This chapter covers the theoretical foundations and essential calculations that every construction electrician must master for the Red Seal exam. Electrical theory is not an academic exercise—it is applied daily on job sites, from conductor sizing to load verification. A solid understanding of these principles will not only help you pass the exam but also make safe, code-compliant decisions in accordance with the Canadian Electrical Code, Part I (CE Code) .
Fundamental Laws of Electricity
Ohm's Law
Ohm's Law establishes the relationship between voltage (E), current (I), and resistance (R) in a circuit. It is expressed as follows:
E = I × R
Where:
E = voltage in volts (V)
I = current in amperes (A)
R = resistance in ohms (Ω)
This relationship can be rearranged to calculate any of the three variables:
Variable to Find
Formula
Voltage (E)
E = I × R
Current (I)
I = E ÷ R
Resistance (R)
R = E ÷ I
Application Example: A resistive load of 12 Ω is supplied at 120 V. The current is calculated as follows: I = 120 V ÷ 12 Ω = 10 A.
Power Law (Joule's Law)
Electrical power (P) represents the rate at which electrical energy is converted into another form of energy (heat, light, motion). The fundamental formula is:
P = E × I
Where P is in watts (W). By combining with Ohm's Law, the following derived formulas are obtained:
Quantity to Find
Formula
Power (P)
P = E × I
Power (with R)
P = I² × R
Power (with R and E)
P = E² ÷ R
Current (with P and E)
I = P ÷ E
Voltage (with P and I)
E = P ÷ I
Example: A 3,000 W water heater is connected to a 240 V circuit. The current is: I = 3,000 W ÷ 240 V = 12.5 A.
Kirchhoff's Laws
Two fundamental rules govern the behaviour of circuits:
First Law (Current Law / Junction Rule): The sum of currents entering a node equals the sum of currents leaving it. In other words, current is never lost—it divides.
Second Law (Voltage Law / Loop Rule): The algebraic sum of voltages in a closed loop is zero. The voltage supplied by the source is entirely consumed within the circuit.
These laws are essential for analysing series circuits, parallel circuits, and combination circuits.
Series, Parallel, and Combination Circuits
Series Circuit
In a series circuit, components are connected end-to-end, forming a single path for current.
Characteristics:
The current is the same through all components: I_total = I₁ = I₂ = I₃
The total voltage is the sum of the individual voltages: E_total = E₁ + E₂ + E₃
The total resistance is the sum of the resistances: R_total = R₁ + R₂ + R₃
Calculating Total Resistance:
R_total = R₁ + R₂ + R₃ + ... + Rₙ
Example: Three resistors of 4 Ω, 6 Ω, and 10 Ω are connected in series to a 120 V source.
R_total = 4 + 6 + 10 = 20 Ω
I = 120 V ÷ 20 Ω = 6 A
Voltage drop across R₁: E₁ = 6 A × 4 Ω = 24 V
Parallel Circuit
In a parallel circuit, components are connected between two common points, providing multiple paths for current.
Characteristics:
The voltage is the same across each branch: E_total = E₁ = E₂ = E₃
The total current is the sum of the branch currents: I_total = I₁ + I₂ + I₃
The total resistance is always less than the smallest resistance in the circuit
Calculating Total Resistance:
1/R_total = 1/R₁ + 1/R₂ + 1/R₃ + ... + 1/Rₙ
For only two resistors, the simplified formula can be used:
R_total = (R₁ × R₂) ÷ (R₁ + R₂)
Example: Two resistors of 10 Ω and 20 Ω are connected in parallel to a 120 V source.
Example: A 10 kW, 600 V three-phase motor with a power factor of 0.85. The line current is:
I = P ÷ (√3 × E × cos φ) = 10,000 W ÷ (1.732 × 600 V × 0.85) = 11.3 A
Electrical Energy
Energy (W) is the power consumed over a given period:
W = P × t
Where:
W = energy in watt-hours (Wh) or kilowatt-hours (kWh)
P = power in watts (W) or kilowatts (kW)
t = time in hours (h)
Example: A 1,500 W heater operates 8 hours per day. The energy consumed is: 1.5 kW × 8 h = 12 kWh per day.
Power Factor
Definition and Importance
The power factor (PF) is the ratio of real power (W) to apparent power (VA):
PF = P ÷ S = cos φ
A low power factor (below 0.90) indicates poor energy utilization. Inductive loads (motors, ballasts) create a phase shift that increases current without increasing useful power.
Power Factor Correction
Correction is achieved by adding capacitors in parallel with the load. The capacitive reactive power (VAR) compensates for the inductive reactive power.
Calculating Reactive Power:
Q = √(S² - P²)
Where Q is in reactive volt-amperes (VAR).
Example: An installation consumes 50 kW with a PF of 0.75. Correction to 0.95 is desired.
Voltage drop is the difference in voltage between the source and the load, caused by the resistance of the conductors. Excessive drop leads to poor equipment performance (dim lighting, overheated motors).
Code Requirements
The Canadian Electrical Code, Part I (CE Code) requires that the total voltage drop from the source to the load not exceed 3% for utilization circuits and 5% for the entire installation (service + branch circuit). These values are recommended in Rule 8-200 and related sections.
Calculating Voltage Drop
For a single-phase circuit:
Voltage drop (V) = (2 × L × I × R) ÷ 1,000
Where:
L = length of the conductor in metres (round trip)
I = current in amperes
R = resistance of the conductor in ohms per 1,000 metres (Ω/km)
For a three-phase circuit:
Voltage drop (V) = (√3 × L × I × R) ÷ 1,000
Table of Copper Conductor Resistance (Ω/km at 75 °C):
Size (AWG)
Resistance (Ω/km)
14
10.2
12
6.4
10
4.0
8
2.5
6
1.6
4
1.0
2
0.63
1
0.50
1/0
0.40
Example: A 120 V, 15 A single-phase circuit using No. 12 AWG copper conductors over a length of 30 m.
R = 6.4 Ω/km
Voltage drop = (2 × 30 m × 15 A × 6.4 Ω/km) ÷ 1,000 = 5.76 V
Percentage: 5.76 V ÷ 120 V × 100 = 4.8% — exceeds 3%, the conductor size must be increased
Circuit Calculations and Conductor Sizing
Conductor Ampacity
Table 2 of the CE Code provides ampacities for copper and aluminum conductors based on size and insulation temperature rating. Correction factors apply for:
Ambient temperature (Tables 5A, 5B, 5C)
Grouping of conductors (Table 5C)
The number of current-carrying conductors in the same conduit
Rule 4-004: The ampacity must be determined based on the actual installation conditions.
Load Calculation
Rule 8-200 of the CE Code defines the minimum load calculation for installations. Loads are calculated based on:
Building floor area (general lighting: 80 VA/m² for dwellings)
Receptacle circuits (12 A per 120 V circuit for dwellings)
Specific loads (range, dryer, water heater, etc.)
Demand factors (reduction for large loads)
Example of a dwelling load calculation:
Floor area of 200 m²: 200 × 80 VA = 16,000 VA
Two receptacle circuits at 12 A: 2 × 12 A × 120 V = 2,880 VA
Range (Table 8-4): 8,000 VA
Water heater: 3,000 VA
Dryer: 5,000 VA
Total: 34,880 VA
The total load divided by the voltage (240 V) gives the service current: 34,880 ÷ 240 = 145.3 A. A 150 A or 200 A service would be selected based on requirements.
Rule 10-600: Bonding and grounding conductors — sizing according to Table 16
Table 16 — Grounding Conductor Sizing (excerpt):
Phase Conductor Size (Copper)
Grounding Conductor Size (Copper)
14
14
12
12
10
10
8
8
6
8
4
8
2
6
1
6
1/0
6
2/0
4
Common Traps to Avoid
205.Confusing power formulas: In AC, do not use P = E × I without accounting for the power factor. For inductive loads, real power is always less than apparent power.
206.Forgetting the √3 factor in three-phase: Three-phase current calculations always require the square root of 3 (1.732). An error by a factor of 1.732 can lead to dangerous undersizing.
207.Neglecting correction factors: Table 2 provides ampacities at 30 °C. If the ambient temperature exceeds 30 °C, or if multiple conductors are grouped, you must apply the correction factors from Tables 5A to 5C.
208.Reversing the rules for motors: The conductor is sized at 125% of the FLC, but overcurrent protection can be higher (up to 250%). Do not confuse these two distinct requirements.
209.Using one-way length instead of round-trip: In single-phase voltage drop calculations, the length must be multiplied by 2 (out and back). For three-phase, use √3 × L.
210.Forgetting the 3% / 5% rule: Voltage drop must be checked systematically. A conductor properly sized for ampacity may be inadequate for voltage drop over long distances.
211.Confusing motor FLC tables: Tables 28-1 to 28-4 of the CE Code provide full-load currents based on motor type (single-phase, three-phase, etc.) and voltage. Always use the correct table.
212.Neglecting continuous loads: For loads that operate for more than 3 hours, the conductor must be sized at 125% of the load (Rule 8-104). This requirement is often overlooked.
213.Error in calculating equivalent parallel resistance: Total resistance in parallel is always less than the smallest resistance. If your calculation gives a higher result, it is an error.
214.Not verifying transformer compliance: Transformer protection rules (Section 26) are specific and differ from ordinary circuits. Always consult Table 26-250.
Summary
Ohm's Law (E = I × R) and the Power Law (P = E × I) are the foundations of all electrical calculations.
Kirchhoff's Laws allow for the analysis of complex circuits: sum of currents at a node = 0; sum of voltages in a loop = 0.
In a series circuit, current is common and voltages add up. In a parallel circuit, voltage is common and currents add up.
Three-phase power is calculated using P = √3 × E × I × cos φ.
The power factor must be corrected to avoid penalties and reduce current.
Voltage drop must not exceed 3% for utilization circuits and 5% overall (Rule 8-200).
Conductor sizing follows a five-step procedure: load current, ampacity, correction factors, voltage drop, protection.
Motor circuits have specific requirements (Chapter 28): conductor at 125% of FLC, separate overcurrent and overload protection.
Transformers follow the transformation ratio E₁/E₂ = N₁/N₂ and the conservation of power.
Grounding and bonding (Section 10) are essential for safety and must be sized according to Table 16.
Exam Tips
Memorize the basic formulas: Ohm's Law, power, voltage drop. Write them down on your scrap paper at the beginning of the exam.
Use the Code tables: The CE Code is provided during the exam. Practice navigating quickly through Tables 2, 5A-5C, 16, 28-1 to 28-4, and 26-250.
Check your units: Always convert kilowatts to watts, metres to kilometres, etc., before performing your calculations.
Make estimates: Before calculating, estimate the order of magnitude of the result. If your answer is unreasonable (e.g., 500 A for a 15 A circuit), review your approach.
Read questions twice: Exam questions often contain traps (line-to-line vs. line-to-neutral voltage, one-way vs. round-trip length, etc.).
Practice with timed exercises: The Red Seal exam is timed. Get used to solving problems in under 2 minutes each.
This chapter provides you with the essential theoretical tools. Mastery comes with practice: redo the examples, vary the parameters, and consult the CE Code for every rule mentioned. Good luck with your preparation!
