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NEET PHYSICSALTERNATING CURRENTEasy

Question

A step-down transformer connected to an AC mains supply of 220 V is made to operate at 11 V, 44 W lamp. Ignoring power losses in the transformer, what is the current in the primary circuit?

A

2 A

B

4 A

C

0.2 A

D

0.4 A

Step-by-Step Solution

For an ideal transformer where power losses are ignored, the power in the primary circuit (PpP_p) is equal to the power in the secondary circuit (PsP_s) . The power consumed by the lamp in the secondary circuit is given as 44 W. Therefore, Pp=VpIp=44P_p = V_p I_p = 44 W. Given the primary voltage (VpV_p) is 220 V, the primary current (IpI_p) can be calculated as: Ip=PsVp=44 W220 V=0.2I_p = \frac{P_s}{V_p} = \frac{44 \text{ W}}{220 \text{ V}} = 0.2 A.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from ALTERNATING CURRENT. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSALTERNATING CURRENTstepdowntransformerconnectedsupplyoperate

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