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JEE Main 2023 (11 Apr, Shift 2) · Inductors & DC Transients (Mixed)

Inductors & DC Transients (Mixed) question from JEE Main 2023 (11 Apr, Shift 2)

Q5JEE Main 2023 · 11 APR · SHIFT 2Medium

Inductors & DC Transients (Mixed)

A coil has an inductance of 2 H2\text{ H} and resistance of 4 Ω4\ \Omega. A 10 V10\text{ V} is applied across the coil. The energy stored in the magnetic field after the current has built up to its equilibrium value will be ___________ ×10−2 J\times 10^{-2}\text{ J}.

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

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Answer: 625

  1. Steady current I=VR=104=2.5 AI = \frac{V}{R} = \frac{10}{4} = 2.5\text{ A}
  2. U=12LI2=12(2)(2.5)2=6.25 JU = \frac{1}{2}LI^2 = \frac{1}{2}(2)(2.5)^2 = 6.25\text{ J}
  3. 6.25 J=625×10−2 J6.25\text{ J} = 625 \times 10^{-2}\text{ J}, so the answer is 625.
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