To solve this problem, we need to use the concept of a transformer and the power relation in an electrical circuit.$\\$Given:$\\$• The primary voltage $V_p = 220$ V$\\$• The secondary voltage $V_s = 11$ V$\\$• The power of the lamp $P = 44$ W$\\$Since the transformer is ideal (ignoring power losses), the power in the primary circuit $P_p$ is equal to the power in the secondary circuit $P_s. \\ P_p = P_s = 44$ W$\\$The power in the secondary circuit is given by $P_s = V_s \cdot I_s,$ where $I_s$ is the current in the secondary circuit.$\\$Rearranging for $I_s: \\ I_s = \frac{P_s}{V_s} = \frac{44}{11} = 4$ A$\\$The transformer equation relates the primary and secondary currents and voltages:$\\ \frac{V_p}{V_s} = \frac{I_s}{I_p} \\$Rearranging for $I_p: \\ I_p = \frac{V_s}{V_p} \cdot I_s \\$Substitute the known values:$\\ I_p = \frac{11}{220} \cdot 4 \\ I_p = \frac{1}{20} \cdot 4 \\ I_p = 0.2$ A$\\$Therefore, the current in the primary circuit is $0.2$ A. This corresponds to Option 1.