63 amperes at 380 volts is how many kilowatts? Accurate power calculation for auto repair and garage

Why is it important for car owners to be able to convert amps to kilowatts?

Have you ever encountered a situation where you need to connect powerful equipment in the garage - for example, a welder, compressor or charger - but the amperage on the breaker or plug is indicated in amperes, and on the device itself the power is in kilowatts? Or vice versa: you know the power of the machine, but you don’t understand which 63A machine will withstand the load at voltage 380V?

This problem is typical for auto electricians and garages. An error in calculations can lead to network overload, circuit breakers tripping at the most inopportune moment, or even a fire. For example, if you connect a 10 kW compressor through a 25A at 380V circuit breaker, it will constantly kick out. And if you install a 63A machine without calculations, you risk burning the wiring. Today we’ll figure out how to accurately translate 63 amps to kilowatts for a three-phase network, what nuances to consider when choosing equipment and where car owners most often make mistakes.

Spoiler: under tension 380V and power factor cosφ = 1 (ideal conditions) 63A correspond to 43.3 kW. But in reality the figure will be different - read why.

Calculation formula: how to convert amperes to kilowatts for 380V

The basic formula relating current strength (I, in amperes), voltage (U, in volts) and power (P, in watts), looks like this:

P = √3 × U × I × cosφ

Where:

• 🔹 √3 ≈ 1,732 — coefficient for a three-phase network (for a single-phase network it is equal to 1).
• 🔹 U = 380V — linear voltage (between phases).
• 🔹 I = 63A - current strength indicated on the machine or equipment.
• 🔹 cosφ — power factor (depending on the type of load).

For most household and garage appliances cosφ ranges from 0.7 to 0.9. For example:

• 🔧 Welding machines: cosφ = 0,7–0,8.
• 🔧 Asynchronous motors (compressors, machine tools): cosφ = 0,8–0,9.
• 🔧 Heating devices (heating elements, stoves): cosφ = 1 (purely active load).

If cosφ unknown, for approximate calculations use the value 0,8.

Why is √3 used in the formula?

In a three-phase network, the voltage between phases (linear) is √3 times greater than the phase voltage (between phase and zero). For example, with a phase voltage of 220V, the linear voltage will be 380V (220 × √3 ≈ 380).

Calculation for 63 amperes at 380V: step-by-step instructions

Let's calculate in practice how many kilowatts the machine can withstand 63A under tension 380V for different types of load.

1. Ideal case: active load (cosφ = 1)

Suitable for heating elements, incandescent lamps, heaters.

P = 1.732 × 380 × 63 × 1 ≈ 43,300 W = 43.3 kW

2. Typical load: cosφ = 0,8

Suitable for most electric motors, welding machines, compressors.

P = 1.732 × 380 × 63 × 0.8 ≈ 34,640 W = 34.6 kW

3. Real example: welding machine with cosφ = 0,7

P = 1.732 × 380 × 63 × 0.7 ≈ 30,310 W = 30.3 kW

As you can see, the difference between ideal and actual conditions reaches 13 kW! This means that a 63A machine, which in theory should withstand 43 kW, in practice may not be able to handle even 35 kW if, for example, an old Soviet machine with low cosφ.

Check the equipment passport for cosφ|

Multiply the calculated power by 1.2–1.3 for reserve|

Make sure the cable size is appropriate for the current (63A requires a minimum of 16mm² copper)|

Remember: motor starting currents are 3–5 times higher than rated currents-->

Table: 63 amps in kilowatts for different power factors

Load type	cosφ	Power (kW) at 380V	Equipment examples
Active	1,0	43,3	Heating elements, heaters, incandescent lamps
Mixed (typical)	0,8	34,6	Welding machines, compressors, machines
Inductive (low cosφ)	0,7	30,3	Old asynchronous motors, transformers
Highly inductive	0,6	25,9	High load engines, cheap welders

Please note: even with the same current 63A actual power may differ by almost 1.7 times depending on the type of equipment. This is critical for garages where they often use homemade or outdated devices with unknown cosφ.

Mistakes made by car owners when calculating power

Even experienced craftsmen sometimes make mistakes that lead to accidents. Here are the most common:

⚠️ Attention: If you connect a three-phase motor through a 63A circuit breaker, but forget about starting currents (they are 3–7 times higher than nominal), the machine will operate when started. Solution: use soft starters or machines with a characteristic D (for example, C63 replace with D63).

Top 5 mistakes:

• 🔴 Ignoring cosφ. Many people use the formula for active load (cosφ=1), and then wonder why the cable is heating up.
• 🔴 Ignoring starting currents. A 15 kW engine can consume up to 50 kW when starting, which will trigger the 63-A circuit breaker.
• 🔴 Incorrect cable section. For 63A you need a cable 16 mm² (copper) or 25 mm² (aluminium). Using a thinner wire will result in overheating.
• 🔴 Connecting single-phase devices to a three-phase machine. For example, if to the machine C63 connect a single-phase load to 380V (for example, 220V), the maximum power will not be 43 kW, but only 13.9 kW (63A × 220V).
• 🔴 Using Class Automata B for engines. Class B triggers when the current exceeds 30–50%, which is not enough for starting currents. Need class machines C or D.

Practical example: a garage owner connected a 22 kW compressor via a machine C63, counting on 43 kW. However, when cosφ=0,7 the actual power was 30 kW, and the starting current exceeded 63A - the machine was triggered every time it was started. Solution: replaced the machine with D80 and added a soft starter.

How to choose a circuit breaker and cable for 63 amps in the garage

Knowing that 63A at 380V corresponds 30–43 kW (depending on cosφ), let's figure out how to choose the right protection and wiring.

1. Selecting a circuit breaker

• 🔹 For active load (heaters, heating elements) automatic is suitable C63.
• 🔹 For engines and compressors it's better to take D63 (so that it does not trigger when starting).
• 🔹 If the total power of the equipment exceeds 35 kW, consider a machine for 80A or 100A.

2. Cable cross-section

For 63A minimum requirements (according to the PUE):

• 🔹 Copper cable: 16 mm² (withstands up to 75A).
• 🔹 Aluminum cable: 25 mm² (withstands up to 65A).

Recommendation: for a garage it is better to take copper with a reserve - 25 mm², especially if the cable is long (more than 20 meters).

3. Overload protection

If you connect several devices through one machine, sum up their power, taking into account cosφ. For example:

• 🔧 Welding machine: 20 kW (cosφ=0,7) → 28.6A.
• 🔧 Compressor: 7.5 kW (cosφ=0,8) → 13.8A.
• 🔧 Heater: 3 kW (cosφ=1) → 4.5A.

Total current: 28.6 + 13.8 + 4.5 = 46.9A. This means that a 63A machine will be enough with a reserve, but if you turn on everything at the same time, it is better to use C63 (it will not work for short-term surges).

Practical example: connecting a welding machine

Let's consider a real problem. Let's say you bought a welding machine Resanta SAI-250 with the following characteristics:

• 🔹 Voltage: 380V.
• 🔹Maximum current: 250A (this is the welding current, not to be confused with the consumption current!).
• 🔹Power: 8.5 kW (indicated in the passport).
• 🔹 cosφ = 0,7 (typical value for welders).

Question: Is a 63A circuit breaker sufficient for this device?

Calculation:

1. By power: P = 8.5 kW, cosφ = 0,7 → current is equal I = P / (√3 × U × cosφ) = 8500 / (1.732 × 380 × 0.7) ≈ 17.5A.
2. But! When welding, the current may briefly reach 250A (this is the current in the arc circuit, not consumption). However current consumption from the network will be lower - about 20–25A (depending on the mode).

Conclusion: Automatic C63 suitable with a large margin. But if you plan to connect more equipment (for example, a compressor), the total current may exceed 63A. In this case it is better to put C80 or D63.

⚠️ Attention: Many cheap welding machines are underrated cosφ (up to 0.5–0.6). In this case, the actual current consumption will be 30–40% higher than the calculated one. Always check the parameters in your passport!

FAQ: Frequently asked questions about 63 amps and 380 volts

❓ Is it possible to connect a 30 kW compressor through a 63A machine at 380V?

Yes, but with reservations:

• 🔹If cosφ ≥ 0,85, then 30 kW corresponds to ~52A (63A is enough).
• 🔹If cosφ = 0,7, then 30 kW = ~61A - the machine will operate at the limit. It's better to take C80.
• 🔹 Take into account the starting currents: for a compressor they can reach 90A (you need a class automatic D).

❓ How many kilowatts will a 63A machine withstand in a single-phase network (220V)?

For a single-phase network, the formula is simplified: P = U × I × cosφ.

• 🔹 When cosφ=1: 220V × 63A = 13.9 kW.
• 🔹 When cosφ=0,8: 220 × 63 × 0,8 = 11.1 kW.

This means that the same machine C63 in a 220V network withstands 3 times less power than in a 380V network!

❓ What cable is needed to connect a 63-amp machine?

Minimum requirements (according to PUE 7):

Material	Section (mm²)	Max. current (A)
Copper	16	75
Copper	25	100
Aluminum	25	65

Recommended for garage copper 25 mm² — this gives a current reserve and reliability.

❓ Why does it turn out to be 43 kW at 63A and 380V, but my 20 kW welding machine knocks out the machine?

The reasons may be:

• 🔹Low cosφ (eg 0.5 instead of 0.7) → actual current is higher.
• 🔹 Starting currents (at the moment the arc is ignited, the current jumps to 100–150A).
• 🔹 Device malfunction (for example, short circuit in a transformer).
• 🔹 Class machine B (triggered at 30–50% overload).

Solution: Use a class automaton D or soft starter.

❓ Is it possible to connect several devices with a total power of 50 kW to a 63-A machine?

No if:

• 🔹 All devices will work simultaneously.
• 🔹 cosφ below 0.9 (for example, at 0.7 the total current will exceed 80A).

Yes, if:

• 🔹 The devices turn on one by one.
• 🔹 Total current taking into account cosφ does not exceed 60A.

Example: 50 kW at cosφ=0,8 → I = 50,000 / (1.732 × 380 × 0.8) ≈ 92A → minimum required 100A.