A 100/5 CT close to its meter
At 60 A the CT gives 3 A. A short lead loop of 0.1 Ω uses 5² × 0.1 = 2.5 VA and the meter 1 VA, so 3.5 VA of the 5 VA rating is used.
Secondary current 3 A Open in the calculator
CT ratio and burden calculator
Check an existing current transformer: its secondary current at the measured load, and whether the meter and the secondary wiring stay within its rated burden.
60through a100CT with a5 Asecondary
gives3on the secondary, 60% of its rated 5 A
Primary current against the rating
60In the range the accuracy class covers
Leads:5.55of 2.5 mm² each way, or0.1loop, and a meter of1on a CT rated5
Burden at the rated 5 A
1.5 VA spare- Meter 1 VA
- Leads 2.5 VA 0.1 Ω loop
- Spare 1.5 VA
Longest lead run each way, copper at 75 °C, within 5 VA
Secondary current
3A
Check the values to see a result.
- Ratio
- 100/5
- Lead burden at rated current
- 2.5 VA
- Connected impedance
- 0.14 Ω
- Longest 2.5 mm² lead run, each way
- 8.878 m
- Pass: Burden within the CT rating3.5 VA connected of 5 VA1.5 VA spare
- Pass: Primary current in the accurate range20% to 120% of the rating60%
Meter and lead burden are added as magnitudes, which errs on the safe side. Never open a CT secondary while the primary carries current.
How it’s calculated
- Is = Ip × Is,rated ÷ Ip,rated=60 ÷ 20=3 A
- VAleads = Is,rated² × Rloop=5² × 0.1=2.5 VA
- VAtotal = VAmeter + VAleads=1 + 2.5=3.5 VA
Is = Ip ÷ (Ip,rated ÷ Is,rated)burden = VAmeter + Is,rated² × Rloop
A 100/5 CT carrying 60 A gives 3 A on its secondary. The meter and leads use 3.5 VA of its 5 VA rated burden.
Tip: With a 5 A secondary, the leads use burden 25 times faster than with 1 A: a long run to the meter is where 1 A CTs earn their place.
How to calculate CT secondary current and burden
A current transformer’s burden is everything connected to its secondary: the meter input, the wiring out and back, and the terminals. Each is a load in VA at the rated secondary current, and together they must stay within the CT’s rated burden, or the CT loses accuracy. The calculator gives the secondary current at your load and checks the burden, including the leads.
Secondary current
Is = Ip × Is,rated ÷ Ip,rated
A 100/5 CT has a ratio of 20. At 60 A in the primary it delivers 60 ÷ 20 = 3 A.
Lead burden
VAleads = Is,rated2 × Rloop
Rloop is the resistance of the whole secondary loop, out and back. It is worked out at the rated secondary current, the point the CT’s rating refers to.
Total burden
VAtotal = VAmeter + VAleads ≤ VArated
Add the meter burden from its datasheet. Adding the two as plain numbers slightly overstates the total, which errs on the safe side.
Because burden goes with the square of the current, a 5 A CT puts 25 times the load on the same leads as a 1 A CT. On long runs between the CT and the meter, 1 A secondaries are the usual answer.
CT burden examples
A 400/5 CT on a long run
The meter is 20 m away on 2.5 mm² copper, a loop of about 0.3 Ω. At 5 A the leads alone use 7.5 VA, so with the meter the burden is 8.5 VA, well over a 5 VA rating. The CT gives 3.125 A at 250 A, but its accuracy class no longer holds.
Secondary current 3.125 A Open in the calculator
The same run with a 1 A CT
With a 400/1 CT the same leads use 1² × 0.3 = 0.3 VA. With a 0.5 VA meter input the total is 0.8 VA, comfortably within 5 VA. At 250 A the secondary carries 0.625 A.
Secondary current 0.625 A Open in the calculator
CT lead burden by loop resistance
The burden the secondary wiring adds at rated current. For the loop resistance, take the conductor resistance per metre times twice the one-way length: 2.5 mm² copper is about 7.4 mΩ per metre at 20 °C, and the conductor resistance calculator works out others.
| Loop resistance (Ω) | 5 A secondary (VA) | 1 A secondary (VA) |
|---|---|---|
| 0.02 | 0.5 | 0.02 |
| 0.05 | 1.25 | 0.05 |
| 0.1 | 2.5 | 0.1 |
| 0.15 | 3.75 | 0.15 |
| 0.2 | 5 | 0.2 |
| 0.3 | 7.5 | 0.3 |
| 0.5 | 12.5 | 0.5 |
| 0.75 | 18.75 | 0.75 |
| 1 | 25 | 1 |
| 1.5 | 37.5 | 1.5 |
| 2 | 50 | 2 |
Questions about CT burden
What is CT burden?
The load on the CT’s secondary, in VA at the rated secondary current. It includes the meter or relay input, the wiring and the connections. A CT meets its accuracy class only while the burden is within its rating.
Should I use a 1 A or a 5 A CT?
For short runs inside a panel, either works. For long runs, 1 A: the wiring burden is 25 times lower for the same cable. Match the secondary to the meter input.
What happens if the burden is too high?
The CT needs more voltage to drive its current through the load. Its ratio and phase errors grow, and at high currents it can saturate and under-read.
Why must a CT secondary never be left open?
A current-output CT forces its current into whatever is connected. With the secondary open and the primary carrying current, the voltage across the terminals can rise to dangerous levels and damage the CT. Short the secondary with the shorting link before disconnecting the meter.
Limits of this result
- For conventional current-output CTs with an external burden. This tool is not for voltage-output CTs or Rogowski coils.
- Ratio and rated burden do not establish accuracy class, saturation, fault duty, insulation rating or aperture fit.
- Enter the resistance of the complete outgoing-and-return secondary lead loop at its operating temperature.
- A cable run uses the IEC 60228 maximum resistance of stranded copper (class 2) at the chosen lead temperature, so a real cable usually uses slightly less burden than shown.
- Enter instrument burden at the selected rated secondary current. For unchanged impedance, burden at another current scales with the square of current.
- Adding instrument VA magnitude and resistive lead VA is a conservative arithmetic sum, not an exact complex-impedance calculation.
Measure it continuously
A ZEM ships with its CTs connected and calibrated to the meter, and Class 0.5S describes the whole measurement, so there is no burden to work out on site.
