Q. 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)

Correct Answer: 314

Solution

Power dissipated due to resistance is given by:

P = I_rms² R

Substituting the given values:

32 = (2)² R ⇒ R = 8 Ω

Magnetic energy stored in an inductor is:

U = ½ L I₀²

For AC current:

I₀ = √2 I_rms = 2√2 A

Substituting:

16 = ½ L (2√2)² = 4L

L = 4 H

Inductive reactance is:

X_L = 2πfL = 2 × 3.14 × 50 × 4 = 1256 Ω

Hence, the ratio of inductive reactance to resistance is:

X_L / R = 1256 / 4 = 314

Therefore, the required answer is 314.

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