Cahier technique no 195 - Schneider Electric

3 Equivalence of the various possible definitions of the same CTIn many cases, you need to know how to “juggle”with the various CT characteristics; ratio, power,class, ALF. The reason for this is not only inorder to get out of a tricky position, but also to beable to use standard CTs that are available, lesscostly and tested.This section thus aims to show how CTcharacteristics can be manipulated. First,however, it should be pointed out that the onlyCT constants are its magnetising curve andresistance and, naturally, its ratio.3.1 How to move from P n1 -5Pk 1 to P n2 -5Pk 2V s1and R ct are fixed.⎛V R P n ⎞ ⎛k R P n ⎞12s = ct +n ct k1 ⎜ ⎟ 1I= ⎜ + ⎟ 2 I2 2⎝ I ⎠ ⎝ I ⎠n⎛R P n ⎞i= ⎜ ct + ⎟ k I2⎝ I ⎠ni n2Knowing that P i = R ct I n(internal ohmic lossesof the CT), we obtain:(P i + Pn) k 1 = (P i + P n ) k 2 = (P i + P k1 2n )3 3.Sometimes, some people ignore P i : this is aserious error as P i can be roughly of the samevalue, if not higher, than P n .nnc if P n2 is imposed, we shall obtain:k 2 =2( Pi+ P n )( P ) k or k ( R P )1ct In+ n11 2 =( P ) k21Pi+ nRctIn+ n2c if k 2 is imposed, we shall obtain:kP n2 =1+ ⎛ ⎞⎝ ⎜ kP1n - 1k k⎟ P1i⎠or else:2k2P n = 1 + ⎛ ⎞2n⎝ ⎜ kP 1n - 1k k⎟ RctI122 ⎠223.2 How to move from P n1 -5Pk 1 to P n2 -10Pk 2We have:VV⎛ P= ⎜Rct +⎝ In1s 1 2n⎛ P= ⎜Rct +⎝ In2s 2 2nBut:V1.6= Vs19 .s 1 2⎞⎟ k⎠c if P n2 is imposed:⎞⎟ k⎠I1 nI2 nc if k 2 is imposed, we shall have:1.9 k 19P n = + ⎛ ⎞12⎝ ⎜ . kP1n- 1 ⎟ R1ct1.6 k216 . k2⎠If you wish to move from a 10P to a 5P definition,the above expressions apply: just reverse theinduction ratio.I2nk 2k 2==21.9( RctIn+ P )n11.6 ( P ) k or21RctIn+n21.9( Pi+ P )n11.6 ( P ) k 1Pi+n2Cahier Technique Schneider Electric no. 195 / p.15