JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Alternating Current

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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 ______ .
(

π = 3.14

)

Answer: 314

Solution

The energy stored in the inductor is

\frac{1}{2} L i_{r m s}^{2} = 16

J, so with

i_{r m s} = 2

L = \frac{2 \times 16}{4} = 8

H. The thermal power dissipated is

i_{r m s}^{2} R = 32

W, giving

R = \frac{32}{4} = 8 Ω

. The inductive reactance is

X_{L} = ω L = 2 π f L = 2 \times 3.14 \times 50 \times 8 = 2512 Ω

. The ratio is

\frac{X _{L}}{R} = \frac{2512}{8} = 314

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Integer answer

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