OpenSpecEngineering

Checks whether a copper or aluminum building wire can carry a load, and how much voltage it loses on the way. Check a gauge you have chosen, or let the page find the smallest gauge, up to 4/0 AWG, that meets both the ampacity and a voltage-drop limit. Electricians and designers use it to size branch circuits and feeders under the National Electrical Code (NEC) [1].

Ampacity

I_allow = min(I_table·k_T·k_N, I_term)

I_table is the ampacity from NEC Table 310.16 for the conductor’s material and insulation temperature rating. k_T is the ambient correction factor of Table 310.15(B)(1)(1), and k_N the adjustment factor of Table 310.15(C)(1) for more than three current-carrying conductors together [1]. I_term is the Table 310.16 value in the column of the terminal rating. Under NEC 110.14(C) a 90 °C wire may be derated from its 90 °C value but can’t be used above the terminals’ column [1]. In check mode, 14 to 10 AWG also show the largest breaker or fuse NEC 240.4(D) allows for them.

Voltage drop

V_d = 2·K·I·L / A

I is the load current in amperes, L the one-way length in feet and A the conductor area in circular mils from NEC Chapter 9 Table 8. K is the resistivity at 75 °C: 12.9 Ω·cmil/ft for copper and 21.2 for aluminum, from the same table [1]. The 2 counts the current going out and coming back. The percentage is V_d divided by the source voltage.

Limits and common mistakes

• The drop formula is for a single-phase or DC circuit. For a balanced three-phase circuit, multiply the drop by 0.866 (√3/2). Reactance, which the page ignores, adds to the drop on large conductors in steel conduit.
• Count only current-carrying conductors. A neutral carrying only the imbalance and the equipment ground don’t count.
• Size for 125 % of a continuous load (3 hours or more). The page uses the current you enter as it is.

Worked example

A 240 V, 40 A load sits 150 ft from its panel. The run uses copper THWN-2 in a conduit with three current-carrying wires, through a space that reaches 35 °C. The breaker and the load have 75 °C terminals. The drop must stay within 3 %.

1. Choose Mode Find minimum gauge, Conductor material Copper and Insulation temperature rating 90 °C (e.g. THHN, THWN-2, XHHW-2).
2. Enter Load current (I) 40 A, One-way run length (L) 150 ft, Source voltage (V) 240 V and Max allowable voltage drop (%) 3.
3. Choose Ambient temperature (NEC Table 310.15(B)(1)(1)) 31–35 °C (87–95 °F), Current-carrying conductors in the raceway or cable 1–3 (× 1) and Terminal temperature rating 75 °C. Press Solve. The Solution shows: Minimum gauge: 6 AWG
   Ampacity (Table 310.16, 90 °C copper): 75 A
   Ambient temperature correction factor: 0.96
   Adjustment factor for the number of conductors: 1
   Allowable ampacity (after factors, limited to the 75 °C column for the terminals): 65 A
   Voltage drop: 5.89939 V
   Voltage drop (%): 2.45808 %

Check: 75 × 0.96 = 72 A, held to the 65 A of the 75 °C column. 8 AWG would carry the current (50 A) but drops 2 × 12.9 × 40 × 150 / 16,510 = 9.38 V, or 3.9 %. The voltage drop, not the ampacity, sets the size.