Current transformer load

There are a number of things that are unclear in your post, such as where you are measuring the 4A and is your burden (marked in VA) actually 0.5VA or is this actually your current transformer accuracy (marked as a unit-less number, 0.5 =5%).

The following link covers the basics of current transformers quite well and should help.

http://en.wikipedia.org/wiki/Current_transformer

Additionally there are other threads on CR4 covering current transformers in more depth.

0.5 is the burden for the first CT. It is a GE and in the paperwork it says burden 0.5 without unit(s). The 4 amps is measured in the secondary of the current transformer, probably measuring 400 Amps primary. Question 1 is: is the current 400 +4 ? Question 2 is: the secondary wire (the one with 4 amps) is send through a different current transformer with 1:3 windings. The current through these 3 windings when shorted, shows also 4 amps. So what is supplying what?

Question 1 is: is the current 400 +4 ?

No, a 500:5 CT will give 4A on the secondary winding when there is 400A on the primary. The fundamental principle of a current transformer is the same as a voltage transformer, that the output is a ratio of the input due to the transformer turns ratio.

Question 2 is: the secondary wire (the one with 4 amps) is send through a different current transformer with 1:3 windings. The current through these 3 windings when shorted, shows also 4 amps. So what is supplying what?

This is still a little confusing as I read this as you have connected a 1:3 ratio CT on to the output of a 500:5 CT. From a practical point of view you should be connecting a 5:1 to a 500:5 so that the output of the 500:5 CT matches the input of the 5:1 CT. Are you trying to convert a 500:5 CT to give a 500:1 ratio or (even more confusing) trying to convert a 500:5 CT to give a 500:15?

The current through these 3 windings when shorted, shows also 4 amps. So what is supplying what?

Ratios will be correct unless the CT burden(s) have been exceeded, the CT current ratings have been exceeded (causing saturation) or the current waveform is non-sinusoidal (due to harmonics). A standard hand-held averaging (not true RMS) current clamp meter wont pick up any of these problems.

Is this an actual application or a general interest (or homework question)? Can you supply more information on the application.

No homework Jack, but work in the plant. We are trying to make separate current measurements to different stations for loads monitoring and control purposes. (kneaders)

Will the current, delivered by the main CT be divided to the different secondary current transformers.

Like the 4 Amp output, will it be split into the demands for the different measurement points?

And will it be proportional? Will the 4 amps still be measurable at the source too, or is it less? Thanks for your reply.

Generally you can add additional CTs to the secondary of a single CT (and get them to output the same current depending on the turns ratio of the CTs) but you need to make sure the load of the secondary CTs does not exceed the burden of the primary CT (and also that the load of the kneader control station CT input(s) does not exceed the burden of the secondary CT(s)), otherwise the current signals wont be proportional.

To do this you really need to know what load your kneader control station CT input is (although you can connect up the CTs and monitor the primary current and split secondary currents to see if the ratio remains the same, although this has its own problems in trying to measure this if the primary current is varying a lot).

If the combined load of the additional CTs is to much for your existing main CT then you could either add an additional main CT or add electronic signal converters (current transducers) that will provide isolated current signals to your equipment (generally with a 4-20mA or other common process output) that should be compatible with your control systems.

http://www.wisegeek.com/what-are-current-transducers.htm