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

A coil of inductance 1 H 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 an instant 15 ms after the circuit is switched on. (Given ln2=0.693, e-32=0.25)

  1. t=10 ms;U=2 mJ
  2. t=10 ms;U=1 mJ
  3. t=7 ms;U=1 mJ
  4. t=7 ms;U=2 mJ

Solution

Given circuit is R-L growth circuit

The current is the circuit is given by, 

i=ER1-e-tτ

For current to be half of the peak value,

i=E2R=ER1-e-tτ

Solving t=τln2

t=LRln2=11000.693=0.00693≃7 ms

At time t=15 ms,

i15 ms=ER1-e-1510

⇒i=61001-14=34×6100=0.045 A

⇒U=12Li2

by solving we get U=1 mJ.

Asked in: JEE Main 2022 (29 Jul Shift 1)

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