A coil of inductance 1H and resistance 100 Ω is connected to a battery of 6.V. Determine approximately:

(A)The time elapsed before the current acquires half of its steady – state value.

(B)The energy stored in the magnetic field associated with the coil at instant 15ms after the circuit is switched on. (Given In 2 = 0.693, e3/2 = 0.25)

Option 1 -

t = 10ms; U = 2mJ

Option 2 -

t = 10ms; U = 1mJ

Option 3 -

 t = 7ms; U = 1mJ

Option 4 -

 t = 7ms; U = 2mJ

0 3 Views | Posted 2 months ago
Asked by Shiksha User

  • 1 Answer

  • V

    Answered by

    Vishal Baghel | Contributor-Level 10

    2 months ago
    Correct Option - 3


    Detailed Solution:

    L = 1H, R = 100 Ω

    As,i=i0et/τ

    Fori=i02i02=i0et/τ

    ln2=t/τ

    i=6100e151/100=.06e1500×103=0.06e1.5=0.06×0.25=.0.015A

    So, U=12Li2=12×1×(15×103)2=12(225)×106=112.5×106J=0.1125mJ

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