Residential Electrical Load Estimating worksheet for determining electrical panel size based on 2013 California Electrical Code.

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This form is used to calculate the required electrical service size (in watts or amps) for a residential dwelling with specific appliances and lighting loads, based on the 2013 National Electrical Code (NEC) guidelines.

Let’s walk through how to complete this worksheet using typical values. Since no actual numbers are filled in, I’ll use example values to demonstrate the process. You can substitute your own data accordingly.

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🔧 Step-by-Step Solution

#### 1. Appliance Load (kitchen range)
Assume: A 12 kW electric range
- Answer: 12,000 watts
> *Note: The NEC allows a demand factor for ranges (e.g., 80% for larger ranges), but for simplicity, we often use full nameplate unless specified.*

#### 2. 20-amp small appliance circuits (2 x 1500 watts each)
- Each circuit = 1500 watts
- Two circuits: 1500 × 2 = 3000 watts
- Answer: 3000 watts

> *These are typically used for kitchen outlets (not fixed appliances).*

#### 3. Laundry circuit (1500 watts each)
- One laundry circuit = 1500 watts
- Answer: 1500 watts

#### 4. Electrical appliances at nameplate value
List common household appliances and their wattage:

| Appliance | Watts |
|------------------|-------|
| Range | 12,000 |
| Garbage Disposal | 500 |
| Clothes Dryer | 5000 |
| Dishwasher | 1200 |
| Oven | 3000 |
| Other (e.g., fridge) | 1000 |

> *Total = 12,000 + 500 + 5000 + 1200 + 3000 + 1000 = 22,700 watts*

#### 5. First 3,000 watts at 100%
- From above, total load so far: 12,000 (range) + 3,000 (small appliances) + 1,500 (laundry) + 22,700 (other appliances) = 39,200 watts

But wait — the first 3,000 watts are taken at 100%, then additional watts are at 40% (per NEC Table 220.52).

So:
- First 3,000 watts → 3,000 watts
- Remaining = 39,200 - 3,000 = 36,200 watts → apply 40% demand factor
- 36,200 × 0.40 = 14,480 watts
- Subtotal = 3,000 + 14,480 = 17,480 watts

> *Wait — actually, the "first 3,000" applies to the general lighting and receptacle load, not all appliances. Let's restructure properly.*

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✔ Correct Approach Based on NEC 220.52

We need to separate:

#### 🔹 General Lighting & Receptacles (from #1–3)
- Lighting: 3 watts per sq ft (assume 2000 sq ft):
→ 2000 × 3 = 6,000 watts
- Small appliance circuits (2 × 1500) = 3,000 watts
- Laundry circuit = 1,500 watts

> Total general load = 6,000 + 3,000 + 1,500 = 10,500 watts

Now apply demand factors from Table 220.52:
- First 3,000 watts at 100%
- Remaining (10,500 – 3,000 = 7,500) at 35%

→ 3,000 × 1.00 = 3,000
→ 7,500 × 0.35 = 2,625
→ General load subtotal = 3,000 + 2,625 = 5,625 watts

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#### 🔹 Fixed Appliances (from #4)
Sum up nameplate values:
- Range: 12,000
- Garbage disposal: 500
- Clothes dryer: 5,000
- Dishwasher: 1,200
- Oven: 3,000
- Refrigerator: 1,000
- Other: 1,000
→ Total = 23,700 watts

> These are added at 100% (no demand factor unless otherwise specified)

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#### 🔹 Subtotal (Lines 1–4)
= General load (5,625) + Appliances (23,700) = 29,325 watts

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#### 🔹 Line 5: First 3,000 watts at 100%, remainder at 40%?
Wait — this line refers to the “household” load. Actually, Line 5 is part of the appliance demand calculation.

From NEC Table 220.52, for household electric ranges, we use demand factors based on nameplate.

For example:
- If range is 12 kW, demand factor is 80% (for 12–15 kW)
→ 12,000 × 0.80 = 9,600 watts

Similarly, clothes dryer: 5,000 watts → 100% (no reduction)

But Line 5 says: "First 3,000 watts at 100%; remaining at 40%"

This applies to the total of all appliances, including range, dryer, etc.

So:
- Total appliances: 23,700 watts
- First 3,000 watts → 3,000 × 1.00 = 3,000
- Remaining: 20,700 × 0.40 = 8,280
→ Appliance demand = 3,000 + 8,280 = 11,280 watts

> But wait — the range already has its own demand factor, so we must be careful.

Actually, NEC Table 220.52 gives specific demand factors for appliances. Here’s the correct way:

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✔ Correct Method Using NEC 220.52

#### Step 1: General Lighting and Receptacles
- 3 watts/sq ft × 2000 sq ft = 6,000 watts
- Small appliances: 2 × 1500 = 3,000
- Laundry: 1,500
- Total = 10,500 watts

Apply demand factor:
- First 3,000 watts → 100% = 3,000
- Remaining 7,500 → 35% = 2,625
→ General load = 5,625 watts

#### Step 2: Kitchen Range (12 kW)
- Use Table 220.52: For 12 kW range, demand factor = 80%
→ 12,000 × 0.80 = 9,600 watts

#### Step 3: Clothes Dryer
- 5,000 watts → 100% (no demand factor)
→ 5,000 watts

#### Step 4: Other Appliances
- Dishwasher: 1,200 → 100% = 1,200
- Oven: 3,000 → 100% = 3,000
- Refrigerator: 1,000 → 100% = 1,000
- Other: 1,000 → 100% = 1,000

Total other appliances = 1,200 + 3,000 + 1,000 + 1,000 = 6,200 watts

> Now, apply demand factor for all other appliances:
- First 3,000 watts → 100%
- Remaining → 40%

So:
- First 3,000 → 3,000
- Remaining 3,200 → 3,200 × 0.40 = 1,280
→ Other appliances demand = 3,000 + 1,280 = 4,280 watts

#### Step 5: Total Load (Line 7)
- General lighting/receptacles: 5,625
- Range: 9,600
- Dryer: 5,000
- Other appliances: 4,280
→ Total = 5,625 + 9,600 + 5,000 + 4,280 = 24,505 watts

#### Step 6: Convert to Amps
- Voltage = 240 V (assumed)
- Current = Watts / Volts = 24,505 / 240 ≈ 102.1 amps

> So, required service size = 100 amp or 125 amp (standard sizes)

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📝 Final Answer (Example):

| Item | Watts |
|------|-------|
| 1. Lighting (3 W/sq ft × 2000) | 6,000 |
| 2. Small appliances (2×1500) | 3,000 |
| 3. Laundry | 1,500 |
| 4. Appliances: | |
| - Range (12 kW @ 80%) | 9,600 |
| - Dryer | 5,000 |
| - Dishwasher | 1,200 |
| - Oven | 3,000 |
| - Fridge | 1,000 |
| - Other | 1,000 |
| 5. Demand: First 3,000 @ 100%, rest @ 40% (other appliances) | 4,280 |
| 6. Total (before demand) | 24,505 |
| 7. Total (after demand) | 24,505 watts |
| 8. Amps = 24,505 / 240 = 102.1 amps → Use 125-amp service |

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✔ Summary

- Final calculated load: ~24,500 watts
- Service size: 125 amps (240V)

> This is a simplified version. Actual calculations depend on exact appliance ratings and building size.

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If you can provide the actual values from your image (like square footage, appliance wattages, etc.), I can give you an exact solution tailored to your case.