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Physics Question 8 – NEET-UG 2023

A 12 V, 60 W lamp is connected to the secondary of a step down transformer, whose primary is connected to ac mains of 220 V. Assuming the transformer to be ideal, what is the current in the primary winding?

For an ideal transformer, there is no power loss, meaning the power in the primary winding is equal to the power in the secondary winding.

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Ninja StrategyPower Conservation & Magnitude Estimation

Recognize that for an ideal transformer, power is conserved. Since the primary voltage is much higher than the secondary voltage, the primary current must be much lower than the secondary current. Estimate the secondary current (60 W/12 V=5 A) and then look for a primary current that is significantly smaller, eliminating options that are too large.

Step 1: Determine Power in Primary Winding✦ Active

The lamp is connected to the secondary winding. The power consumed by the lamp is given as Ps=60 W. Since the transformer is ideal, there is no power loss. Therefore, the power in the primary winding (Pp) is equal to the power in the secondary winding (Ps).

Pp=Ps=60 W
💡 Teacher's Secret Hint

Remember that for an ideal transformer, efficiency is 100%, so input power equals output power.

Step 2: Calculate Current in Primary Winding○ Expand

The primary winding is connected to ac mains of Vp=220 V. We can use the power formula P=V×I to find the current in the primary winding (Ip).

Pp=Vp×Ip 60 W=220 V×Ip Ip=60220 A
💡 Teacher's Secret Hint

Ensure you use the correct voltage (Vp) for the primary current calculation.

Step 3: Simplify and Select Option○ Expand

Simplifying the fraction gives the primary current:

Ip=622=311 A Ip0.2727 A

Rounding to two decimal places, the current in the primary winding is approximately 0.27 A, which corresponds to option (1).

💡 Teacher's Secret Hint

Always check the units and round to an appropriate number of significant figures based on the options.

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