redzilla
All tools
Power

Load & Breaker

Enter the load (W or VA), voltage and number of phases and I compute the current, the minimum breaker by the 80% continuous-load rule and the recommended standard rating (IEC/NEC).

VA is apparent power (maps directly to current). If you enter W I divide it by the PF: VA = W / PF.

System

Single phase: line-to-neutral voltage (e.g. 120 / 230 V).

Only used when the load is in W. Resistive load (heater, incandescent light) = 1; mixed ≈ 0.9; motors ≈ 0.8.

Load type

A continuous load runs 3 hours or more at a stretch. It triggers the 80% rule: breaker ≥ I / 0.8. Non-continuous uses the current as is.

Rating series

IEC: 6 · 10 · 13 · 16 · 20 · 25 · 32 · 40 · 50 · 63 A… NEC: 15 · 20 · 25 · 30 · 35 · 40 · 50 · 60 A…

Examples
redzilla.cl — breaker
 
Current
Min breaker
Rating
VA
Load current
amperes
Minimum breaker
80% rule
Recommended rating
standard
Breaker usage
current / rating
Breaker utilization

The breaker must sit above the load current with headroom.

Series ratings

RatingMax continuous load (80%)Status
How it is computed · current, 80% and rating

1. If the load comes in W I turn it into apparent power by dividing by the power factor: VA = W / PF. If it already comes in VA, it goes straight through.

2. Load current. Single phase: I = VA / V. Three phase (V = line-to-line): I = VA / (√3 · V).

3. The 80% rule for a continuous load (runs ≥ 3 h): the breaker must not carry more than 80% of its rating, so breaker ≥ I / 0.8. For a non-continuous load breaker ≥ I is enough.

4. I pick the next standard rating above that minimum, from the IEC 60898 or NEC 240.6 series. Also check the conductor can carry that current.

Runs locally in your browser · no sign-up · nothing leaves your browser. Design reference; verify against your local code and the conductor.

How it works

The calculator sizes the thermal-magnetic breaker for a load: it converts power into current and suggests the standard rating. If the load is entered in W, it first converts it to apparent power with VA = W ÷ PF; the current is I = VA ÷ V for single-phase or I = VA ÷ (√3 × V) for three-phase, with V line-to-line. Typical power factors are 1.0 for resistive loads, 0.9 for mixed loads and 0.8 for motors.

For continuous loads (operating 3 hours or more) it applies the NEC 80 % rule: the breaker must not be loaded beyond 80 % of its rating, so breaker ≥ I ÷ 0.8; for non-continuous loads breaker ≥ I is enough. It then picks the next standard rating from the IEC 60898 series (6, 10, 13, 16, 20, 25, 32, 40, 50, 63 A…) or NEC 240.6 (15, 20, 25, 30, 35, 40, 50, 60 A…) and shows the utilization percentage. It is a design reference: the conductor must be verified separately and local code prevails.

Example: breaker for a continuous 3 kVA load at 230 V single-phase (IEC series)

  1. Current: 3,000 VA ÷ 230 V = 13.0 A.
  2. Continuous load, 80 % rule: 13.0 ÷ 0.8 = 16.3 A minimum.
  3. Next IEC rating above 16.3 A: 20 A.
  4. Resulting utilization: 13.0 ÷ 20 = 65 % of the rating, with proper headroom.

Frequently asked questions

What breaker do I need for a 2,200 W load at 230 V?
With PF 0.9 that is 2,444 VA and 10.6 A of current. If the load is continuous, the 80 % rule minimum is 13.3 A and the next IEC rating is 16 A. If it is non-continuous you only need to cover 10.6 A, and a 13 A (or 16 A) breaker from the IEC series works. The calculator applies this adjustment automatically based on the load type.
What is the 80 % rule and when does it apply?
It is the NEC criterion for continuous loads (those operating 3 hours or more without interruption, such as lighting, servers or HVAC): the load current must not exceed 80 % of the breaker rating, because the breaker and conductors heat up under sustained operation. It is equivalent to sizing the breaker at no less than 125 % of the continuous current.
What is the difference between calculating with W and with VA?
VA is apparent power and converts directly into current; W is real power and must be divided by the power factor first (VA = W ÷ PF). With motor loads (PF ≈ 0.8), 15 kW is actually 18.75 kVA: using the watts uncorrected would underestimate the current by 25 % and the breaker would come up short.
Does the suggested breaker also protect the cable?
Not necessarily: the breaker here is chosen from the load current, but the conductor must have an ampacity equal to or greater than the breaker rating. Check the wire gauge against an ampacity table (such as NEC 310.16) and your local electrical code; the calculation runs 100 % in your browser and is a design reference only.
Was this tool useful?
Disclaimer We take great care to keep every tool accurate and review it thoroughly; even so, we can't guarantee it is free of errors or take responsibility for how the results are used. We recommend double-checking anything critical.
Found an error? Let us know →