Voltage drop: master Rule 8-102 to earn marks on the Red Seal exam

Why voltage drop is a guaranteed question for electricians

Roughly a quarter of the questions on the Red Seal (Interprovincial) electrician exam are calculations, and voltage drop is one of the most testable of them all: it follows a fixed method, uses only two formulas, and punishes the habit of sizing conductors for ampacity alone. The classic exam scenario is a long circuit where the conductor is perfectly legal under Table 2, yet the voltage arriving at the load quietly breaks Rule 8-102. Candidates who know how to check, and how to fix, the drop pick up marks that others leave on the table.

Rule 8-102: the 3% / 5% limits

Rule 8-102 of the Canadian Electrical Code, Part I, sets two limits:

• 3% maximum voltage drop in any feeder or branch circuit;
• 5% maximum total from the supply side of the consumer's service to the point of utilization.

Convert the percentages into volts before comparing: a 120 V branch circuit may drop 3.6 V; a 240 V circuit, 7.2 V; a 347 V lighting circuit, 10.4 V. The 5% totals are 6 V, 12 V and 17.35 V respectively.

The two formulas you must know cold

Both formulas use the same pieces: I = current (A), L = one-way length (m), and R = resistance per kilometre from Table D3 of Appendix D (Ω/km).

• Single phase: VD = 2 × I × L × R ÷ 1000 (the factor 2 covers the return path)
• Three phase: VD = √3 × I × L × R ÷ 1000

Divide the result by the nominal voltage and multiply by 100 to get the percentage drop.

Worked example 1 — single-phase branch circuit

A 120 V, 15 A general-purpose circuit feeds receptacles 25 m away through #12 AWG copper (R = 6.50 Ω/km at 75 °C).

VD = 2 × 15 × 25 × 6.50 ÷ 1000 = 4.88 V = 4.06% — over the 3% limit (3.6 V). The #12 is fine for ampacity (rated 20 A), yet it fails on voltage drop. Upsize to #10 AWG (R = 4.07 Ω/km): VD = 2 × 15 × 25 × 4.07 ÷ 1000 = 3.05 V = 2.54% — compliant. Rule of thumb: on a 15 A load, #12 runs out of room at roughly 18 m.

Worked example 2 — three-phase feeder

A 208 V three-phase feeder supplies a panel 100 m away with 100 A. Try #1 AWG copper (R = 0.524 Ω/km): ampacity is fine at 130 A. VD = 1.732 × 100 × 100 × 0.524 ÷ 1000 = 9.08 V = 4.36% — the feeder limit is 3% (6.24 V), so this fails. Move up to 2/0 AWG (R = 0.335 Ω/km): VD = 5.80 V = 2.79% — compliant. The trap? The conductor was never undersized for current; it was undersized for distance.

The five traps that cost marks

• Using round-trip length: L is the one-way distance; the factor 2 already accounts for the return conductor.
• Mixing units: R is in Ω/km and L in metres — feed feet into the formula and every answer is wrong.
• Wrong phase factor: 2 for single-phase, 1.732 for three-phase.
• Skipping the 125% step: for continuous loads and motor-driven equipment, the minimum conductor ampacity is 125% of the load (Rules 8-104(3)(a) and (6)(a)) — verify that first, then run the voltage-drop check with the actual circuit current.
• Comparing against the wrong number: 3% of 240 V is 7.2 V — a 6 V drop passes, an 8 V drop fails, even if the equipment "seems to work".

How to bank these marks on exam day

Voltage-drop questions are mechanical: identify the system (single or three phase), pull R from Table D3, apply the formula, compare against 3% or 5%. Work through long-run feeder and branch examples until the sequence is automatic, then practise on timed Red Seal-style simulations where the trap answers (ampacity-only sizing, one-way confusion, imperial units) are baked in. Master Rule 8-102 and a whole family of calculation marks becomes yours.