Explanation:
For a transformer,
\(\dfrac{I_{p}}{I_{s}}=\dfrac{N_{s}}{N_{p}}\) or \(I_{s}=\left(\dfrac{N_{p}}{N_{s}}\right) I_{p}\)
where \(I_{p}, N_{p}\) be the current and number of turns in the primary coil and \(I_{s}, N_{s}\) be corresponding quantities in the secondary coil.
Here, \({I_p} = 3A,\frac{{{N_s}}}{{{N_p}}} = \frac{3}{2}.\)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\because {\text{Turns}}\,{\text{ratio}} = \frac{{{N_s}}}{{{N_p}}}} \right)\)
\(\therefore \quad {I_s} = \left( {\frac{2}{3}} \right)(3\;A) = 2\;A\)
Hence the current through load resistance is \(2\;A\).