NEC Conductor Ampacity and Derating Practice Questions

The code book works this exact problem in Annex D, Example D3(a), and it says the method in one sentence: the conductors must independently meet the requirements for (1) terminations and (2) conditions of use throughout the raceway run. Do both, then take whichever is worse.

The correction and adjustment factors are applied to the conductor's own temperature rating. The wire is THHN, so that is the 90 °C column of Table 310.16:

125 °F falls in the 123–131 °F row → 90 °C factor = 0.76. (The tables assume 86 °F, so a hot space derates you.)

Six current-carrying conductors is the 4–6 band → 0.80.

The lugs are 75 °C, so the terminations are judged on the 75 °C column: #10 = 35 A (110.14(C)). That is the second of the two independent checks — it is not something you derate.

Annex D, Example D3(a) sizes a feeder exactly this way: the termination is taken from the 75 °C column, the raceway run is worked from the 90 °C column with the factors applied, and the book then says “the worst case is given by the raceway conditions.” It even checks the finished conductor the same way — “195 A × 0.96 × 0.7 = 131 A.”

1. Derating the 75 °C column. That gives 35 × 0.67 × 0.80 = 19 A. The 75 °C column is the termination check, not the starting point for the factors.
2. Doing only one of the two checks. They are independent — sometimes the terminations win, sometimes the raceway does.
3. Applying one factor, rounding, then the other. Apply both to the base at the same time.
4. Counting every wire. Count current-carrying conductors only; grounds never count (310.15(F)).

Multiwire branch circuit — neutral

The shared neutral carries only the difference between the two hot legs: Neutral = | Leg A − Leg B |. Balanced legs cancel to zero; the leftover unbalance is what flows.

1. Adding the legs — the neutral carries the difference, not the sum.
2. Forgetting that a balanced load gives 0 A on the neutral.
3. Fusing or switching a shared neutral — never do it.

On a single-phase 3-wire system, the neutral carries the difference between the two legs.

Multiwire branch circuit — balanced neutral

The shared neutral carries only the difference between the two hot legs: Neutral = | Leg A − Leg B |. Balanced legs cancel to zero; the leftover unbalance is what flows.

1. Adding the legs — the neutral carries the difference, not the sum.
2. Forgetting that a balanced load gives 0 A on the neutral.
3. Fusing or switching a shared neutral — never do it.

On a single-phase, 3-wire (multiwire) circuit the two hot legs share one neutral. The neutral only carries the difference between the legs — the part that isn't balanced.

Picture the two hots pushing current in opposite directions through the neutral. Equal legs cancel to zero; the leftover (the unbalance) is what flows on the neutral. Balanced 40/40 → 0 A; 40/30 → 10 A.

1. Adding the legs (40 + 30 = 70). The neutral carries the difference, not the sum.
2. Forgetting a balanced load gives a 0 A neutral.
3. This is why you never fuse or switch a shared neutral — opening it can overload the branch.

The terminal rating sets which Table 310.16 column you use (110.14(C)). The terminals are rated 75 °C.

110.14(C) · Table 310.16 (75 °C column)

Use the column that matches the terminal rating (110.14(C)) — 60 °C or 75 °C for most gear, never the 90 °C column for the final pick. Go down that column to the first conductor whose ampacity meets or beats the load.

1. Reading the 90 °C column for the final size — that column is only for derating math.
2. Ignoring the terminal rating — the lug, not the insulation, sets the column.
3. Picking a wire rated below the required amps.

Ambient correction adjusts a conductor's ampacity when the surrounding air is hotter or cooler than the 30 °C (86 °F) base.

310.15(B)(1) / Table 310.15(B)(1)(1) — ambient temperature correction

Start at the conductor's 90 °C ampacity (Table 310.16), then multiply by the ambient-temperature correction (Table 310.15(B)(1)) and the more-than-3-conductors adjustment (Table 310.15(C)(1)). The final result still cannot exceed the terminal column's ampacity.

1. Starting the math from the 75 °C ampacity instead of 90 °C.
2. Applying only one factor when both ambient and fill apply.
3. Counting the neutral or ground as current-carrying when it is not.

When more than 3 current-carrying conductors share a raceway, reduce ampacity per Table 310.15(C)(1).

310.15(C)(1) — conductor fill adjustment

Start at the conductor's 90 °C ampacity (Table 310.16), then multiply by the ambient-temperature correction (Table 310.15(B)(1)) and the more-than-3-conductors adjustment (Table 310.15(C)(1)). The final result still cannot exceed the terminal column's ampacity.

1. Starting the math from the 75 °C ampacity instead of 90 °C.
2. Applying only one factor when both ambient and fill apply.
3. Counting the neutral or ground as current-carrying when it is not.

Apply BOTH adjustments - ambient heat and conductor fill.

310.15(B)(1) & 310.15(C)(1)

Start at the conductor's 90 °C ampacity (Table 310.16), then multiply by the ambient-temperature correction (Table 310.15(B)(1)) and the more-than-3-conductors adjustment (Table 310.15(C)(1)). The final result still cannot exceed the terminal column's ampacity.

1. Starting the math from the 75 °C ampacity instead of 90 °C.
2. Applying only one factor when both ambient and fill apply.
3. Counting the neutral or ground as current-carrying when it is not.

≈ 40 A.

NEC 310.15(B) & (C)

Start at the conductor's 90 °C ampacity (Table 310.16), then multiply by the ambient-temperature correction (Table 310.15(B)(1)) and the more-than-3-conductors adjustment (Table 310.15(C)(1)). The final result still cannot exceed the terminal column's ampacity.

1. Starting the math from the 75 °C ampacity instead of 90 °C.
2. Applying only one factor when both ambient and fill apply.
3. Counting the neutral or ground as current-carrying when it is not.