The secondary current, IS$,$ in an iron core transformer equals $1\mathrm{.}8$A$.$ If the turns ratio, NP$/$NS$,$ equals $3$:$1,$ How much is the primary current, IP$?$

Group of answer choices

IP$=1\mathrm{.}8$A

IP$=600$mA

IP$=5\mathrm{.}4$A

none of the above.