Verified Solution • 2025–26 Session moderate

QJEEM07Ph93cc: JEE Main 2007 Physics - Class 12 JEEM07Ph93cc Solution: An ideal coil of is connected in series with a resistance of…

An ideal coil of 10H10H is connected in series with a resistance of 5Ω5\Omega and a battery of 5V5V . 22 second after the connection is made, the current flowing in ampere in the circuit is

Multiple Choice Options

  • A (1−e−1)\left( {1 - {e^{ - 1}}} \right)
  • B (1−e)\left( {1 - e} \right)
  • C ee
  • D e−1{{e^{ - 1}}}

Step-by-Step Verified Solution

KEY CONCEPT : I=I0(1−e−RLt)I = {I_0}\left( {1 - {e^{ - {R \over L}t}}} \right)

(When current is in growth in LRLR circuit)

=ER(1−e−RLt)= {E \over R}\left( {1 - {e^{ - {R \over L}t}}} \right)

=55(1−e−510×2)= {5 \over 5}\left( {1 - {e^{ - {5 \over {10}} \times 2}}} \right)

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

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