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NEET PHYSICSALTERNATING CURRENTMedium

Question

A 220V input is supplied to a transformer. The output circuit draws a current of 2.0A at 440V. If the efficiency of the transformer is 80%, the current drawn by the primary windings of the transformer is

A

3.6A

B

2.8A

C

2.5A

D

5.0A

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}). Formula: η=PoutPin=VsIsVpIp\eta = \frac{P_{out}}{P_{in}} = \frac{V_s I_s}{V_p I_p} . Given: Input Voltage (VpV_p) = 220 V Output Voltage (VsV_s) = 440 V Output Current (IsI_s) = 2.0 A Efficiency (η\eta) = 80% = 0.8

Calculation:

  1. Calculate Output Power: Pout=VsIs=440 V×2.0 A=880 WP_{out} = V_s I_s = 440 \text{ V} \times 2.0 \text{ A} = 880 \text{ W}.
  2. Calculate Input Power: Pin=Poutη=8800.8=1100 WP_{in} = \frac{P_{out}}{\eta} = \frac{880}{0.8} = 1100 \text{ W}.
  3. Calculate Primary Current: Since Pin=VpIpP_{in} = V_p I_p, then Ip=PinVp=1100220=5.0 AI_p = \frac{P_{in}}{V_p} = \frac{1100}{220} = 5.0 \text{ 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 CURRENTsuppliedtransformeroutputcircuitcurrent

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