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



  As far as i read about CT, their secondary current is independent on
  load impedance.


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.