Selecting a circuit breaker based on wattage is a critical step when designing the electrical wiring in your home, garage or workshop. An error in calculations can lead to two extremes: the machine will operate at the slightest load (false shutdowns) or, more dangerously, will not respond to a short circuit, causing a fire. In this article we will analyze universal calculation formula taking into account demand factors, temperature and load type, which is used by professional electricians, but we adapt it for independent use.

Many people mistakenly believe that it is enough to divide the total power of the devices by the voltage - but this only works under ideal conditions. In reality it is necessary to take into account starting currents (for example, for a refrigerator or compressor they are 5–7 times higher than the nominal value), cable length (voltage loss at 10 meters can reach 5%), and also machine type (trigger characteristics B, C or D). We will show you how to avoid common mistakes and choose a device that will last for decades without false alarms.

Why you can’t choose a slot machine “with a reserve”

The most common mistake is buying a machine with a denomination significantly higher than calculated, “just in case.” For example, for a load of 5 kW, set 32A instead of 25A. It would seem that this provides a margin of safety, but in practice it leads to:

  • 🔥 Cable overheating. The machine will not turn off if the permissible load on the wire is exceeded, which is fraught with melting of the insulation and fire. For example, cable VVGng 3×2.5 can withstand 25A, and a 32A automatic will allow it to warm up for hours.
  • ⚡ Damage to equipment. Sensitive electronics (such as control units for electric vehicle charging stations) can fail due to prolonged overloads.
  • 💸 Increased cost of the system. Overrated machines cost more and require thicker (and more expensive) cables to install.

Exception - introductory machines, where the current reserve is justified for selectivity (so that during a short circuit only the problem area is triggered, and not the entire network). But even here the reserve should not exceed 25–30% from the calculated value.

⚠️ Attention: If your house or garage already has oversized circuit breakers installed, check the cable cross-section. With a probability of 80% it does not correspond to the current of the machine. In this case, it is necessary to replace the wiring or reduce the rating of the switch.

Formula for calculating machine power: basic algorithm

Basic formula for a single-phase network (220V):

I = (P × Ks) / (U × cosφ)

Where:

  • I — load current in amperes (using it we select the rating of the machine);
  • P — total power of all devices in watts (W);
  • Ks — demand coefficient (takes into account that not all devices operate simultaneously);
  • U — mains voltage (220V for single-phase, 380V for three-phase);
  • cosφ — power factor (for household appliances usually 0.95, for engines 0.8).

For a three-phase network (380V) the formula is modified:

I = (P × Ks) / (√3 × U × cosφ)

Where √3 ≈ 1.73 — root of three (phase multiplier).

📊 What type of network do you have at home?
Single phase (220V)
Three-phase (380V)
I don't know
Other

Demand coefficient (Kc): why without it the calculation is incorrect

The demand factor takes into account that not all electrical appliances work at the same time. For example, in your garage you might have a 5 kW welder, a 2 kW compressor and a 3 kW heater. But it's unlikely that you'll use them all at once. Typical values Ks:

Type of room/load Demand factor (Ks)
Apartment (lighting, sockets) 0.6–0.8
Private house (without electric stove) 0.7–0.9
Garage with tools 0.5–0.7
Workshop (welding station, machines) 0.4–0.6
Office (computers, office equipment) 0.8–0.9

Example: in the garage installed:

  • Welding inverter - 4 kW;
  • Compressor - 2 kW;
  • Heater - 1.5 kW;
  • Lighting - 0.5 kW.

Total power: 4 + 2 + 1.5 + 0.5 = 8 kW. But given Ks = 0.6 (garage) actual load: 8 × 0.6 = 4.8 kW. This is almost 2 times less!

💡

If there are devices in the room with high starting currents (for example, a refrigerator or a pump), increase the rated current by 20–30% or select a machine with the characteristic D instead of C.

Starting currents: why the machine is triggered when the equipment is turned on

Many people are faced with a situation where the machine crashes on startup washing machine, compressor or welding machine - but it works fine at normal load. Reason - starting current, which is 3–7 times higher than the nominal one. For example:

  • 🔌 Refrigerator (nominal 300 W) at startup consumes up to 1.5–2 kW;
  • 🔧 Compressor (1.5 kW) - up to 7–10 kW in the first seconds;
  • 🔥 Welding machine (5 kW) - up to 20–25 kW when igniting the arc.

Solutions:

  1. Select machine with characteristic D (for starting currents) instead C;
  2. Increase the nominal value of the machine by 1–2 steps (for example, from 16A to 20A), but only if the cable cross-section allows;
  3. Use soft starters (for compressors, pumps).
⚠️ Attention: If the machine is triggered when the device is turned on, do not replace it with a more powerful onewithout checking the cable cross-section. For example, for VVGng 3×1.5 maximum current - 16A. Installing a 25A machine will lead to overheating of the wire!
What happens if you ignore inrush currents?

If you ignore the starting currents, the machine will trigger falsely, which will lead to:

1. Wear of the switch mechanism (operation life is limited).

2. Risk of “sticking” of contacts during frequent shutdowns under load.

3. Possible failure of sensitive electronics (for example, control units of charging stations for electric vehicles) due to voltage surges when the machine is triggered.

Examples of machine calculations for different cases

Let's look at real cases with step-by-step calculations.

Example 1: Apartment with an electric stove (single-phase network, 220V)

Devices:

  • Electric stove - 7 kW;
  • Washing machine - 2 kW;
  • Refrigerator - 0.5 kW;
  • Microwave - 1 kW;
  • Lighting - 0.5 kW.

Calculation:

  1. Total power: 7 + 2 + 0.5 + 1 + 0.5 = 11 kW;
  2. Demand coefficient for an apartment: Ks = 0.8;
  3. Real power: 11 × 0.8 = 8.8 kW;
  4. Current: I = (8.8 × 1000) / (220 × 0.95) ≈ 42.5 A;
  5. Select the machine: 40A (nearest standard denomination).

Nuances:

  • For an electric stove, a separate line with an automatic machine is recommended 32A and cable VVGng 3×6;
  • The remaining devices can be divided into 2 groups according to 16A each.

Example 2: Garage with three-phase network (380V)

Devices:

  • Welding machine - 6 kW;
  • Compressor - 3 kW;
  • Heater - 2 kW;
  • Lighting - 0.5 kW.

Calculation:

  1. Total power: 6 + 3 + 2 + 0.5 = 11.5 kW;
  2. Demand factor for garage: Ks = 0.6;
  3. Real power: 11.5 × 0.6 = 6.9 kW;
  4. Current: I = (6.9 × 1000) / (1.73 × 380 × 0.8) ≈ 13.5 A;
  5. Select the machine: 16A (characteristic D due to the welding machine).

Nuances:

  • We connect the welding machine through a separate machine 25A with cable VVGng 5×4;
  • For the compressor we use soft starterto reduce the starting current.

1. Calculated the total power of all devices (including rarely used ones)

2. Apply the demand coefficient for your type of premises

3. Taken into account starting currents (if there are engines, compressors, welding machines)

4. I checked the cross-section of the cable - it should withstand the current of the machine

5. Selected response characteristic (B, C or D) depending on the type of load-->

Table for selecting a machine by power (for quick selection)

To simplify the calculations, use a ready-made table. Values are given for single-phase network 220V with cosφ = 0.95 and Ks = 0.8 (typical apartment). For a three-phase network, the current must be divided by √3 ≈ 1.73.

Load power (kW) Current (A) Recommended machine (A) Minimum cable cross-section (mm²)
Up to 3.5 16 16 1.5
3.5–5.5 25 20 or 25 2.5
5.5–7.5 35 25 or 32 4
7.5–10 45 40 6
10–15 68 50 or 63 10

Example of using a table: for load 6 kW the current will be ~27A. The nearest machine is 32A, but the cable should not be thinner 4 mm².

💡

If the load power falls between two values in the table (for example, 4 kW), always choose a circuit breaker with a higher rating (in this case 25A instead of 20A), but at the same time increase the cable cross-section to 2.5 mm².

Typical mistakes when choosing a machine and how to avoid them

Even experienced electricians sometimes make mistakes. Here are the most dangerous of them:

  1. Ignoring ambient temperature. The machines are designed to operate at +30°C. In a hot garage or near a heater, their rating is reduced by 10-15%. Solution: choose a machine with a reserve or install it in a ventilated panel.
  2. Connecting an aluminum cable to the machine. Aluminum “flows” over time, weakening the contact. This leads to heat and fire. Solution: use copper cables or special terminals for aluminum.
  3. Installing a machine without an RCD. The machine only protects against short circuits and overloads, but not against current leakage (for example, in the event of an insulation breakdown on the body of a washing machine). Solution: combine the machine with RCD or use differential machine.
  4. Wrong choice of characteristic (B, C, D). For example, automatic B16 will work with current 16–24A, and C16 - at 32–48A. Suitable for lighting B, for sockets - C, for engines - D.

Check yourself: if there are machine guns in your dashboard one characteristic (for example, all C), most likely some of them were chosen incorrectly.

FAQ: Frequently asked questions about calculating slot machines

Is it possible to install a machine with a higher denomination if the cable allows it?

Technically possible, but inappropriate. The machine must protect the weakest link in a circuit - usually a cable or socket. For example, if you have a socket on 16A, and the cable can withstand 25A, the machine must be 16A, otherwise the socket will melt if overloaded.

How to calculate an automatic machine for a three-phase motor?

For a three-phase motor, use the formula:

I = P / (√3 × U × cosφ × η)

Where η — Engine efficiency (usually 0.8–0.9). For example, for an engine 5.5 kW (cosφ = 0.8, η = 0.85):

I = 5500 / (1.73 × 380 × 0.8 × 0.85) ≈ 12.3 A

Choose a machine at 16A (characteristic D).

What to do if the machine goes off for no apparent reason?

Causes and solutions:

  • 🔍 Check cable size — if it is too low, replace the wire;
  • 🔌 Eliminate current leakage (use a megohmmeter or call an electrician);
  • 🌡️ Check the temperature in the panel — at +40°C the machine may operate falsely;
  • 🔄 Replace the machine - it may be faulty (especially if it is a cheap model).
Do I need to change machines when replacing wiring?

Yes, necessarily. Old machines could be designed for aluminum wiring with a smaller cross-section. For example, if you replaced aluminum 2.5 mm² for copper 2.5 mm², the current load will increase with 19A up to 27A - old machine gun 16A will become a bottleneck.

What kind of machine should I put into the house?

The nominal value of the input machine is determined allocated power to the home (indicated in the technical specifications from the energy sales company). Typical values:

  • 5 kW - automatic 25A;
  • 10 kW — 50A;
  • 15 kW — 63A.

The characteristic must be C or D (for protection against short circuit at the input). Input cable cross-section - not less than 10 mm² for copper.