Convert 5 amp ct to 333mv

..."A burden resistor connected across the secondary produces an output voltage proportional to the resistor value, based on the amount of current flowing through it. With our 1:10 turns ratio transformer that produces a 10:1 current ratio, a burden resistor can be selected to produce the voltage we want. If 1A on the primary produces 0.1A on the secondary, then by Ohm's law, 0.1 times the burden resistor will result in an output voltage per amp."...

https://www.powerelectronics.com/content/article/21863556/exploring-current-transformer-applications#:~:text=A%20burden%20resistor%20connected%20across,produce%20the%20voltage%20we%20want.

https://www.foxtamcontrols.co.uk/products/5-amp-1-amp-current-transformer-secondary-to-333mvac-converter

Is this relaylike-gadget just a glorified resistor, or are there more electronics involved?

It's basically code compliant for several standards and Din rail mounted which is necessary in some applications...

"Currently installed" suggests the CT already has a load.

1. If you just parallel the 0.0666 ohm with existing load, only some of the current will flow through your new load.
2. The correct connection for multiple loads is to put them in series with each other and the CT 5 amp secondary.
3. The series loads require more voltage from the CT : CT rating is in VA = volts x amps e.g. a 5 amp 5 VA CT can deliver 5 VA/5 amps = 1 volt at 5 amps within its rated error. It may be able to give more VA without overheating, but with more error.
4. Measure the volts across the CT output terminals with the existing load and correct to 5 amps [maybe existing current load is 2.5 amps, so you have to correct voltage reading by increasing in ratio 5/2.5]: you will know the existing "burden" by calculating volts x 5 amps = VA.
5. What markings or ratings does the CT have on it??

They are in use right now but will only be connected to new meter. No parrallel connections. Some ratios are 600:5, 300:5, 200:5. Kind of all over the map.( The new meter does have a different multiplier for each ct.)

Going back to your original question/post, it is almost as simple as loading the CT with 0.0666 ohm.

CTs are rated for a certain accuracy, according to the standard they are made to, the lower the fraction of rating the greater the error as % of current flowing. Usually, error is not specified below a fraction of rating, like 30%. The actual burden VA and its power factor affect the error, but a low resistive burden gives good accuracy, since the error rated is usually with the maximum burden and less inside the limit "map". You have to decide what you mean by accurate!

Your full-load burden is 5 amps x 0.333 volts = 1.66 VA, plus the cable burden which could be a lot more at 5 amps with a long cable run. A 6 square millimetre copper cable gives 7.3 mV per amp per metre run @ 70'C conductor. If run length is 25 metres then circuit volts drop is 7.3 x 5 amps x 25 metres = 0.92 volts, then x 5 amps = 4.6 VA.

0.0666 ohms may not be a stock value but the supplier of your Power Quality?? instrument probably has a stock of shunts for 5 amp CTs or knows of a supplier. Shunts themselves have a resistance tolerance which adds to overall error.

There is one problem - if the resistor [shunt] cannot stand the current/voltage/duration applied if your system has a short circuit, it will open circuit - this will not stop the current, it will divert into the PQ instrument & burn that out. The PQ instrument will have a maximum voltage or voltage /time limit beyond which it becomes damaged or out of calibration. Damage could extend to catching fire.

You will have to find out the short-circuit current and time from the supply voltage/impedance and the protective fuses or breakers installed. Data may be in the existing drawings or parts lists - if you knew the CT type, rating & manufacture/standard, then tolerable current/time are findable ( but it really depends on whether the CT was correctly sized for your installation - maybe it has been re-located?).

67model