# Is Current Transformer secondary current independent of load impedance? and How to find voltage of CT Secondary side?

i am trying to figure out a CT (Current Transformer) which is installed in my device of analysis. This CT has Two wires passing through its center hole (one is phase/red wire of single Phase AC supply and the other is neutral).

I don't know the specifications of it, because there is no marking over it (probably locally made) nor i have a datasheet or something.

So, i applied different values of ac current through it and noted down the current at secondary of this CT using Multi-meter's Ammeter Function. I came to know that, its turn ratio is Ns/Np = 0.01. It means for current flowing through one primary conductor is 10A, the secondary side output current of 0.1A.

The problems, i am facing are these;

1- When the current of 10A is passed through each pf primary conductors (phase and neutral passing through center), and the current is in same direction (in CT, direction of current is same in both conductors). The output secondary current is approx. zero. If the current is in opposite direction, the secondary current is 0.2A. How is this possible? In case of different direction, the resulting flux should cancel each other and no secondary current should be present and vice versa? But i am getting opposite behavior?? is there any possible internal structure that makes it work??

2- As far as i read about CT, their secondary current is independent on load impedance. It means if primary current is fixed, secondary should have been fixed no matter impedance of load is.right? But if is simulate this in Orcade PSPICE, changing the load impedance changes the value of secondary current although primary current source is fixed at some value.??

3- I know the secondary current based on primary current and turn ratio. But since CT;s are used to sense current, one actually sense the voltage at secondary of CT whose value is corresponding to primary current. But how do i know this secondary voltage at CT? Because at the output of CT, a bridge rectifier is attached and some RC circuit, so to find the ripple factors, values of ripple voltages, i need to know what voltage is appearing at secondary of CT. How do i find that?? • Sort your issues out with 10 amps in the primary - this is a misunderstanding and doesn't happen that way - you must be mistaking what is actually happening. Try looping 10 amps on one wire through then loop it back in the opposite direction. – Andy aka Sep 26 '18 at 11:45
• There is a power source, single Phase (e.g. phase and neutral). first the phase wire from source is passed through CT and then to load, then the return wire from load (neutral) also passed from CT and goes back to source. so when load of 10A is turned on, 10 passes through Source-to-CT-to-load, and then from load-to-CT-to-Source. I hope this make it clear.. – BetaEngineer Sep 26 '18 at 11:50
• Show a picture. – Andy aka Sep 26 '18 at 11:57
• Look for an online tutorial. lecture notes, application sheet or something like that. This site is more for single-issue questions and explaining individual points that you have difficulty with while studying the materials mentioned. – Charles Cowie Sep 26 '18 at 13:03

1. Either you are doing something wrong or you are mistaken in some way or another. If there are two wires passing through a CT and they both carry identical current amplitude wave forms of opposite phase then there is no net magnetic field and there can be no induced voltage in the secondary.

That isn't true when taken to extremes. What you have to remember is that the burden resistor on the secondary (maybe 1 ohm for example) becomes reflected to the primary by the turns ratio squared. So, if the turns ratio is 1:100, the primary looks like a resistance of 0.1 milli ohms (i.e. divided by 10,000).

But there is also the magnetization inductance and this appears in parallel with that reflected impedance of 0.1 milli ohms. There is only 1 turn (usually) in the primary and this might have a net inductance of 10 uH (made so by the ferrous core). At 50 Hz (for example), 10 uH has an impedance of 3.14 milli ohms i.e. 30 odd times more impedance than the reflected burden.

This means that the magnetization impedance can be ignored but, if you raise the burden resistor too high, that magnetization inductance is going to produce an error. Ultimately, if you remove the burden from the secondary all you are left with is the magnetization inductance and, if you are passing (say) 10 amps, it will develop a volt-drop of 31.4 mV and the secondary voltage will by 100 x higher at 3.41 V (but no current passes).

It behaves like a voltage transformer but with vastly different values for magnetization inductance.

But if is simulate this in Orcade PSPICE, changing the load impedance changes the value of secondary current although primary current source is fixed at some value

That's because of the magnetization inductance becomes significant in terms of an impedance in parallel with the reflected burden resistor value.

I know the secondary current based on primary current and turn ratio

No, not in extremes as I hope I have shown.