Q 12-07-039JEE MainJEE Main 2026 (28 Jan, Shift 2)Medium
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 ______. ($\pi = 3.14$)
Numerical value type. Enter your answer.
Answer: 314
Taking the stored energy with the rms current, $\frac12LI_{\text{rms}}^2 = 16$:
$$L = \frac{2\times16}{2^2} = 8\ \text{H}$$
Power dissipated: $I_{\text{rms}}^2R = 32 \Rightarrow R = \dfrac{32}{4} = 8\ \Omega$.
$$X_L = 2\pi fL = 2\times3.14\times50\times8 = 2512\ \Omega$$
$$\frac{X_L}{R} = \frac{2512}{8} = 314$$
Solution by Sreeraj P, M.Sc Physics