Physics · Alternating Current

JEE Main 2026 — 28 January, Evening Shift — Question 48

An inductor stores 16 J of magnetic field energy and dissipates 32 W of thermal energy due to its resistance when an a.c. current of 2 A (rms) and frequency 50 Hz flows through it. The ratio of inductive reactance to its resistance is $\_\_\_\_$ . ( $\pi=3.14$ )

Answer: 314

Numerical answer — enter this value.

Step-by-step solution

$\frac{1}{2} \mathrm{Li}_{\text {rms }}^{2}=16 \Rightarrow \mathrm{~L}=8$ $i^{2} R=32 \Rightarrow R=8$ $\mathrm{x}_{\mathrm{L}}=\omega \mathrm{L} \Rightarrow 2 \times 3.14 \times 50 \times 8$ $\Rightarrow 800 \times 3.14$ $\mathrm{R}=8$ $\frac{\mathrm{x}_{\mathrm{L}}}{\mathrm{R}}=314$

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Exam: JEE Main 2026
Paper: 28 January, Evening Shift
Subject: Physics
Chapter: Alternating Current
Topic: Series L-R, R-C, L-C Circuits with AC Source