Physics · Electromagnetic Induction

JEE Main 2024 — 27 January, Shift 2 — Question 41

Primary side of a transformer is connected to $230 \mathrm{~V}, 50 \mathrm{~Hz}$ supply. Turns ratio of primary to secondary winding is $10: 1$ . Load resistance connected to secondary side is $46 \Omega$ . The power consumed in it is :

Option A:
12.5 W
Option B:
10.0 W
Option C:
11.5 W
Correct
Option D:
12.0 W

Answer: C

Step-by-step solution

$\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}$

$\frac{230}{\mathrm{~V}_{2}}=\frac{10}{1}$

$\mathrm{V}_{2}=23 \mathrm{~V}$

Power consumed $=\frac{\mathrm{V}_{2}^{2}}{\mathrm{R}}$

$=\frac{23 \times 23}{46}=11.5 \mathrm{~W}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 27 January, Shift 2
Subject: Physics
Chapter: Electromagnetic Induction
Topic: Eddy Currents, AC Generator and Transformers