$
- Secondary (Star-connected): In Star, phase voltage is line voltage divided by $\sqrt{3}$: $$V_{s\_phase} = \frac{V_{s\_line}}{\sqrt{3}} = \frac{400}{\sqrt{3}} \approx 230.94\text{ V}$$
- Answer 2b: Secondary phase voltage is $230.94\text{ V}$
Step 3: Calculate the phase turns ratio ($\frac{N_p}{N_s}$) Using the ratio of phase voltages: $$\frac{N_p}{N_s} = \frac{V_{p\_phase}}{V_{s\_phase}} = \frac{10,000\text{ V}}{230.94\text{ V}} \approx 43.3$$ * Answer 3: The phase turns ratio is $43.3:1$ (or $43.3$).
Step 4: Calculate the primary phase current ($I_{p\_phase}$) * Method A (Using Delta connection current rules): $$I_{p\_phase} = \frac{I_{p\_line}}{\sqrt{3}} = \frac{8.66\text{ A}}{\sqrt{3}} \approx 5.0\text{ A}$$
- Method B (Using phase turns ratio): Since secondary is Star, secondary phase current equals line current ($I_{s\_phase} = I_{s\_line} = 216.51\text{ A}$): $$I_{p\_phase} = I_{s\_phase} \times \frac{N_s}{N_p} = 216.51 \times \frac{1}{43.3} \approx 5.0\text{ A}$$ Both methods yield the same result.
- Answer 4: The primary phase current is $5.0\text{ A}$.
Phase 6 Electrical Mathematics: Revision Guide & Question Bank
This document serves as your guide to mastering the advanced mathematical applications in Phase 6 of the Irish Electrical Apprenticeship.
Key Revision Concepts
1. Parallel AC Circuits
In a parallel AC circuit, the voltage ($V$) is the reference vector and is common to all branches. The total current ($I_T$) is the vector sum of the individual branch currents:
| Parameter | Formula | Description & Phase Angle |
|---|---|---|
| Active Current ($I_R$) | $I_R = \frac{V}{R}$ | Current through resistors (in phase with voltage, $0^\circ$) |
| Inductive Current ($I_L$) | $I_L = \frac{V}{X_L}$ | Current through inductors (lags voltage by $90^\circ$, $-90^\circ$) |
| Capacitive Current ($I_C$) | $I_C = \frac{V}{X_C}$ | Current through capacitors (leads voltage by $90^\circ$, $+90^\circ$) |
| Total Current ($I_T$) | $I_T = \sqrt{I_R^2 + (I_C - I_L)^2}$ | Vector combination of horizontal and vertical currents |
| Circuit Power Factor | $\cos\phi = \frac{I_R}{T}$ | Ratio of active current to total current |
2. Power Factor Correction
Inductive loads (like motors and transformers) cause a lagging power factor, drawing more current than necessary from the grid. To correct this, capacitor banks are connected in parallel.
| Parameter | Formula | Description & Variable Details |
|---|---|---|
| Required Capacitor Power ($Q_c$) | $Q_c = P(\tan\phi_1 - \tan\phi_2)$ | $P$ = active power, $\phi_1$ = original angle, $\phi_2$ = targeted angle ($\text{VAr}$ or $\text{kVAr}$) |
| Capacitor Current ($I_c$, Star/1-$\phi$) | $I_c = \frac{Q_c}{V_{phase}}$ | For single-phase or three-phase Star connections |
| Capacitor Current ($I_c$, Delta) | $I_c = \frac{Q_c}{\sqrt{3} \cdot V_L}$ | For three-phase Delta-connected capacitor banks |
| Required Capacitance ($C$) | $C = \frac{Q_c}{2\pi f V^2}$ | $f$ = frequency, $V$ = voltage directly across capacitor ($\text{F}$) |
3. Advanced Three-Phase Power & Measurement
| Parameter | Formula | Description & Units |
|---|---|---|
| Active Power ($P$) | $P = \sqrt{3} \times V_L \times I_L \times \cos\phi$ | True/active work-producing power ($\text{W}$ or $\text{kW}$) |
| Apparent Power ($S$) | $S = \sqrt{3} \times V_L \times I_L$ | Total vector power supplied ($\text{VA}$ or $\text{kVA}$) |
| Reactive Power ($Q$) | $Q = \sqrt{3} \times V_L \times I_L \times \sin\phi$ | Field/magnetizing stored power ($\text{VAr}$ or $\text{kVAr}$) |
| Two-Wattmeter Total Power ($P_T$) | $P_T = W_1 + W_2$ | Direct sum of wattmeter readings (one may be negative if $\text{PF} < 0.5$) | | Two-Wattmeter Phase Angle ($\phi$) | $\tan\phi = \sqrt{3} \frac{W_1 - W_2}{W_1 + W_2}$ | Enables power factor calculation: find $\phi$, then compute $\cos\phi$ |
4. Induction Motor Formulas
Induction motors run at a rotor shaft speed slightly below the synchronous speed of the rotating stator magnetic field.
| Parameter | Formula | Description & Units |
|---|---|---|
| Synchronous Speed ($N_s$) | $N_s = \frac{120 \times f}{p}$ | Speed of rotating stator magnetic field (RPM) |
| Fractional Slip ($s$) | $s = \frac{N_s - N_r}{N_s}$ | Relative speed difference as a decimal fraction |
| Percentage Slip ($s\%$) | $s\% = \frac{N_s - N_r}{N_s} \times 100$ | Relative speed difference as a percentage |
| Rotor Frequency ($f_r$) | $f_r = s \times f$ | Frequency of induced currents in rotor windings ($\text{Hz}$) |
5. Cable Sizing & Regulations (I.S. 10101:2020)
Apprentices must size cables to satisfy load capacity, voltage drop, and fault loop safety rules.
| Sizing Step / Rule | Formula | Variable Description & Meanings |
|---|---|---|
| 1. Design Current ($I_b$, 1-$\phi$) | $I_b = \frac{P}{V \cos\phi}$ | Design current for single-phase circuits |
| 1. Design Current ($I_b$, 3-$\phi$) | $I_b = \frac{P}{\sqrt{3} \cdot V_L \cos\phi}$ | Design current for balanced three-phase circuits |
| 2. Device Rating ($I_n$) | $I_n \ge I_b$ | Select nominal rating of protective device |
| 3. Adjusted Capacity ($I_t$) | $I_t \ge \frac{I_n}{C_a \cdot C_g \cdot C_i \cdot C_f}$ | Tabulated capacity adjusted for ambient ($C_a$), grouping ($C_g$), insulation ($C_i$), fuse type ($C_f$) |
| 4. Volt Drop ($V_{d}$, 3-$\phi$) | $V_d = \frac{\sqrt{3} \times I_b \times L \times (mV/A/m)}{1000}$ | Check voltage drop limit (lighting $3\%$, power $5\%$) |
| 5. Loop Impedance ($Z_s$) | $Z_s = Z_e + \frac{(R_1 + R_2) \times L \times C_T}{1000}$ | Earth loop impedance ($Z_e$ = external, $R_1+R_2$ = resistance/m, $C_T$ = temp correction) |
6. Neutral Current Calculations in Unbalanced Star Systems
In star systems, neutral current ($\vec{I}_N$) is the vector sum of the phase line currents. It is zero in balanced systems, but flows in unbalanced networks.
| System State | Vector Equation | Rectangular Component Calculations |
|---|---|---|
| Balanced Star | $\vec{I}_N = \vec{I}_{L1} + \vec{I}_{L2} + \vec{I}_{L3} = 0$ | No neutral current flows |
| Unbalanced Star | $\vec{I}_N = \vec{I}_{L1} + \vec{I}_{L2} + \vec{I}_{L3}$ | Vector sum of unbalanced line currents |
Resolving Line Currents (taking $I_{L1}$ as reference $0^\circ$)
- Phase L1 Current: $\vec{I}_{L1} = I_{L1} \angle 0^\circ = I_{L1} + j0$
- Phase L2 Current: $\vec{I}_{L2} = I_{L2} \angle -120^\circ = -0.5 I_{L2} - j0.866 I_{L2}$
- Phase L3 Current: $\vec{I}_{L3} = I_{L3} \angle 120^\circ = -0.5 I_{L3} + j0.866 I_{L3}$
Component & Magnitude Calculations
| Step | Component Formula | Description |
|---|---|---|
| Horizontal Sum ($I_{Nx}$) | $I_{Nx} = I_{L1} - 0.5(I_{L2} + I_{L3})$ | Combined active components along x-axis |
| Vertical Sum ($I_{Ny}$) | $I_{Ny} = 0.866(I_{L3} - I_{L2})$ | Combined reactive components along y-axis |
| Total Neutral Current ($I_N$) | $I_N = \sqrt{I_{Nx}^2 + I_{Ny}^2}$ | Resultant magnitude via Pythagoras |
Worked Example Questions
Question 1: Parallel AC Circuit
Scenario: A parallel AC circuit connected across a $230\text{V}, 50\text{Hz}$ single-phase supply contains: * A branch with a heating element drawing a resistive current of $8\text{ A}$. * A branch with a coil drawing $10\text{ A}$ of current with an inductive phase angle of $60^\circ$ lagging. 1. Express the currents in rectangular (vector) components. 2. Calculate the total circuit current ($I_T$). 3. Calculate the overall circuit power factor ($\cos\phi_T$).
Step-by-Step Solution:
Step 1: Resolve branch currents into active and reactive components * Branch 1 (Resistor, $I_1$): * Current is completely in-phase (resistive): $$I_{1\_active} = 8\text{ A}$$ $$I_{1\_reactive} = 0\text{ A}$$ * Branch 2 (Coil, $I_2$): * Current ($10\text{ A}$) lags by $60^\circ$ ($\phi = -60^\circ$): $$I_{2\_active} = I_2 \cos(60^\circ) = 10 \times 0.5 = 5\text{ A}$$ $$I_{2\_reactive} = -I_2 \sin(60^\circ) = -10 \times 0.866 = -8.66\text{ A}$$
Step 2: Sum the components to find total current components * Total Active Current ($I_{active\_T}$): $$I_{active\_T} = I_{1\_active} + I_{2\_active} = 8\text{ A} + 5\text{ A} = 13\text{ A}$$ * Total Reactive Current ($I_{reactive\_T}$): $$I_{reactive\_T} = I_{1\_reactive} + I_{2\_reactive} = 0\text{ A} - 8.66\text{ A} = -8.66\text{ A}$$
Step 3: Calculate total current magnitude ($I_T$) Using Pythagoras: $$I_T = \sqrt{(I_{active\_T})^2 + (I_{reactive\_T})^2}$$ $$I_T = \sqrt{13^2 + (-8.66)^2} = \sqrt{169 + 75} = \sqrt{244} \approx 15.62\text{ A}$$ * Answer 2: The total current is $15.62\text{ A}$
Step 4: Calculate the total power factor ($\cos\phi_T$) $$\cos\phi_T = \frac{I_{active\_T}}{I_T} = \frac{13\text{ A}}{15.62\text{ A}} \approx 0.832$$ Since the net reactive current is negative (inductive), the power factor is lagging. * Answer 3: Power factor is $0.832$ (lagging)
Question 2: Power Factor Correction
Scenario: A small factory draws a three-phase load of $120\text{ kW}$ at a power factor of $0.62$ lagging from a $400\text{V}, 50\text{Hz}$ supply. Calculate the total capacitor rating in kVAr and the required capacitance ($C$) of each capacitor in microfarads ($\mu\text{F}$) if they are connected in Delta to improve the power factor to $0.95$ lagging.
Step-by-Step Solution:
Step 1: Calculate the phase angles * Original phase angle ($\phi_1$): $$\phi_1 = \arccos(0.62) \approx 51.68^\circ \implies \tan(51.68^\circ) \approx 1.265$$ * Target phase angle ($\phi_2$): $$\phi_2 = \arccos(0.95) \approx 18.19^\circ \implies \tan(18.19^\circ) \approx 0.329$$
Step 2: Calculate the required kVAr capacity ($Q_c$) $$Q_c = P(\tan\phi_1 - \tan\phi_2)$$ $$Q_c = 120\text{ kW} \times (1.265 - 0.329)$$ $$Q_c = 120 \times 0.936 = 112.32\text{ kVAr}$$ * Answer 1: The total capacitor bank rating must be $112.3\text{ kVAr}$
Step 3: Calculate the capacitance per phase in Delta For a Delta connection, three capacitor units share the load. * Capacitor rating per phase ($Q_{c\_phase}$): $$Q_{c\_phase} = \frac{Q_c}{3} = \frac{112.32\text{ kVAr}}{3} = 37.44\text{ kVAr} = 37,440\text{ VAr}$$ * In Delta, the voltage across each capacitor is the full line voltage ($V_L = 400\text{V}$). Apply the capacitance formula: $$C = \frac{Q_{c\_phase}}{2\pi f V^2}$$ $$C = \frac{37,440}{2 \times \pi \times 50 \times 400^2}$$ $$C = \frac{37,440}{314.16 \times 160,000} = \frac{37,440}{50,265,600} \approx 0.0007448\text{ Farads}$$ * Convert to microfarads ($\mu\text{F}$): $$C = 0.0007448 \times 10^6 \approx 744.8\ \mu\text{F}$$ * Answer 2: The required capacitance per phase is $744.8\ \mu\text{F}$
Question 3: Two-Wattmeter Power Measurement
Scenario: A three-phase induction motor running on a balanced $400\text{V}$ supply has its input power measured using two wattmeters. The wattmeters register $14.2\text{ kW}$ and $6.8\text{ kW}$ respectively. Calculate: 1. The total active power ($P_T$) drawn by the motor. 2. The power factor ($\cos\phi$) of the motor.
Step-by-Step Solution:
Step 1: Calculate total active power ($P_T$) $$P_T = W_1 + W_2 = 14.2\text{ kW} + 6.8\text{ kW} = 21.0\text{ kW}$$ * Answer 1: Total active power is $21.0\text{ kW}$
Step 2: Calculate the phase angle ($\phi$) Using the two-wattmeter formula: $$\tan\phi = \sqrt{3} \frac{W_1 - W_2}{W_1 + W_2}$$ $$\tan\phi = \sqrt{3} \times \frac{14.2 - 6.8}{14.2 + 6.8} = \sqrt{3} \times \frac{7.4}{21.0}$$ $$\tan\phi = 1.732 \times 0.3524 \approx 0.6104$$ Now find $\phi$: $$\phi = \arctan(0.6104) \approx 31.4^\circ$$
Step 3: Calculate the power factor ($\cos\phi$) $$\cos\phi = \cos(31.4^\circ) \approx 0.854$$ * Answer 2: The power factor is $0.854$ (lagging)
Question 4: Induction Motors
Scenario: A 3-phase, 6-pole induction motor is connected to a $400\text{V}, 50\text{Hz}$ supply. The rotor speed at full load is measured to be $960\text{ RPM}$. Calculate: 1. The synchronous speed ($N_s$). 2. The fractional slip ($s$) and percentage slip ($s\%$). 3. The frequency of the rotor current ($f_r$).
Step-by-Step Solution:
Step 1: Calculate Synchronous Speed ($N_s$) $$N_s = \frac{120 \times f}{p} = \frac{120 \times 50}{6} = \frac{6000}{6} = 1000\text{ RPM}$$ * Answer 1: Synchronous speed is $1000\text{ RPM}$
Step 2: Calculate Slip ($s$) * Fractional slip ($s$): $$s = \frac{N_s - N_r}{N_s} = \frac{1000 - 960}{1000} = \frac{40}{1000} = 0.04$$ * Percentage slip ($s\%$): $$s\% = 0.04 \times 100 = 4\%$$ * Answer 2: Slip is $0.04$ (or $4\%$)
Step 3: Calculate Rotor Frequency ($f_r$) $$f_r = s \times f = 0.04 \times 50\text{ Hz} = 2.0\text{ Hz}$$ * Answer 3: Rotor frequency is $2.0\text{ Hz}$ (under normal running conditions, rotor frequency is always very low).
Question 5: Cable Sizing Calculations (I.S. 10101)
Scenario: A three-phase $400\text{V}$ balanced commercial heating unit rated at $48\text{ kW}$ is installed. The power factor is unity ($\cos\phi = 1.0$). The cable will be run in a trunking (Ref Method B) along with 3 other loaded circuits. The ambient temperature is $30^\circ\text{C}$ (correction factor $C_a = 1.0$), and grouping correction factor $C_g = 0.65$. The protection will be a standard MCB. 1. Calculate the design current ($I_b$). 2. Select an appropriate MCB rating ($I_n$). 3. Calculate the minimum required tabulated current capacity ($I_t$) for the cable.
Step-by-Step Solution:
Step 1: Calculate Design Current ($I_b$) Using the three-phase active power formula: $$P = \sqrt{3} \times V_L \times I_L \times \cos\phi \implies I_b = \frac{P}{\sqrt{3} \cdot V_L \cdot \cos\phi}$$ $$I_b = \frac{48,000}{\sqrt{3} \times 400 \times 1.0} = \frac{48,000}{692.82} \approx 69.28\text{ A}$$ * Answer 1: Design current $I_b = 69.3\text{ A}$
Step 2: Select protective device rating ($I_n$) The protective device must satisfy the rule $I_n \ge I_b$. Standard MCB ratings are: 50A, 63A, 80A, 100A... Since $I_b = 69.3\text{ A}$, the next standard MCB size up is $80\text{ A}$. * Answer 2: Selected MCB size is $80\text{ A}$
Step 3: Calculate minimum current rating ($I_t$) Apply correction factors to the device rating ($I_n = 80\text{ A}$): $$I_t \ge \frac{I_n}{C_a \cdot C_g}$$ $$I_t \ge \frac{80}{1.0 \times 0.65} = \frac{80}{0.65} \approx 123.08\text{ A}$$ * Answer 3: The selected cable must be rated to carry at least $123.1\text{ A}$ under tabulated conditions in the standard tables.
Question 6: Voltage Drop & Loop Impedance Verification
Scenario: Following Question 5, a $35\text{ mm}^2$ multi-core copper XLPE cable is selected, which has a tabulated rating of $125\text{ A}$ (Ref Method B). The route length is $85\text{ meters}$. From tables: * Volt drop factor $(mV/A/m) = 1.15$ * Conductor resistance per meter at $20^\circ\text{C}$: $(R_1 + R_2) = 1.05\text{ m}\Omega/\text{m}$. * External loop impedance $Z_e = 0.22\ \Omega$. 1. Calculate the voltage drop ($V_d$) on full load. Verify if it is within the regulatory limit of $5\%$ ($20\text{V}$ drop for $400\text{V}$ system). 2. Calculate the Earth fault loop impedance ($Z_s$) at operating temperature (multiplying loop resistance by temperature factor $C_T = 1.2$).
Step-by-Step Solution:
Step 1: Calculate Voltage Drop ($V_d$) Using the three-phase volt drop formula: $$V_d = \frac{\sqrt{3} \times I_b \times L \times (mV/A/m)}{1000}$$ Substitute the parameters: $$V_d = \frac{\sqrt{3} \times 69.28 \times 85 \times 1.15}{1000}$$ $$V_d = \frac{1.732 \times 69.28 \times 85 \times 1.15}{1000} = \frac{117,301}{1000} \approx 11.73\text{ Volts}$$ * Maximum allowed drop = $5\%$ of $400\text{V} = 20\text{V}$. * Our calculated drop is $11.73\text{V}$, which complies with the regulations. * Answer 1: Voltage drop is $11.73\text{ V}$ (Complies).
Step 2: Calculate Earth Loop Impedance ($Z_s$) * Conductor loop resistance ($R_{loop}$): $$R_{loop} = \frac{(R_1 + R_2) \times L \times C_T}{1000}$$ $$R_{loop} = \frac{1.05\text{ m}\Omega/\text{m} \times 85\text{ m} \times 1.2}{1000} = \frac{107.1}{1000} = 0.107\ \Omega$$ * Total Loop Impedance ($Z_s$): $$Z_s = Z_e + R_{loop} = 0.22\ \Omega + 0.107\ \Omega = 0.327\ \Omega$$ * Answer 2: The total Earth loop impedance $Z_s$ is $0.327\ \Omega$.
Question 7: Unbalanced Three-Phase Star System
Scenario: A $400\text{V}$ (line voltage), three-phase, 4-wire star-connected system supplies three separate single-phase resistive heating loads: * Load 1 connected to L1: $46\text{ kW}$ * Load 2 connected to L2: $34.5\text{ kW}$ * Load 3 connected to L3: $23\text{ kW}$ 1. Calculate the line currents ($I_{L1}, I_{L2}, I_{L3}$) drawn by each load. 2. Calculate the magnitude and phase angle of the current flowing in the neutral conductor ($I_N$).
Step-by-Step Solution:
Step 1: Calculate the line currents For a star-connected system, the phase voltage is: $$V_{phase} = \frac{V_L}{\sqrt{3}} = \frac{400\text{ V}}{\sqrt{3}} \approx 230\text{ V}$$
Since the loads are purely resistive ($\cos\phi = 1$), calculate the currents: * Current in Line 1 ($I_{L1}$): $$I_{L1} = \frac{P_1}{V_{phase}} = \frac{46,000\text{ W}}{230\text{ V}} = 200\text{ A}$$ * Current in Line 2 ($I_{L2}$): $$I_{L2} = \frac{P_2}{V_{phase}} = \frac{34,500\text{ W}}{230\text{ V}} = 150\text{ A}$$ * Current in Line 3 ($I_{L3}$): $$I_{L3} = \frac{P_3}{V_{phase}} = \frac{23,000\text{ W}}{230\text{ V}} = 100\text{ A}$$ * Answer 1: Line currents are $I_{L1} = 200\text{ A}$, $I_{L2} = 150\text{ A}$, and $I_{L3} = 100\text{ A}$.
Step 2: Calculate the neutral current vector components Express each current as a phasor in rectangular coordinates: * $\vec{I}_{L1} = 200 \angle 0^\circ = 200 + j0\text{ A}$ * $\vec{I}_{L2} = 150 \angle -120^\circ = 150\cos(-120^\circ) + j150\sin(-120^\circ) = -75 - j129.9\text{ A}$ * $\vec{I}_{L3} = 100 \angle 120^\circ = 100\cos(120^\circ) + j100\sin(120^\circ) = -50 + j86.6\text{ A}$
Sum the horizontal components ($I_{Nx}$): $$I_{Nx} = 200 - 75 - 50 = 75\text{ A}$$
Sum the vertical components ($I_{Ny}$): $$I_{Ny} = 0 - 129.9 + 86.6 = -43.3\text{ A}$$
Step 3: Calculate the magnitude and phase angle of the neutral current ($I_N$) * Magnitude ($I_N$): $$I_N = \sqrt{I_{Nx}^2 + I_{Ny}^2} = \sqrt{75^2 + (-43.3)^2} = \sqrt{5625 + 1874.89} = \sqrt{7499.89} \approx 86.6\text{ A}$$
-
Phase Angle ($\theta_N$): $$\theta_N = \arctan\left(\frac{I_{Ny}}{I_{Nx}}\right) = \arctan\left(\frac{-43.3}{75}\right) = \arctan(-0.5773) \approx -30^\circ\ (\text{or } 330^\circ)$$
-
Answer 2: The neutral current is $86.6\text{ A}$ flowing at a phase angle of $-30^\circ$ relative to $I_{L1}$.
Curated Learning Resources for Irish Electrical Apprentices
Passing the math-heavy science exams in Phase 2, 4, and 6 requires utilizing the best study aids available. Below is a curated list of top-tier websites, video channels, textbooks, and support programs tailored to the Irish SOLAS curriculum and I.S. 10101 standards.
YouTube Channels (Highly Recommended)
- Joe Robinson (Electrical Training)
- Why it's great: The absolute gold standard for Irish electrical apprentices. Joe provides detailed walkthroughs of past exam papers, breaking down complex AC science, Star/Delta conversions, and motor calculations.
- Focus: Phase 4 & 6 Science revision.
- John Clare (Wiring Rules)
- Why it's great: John is a renowned expert on the Irish National Rules. He covers practical installation testing, cable calculations, and loop impedance in line with I.S. 10101.
- Focus: Practical science, regulations, and Phase 6 cable calculations.
- Dave Hergathy (Electrical Science)
- Why it's great: Excellent whiteboard demonstrations of basic Ohm's Law, parallel resistors, and power calculations.
- Focus: Phase 2 Science foundations.
- The Engineering Mindset
- Why it's great: High-quality, animated explanations of three-phase power, transformers, AC waveforms, and reactive power. Excellent for visualizing the concepts behind the math.
- Focus: Core theory for all phases.
Key Reference Books & Standards
- "A Practical Guide to the National Rules for Electrical Installations (I.S. 10101:2020)" by John Clare
- Why you need it: An essential companion book that simplifies the massive NSAI regulations book. It contains practical calculation examples for voltage drop, earth fault loop impedance, and correction factors.
- "Electrical Installation Calculations: Basic & Advanced" by A.J. Watkins and Christopher Kitcher
- Why you need it: While written for the UK City & Guilds curriculum, the math is identical. It contains hundreds of practice exercises with answers covering all aspects of cable sizing, motors, and AC circuits.
- "I.S. 10101:2020 National Rules for Electrical Installations" by NSAI
- Why you need it: The official rulebook. You should learn how to navigate its annexes and tables (especially Table A, B, and C for current capacity and correction factors) during your Phase 4 and Phase 6 college blocks.
Websites & Forums
- Safe Electric Ireland (safeelectric.ie)
- The regulatory body for electrical contractors in Ireland. Features free technical downloads, guides on the transition to I.S. 10101, and details on RCD and AFDD requirements.
- Boards.ie - Electrical Forum (boards.ie/categories/electrical)
- A very active Irish community forum. Many apprentices post questions about difficult college homework, and experienced Irish sparks and lecturers provide step-by-step guidance.
- Apprenticeship Ireland (apprenticeship.ie)
- The official SOLAS portal. Provides general information on training durations, employer requirements, and curriculum structures.
Educational Support & Grinds
- ETB Learner Support Centres
- If you are struggling during Phase 2, most Education and Training Boards (ETBs) offer free math support drop-in clinics. Ask your tutor or training coordinator for details.
- College Math Tutoring (TU Dublin, TUS, SETU, etc.)
- During Phase 4 and 6, technical universities run free "Maths Support Centres" where tutors can help you with algebraic transposition, trigonometry, and complex numbers.
- Private Revision Courses
- Various private electrical training schools in Ireland (e.g., John Clare Training, C²S Consulting, and independent tutors) run intensive 1-day or weekend exam prep seminars leading up to college science exams.
Useful Conversions and Metric Prefixes
This reference section compiles the most common mathematical conversions, metric prefixes, and unit scaling rules required for electrical science calculations in Phase 2, Phase 4, and Phase 6.
🔢 Metric Prefixes and Multipliers
Electrical engineering uses a wide range of values—from tiny currents in micro-electronics to massive power in megawatts. Always scale values back to their base units (Volts, Amperes, Ohms, Farads, Henrys, Watts) before inserting them into formulas.
| Prefix | Symbol | Power of 10 | Numeric Value | Practical Electrical Example |
|---|---|---|---|---|
| Mega | $M$ | $\times 10^6$ | $1,000,000$ | $1.2\text{ M}\Omega = 1,200,000\ \Omega$ (Insulation resistance) |
| Kilo | $k$ | $\times 10^3$ | $1,000$ | $4.5\text{ kW} = 4,500\text{ W}$ (Oven power load) |
| Unit | — | $10^0$ | $1$ | $230\text{ V}$, $16\text{ A}$, $50\text{ Hz}$ (Standard grid values) |
| Milli | $m$ | $\times 10^{-3}$ | $0.001$ | $30\text{ mA} = 0.03\text{ A}$ (RCD trip current threshold) |
| Micro | $\mu$ | $\times 10^{-6}$ | $0.000001$ | $47\ \mu\text{F} = 0.000047\text{ F}$ (PFC capacitor value) |
| Nano | $n$ | $\times 10^{-9}$ | $0.000000001$ | $17.2\text{ n}\Omega\cdot\text{m} = 1.72 \times 10^{-8}\ \Omega\cdot\text{m}$ (Copper resistivity) |
💡 Step-by-Step Prefix Scaling Examples
- Milliamperes ($\text{mA}$) to Amperes ($\text{A}$): $$I = 150\text{ mA} = 150 \times 10^{-3}\text{ A} = 0.15\text{ A}$$
- Kilohms ($\text{k}\Omega$) to Ohms ($\Omega$): $$R = 5.6\text{ k}\Omega = 5.6 \times 10^3\ \Omega = 5,600\ \Omega$$
- Microfarads ($\mu\text{F}$) to Farads ($\text{F}$): $$C = 22\ \mu\text{F} = 22 \times 10^{-6}\text{ F} = 0.000022\text{ F}$$
- Millihenrys ($\text{mH}$) to Henrys ($\text{H}$): $$L = 120\text{ mH} = 120 \times 10^{-3}\text{ H} = 0.12\text{ H}$$
⚙️ Power and Mechanical Conversions
These conversions are essential when calculating motor efficiency, mechanical output power, and electrical input power.
1. Horsepower (HP) and Kilowatts (kW)
Motors are often rated in Horsepower (mechanical output power). You must convert this to Watts or kW to calculate current draw. * Horsepower to Watts: $$1\text{ HP} = 746\text{ Watts} = 0.746\text{ kW}$$ $$\text{Power (Watts)} = \text{HP} \times 746$$ * Kilowatts to Horsepower: $$1\text{ kW} \approx 1.34\text{ HP}$$ $$\text{Power (HP)} = \frac{\text{Power (kW)}}{0.746}$$
2. Energy: Kilowatt-Hours (kWh) and Joules (J)
- Kilowatt-Hour is the standard commercial unit of electrical energy.
- Joule is the SI unit of energy ($1\text{ Joule} = 1\text{ Watt-second}$). $$1\text{ kWh} = 1,000\text{ W} \times 3,600\text{ seconds} = 3,600,000\text{ Joules} = 3.6\text{ MJ}$$
📏 Conductor and Resistivity Conversions
1. Conductor Cross-Sectional Area ($\text{mm}^2$ to $\text{m}^2$)
When using the resistivity formula ($R = \rho \frac{l}{A}$), the area ($A$) must be converted from square millimeters to square meters. * Conversion Factor: Multiply by $10^{-6}$ (or divide by $1,000,000$). $$2.5\text{ mm}^2 = 2.5 \times 10^{-6}\text{ m}^2 = 0.0000025\text{ m}^2$$ $$16\text{ mm}^2 = 16 \times 10^{-6}\text{ m}^2 = 0.000016\text{ m}^2$$
2. Cable Loop Resistance ($\text{m}\Omega/\text{m}$ to $\Omega$)
Conductor tables in I.S. 10101 express $(R_1 + R_2)$ loop resistance in milliohms per meter ($\text{m}\Omega/\text{m}$). * Conversion Factor: Divide the result by $1,000$ to get the value in Ohms ($\Omega$). $$\text{Loop Resistance } (R) = \frac{(R_1 + R_2) \times \text{Length (m)}}{1000} \quad (\Omega)$$
📐 Trigonometric and Angular Conversions
Trigonometric calculations are central to AC circuit analysis (calculating phase angles, impedance triangles, and power factor).
1. Degrees and Radians
Calculators must be toggled between Degrees ($^\circ$) and Radians ($\text{rad}$) depending on the formula. * Degrees to Radians: $$\text{Angle (Radians)} = \text{Angle (Degrees)} \times \frac{\pi}{180}$$ $$\text{Example: } 90^\circ = 90 \times \frac{\pi}{180} = \frac{\pi}{2} \approx 1.571\text{ rad}$$ * Radians to Degrees: $$\text{Angle (Degrees)} = \text{Angle (Radians)} \times \frac{180}{\pi}$$ $$\text{Example: } 1\text{ rad} = 1 \times \frac{180}{\pi} \approx 57.3^\circ$$
IMPORTANT: Calculator Mode Tip: Use Degree mode when calculating power factors ($\cos\phi$) or impedance triangles using right-angled trigonometry. Switch to Radian mode only when calculating instantaneous values using the time-based sine equation: $v = V_{peak} \sin(2\pi f t)$.
Interactive Formula Calculator
Select a mathematical formula from the list below to solve for any variable in real-time. Enter the known parameters, and the hub will automatically transpose the equation and compute the result.
Select Formula
- Ohm's Law ($V = I \times R$)
- Power Law ($P = V \times I$)
- Resistivity ($R = \rho \frac{l}{A}$)
- Temperature Coefficient ($R_\theta = R_0(1 + \alpha\theta)$)
- Inductive Reactance ($X_L = 2\pi f L$)
- Capacitive Reactance ($X_C = \frac{1}{2\pi f C}$)
- Synchronous Speed ($N_s = \frac{120 f}{p}$)
- Motor Percentage Slip ($s\% = \frac{N_s - N_r}{N_s} \times 100$)
- Single-Phase Volt Drop ($V_d = \frac{mV \cdot I \cdot L}{1000}$)
Interactive Revision Exam
Test your electrical mathematics knowledge with this interactive revision test. The deck consists of 40 exam-style questions spanning Phase 2, Phase 4, and Phase 6 of the Irish SOLAS curriculum.
Full 40-Question Exam
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Phasor & Impedance Vector Visualizer
Explore single-phase AC theory by manipulating resistance ($R$) and reactance ($X$) in real time. Drag the phasor vector directly on the complex plane coordinate canvas, or use the sliders to observe changes in total impedance ($Z$), phase angle ($\phi$), and power factor ($\cos\phi$).