It is more useful to think of a transformer in terms of Voltage rather than current.

Here's a link that may help you

http://www.tpub.com/neets/book2/5f.htm

The ratio is reduced proportionally by the coupling coefficient.

Let's take the simplest case of a 1:1 transformer. Assume we have identical windings wound on a common core. Thus, Np = Ns and Lp = Ls. The mutual inductance (M) between the two windings is a function of the coupling coefficient:

M = k*sqrt(Lp*Ls)

The ideal transformer equation can be rewritten to take into account lower than unity coupling as follows:

Is = M*Ip/Ls

If Np = Ns, then Lp = Ls, and substituting for M (adjusted for non-unity coupling coefficient):

Is = k*sqrt(Lp*Ls)*Ip/Ls = k*I1

If k = 0.5, then we get an output current that is half that within the primary. This is expected since only half of the total primary flux links the secondary (and vice-versa).

You can also compute the above relationship for the more general case where Np <> Ns by remembering that the individual winding inductances will scale as turns squared. So, if Ns = 2*Np, then Ls = 4*Lp, or Lp = 0.25Ls. In this case:

Is = k*sqrt(Lp*Ls)*Ip/Ls = k*sqrt(0.25*Ls*Ls)*Ip/Ls = 0.5*k*Ip

Again, you can see that the secondary current is the ideal current ratio reduced proportionally by the coupling coefficient.

BertHickman,thanks for the great mathematical derivation and the true results you obtained, but how you got the relation:

Is=M*Ip/Ls

If as you say: Is=N1/N2*k*Ip

How come the following basic transformer equations can be maintained :

1- Np*Ip = Ns*Is

2- Ep*Ip = Es*Is

I think that the current ratio Is/Ip should equal to Np/Ns and k has no effect on that ratio because each of the two windings has its own leakage reactance and K1=K2