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Alternating Current question

2023 · 11 Apr · Shift 2 · Q69
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Alternating Current question

2023 · 11 Apr · Shift 2 · Q69

JEE MainPhysicsAlternating CurrentNumerical+4 / −1
A coil has an inductance of 2H2 \mathrm{H}2H and resistance of 4 Ω4 ~\Omega4 Ω. A 10 V10 \mathrm{~V}10 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 ‾\underline{\hspace{2cm}}​×10−2 J\times 10^{-2} \mathrm{~J}×10−2 J.
Numerical answer
View written solutionFree

Correct answer: 625

  1. Given data
  • Inductance: L=2 HL = 2\,\text{H}L=2H
  • Resistance: R=4 ΩR = 4\,\OmegaR=4Ω
  • Applied voltage: V=10 VV = 10\,\text{V}V=10V

We need the energy stored in the magnetic field after the current reaches equilibrium.

  1. Current at equilibrium in an RL circuit

At steady state, the inductor behaves like a short circuit for DC, so the current is determined only by the resistance:

I0=VR=104=2.5 AI_0 = \frac{V}{R} = \frac{10}{4} = 2.5\,\text{A}I0​=RV​=410​=2.5A

  1. Energy stored in an inductor

The magnetic energy stored in an inductor is

U=12LI2U = \frac{1}{2} L I^2U=21​LI2

Substitute the values:

U=12⋅2⋅(2.5)2U = \frac{1}{2} \cdot 2 \cdot (2.5)^2U=21​⋅2⋅(2.5)2

U=1⋅6.25=6.25 JU = 1 \cdot 6.25 = 6.25\,\text{J}U=1⋅6.25=6.25J

  1. Match with the required format

The question asks for

‾×10−2 J\underline{\hspace{2cm}} \times 10^{-2}\,\text{J}​×10−2J

So write 6.25 J6.25\,\text{J}6.25J as

6.25 J=625×10−2 J6.25\,\text{J} = 625 \times 10^{-2}\,\text{J}6.25J=625×10−2J

Hence the required integer is

625\boxed{625}625​

  1. Comparison with stored correct answer
  • Derived answer: 625625625
  • Stored correct answer: 625625625

They match.

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