The current drawn by the primary coil of an ideal transformer, which steps up $22 \mathrm{~V}$ into $220 \mathrm{~V}$, to operate a device having a load resistance $110 \Omega$ is:

For an ideal transformer, the power input equals the power output. Let $V_p$ and $I_p$ be the primary voltage and current, and $V_s$ and $I_s$ be the secondary voltage and current.

Given $V_p = 22 \text{ V}$ and $V_s = 220 \text{ V}$.

The load resistance is $R = 110 \Omega$. The secondary current $I_s$ is given by Ohm's law:

$I_s = \dfrac{V_s}{R} = \dfrac{220 \text{ V}}{110 \Omega} = 2 \text{ A}$.

Since the transformer is ideal, the power balance equation is $V_p I_p = V_s I_s$.

Substituting the known values:

$22 \text{ V} \times I_p = 220 \text{ V} \times 2 \text{ A}$

$22 I_p = 440$

$I_p = \dfrac{440}{22} = 20 \text{ A}$.

Answer: 20 A