The current transformer rate (commonly called the CT ratio) is the fixed proportional relationship between the high primary AC current flowing through the main conductor and the safe, stepped-down secondary current delivered to a meter or relay. When you install a 400A feeder but your digital power meter maxes out at 5A, the CT rate is the scaling factor that bridges that gap. In a real installation, this component changes a dangerous, high-energy AC circuit into a standardized, galvanically isolated low-voltage signal, protecting both your measurement equipment and the human reading it.
Think of the CT rate like a mechanical gear reduction on a winch: you trade high speed (primary current) for high torque (secondary voltage potential), keeping the overall power balanced while making the output safe to handle. However, unlike a simple voltage transformer, a CT acts as a constant current source, which introduces unique hazards and sizing rules we will cover below.
Standard Current Transformer Rate Table
Before pulling wire or programming a power meter, you need to select the right CT rate. The table below outlines standard primary-to-secondary ratios governed by IEEE C57.13 and IEC 61869-2 standards. The secondary output is almost universally standardized to either 5A or 1A to match the input terminals of commercial metering and protection relays.
| Primary Current Rating (A) | Secondary Output (A) | CT Rate (Ratio) | Typical Application & Conductor Size |
|---|---|---|---|
| 100A | 5A | 100:5 (20:1) | Subpanels, small motor starters (1/0 AWG) |
| 200A | 5A | 200:5 (40:1) | Commercial lighting panels (3/0 AWG) |
| 400A | 5A | 400:5 (80:1) | Large HVAC disconnects, main feeders (600 kcmil) |
| 800A | 5A | 800:5 (160:1) | Switchgear, heavy industrial busbars |
| 1200A | 1A | 1200:1 (1200:1) | Utility metering, long-distance relay runs |
| 2000A | 5A | 2000:5 (400:1) | Main service entrance, utility substations |
Worked Numeric Example: Calculating Secondary Output
Let’s move from the spec sheet to the workbench. Suppose you are monitoring a 400A main breaker feeding a workshop. The actual continuous load drawing through the busbar is 250A. You have installed a standard split-core CT with a 400:5 rate around the primary conductor, and the secondary wires are terminated into an Emporia Vue or similar energy monitor.
What is the actual current flowing through the secondary wires back to the meter?
The Formula:
Secondary Current (Is) = Primary Current (Ip) × (Secondary Rate / Primary Rate)
The Math:
Is = 250A × (5 / 400)
Is = 250A × 0.0125
Is = 3.125A
What if the workshop turns on a massive CNC machine and the primary current spikes to 450A? The secondary current will rise to 5.625A. Most 5A-nominal metering circuits can handle a 20% continuous overload, but sustained over-ranging will saturate the CT core, causing the meter to read inaccurately low—a phenomenon detailed in Schweitzer Engineering Laboratories (SEL) technical papers on CT saturation.
Where You Meet This in Practice (And Common Confusions)
You will encounter current transformer rates in three main environments: revenue metering (utility billing), energy monitoring (DIY solar or shop subpanels), and protection relaying (tripping a breaker during a short circuit). However, when sizing a CT, hobbyists and junior technicians frequently confuse the ratio with two other critical specifications:
Confusion 1: CT Rate vs. CT Burden (VA)
The rate tells you the current scaling. The burden tells you how much load (in Volt-Amps) the CT can drive through the secondary wiring and meter impedance without losing accuracy. If you run 14 AWG wire 100 feet to a meter with a high internal resistance, you might exceed a 2.5VA burden rating, causing the core to saturate and the meter to under-report your power usage. Always calculate the total loop resistance and ensure it falls below the CT's VA rating.
Confusion 2: Accuracy Class (Metering vs. Protection)
A 200:5 CT might be stamped with '0.5' or '5P20'. These are not rates; they are accuracy classes defined by instrument transformer theory. A '0.5' class is highly accurate up to 120% of the rated current, perfect for billing meters. A '5P20' class is less precise at normal loads but is designed to remain linear up to 20 times the rated current (e.g., 4000A on a 200:5 CT) so a protection relay can accurately detect a massive short circuit and trip the breaker.
Confusion 3: Assuming CTs Work on DC
Current transformers operate strictly on the principle of alternating magnetic flux. They cannot measure Direct Current. If you pass 100A of DC from a solar battery bank through a 100:5 CT, the secondary output will be exactly 0A. For DC measurement, you must use a Hall-effect sensor or a shunt resistor.
FAQ: Sizing and Selecting the Right CT
Q: Can I use a 400:5 CT to measure a small 20A lighting circuit?
A: Technically yes, but practically no. At 20A primary, a 400:5 CT outputs only 0.25A on the secondary. Most standard panel meters lose accuracy below 10% of their nominal input range. You would be operating at 5% of the meter's scale, resulting in massive measurement errors. Always size the CT rate so your expected normal load falls between 50% and 80% of the primary rating.
Q: Does the physical size of the CT window matter if the rate is correct?
A: Yes. The window must accommodate the physical wire gauge and insulation thickness. If you try to force three parallel runs of 500 kcmil THHN through a window meant for a single 4/0 AWG wire, you will damage the conductor insulation or fail to close the split-core halves completely. An air gap in a split-core CT introduces massive phase angle errors and destroys the stated accuracy class.
Q: What happens if I wire the secondary leads backward (X1 and X2 swapped)?
A: For a simple analog ammeter, nothing changes; the needle will still deflect correctly because it reads AC RMS magnitude. However, for a digital wattmeter or a protection relay measuring power factor and directional fault current, swapping the polarity will invert the phase angle by 180 degrees. Your wattmeter will read negative power (assuming you are exporting to the grid), and a directional overcurrent relay might fail to trip during a fault.






