Chapter V

Transformers and Power Distribution

Red Seal Practice study guide with diagrams.

Transformers and Power Distribution

Chapter Introduction

This chapter covers all the knowledge required for the Red Seal exam concerning transformers and electrical power distribution. You will find the fundamental principles, common configurations, essential calculations, the requirements of the Canadian Electrical Code (CE Code), and typical pitfalls to avoid. Mastering this content is essential, as questions on this topic represent a significant portion of the exam.


Fundamental Principles of Transformers

Transformers — Electromagnetic induction and transformation ratio Transformers — Electromagnetic induction and transformation ratio Simplified diagram Core (laminated) Primary V₁, I₁ N₁ turns Secondary V₂, I₂ N₂ turns ● Alternating magnetic flux (Φ) Transformation ratio V₁ / V₂ = N₁ / N₂ = a Voltage (Volts) Turns (windings) a = ratio (transformation ratio) Calculation example Given data: N₁ = 500 turns, N₂ = 100 turns V₁ = 600 V (primary side) Solution: a = N₁ / N₂ = 500 / 100 = 5 V₂ = V₁ / a = 600 / 5 = 120 V V₂ = 120 Volts Step-down transformer (step-down transformer) Electromagnetic induction requires alternating current (AC) — direct current (DC) produces no induction.

Faraday's Law and Electromagnetic Induction

A transformer operates on the principle of electromagnetic induction discovered by Michael Faraday. When an alternating current flows through the primary winding, it creates a varying magnetic flux in the ferromagnetic core. This varying flux induces an electromotive force (EMF) in the secondary winding.

The fundamental relationship is expressed by Faraday's Law:

E = N × (dΦ/dt)

Where:

E = induced electromotive force (volts)
N = number of turns
dΦ/dt = rate of change of magnetic flux over time

Transformation Ratio

The transformation ratio (a) is the ratio between the number of turns in the primary and that of the secondary:

a = N₁/N₂ = V₁/V₂ = I₂/I₁

For an ideal transformer (no losses):

V₁/V₂ = N₁/N₂
I₁/I₂ = N₂/N₁
V₁ × I₁ = V₂ × I₂ (power is conserved)

Calculation Example: A transformer has 480 turns on the primary and 120 turns on the secondary. If the primary voltage is 600 V, what is the secondary voltage?

V₂ = V₁ × (N₂/N₁) = 600 × (120/480) = 600 × 0.25 = 150 V

Types of Transformers by Application

TypePrimary ApplicationCharacteristic
PowerSubstations, distribution> 500 kVA, high voltage
DistributionNeighborhood and industrial supply5 kVA to 500 kVA
IsolationGalvanic separation1:1 ratio
InstrumentationCTs, PTs for measurement and protectionHigh accuracy
AutotransformerModerate voltage variationSingle winding
Three-phaseIndustrial networks3 cores or 3 single-phase units

Three-Phase Transformers

Connection Configurations

Three-phase transformers can be connected in various configurations. Each one has specific advantages and disadvantages.

Wye-Wye (Y-Y)

Advantages: economical, lightweight
Disadvantages: harmonic issues (3rd harmonic), possible unbalance
The primary neutral must be grounded

Delta-Wye (Δ-Y)

Most common configuration for distribution
30° phase shift between primary and secondary
Neutral is available on the secondary
Good stability under unbalanced load

Wye-Delta (Y-Δ)

Used for large generating stations (voltage step-up)
Also has a 30° phase shift
The delta traps 3rd order harmonics

Delta-Delta (Δ-Δ)

No phase shift
Can operate in open delta (V-V) configuration if one winding is faulty
No neutral available

Clock Hour Index

The clock hour index indicates the phase shift between primary and secondary voltages. It is expressed in multiples of 30°. For example, a Δ-Y connection with a 30° phase shift is designated Dyn1 or Dyn11 depending on the direction of the phase shift.


Power and Current Calculations

Apparent, Active, and Reactive Power

Apparent power: S = V × I (VA, kVA, MVA)
Active power: P = V × I × cos φ (W, kW, MW)
Reactive power: Q = V × I × sin φ (VAR, kVAR, MVAR)

For a balanced three-phase system:

S = √3 × V_L-L × I_L
P = √3 × V_L-L × I_L × cos φ
Q = √3 × V_L-L × I_L × sin φ

Calculating Rated Current

Example: A three-phase transformer rated 750 kVA, 600 V/208 V, Δ-Y connection. Calculate the rated current on the secondary.

I₂ = S / (√3 × V₂) = 750,000 / (1.732 × 208) = 750,000 / 360.3 = 2,082 A

Transformer Efficiency

η = (P_output / P_input) × 100%

Losses are divided into:

Copper losses (P = I² × R) — proportional to the square of the current
Iron losses (hysteresis and eddy currents) — constant at fixed voltage

Maximum efficiency occurs when copper losses equal iron losses.


Grounding and Bonding

Canadian Electrical Code Requirements

The Canadian Electrical Code, Part I (CE Code) imposes specific rules for the grounding of transformers.

Rule 10-204 — Grounding of Transformers

The metal enclosure and core of every transformer must be grounded in accordance with the rules in Section 10. The secondary neutral must be connected to ground at the service point or at the transformer itself.

Rule 10-206 — Grounding Conductor

The grounding conductor must be:

Copper or aluminum
Of minimum size according to Table 16 of the CE Code
Run without splices, unless an accessible junction box is used

Rule 10-208 — Grounding Electrode

The electrode must have a resistance of 25 Ω or less. If this value cannot be achieved, an additional electrode must be installed.

Grounding System Schemes (System Grounding Types)

Grounding systems are classified by three letters:

First letter: relationship to ground (T = directly grounded, I = isolated)
Second letter: exposed conductive parts (T = connected to ground, N = connected to neutral)
Third letter: neutral and protective conductor arrangement (S = separate, C = combined)
SchemeDescriptionTypical Use
TN-SNeutral and PE separateCommercial buildings
TN-CNeutral and PE combined (PEN)Older industrial networks
TN-C-SCombined upstream, separate downstreamModern distribution
TTNeutral grounded, exposed parts groundedPublic networks
ITNeutral isolated, exposed parts groundedHospitals, service continuity

Transformer Protection

Overcurrent Protection

CE Code Rule 26-248

This rule requires that every transformer be protected against overcurrents on the primary side. The protective device must be set:

At 125% of the primary rated current for transformers 600 V and less
At 150% for transformers over 600 V

If the exact setting is not available, you may round up to the next standard size.

Secondary Protection

Secondary protection is required when:

The secondary voltage exceeds 750 V
The secondary current exceeds 9 A and the voltage exceeds 150 V

Protection Against Internal Faults

Buchholz relay: detects gas accumulation in oil-filled transformers
Pressure relay: detects sudden pressure changes
Differential relay: compares primary and secondary currents
Temperature sensors: monitor winding heating

Instrument Transformers

Current Transformers (CTs)

Current transformers reduce high currents to measurable values (typically 5 A or 1 A on the secondary).

Important characteristics:

The primary is connected in series with the circuit
The secondary must never be opened under load (dangerous voltage)
The ratio is expressed as 400:5, 600:5, etc.
The accuracy class is indicated (0.3, 0.6, 1.2, etc.)

Potential Transformers (PTs)

Potential transformers reduce high voltages to measurable values (typically 120 V or 69.3 V on the secondary).

The primary is connected in parallel with the circuit
The secondary must be protected by a fuse
Accuracy is typically 0.3% or 0.6%

Power Distribution in Industrial Buildings

Typical Distribution Systems

VoltageUseConfiguration
600 VIndustrial motors, heavy equipmentThree-phase, 3 or 4 wire
480 VCommercial and industrial equipmentThree-phase, 3 or 4 wire
347/600 VLighting and motorsThree-phase, 4 wire
208/120 VReceptacles, lighting, small equipmentThree-phase, 4 wire
240/120 VResidential, small commercialSingle-phase, 3 wire

Rule 8-200 — Load Calculations

Rule 8-200 of the CE Code specifies minimum demands for load calculations. Demand factors apply according to the type of load:

Type of LoadDemand Factor
General lighting100% of the first 100 kVA + 70% of the remainder
Receptacles100% of the first 10 kVA + 50% of the remainder
Motors100% of the largest + 25% of the others
Heating100% of the connected load

Conductors and Raceways

Rule 4-004 — Ampacity of Conductors

The ampacity of conductors is determined according to Tables 1 to 4 of the CE Code, based on:

Ambient temperature
Number of conductors in the raceway
Type of insulation (TW, THW, XHHW, etc.)

Correction Factors

Temperature: Table 5A of the CE Code (correction factors)
Grouping: Table 5C of the CE Code (correction factors for more than 3 conductors)

Harmonics and Power Quality

Sources of Harmonics

Non-linear loads (variable frequency drives, switching power supplies, electronic lighting) generate harmonic currents that:

Overheat transformers
Cause heating in the neutral
Deteriorate power quality

K-Factor Transformers

K-factor transformers are designed to withstand harmonic currents. The K-factor indicates the transformer's ability to handle harmonics:

K-FactorApplication
K-4Light loads, lighting
K-13Mixed loads, offices
K-20Heavy loads, data centers
K-30Very heavy loads, industrial

Rule 26-250 — Transformers and Harmonics

The CE Code requires that the neutral of 3-phase, 4-wire systems supplying non-linear loads be sized to handle the resulting neutral current. In some cases, the neutral must be full size or even oversized.


Testing and Commissioning

Tests Prior to Energization

149.Insulation test (megohmmeter): measure the insulation resistance between each winding and ground, and between primary and secondary
Typical minimum value: 1 MΩ per kV of rated voltage
151.Transformation ratio test: verify the V₁/V₂ ratio at no load
152.Polarity test: verify terminal markings
153.Continuity test: verify winding continuity
154.Ohmic resistance test: measure winding resistance

Energized Tests

Applied voltage test: apply 2 × V_rated + 1,000 V for 60 seconds
Induction test: apply a voltage higher than rated to verify inter-turn insulation

Energization Procedure

159.Verify all connections and terminal torque
160.Verify grounding of the enclosure and neutral
161.Remove temporary grounds
162.Close the primary disconnecting means
163.Measure secondary voltages (should be balanced within ± 2%)
164.Verify phase rotation
165.Apply load gradually

Pitfalls to Avoid

168.Confusing transformation ratios: N₁/N₂ = V₁/V₂, but I₁/I₂ = N₂/N₁. Always check the orientation of the ratio.
169.Forgetting the √3 factor in three-phase calculations: S = √3 × V × I, not V × I.
170.Opening the secondary of a CT under load: this creates an extremely dangerous voltage. The secondary of a CT must always be short-circuited before removing any device.
171.Neglecting correction factors: temperature and grouping change conductor ampacity. A conductor properly sized at 30 °C may be inadequate at 40 °C.
172.Confusing connection configurations: a Δ-Y transformer has a 30° phase shift, not 0°. Check the clock hour index.
173.Forgetting the 125% rule for primary transformer protection (Rule 26-248).
174.Using the wrong demand factor according to Rule 8-200. Factors vary by load type.
175.Ignoring harmonics when sizing the neutral. Non-linear loads can produce neutral currents higher than phase currents.
176.Not checking polarity when paralleling transformers. Opposite polarities create a short circuit.
177.Confusing copper and iron losses: copper losses vary with the square of the current; iron losses are constant at fixed voltage.

Summary

The transformation ratio is fundamental: a = N₁/N₂ = V₁/V₂ = I₂/I₁
Three-phase transformers use Y-Y, Δ-Y, Y-Δ, Δ-Δ connections, each with its advantages and disadvantages
Three-phase power is calculated using S = √3 × V × I
The Canadian Electrical Code, Part I imposes specific rules for protection (Rule 26-248), grounding (Section 10), and load calculations (Rule 8-200)
CTs must never be opened under load; PTs must be protected by fuses
Harmonics require K-factor transformers and oversized neutrals
Commissioning tests include: insulation, transformation ratio, polarity, continuity, and applied voltage
Correction factors (temperature, grouping) are essential for conductor sizing

Review Questions

190.A single-phase transformer rated 25 kVA, 600 V/120 V, supplies a load of 18 kW with a power factor of 0.85. What is the current on the secondary?
191.A three-phase transformer rated 500 kVA, 13.8 kV/600 V, Δ-Y connection, is protected on the primary. What is the minimum size of the protective device according to Rule 26-248?
192.What is the fundamental difference between a current transformer and a potential transformer?
193.A distribution transformer supplies a data center with 40% non-linear loads. What minimum K-factor do you recommend?
194.What checks do you perform before energizing a newly installed transformer?

This chapter covers the essential concepts for the Red Seal exam. Make sure you master the calculations, CE Code rules, and safety procedures before moving on to the exam.

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