5.3 Service & Feeder Sizing Calculations

Pulling It Together: Article 220 into 230 and 215

The exam will give you a list of loads and ask for the minimum service or feeder size. The workflow is always:

1. Calculate the total load using Article 220 standard method (or optional method 220.82 for dwellings 100 A+).
2. Apply the 125% continuous-load factor where applicable (230.42 / 215.2).
3. Select the conductor from Table 310.16 using the correct temperature column (110.14(C)(1)).
4. Select the OCPD and disconnect from 240.6 standard sizes, not less than 230.79 minimums.
5. Size the neutral per 220.61 with permitted reductions.
6. Add the largest motor contribution per 430.62 (25% of FLC).

Dwelling Load Calculation (Standard Method, Article 220 Part III)

Step 1: General Lighting

General lighting = 3 VA per sq ft for dwellings (Table 220.12). Add the small-appliance circuits (2 × 1500 VA = 3000 VA per 220.52(A)) and the laundry circuit (1500 VA per 220.52(B)).

Apply Table 220.42 demand factors to the combined general lighting + small-appliance + laundry total:

Step 2: Appliances

• Electric range (220.55, Table 220.55): Use the demand factors in Table 220.55. For a single range rated 12 kW or less, the demand is 8 kW (Column C).
• Clothes dryer (220.54): Use 5,000 VA or the nameplate, whichever is larger. With 5+ dryers in a building, Table 220.54 demand factors apply.
• Fastened-in-place appliances (220.53): Four or more appliances rated 1/4 hp or 500 W+ may use a 75% demand factor on the total connected load. Excludes ranges, dryers, space heating, and A/C.
• Space heating (220.51): 100% connected, no demand factor.
• A/C vs heating (220.50): Use the larger of A/C or heating load, not both.

Step 3: Neutral Sizing (220.61)

The neutral carries the maximum unbalanced load between the neutral and any one ungrounded conductor. Line-to-line 240 V loads do not count toward the neutral.

Permitted reductions:

• 220.61(B)(1): Ranges, wall ovens, cooktops, and clothes dryers may use 70% of the demand load for the neutral.
• 220.61(B)(2): The portion of unbalanced load exceeding 200 A may be reduced to 70% for 3-wire single-phase or 4-wire 3-phase systems.

Prohibited reductions (220.61(C)):

• 3-wire circuits from 4-wire wye systems (no neutral reduction).
• Nonlinear loads on 3-phase systems (harmonic content).

Step 4: Largest Motor (430.62)

The service or feeder conductor ampacity must include 25% of the largest motor full-load current in addition to the calculated load. This accounts for motor inrush without requiring every motor to be fully sized.

Worked Example: Single-Family Dwelling Service

A 2,400 sq ft dwelling has: 12 kW electric range, 5 kW clothes dryer, 4 fastened-in-place appliances totaling 9,000 VA (dishwasher, disposal, water heater, furnace blower), a 5 hp 240 V air-conditioner motor with 24 A FLC, and electric space heating of 10,000 VA.

General Lighting Load

• 2,400 sq ft × 3 VA = 7,200 VA
• Small-appliance: 2 × 1,500 = 3,000 VA
• Laundry: 1,500 VA
• Subtotal: 11,700 VA

Apply Table 220.42:

• First 3,000 VA × 100% = 3,000 VA
• Remaining 8,700 VA × 35% = 3,045 VA
• Demand lighting load: 6,045 VA ≈ 6.05 kVA

Appliances

• Range: 12 kW ≤ 12 kW, use 8 kW (Table 220.55 Column C).
• Dryer: 5,000 VA (220.54 minimum).
• Fastened-in-place: 4 appliances × 9,000 VA × 75% (220.53) = 6,750 VA.
• Space heating: 10,000 VA (220.51, no demand factor).
• A/C vs heating: heating 10,000 VA > A/C 24 A × 240 V = 5,760 VA. Use 10,000 VA.

Total Calculated Load

6,045 + 8,000 + 5,000 + 6,750 + 10,000 = 35,795 VA ≈ 35.8 kVA ≈ 149 A at 240 V.

Add largest motor (430.62): 24 A × 240 V × 0.25 = 1,440 VA. New total = 37,235 VA ≈ 155 A.

Service Conductor and Disconnect

• 230.79(C): One-family dwelling minimum is 100 A; calculated load is 155 A. Use a 175 A or 200 A disconnect. Choose 200 A.
• 230.42(A): Conductor ampacity must be not less than 200 A. From Table 310.16, 75 °C column: 2/0 AWG copper at 175 A (too small); 3/0 AWG copper at 200 A. 3/0 AWG Cu or 4/0 AWG aluminum.
• OCPD: 200 A main breaker (240.6).

Neutral Sizing (220.61)

Unbalanced load = general lighting + range (8 kW × 70%) + dryer (5 kW × 70%) + fastened appliances + largest motor.

6,045 + (8,000 × 0.70) + (5,000 × 0.70) + 6,750 + 1,440 = 6,045 + 5,600 + 3,500 + 6,750 + 1,440 = 23,335 VA ≈ 97 A at 240 V.

From Table 310.16, 75 °C column: 4 AWG copper at 85 A (too small); 3 AWG copper at 100 A. Use 3 AWG Cu neutral (or 2 AWG Al).

Exam Traps

• 220.42 first 3000 VA: The 100% tier applies to the combined lighting + small-appliance + laundry load, not just lighting. A common error is applying 35% to the entire subtotal.
• Table 220.55 Column C: For one range ≤ 12 kW, the demand is 8 kW, not the nameplate. The nameplate only applies if it exceeds 12 kW; then add 5% per kW above 12 kW (Column A or B).
• A/C vs heating: Compare at 100%; do not add both. The 65% heating factor only applies in the optional method (220.82).
• 430.62 motor add: Applies to the service or feeder conductor, not to branch-circuit conductors.
• 230.79(A) vs (C): For a single dwelling, the 100 A 3-wire minimum in (C) overrides the 60 A "all other" minimum in (D).
• Neutral reduction: 70% for ranges/dryers applies to the demand load, not the nameplate; the 200 A excess reduction is separate.