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Physics · Electromagnetic Induction

JEE Main 2024 — 1 February, Shift 2 — Question 48

A transformer has an efficiency of 80%80 \% and works at 10 V and 4 kW . If the secondary voltage is 240 V , then the current in the secondary coil is :

  1. Option A:

    1.59 A

  2. Option B:

    13.33 A

    Correct
  3. Option C:

    1.33 A

  4. Option D:

    15.1 A

Answer: B

Step-by-step solution

Efficiency =ESISEPIP=\frac{E_{S} I_{S}}{E_{P} I_{P}} 0.8=240IS40000.8=\frac{240 \mathrm{I}_{\mathrm{S}}}{4000} IS=3200240=13.33 A\mathrm{I}_{\mathrm{S}}=\frac{3200}{240}=13.33 \mathrm{~A}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
1 February, Shift 2
Subject
Physics
Chapter
Electromagnetic Induction
Topic
Eddy Currents, AC Generator and Transformers

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