Solution:
Soltion
Here, Ip=?,Ep=220V,Es=22V
Rs=220ohm
$$\displaystyle{I}_{{{s}}}=\frac{{{E}_{{{s}}}}}{{{R}_{{{s}}}}}=\frac{{{22}}}{{{220}}}={1.0}{A}$$
In an an ideal transformer,
$$\displaystyle{I}_{{{s}}}=\frac{{{E}_{{{s}}}}}{{{R}_{{{s}}}}}=\frac{{{22}}}{{{220}}}={1.0}{A}$$
$$\displaystyle\therefore{I}_{{{p}}}=\frac{{{E}_{{{s}}}}}{{{E}_{{{p}}}}}\times{I}_{{{s}}}=\frac{{{22}\times{0.1}}}{{{220}}}={10}^{{-{2}}}{A}$$