For an ideal transformer where power losses are ignored, the power in the primary circuit ($P_p$) is equal to the power in the secondary circuit ($P_s$) . The power consumed by the lamp in the secondary circuit is given as 44 W. Therefore, $P_p = V_p I_p = 44$ W. Given the primary voltage ($V_p$) is 220 V, the primary current ($I_p$) can be calculated as: $I_p = \frac{P_s}{V_p} = \frac{44 \text{ W}}{220 \text{ V}} = 0.2$ A.