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NEET PHYSICSALTERNATING CURRENTEasy

Question

A transformer is used to light a 100 W and 110 V lamp from a 220 V main. If the main current is 0.5 A, the efficiency of the transformer is approximately:

A

30%

B

50%

C

90%

D

10%

Step-by-Step Solution

The efficiency (η\eta) of a transformer is defined as the ratio of output power (PoutP_{out}) to input power (PinP_{in}).

  1. Output Power: The lamp consumes 100 W. So, Pout=100P_{out} = 100 W.
  2. Input Power: The power drawn from the mains is given by the product of the primary voltage (VpV_p) and the primary current (IpI_p). Pin=VpIp=220 V×0.5 A=110P_{in} = V_p I_p = 220 \text{ V} \times 0.5 \text{ A} = 110 W .
  3. Efficiency: η=PoutPin×100%=100110×100%≈90.9%\eta = \frac{P_{out}}{P_{in}} \times 100\% = \frac{100}{110} \times 100\% \approx 90.9\%. The closest approximation among the options is 90%.

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 CURRENTtransformercurrentefficiencytransformerapproximately

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