Load Calculation
How to calculate the total current demand of a circuit to correctly size wire, fuses, and supply capacity.
Why This Matters
Every wire, fuse, and power source has a maximum current it can carry before damage. Wire that is too small for its load heats up — insulation melts, fires start, and circuits fail silently inside walls. A fuse or breaker that is too large allows dangerous overloads to persist unchecked.
Load calculation is the arithmetic that ties together your power source capacity, your wire sizes, your overcurrent protection, and your connected equipment. Done correctly before wiring begins, it prevents the most common and dangerous electrical mistakes. Done incorrectly — or not at all — it produces installations that work initially but fail catastrophically under full load.
In a rebuilding context where you have one generator or one battery bank supplying multiple critical loads, getting this right matters even more. You cannot afford to rewire after a failure.
Fundamental Relationships
All load calculations rest on Ohm’s Law and the power equation:
| Formula | Meaning |
|---|---|
| V = I × R | Voltage = Current × Resistance |
| P = V × I | Power (watts) = Voltage × Current |
| I = P ÷ V | Current (amps) = Power ÷ Voltage |
Example: A 100-watt lamp on a 12 V system draws: I = 100 ÷ 12 = 8.3 A
The same lamp on a 120 V system draws: I = 100 ÷ 120 = 0.83 A
Higher voltage systems carry the same power at much lower current — this is why long-distance distribution uses high voltage.
Step 1: List All Loads
Inventory every device that will connect to the circuit:
| Device | Watts | Voltage | Current (I = P/V) |
|---|---|---|---|
| LED light × 4 | 10 W each | 12 V | 3.3 A |
| Water pump | 150 W | 12 V | 12.5 A |
| Radio/communications | 20 W | 12 V | 1.7 A |
| Battery charger | 60 W | 12 V | 5.0 A |
| Total | 300 W | — | 22.5 A |
Motor Starting Current
Electric motors draw 3–7× their rated running current for 0.5–2 seconds at startup. A pump rated at 12.5 A running current may draw 50–75 A at startup. Size fuses and wire for the starting surge, or use a slow-blow fuse rated for running current but tolerant of brief surges.
Step 2: Apply a Demand Factor
Not all loads operate simultaneously. A demand factor accounts for this:
- Critical infrastructure (lights, communications): 100% demand — always assume on
- Motors, pumps: 100% of largest motor + 50% of remaining motors
- General loads: 80% of total if realistic simultaneous use is lower
Conservative approach for a rebuilding context: use 100% of all loads. Oversizing capacity is almost always the right choice when reliability matters.
Step 3: Add a Safety Margin
Multiply total calculated current by 1.25 (25% headroom). This accounts for:
- Inaccurate nameplate ratings
- Aging equipment drawing more current
- Future load additions
- Voltage drop that increases current at lower voltages
Example continued: 22.5 A × 1.25 = 28.1 A minimum circuit capacity required
Round up to the next standard fuse size: 30 A
Step 4: Verify Supply Capacity
Your power source must supply the total load plus margin:
- Generator: Check rated continuous watts, not peak. A 500 W generator can sustain 500 W continuously; peak ratings are short-duration.
- Battery bank: Check amp-hour (Ah) capacity and maximum discharge rate (C-rate). A 100 Ah battery at 12 V with a 1C maximum discharge rate can deliver 100 A for 1 hour — but repeated deep discharges degrade it. For reliable operation, draw no more than 20 A from a 100 Ah battery (C/5 rate).
- Solar array: Rated watts at peak sun. In practice, plan on 4–5 peak sun hours per day in temperate climates.
Step 5: Calculate Voltage Drop
Long wire runs lose voltage due to wire resistance. Excessive voltage drop causes:
- Lights to dim
- Motors to run hot and underpowered
- Electronics to malfunction
Voltage drop formula: V_drop = I × R_wire
Where R_wire = (wire resistance per meter) × (2 × run length in meters) — factor of 2 for the return conductor.
Maximum acceptable voltage drop: 3% for lighting, 5% for motors and heating loads.
Example: 20 A load, 10 m run, 4 mm² copper wire (resistance 4.61 mΩ/m): V_drop = 20 × (4.61 × 10⁻³ × 20) = 20 × 0.092 = 1.84 V drop on a 12 V system = 15.3% — unacceptable.
Solution: use 16 mm² wire, reducing V_drop to 0.46 V (3.8%) — acceptable.
Common Load Calculation Mistakes
| Mistake | Consequence | Fix |
|---|---|---|
| Ignoring motor starting current | Fuse blows at startup, or fuse is oversized and allows dangerous overload | Calculate starting current; use slow-blow fuse |
| Forgetting return conductor in voltage drop calc | Calculated drop is half actual | Always multiply wire length by 2 |
| Using peak load as continuous load | Wire and fuse sized correctly but supply is undersized | Distinguish continuous from peak |
| No safety margin | Any load addition overloads circuit | Always design to 80% of capacity maximum |
| Ignoring parallel loads on shared supply | Individual circuits correct but supply overloaded | Sum all circuit loads before sizing supply |
Quick Reference: Current at Common Voltages
| Power (W) | 12 V | 24 V | 120 V | 240 V |
|---|---|---|---|---|
| 50 W | 4.2 A | 2.1 A | 0.4 A | 0.2 A |
| 100 W | 8.3 A | 4.2 A | 0.8 A | 0.4 A |
| 500 W | 41.7 A | 20.8 A | 4.2 A | 2.1 A |
| 1,000 W | 83.3 A | 41.7 A | 8.3 A | 4.2 A |
This table makes immediately clear why 12 V systems require much heavier wire than 120 V systems for the same power. Every time you double the voltage, you halve the current and can use wire four times lighter by cross-section.