QNEET-2023-Ph-225: NEET 2023 Physics - Class 12 NEET-2023-Ph-225 Solution: A 12 V 60 W lamp is connected to the secondary of a step down…

To find the current in the primary winding of an ideal step-down transformer, we first understand that an ideal transformer's power input (primary side) equals its power output (secondary side). This is because an ideal transformer is assumed to have 100% efficiency, meaning there are no losses in the transformation process. The power of the lamp (which is connected to the secondary side of the transformer) is given as 60 W , and it operates at 12 V . The voltage on the primary side of the transformer is 220 V . Using the power formula P = I V , where P is the power in watts, I is the current in amperes, and V is the voltage in volts, we equate the power on both sides of the transformer because of its ideal nature, i.e., P primary = P secondary . Hence, 60 W = 220 V _ I primary . Solving for I primary gives: I primary = 60 W 220 V = 0.2727 . . . � 0.27 A . Therefore, the current in the primary winding of the transformer is approximately 0.27 A .