3.2 Conductors, Resistance, and Voltage Drop

Why This Matters for the Texas Journeyman Exam

Voltage-drop calculations appear directly on the Calculations part of the TDLR/PSI exam and indirectly on the Knowledge part when questions ask about conductor upsizing, feeder sizing, and the NEC Informational Notes that recommend voltage-drop limits. A candidate who can quickly pull a resistance value from NEC Chapter 9, Table 8, plug it into the correct formula, and compare the result to 3% or 5% will pick up several points that candidates who guess on conductor size will miss.

Conductor Resistance — NEC Chapter 9, Table 8

NEC Chapter 9, Table 8 ("Conductor Properties") lists the cross-sectional area, stranding, and DC resistance at 75 °C for copper (coated and uncoated) and aluminum conductors. The resistance column is given in ohms per 1,000 feet (Ω/kft) and is the value you plug into the voltage-drop formulas.

Selected copper resistances at 75 °C (Ω per 1,000 ft):

Two patterns to remember:

1. Larger conductors have lower resistance. Resistance is inversely proportional to cross-sectional area; doubling the kcmil halves the resistance.
2. Aluminum has ~1.6× the resistance of copper at the same size. To match copper voltage drop, aluminum usually needs to be one AWG size larger.

Voltage-Drop Formulas

The exam uses two formulas, depending on the phase configuration. L is the one-way length of the circuit (in feet), I is the load current (in amperes), and R is the Table 8 resistance (in Ω per 1,000 ft).

The 1.732 factor (√3) reflects that in a balanced three-phase load, current flows through two conductors at any instant — the vector sum of the two phase currents is √3 times one phase current. The 2 in the single-phase formula is the round-trip factor: current travels out on one conductor and returns on the other, so the total conductor length is twice the one-way run.

Formula Derivation (Intuition)

Ohm's Law says the voltage dropped on one conductor is I × R_conductor. For a single-phase circuit the current uses two conductors (hot and neutral, or hot and hot), so the total loop drop is 2 × I × R. For three-phase, the vector combination of two energized conductors yields √3 ≈ 1.732 times the single-conductor drop. The ÷ 1,000 converts the per-1,000-ft Table 8 value to per-foot resistance.

NEC Voltage-Drop Limits — Informational Notes

The NEC states voltage-drop guidance in Informational Notes, which are not mandatory Code but describe best practice and are routinely tested:

• NEC 210.19(A) Informational Note No. 4 — Branch-circuit conductors should be sized so that the voltage drop does not exceed 3%, and the combined voltage drop of feeder plus branch circuit should not exceed 5%.
• NEC 215.2(A) Informational Note No. 2 — Feeder conductors should be sized so that the voltage drop does not exceed 3%, with the same 5% combined recommendation.

Reasonable design practice (and exam convention) is to treat 3% as the target for an individual feeder or branch circuit and 5% as the maximum for the total run from service to load.

Worked Example — Size Conductor for 250 A at 200 ft (Three-Phase, 480 V, Copper)

Problem: A 250 A continuous load is supplied by a 480 V three-phase feeder, 200 ft one-way, copper conductors. Size the conductor so voltage drop stays within 3%.

Step 1 — Find the maximum allowable voltage drop

Max VD = 480 V × 0.03 = 14.4 V

Step 2 — Solve for the maximum allowable resistance

Rearrange the three-phase formula:

R = VD × 1,000 ÷ (1.732 × I × L)

R = 14.4 × 1,000 ÷ (1.732 × 250 × 200)

R = 14,400 ÷ 86,600 = 0.1663 Ω/kft

Any copper conductor with Table 8 resistance ≤ 0.1663 Ω/kft satisfies the voltage-drop limit.

Step 3 — Check ampacity (NEC Table 310.16)

Before accepting a conductor based on voltage drop alone, verify it can carry the load. From NEC Table 310.16 (75 °C column, copper):

• 4/0 AWG Cu = 230 A — not enough for a 250 A load.
• 250 kcmil Cu = 255 A — meets 250 A.

Step 4 — Verify voltage drop with 250 kcmil

From Chapter 9 Table 8, 250 kcmil Cu R = 0.0515 Ω/kft.

VD = 1.732 × 250 × 0.0515 × 200 ÷ 1,000

VD = 1.732 × 250 × 0.0103 = 4.46 V

%VD = 4.46 ÷ 480 = 0.93% — well within 3%.

Step 5 — Conclusion

250 kcmil copper satisfies both the ampacity requirement (255 A ≥ 250 A) and the voltage-drop recommendation (0.93% < 3%). At 200 ft, ampacity governs; voltage drop is not the limiting factor.

What Happens at Longer Distances?

At 500 ft with 250 kcmil: VD = 1.732 × 250 × 0.0515 × 0.5 = 11.15 V → 2.32% (still under 3%).

At 700 ft with 250 kcmil: VD = 1.732 × 250 × 0.0515 × 0.7 = 15.6 V → 3.25% (exceeds 3%). Upsize to 300 kcmil (R = 0.0429): VD = 1.732 × 250 × 0.0429 × 0.7 = 13.0 V → 2.71% (within 3%).

This is the pattern the exam tests: compute the drop, compare to 3% (or 5%), and upsize if needed.

Worked Example — Single-Phase Voltage Drop

A 120 V, 15 A branch circuit runs 100 ft one-way using #12 copper (R = 1.93 Ω/kft). Find the voltage drop and percentage.

VD = 2 × 15 × 1.93 × 100 ÷ 1,000 = 5.79 V

%VD = 5.79 ÷ 120 = 4.83% — exceeds the 3% branch-circuit recommendation.

To bring it under 3% (max 3.6 V), solve for R: R = 3.6 × 1,000 ÷ (2 × 15 × 100) = 1.2 Ω/kft. #10 Cu (R = 1.21) is just over; #8 Cu (R = 0.764) gives VD = 2 × 15 × 0.764 × 0.1 = 2.29 V → 1.91%. So upsizing from #12 to #8 brings a 100 ft, 15 A circuit into compliance.

Exam Tips for Voltage-Drop Questions

1. Identify the phase configuration first. Three-phase → 1.732; single-phase → 2. Mixing these up is the most common calculation error.
2. Use the one-way length, not round-trip. The formula factor already accounts for return path.
3. Pull R from Chapter 9 Table 8, not Table 310.16. Table 310.16 gives ampacity, not resistance.
4. Check ampacity first, then voltage drop. If a conductor is too small for the load, voltage drop doesn't matter.
5. Remember the limits are Informational Notes. A 3.2% drop is not a Code violation — but the exam treats 3% as the design threshold.
6. Aluminum runs need ~1.6× the copper area, so go one size up to match.