Electrical Theory and Calculations

Introduction

This chapter covers the theoretical foundations and essential calculations that every industrial electrician must master for the Red Seal exam. Electrical theory is not an academic exercise: it is the foundation of every diagnostic, every installation, and every commissioning. You will need to apply these principles to size conductors, protect circuits, correct power factor, and analyze alternating current (AC) and direct current (DC) circuits. Exam questions focus less on memorizing formulas than on applying them correctly in realistic industrial contexts.

Fundamental Laws and Units

The International System (SI) and Derived Units

You must know the base units and their symbols, as well as the metric prefixes commonly used in industry (kilo, mega, milli, micro). Conversion errors are a frequent cause of failure.

Quantity	Unit	Symbol
Voltage	Volt	V
Current	Ampere	A
Resistance	Ohm	Ω
Active power	Watt	W
Reactive power	Volt-ampere reactive	var
Apparent power	Volt-ampere	VA
Frequency	Hertz	Hz
Energy	Joule or watt-hour	J or Wh

Prefixes to remember: kilo (k = 10³), mega (M = 10⁶), giga (G = 10⁹), milli (m = 10⁻³), micro (µ = 10⁻⁶), nano (n = 10⁻⁹).

Ohm's Law and Power in Direct Current

Ohm's Law establishes the relationship between voltage (U), current (I), and resistance (R): U = R × I. In direct current, power is calculated by P = U × I. By combining these two relationships, you obtain the derived forms:

P = I² × R

P = U² / R

Example: A heating element of 20 Ω is supplied at 240 V DC. The current is I = 240 / 20 = 12 A. The power is P = 240 × 12 = 2,880 W, or 2.88 kW.

Kirchhoff's Laws

Two fundamental laws govern circuit analysis:

16.Current Law (Junction Rule): the algebraic sum of currents entering a node equals the sum of currents leaving it. In practice, ΣI = 0.

17.Voltage Law (Loop Rule): the algebraic sum of voltages around a closed loop is zero. In practice, ΣU = 0.

These laws apply equally to DC and AC, provided you use vector quantities (phasors) in AC.

Series and Parallel Circuits

Resistors in Series

In a series circuit, the current is identical through all elements. The total resistance is the sum of the individual resistances: R_total = R₁ + R₂ + R₃ + …. The total voltage divides proportionally across the resistances (voltage divider).

Resistors in Parallel

In a parallel circuit, the voltage is identical across each branch. The total conductance (inverse of resistance) is the sum of the conductances: 1/R_total = 1/R₁ + 1/R₂ + 1/R₃ + …. For two resistors only, you use the simplified formula: R_total = (R₁ × R₂) / (R₁ + R₂).

Common trap: For three or more resistors, the product-over-sum formula only works for two resistors. Always use the reciprocal formula for three or more elements.

Mixed Circuits

For a mixed circuit, simplify progressively: identify series and parallel groups, calculate their equivalents, then combine. This method also applies to circuits with inductors and capacitors, but using complex impedance.

Alternating Current: Fundamental Principles

Characteristic Values of a Sinusoidal Wave

In AC, voltage and current vary sinusoidally. Four values are essential:

Peak value (U_max): maximum amplitude of the wave.

RMS value (U_rms): the equivalent heating value of a DC voltage. For a sine wave, U_rms = U_max / √2 ≈ 0.707 × U_max.

Average value: for a complete sine wave, it is zero. For a half-cycle, it equals 0.637 × U_max.

Peak-to-peak value: the difference between the positive maximum and the negative maximum, i.e., 2 × U_max.

Industrial application: A 600 V AC RMS voltage has a peak value of 600 × √2 ≈ 848 V. This value is critical for selecting the withstand voltage of components and for insulation calculations.

Period and Frequency

Frequency (f) is the number of cycles per second, in hertz (Hz). The period (T) is the inverse of frequency: T = 1/f. In Canada, the standard frequency is 60 Hz, so T = 1/60 ≈ 16.67 ms.

Phase and Phase Shift

Phase shift (φ) is the angle between voltage and current in a circuit. It is expressed in degrees or radians. A zero phase shift means voltage and current are in phase (purely resistive load). A phase shift of +90° indicates a purely inductive load (current lags voltage). A phase shift of -90° indicates a purely capacitive load (current leads voltage).

Impedance and RLC Circuits

Definition of Impedance

Impedance (Z) is the total opposition to the flow of alternating current in a circuit. It combines resistance (R), inductive reactance (X_L), and capacitive reactance (X_C). It is measured in ohms (Ω) and is represented as a complex number: Z = R + j(X_L - X_C).

Reactances

Inductive reactance: X_L = 2πfL, where L is the inductance in henrys (H). It increases with frequency.

Capacitive reactance: X_C = 1 / (2πfC), where C is the capacitance in farads (F). It decreases with frequency.

Calculating Total Impedance

For a series RLC circuit, the total impedance is: Z = √(R² + (X_L - X_C)²). The phase angle is: φ = arctan((X_L - X_C) / R).

Example: A series circuit contains R = 30 Ω, L = 0.1 H, and C = 50 µF, supplied at 60 Hz.

X_L = 2π × 60 × 0.1 = 37.7 Ω

X_C = 1 / (2π × 60 × 50 × 10⁻⁶) = 53.1 Ω

Z = √(30² + (37.7 - 53.1)²) = √(900 + 237.2) = √1137.2 ≈ 33.7 Ω

φ = arctan((37.7 - 53.1) / 30) = arctan(-0.513) ≈ -27.2° (current leads voltage, capacitively dominant circuit)

Resonance

Resonance occurs when X_L = X_C, therefore when f_r = 1 / (2π√(LC)). At resonance, impedance is minimal (equal to R) in series, and current is maximal. In parallel, impedance is maximal. This phenomenon is used in filter circuits and reactive power compensators.

Power in Alternating Current

The Three Powers

In AC, three powers are distinguished:

57.Active power (P): measured in watts (W), this is the power actually converted into work or heat. P = U × I × cos φ.

58.Reactive power (Q): measured in volt-amperes reactive (var), it is exchanged between the source and inductive/capacitive elements. Q = U × I × sin φ.

59.Apparent power (S): measured in volt-amperes (VA), it is the product of voltage and current without accounting for phase shift. S = U × I.

The relationship between these three powers is: S² = P² + Q², or S = √(P² + Q²).

Power Factor

Power factor (PF) is the ratio of active power to apparent power: PF = P / S = cos φ. A unity power factor (1.0) indicates perfect energy utilization. A low PF (for example, 0.7) means the current is higher than necessary for the useful power, resulting in conductor losses and financial penalties.

Industrial example: A 50 kW motor operates with a PF of 0.75 at 600 V. The current is I = P / (U × PF) = 50,000 / (600 × 0.75) = 111.1 A. If the PF is corrected to 0.95, the current becomes I = 50,000 / (600 × 0.95) = 87.7 A. The current reduction is 21%.

Power Factor Correction

Correction is achieved by installing capacitors in parallel with the load. The required capacitive reactive power (Q_C) is: Q_C = P × (tan φ₁ - tan φ₂), where φ₁ is the initial angle and φ₂ the target angle. The required capacitance is: C = Q_C / (2πfU²).

Example: An installation consumes 100 kW with a PF of 0.70. You want to correct to 0.95.

φ₁ = arccos(0.70) = 45.57°, tan φ₁ = 1.02

φ₂ = arccos(0.95) = 18.19°, tan φ₂ = 0.33

Q_C = 100,000 × (1.02 - 0.33) = 69,000 var = 69 kvar

At 600 V and 60 Hz, the capacitance is C = 69,000 / (2π × 60 × 600²) = 69,000 / 135,716,800 ≈ 508 µF.

Three-Phase Circuits

Basic Principles

A three-phase system has three sinusoidal voltages offset by 120° from each other. There are two configurations: wye (Y) and delta (Δ).

Voltage-Current Relationships

Configuration	Phase-to-phase voltage (U_LL)	Phase-to-neutral voltage (U_LN)	Line current (I_L)	Phase current (I_Ph)
Wye (Y)	U_LL = √3 × U_LN	U_LN = U_LL / √3	I_L = I_Ph	I_Ph = I_L
Delta (Δ)	U_LL = U_Ph	U_Ph = U_LL	I_L = √3 × I_Ph	I_Ph = I_L / √3

Values to memorize: √3 ≈ 1.732 and 1/√3 ≈ 0.577.

Example: A 600 V three-phase wye system has a phase-to-neutral voltage of 600 / √3 ≈ 346 V. A 208 V three-phase system has a phase-to-neutral voltage of 120 V.

Three-Phase Power

The total active power is: P = √3 × U_LL × I_L × cos φ. The apparent power is: S = √3 × U_LL × I_L. The reactive power is: Q = √3 × U_LL × I_L × sin φ.

Example: A 25 kW three-phase motor, 600 V, PF = 0.85. The line current is I_L = P / (√3 × U × PF) = 25,000 / (1.732 × 600 × 0.85) = 25,000 / 883.3 ≈ 28.3 A.

Unbalanced Loads

In practice, three-phase loads are not always perfectly balanced. The current in the neutral of a wye system is the vector sum of the three phase currents. For unbalanced loads, the neutral current can be significant. The exact calculation requires vector analysis, but a common approximation for the exam is to consider the worst case where the neutral carries nearly the current of the most heavily loaded phase.

Voltage Drop Calculations

General Formula

Voltage drop (ΔU) in a conductor is caused by its resistance and reactance. For a single-phase circuit: ΔU = 2 × I × L × (R_cos φ + X_sin φ), where L is the total circuit length in meters (out and return), R and X are the resistance and reactance of the conductor per meter. For a three-phase circuit: ΔU = √3 × I × L × (R_cos φ + X_sin φ).

In practice, for conductors of moderate size, reactance is often neglected, and you use: ΔU = 2 × I × L × R (single-phase) or ΔU = √3 × I × L × R (three-phase).

Code Requirements

The Canadian Electrical Code, Part I (CE Code) recommends that voltage drop not exceed 3% for utilization circuits and 5% in total (service + utilization circuit). These values are recommendations unless made mandatory by the authority having jurisdiction. Rule 8-200 of the CE Code addresses the calculation of conductor sizes based on voltage drop.

Example: A 10 kW motor, 600 V three-phase, PF = 0.85, is supplied by a 75 m cable. The current is I = 10,000 / (1.732 × 600 × 0.85) = 11.3 A. The maximum allowable voltage drop is 3% of 600 V = 18 V. The maximum conductor resistance is R = ΔU / (√3 × I × L) = 18 / (1.732 × 11.3 × 75) = 18 / 1468 ≈ 0.0123 Ω/m. By consulting the CE Code tables, you select the appropriate size.

Table of Approximate Resistances (copper conductors at 75 °C)

Size (mm²)	Resistance (Ω/km)
2.5	9.42
4.0	5.90
6.0	3.94
10.0	2.37
16.0	1.48
25.0	0.943
35.0	0.674
50.0	0.471

Circuit Protection and Sizing

Rated Current and Full-Load Current

A motor's rated current is indicated on its nameplate. The full-load current (FLC) can be estimated from the CE Code tables (Appendix B) for standard motors. These tables provide typical values based on horsepower and voltage.

Overcurrent Protection Rules

The CE Code, Part I, Rule 28-200, specifies that branch circuit protectors for motors must be sized at 125% of the full-load current for continuous-duty motors. Overload protectors (thermal relays) are set between 115% and 125% of the rated current, depending on the motor's service factor.

Example: A 20 A continuous-duty motor requires a branch circuit protector of at least 20 × 1.25 = 25 A. You select the next standard size up, which is 30 A, if starting conditions require it.

Conductor Ampacity

Ampacity depends on size, material (copper or aluminum), ambient temperature, number of grouped conductors, and insulation type. The CE Code tables (Tables 1 to 4) provide these values. Correction factors for temperature and grouping are given in Tables 5A to 5D and 5E to 5H.

Rule 4-004: Conductors must be protected against overcurrent in accordance with their ampacity, except for specific exceptions (for example, motor circuits with overload protection).

Practical Circuit Analysis

Step-by-Step Problem-Solving Method

To solve a complex circuit problem on the exam:

104.Identify the circuit type (series, parallel, mixed, three-phase).

105.Determine the known and unknown quantities.

106.Select the appropriate formula.

107.Convert all units to the SI system.

108.Perform calculations with three significant figures of precision.

109.Verify the coherence of results (order of magnitude, units).

Using Phasors

In AC, voltages and currents are phasors. To add currents in a parallel circuit, you must account for phase shifts. The graphical method (Fresnel diagram) is often more intuitive than complex calculation. For the exam, master the conversion between polar and rectangular forms.

Example: Two loads in parallel: load 1: 10 A at PF 0.8 inductive; load 2: 5 A at PF 0.9 capacitive. The total current is the vector sum. Load 1: I₁ = 10∠36.87° (since cos φ = 0.8, φ = 36.87°). Load 2: I₂ = 5∠-25.84° (since cos φ = 0.9, φ = -25.84°). In rectangular coordinates: I₁ = 8 + j6, I₂ = 4.5 - j2.18. Total: I = 12.5 + j3.82. The total current is I = √(12.5² + 3.82²) ≈ 13.1 A, with an angle φ = arctan(3.82/12.5) ≈ 17° (inductive).

Pitfalls to Avoid

Confusing phase-to-neutral and phase-to-phase voltage: In a three-phase system, the phase-to-phase voltage is √3 times greater than the phase-to-neutral voltage. An error here invalidates all power calculations.

Forgetting the √3 factor in three-phase calculations: Three-phase power is √3 × U × I × cos φ, not U × I × cos φ.

Using peak value instead of RMS value: Power and voltage drop calculations always use RMS values, unless otherwise indicated.

Neglecting temperature correction: Conductor ampacity decreases at high temperatures. Failing to apply CE Code correction factors leads to undersized conductors.

Confusing active and apparent power: Current is calculated from apparent power, not active power. A 10 kW motor at PF 0.7 does not draw 10 kVA, but 14.3 kVA.

Forgetting the 125% rule: Motor circuit conductors and protectors must be sized at 125% of the rated current, not 100%.

Rounding too early: Perform calculations with sufficient precision (at least three significant figures) and round only at the end.

Ignoring units: Always verify that units are consistent. For example, capacitance is in farads, not microfarads, in formulas.

Exam Tips

Memorize the constants: √2 ≈ 1.414, √3 ≈ 1.732, 1/√3 ≈ 0.577, π ≈ 3.1416.

Learn the CE Code tables: Ampacity tables and correction factors are provided with the exam, but you must know how to use them quickly.

Practice unit conversion: Questions often mix kW, kVA, and hp. Remember that 1 hp = 746 W.

Check the coherence of answers: A current of 500 A for a 10 kW motor is obviously wrong. Use your judgment to detect gross errors.

Manage your time: Calculation questions are often worth more points. Don't get stuck on a difficult question; move on and come back later.

Summary

Ohm's Law (U = R × I) and Kirchhoff's Laws are the foundations of circuit analysis.

In AC, distinguish between RMS, peak, and average values. Always use RMS values for power calculations.

Impedance combines resistance and reactances: Z = √(R² + (X_L - X_C)²).

The three AC powers are related by S² = P² + Q², with PF = P/S = cos φ.

In three-phase, the relationships between voltages and currents depend on the configuration (wye or delta). Power is P = √3 × U_LL × I_L × cos φ.

Power factor correction uses capacitors in parallel: Q_C = P × (tan φ₁ - tan φ₂).

Voltage drop must not exceed 3% for utilization circuits and 5% in total, according to the CE Code, Part I.

Branch circuit protectors for motors are sized at 125% of the full-load current (Rule 28-200).

Conductors must be sized according to their ampacity, with CE Code correction factors for temperature and grouping.

Regulatory References

Canadian Electrical Code, Part I (CE Code): Rules 4-004 (conductor protection), 8-200 (voltage drop), 28-200 (motor protection).

CSA C22.1: Canadian Electrical Code, Part I (installation).

CSA C22.2: Standards for industrial electrical equipment.

Mastery of these concepts and calculations will enable you not only to pass the exam but also to practice your trade competently and safely. Practice regularly with varied exercises, timing your sessions, to develop the speed and accuracy needed on exam day.

Electrical Theory and Calculations