Alternating Current

JEE Main level, numerical answer, medium

57 of 190

Question 57 of 190

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)

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Answer: 314

Given: Magnetic field energy =16= 16 J, Thermal power dissipated =32= 32 W, Irms=2I_{rms} = 2 A, f=50f = 50 Hz.
From energy stored in inductor:
12LIrms2=16\frac{1}{2}L I_{rms}^2 = 16
12⋅L⋅4=16\frac{1}{2} \cdot L \cdot 4 = 16
L=8 HL = 8 \text{ H}
From thermal power dissipated:
Irms2R=32I_{rms}^2 R = 32
4R=324R = 32
R=8  ΩR = 8 \; \Omega
Inductive reactance:
XL=ωL=2πfL=2×3.14×50×8=2512  ΩX_L = \omega L = 2\pi f L = 2 \times 3.14 \times 50 \times 8 = 2512 \; \Omega
XLR=25128=314\frac{X_L}{R} = \frac{2512}{8} = 314

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