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Electromagnetic Induction question

2007 · Shift 0 · Q70
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Electromagnetic Induction question

2007 · Shift 0 · Q70

JEE MainPhysicsElectromagnetic InductionMCQ+4 / −1
An ideal coil of 10H10H10H is connected in series with a resistance of 5Ω5\Omega5Ω and a battery of 5V5V5V. 222 second after the connection is made, the current flowing in ampere in the circuit is
  1. A
    (1−e−1)\left( {1 - {e^{ - 1}}} \right)(1−e−1)
  2. B
    (1−e)\left( {1 - e} \right)(1−e)
  3. C
    eee
  4. D
    e−1{{e^{ - 1}}}e−1
View written solutionFree

Correct answer: A

  1. Identify the circuit type

    This is a series RLRLRL circuit with:

    • Inductance: L=10 HL = 10\,\text{H}L=10H
    • Resistance: R=5 ΩR = 5\,\OmegaR=5Ω
    • Battery voltage: V=5 VV = 5\,\text{V}V=5V

  2. Use the current growth formula for an RLRLRL circuit

    When a battery is connected to an RLRLRL series circuit, the current grows as

    i(t)=VR(1−e−Rt/L)i(t) = \frac{V}{R}\left(1 - e^{-Rt/L}\right)i(t)=RV​(1−e−Rt/L)

  3. Substitute the given values

    First compute:

    VR=55=1\frac{V}{R} = \frac{5}{5} = 1RV​=55​=1

    And

    RtL=5⋅210=1\frac{Rt}{L} = \frac{5 \cdot 2}{10} = 1LRt​=105⋅2​=1

    Therefore,

    i(2)=1(1−e−1)i(2) = 1\left(1 - e^{-1}\right)i(2)=1(1−e−1)

    So the current after 222 s is

    i=1−e−1 Ai = 1 - e^{-1}\,\text{A}i=1−e−1A

  4. Match with options

    • A: (1−e−1)\left(1 - e^{-1}\right)(1−e−1) ✅
    • B: (1−e)\left(1 - e\right)(1−e) ❌
    • C: eee ❌
    • D: e−1e^{-1}e−1 ❌

  5. Final answer

    The correct option is A.

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