To solve this problem, we need to use the concept of an ideal transformer.$\\$Given:$\\$• Voltage across the primary $V_p = 220 \,$V$\\$• Voltage across the secondary $V_s = 12 \,$V$\\$• Power of the lamp $P = 60 \,$W$\\$For an ideal transformer, the power in the primary equals the power in the secondary:$\\ P_p = P_s \\$Since the lamp is connected to the secondary, the power in the secondary is:$\\ P_s = V_s \cdot I_s \\$Substitute the known values:$\\ 60 = 12 \cdot I_s \\$Solve for $I_s: \\ I_s = \frac{60}{12} = 5 \,$A$\\$Now, using the transformer equation for an ideal transformer:$\\ \frac{V_p}{V_s} = \frac{I_s}{I_p} \\$Substitute the known values:$\\ \frac{220}{12} = \frac{5}{I_p} \\$Rearrange to solve for $I_p: \\ I_p = \frac{5 \cdot 12}{220} \\ I_p = \frac{60}{220} \\ I_p = \frac{3}{11} \\ I_p \approx 0.27 \,$A$\\$Therefore, the current in the primary winding is approximately $0.27 \,$A$. \\$This corresponds to Option 1.