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JEE Main 2026 (28 Jan, Shift 2) · RL and RC Series Circuits

RL and RC Series Circuits question from JEE Main 2026 (28 Jan, Shift 2)

Q7JEE Main 2026 · 28 JAN · SHIFT 2Medium

RL and RC Series Circuits

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\pi = 3.14)

Numerical answer type. Work it out, then open the answer.

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Answer: 314

  1. Stored energy U=12LI2U = \frac{1}{2}LI^2: 16=12L(2)2⇒L=8 H16 = \frac{1}{2} L (2)^2 \Rightarrow L = 8\text{ H}.
  2. Power P=I2RP = I^2R: 32=4R⇒R=8 Ω32 = 4R \Rightarrow R = 8\ \Omega.
  3. XL=2πfL=2×3.14×50×8=2512 ΩX_L = 2\pi f L = 2 \times 3.14 \times 50 \times 8 = 2512\ \Omega
  4. XLR=25128=314\frac{X_L}{R} = \frac{2512}{8} = 314
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