Guide to calculating the electrical load for a low-voltage electrical system

The first step is to determine all the electrical devices that will be connected to the system. For each device, you need to know the power rating, expressed in watts (W) or VA (volt-amperes). This information is usually stated on the product label or in the data sheet.

For example, in a low-voltage lighting system there may be:

• 5 LED lamps of 10 W each.
• 2 halogen spotlights of 50 W each.
• 1 pump for a 100 W fountain.

To calculate the total load in watts, simply add up the power ratings of all devices. Using the example above:

• Total power of LED lamps = 5 × 10 W = 50 W.
• Total power of halogen spotlights = 2 × 50 W = 100 W.
• Pump power = 100 W

Total load = 50 W + 100 W + 100 W = 250 W.

The power factor (cosφ) represents the efficiency with which a load uses electrical energy. For purely resistive loads, such as incandescent lamps or high-quality LEDs, the power factor is close to 1. However, for inductive loads (e.g. motors, transformers or pumps), the power factor can be lower, typically between 0.7 and 0.9.

If the power factor of the devices is not specified, it is prudent to assume an average value, e.g. cosφ = 0.8 for inductive loads.

To convert apparent power to VA (volt-ampere), the formula is used:

Apparent power (VA) = Active power (W) ÷ cosφ

Assuming that the system contains only loads with cosφ = 0.8:

• Apparent power = 250 W ÷ 0.8 = 312.5 VA.

The total current drawn by the system can be calculated by knowing the total electrical load and supply voltage. The formula is:

I = P ÷ V

Where:

• I = current in amperes (A).
• P = total power in watts (W).
• V = supply voltage in volts (V).

For a low-voltage system with a supply voltage of 12 V:

I = 250 W ÷ 12 V = 20.83 A.

If apparent power (VA) is used, the formula becomes:

I = VA ÷ V

In this case:

I = 312.5 VA ÷ 12 V = 26.04 A.

Once the total current has been calculated, you need to size the cables based on:

• Rated Current (A): The cables must be able to withstand the total current without overheating.
• Cable Length (m): The longer the cable, the greater the voltage drop, which must remain within acceptable limits (generally no more than 3% of the total voltage).
• Cable cross-section (mm²): The minimum cross-section can be determined by consulting cable capacity tables as a function of current and length.

For example, for a current of around 26 A, a copper cable with a cross-section of 4 mm² might be sufficient for a length of up to 15 metres. If the length is longer, you may need to increase the cross-section to 6 mm² or more.

The voltage drop can be calculated with the formula:

ΔV = (2 × ρ × I × L) ÷ A

Where:

• ΔV = voltage drop in volts.
• ρ = resistivity of the cable material (copper ≈ 0.0178 Ω·mm²/m).
• I = current in amperes (A).
• L = length of the cable in meters (round trip).
• A = cable cross-section in mm².

For example, for a copper cable with:

• Current = 26 A
• Length = 10 meters (20 meters round trip).
• Cross-section = 4 mm².

ΔV = (2 × 0.0178 × 26 × 20) ÷ 4 = 4.63 V.

The voltage drop is high compared to the nominal 12 V (greater than 3%). Then, you need to increase the cable cross-section to reduce ΔV.

Circuit breakers or fuses must be sized to protect the system from overloads and short circuits. The rated current of the protective device must be at least 125% of the total calculated current.

In our example:

• Total current = 26 A.
• Minimum rated current of the protective device = 26 × 1.25 ≈ 32.5 A.

You could choose a 32A circuit breaker.

After calculating the electrical load, verify that:

1. The total current does not exceed the capacities of the cables and protective devices.
2. The voltage drop is acceptable.
3. Safety margins have been included for any future expansion of the plant.

A low-voltage system powers:

• 10 x 8 W LED lamps (total: 80 W).
• 1 x 120W pump

Voltage = 12 V, cosφ = 0.9.

1. Total load: 80 + 120 = 200 W
2. Apparent power: 200 ÷ 0.9 = 222.2 VA.
3. Total current: 222.2 ÷ 12 = 18.5 A.
4. Cables: Minimum cross-section 2.5 mm², but increase to 4 mm² if the length exceeds 10 meters.
5. Protection device: 25 A circuit breaker

By following these calculations, a safe and efficient system can be designed for lighting or other low-voltage loads.