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When you run cables underground, buried directly in the earth without conduit, you have to size them properly so they:

  1. Carry enough current without overheating.
  2. Stay safe and code-compliant under all conditions (soil, temperature, depth, etc.).

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How We Size Cables for 3-Phase Service (Direct Burial)

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Step-by-Step:

  1. Determine the load
    • This is the total amperage the service will carry.
    • For example: 100A 3-phase service.
  2. Use CEC (Assorted Table D’s in Appendix D)
    • These tables are in the Canadian Electrical Code and give you the ampacity (current-carrying capacity) of different cable sizes.
  3. Consider soil temperature
    • Ampacity tables assume a certain ground temperature (usually 50-60°C).
    • If your soil is hotter, you may need to de-rate the ampacity (lower it).
  4. Factor in length (voltage drop)
    • Long underground runs may cause a voltage drop, meaning the voltage at the far end is lower.
    • Use voltage drop calculations (Table D3) (using cable resistance and reactance) to check that voltage stays within acceptable limits (usually 3–5%).
  5. Choose the smallest conductor that meets all these conditions
    • Use the ampacity table to find the smallest size that can safely carry the load, stay cool enough, and meet voltage drop rules.

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Why All This Matters

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  • Underground cables can’t cool off as easily as cables in open air.
  • The soil traps heat, especially if dry or compacted.
  • Overheated cables =  fire risk, insulation breakdown, early failure. Fun fact: Fire is bad.
  • Getting the size right ensures safety, long life, and efficient power delivery.

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Think of a direct burial cable like a garden hose buried in the ground. If it’s too narrow and buried deep under heavy dirt, the water (current) can’t flow properly, and pressure (heat) builds up. A wider hose (larger cable) solves the problem and keeps things safe.

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So, here's some tips for these questions:

Tables D8-D11 have both Diagrams AND Tables. All of these tables are for Voltages UNDER 5000V.

(If you want more than 5000V, you have to go to Table(s) D17+)

They’re surprisingly easy to understand. Which is a very weird thing to say about the Code book.

Each option has a “Detail”. You can see how many conductors and the configuration, and the spacing info in each detail. You then go to the Table, and find the relevant Detail.

The C of Q Exam is going to tell you exactly which Diagram and “Detail” to go to (don’t worry, you don’t have to guess!)

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So what do we do once we get there?

Well, it depends on what type of configuration we need to deal with.

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What do we do if we have 1 cable per phase?
  • The service is rated 200A, so each phase conductor must be able to safely carry at least 200A.
  • We size our conductor for 200A
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What do we do if we have 2 cables per phase?
  • We need 200A total per phase, but we are using 2 conductors per phase, so each conductor will carry half the current.
  • 200 A / 2 = 100A per conductor

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What do we do if we have 3 cables per phase?
  • 200 A / 3 = 66.7A per conductor

You get the point.

BUT (because there’s always a “but” with code….)

You have to pay attention very closely to what the question is asking/how the question is phrased.

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Sometimes Multiply, and Sometimes Divide:

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It all depends on what the question is asking:

You multiply when the question gives you the amperage per conductor and you need to find the total required ampacity.

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Example Questions

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“The service is 100A, and we’re using 3 cables per phase. What’s the total ampacity required from all cables?”

  • Since you're using 3 parallel cables, and the load is 100A per phase, you multiply:
  • 100A×3=300A total ampacity needed from the cable group

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How Do I Know That a "100A 3-Phase Service" Means 100A Per Phase?

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When a service is described as “100A 3-phase”, it means 100A per line conductor (per phase) — not shared across the 3 phases.

Because service ratings are always based on the capacity of each individual ungrounded (hot) conductor, not the total sum across all phases.

  • A 100A single-phase service = 100A on each hot leg.
  • A 100A 3-phase service = 100A on L1, L2, and L3 — each carrying up to 100A.

So when you're:

  • Sizing conductors ➝ size each phase to carry 100A
  • Calculating load ➝ assume each line is rated for 100A
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Think of It Like This:

Imagine you're wiring a 3-phase panel labeled "100A":

  • You install:
    • A 3-pole 100A breaker
    • 3 conductors (L1, L2, L3)
    • Each of those conductors must carry up to 100A.

So you must size each ungrounded conductor to handle 100A, whether it's one cable per phase, or multiple in parallel.

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Example: 1/10 Difficulty

What is the minimum size RWU 90 conductor that is required for a 800A 347V/600V three-phase service, using the IEEE installation configuration detail 2 of diagram D10? Termination Temperature is 90C.

  • It shows 2 cables per phase
  • 800A/2= 400A
  • Going to the Table: Appendix D
  • Detail 2 → 350 kcmil

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Example: 4/10 Difficulty

What is the minimum size RWU 75 conductor that is required for a 800A 347V/600V three-phase service, using the IEEE installation configuration detail 2 of diagram D10?

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  • It shows 2 cables per phase
  • 800A /2= 400A
  • Note 2 says we have to “de-rate” because the table is based on 90C.
  • 400A * 0.886= 354.4A
  • Going to Table: Appendix D
  • Detail 2 → 300 kcmil

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Example: 10/10 Difficulty

What is the minimum size RWU 75 conductor that is required for a 800A 347V/600V three-phase service with an 80% Power Factor with a length of 150m, using the IEEE installation configuration detail 2 of diagram D10 that stays within 3% voltage drop?

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  • It shows 2 cables per phase
  • 800A /2 = 400A
  • Note 2 says we have to “de-rate” because the table is based on 90C.
  • 400A * 0.886= 354.4A
  • Going to Table: Detail 2: 300 kcmil

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Numbers

  • V = 600V
  • VD = ??
  • K = 0.167
  • f = 1.73 
  • I = 177.2A
  • L = 150m

VD = K*f*I*L / 1000

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Use CEC Table D3 for K

  • 300 kcmil copper, 75°C, Cable, 80% PF
  • K = 0.167 ohms/km

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What are my Amps?

  • Adjusted Total Load (after de-rating) : 354.4A
  • There are 2 cables per phase, and we need to size for each individual conductor:
  • 354.4 A / 2 = 177.2 A

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Apply Voltage Drop Formula

  • VD = 0.167 x 1.73 x 177.2 x 150 / 1000
  • VD= 7.67V

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Calculate Percent Drop

  • VD% = 7.67 * 100 / 600
  • VD% = 1.27%

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It stays. 

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