Question

A coil of inductance $1 H$ and resistance $100 \Omega$ 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 $\operatorname{In} 2=0.693, e ^{-3 / 2}=0.25$ )

• A. t = 10 ms; U = 2 mJ
• B. t = 10 ms; U = 1 mJ
• C. t = 7 ms; U = 1 mJ
• D. t = 7 ms; U = 2 mJ

Answer 1

Answer: (C)

Given circuit is R - L growth circuit

$ i=\frac{E}{R}\left(1-e^{-t / \tau}\right) $
$ i=\frac{E}{2 R}=\frac{E}{R}\left(1-e^{-t / \tau)}\right)$
Solving $t =\tau \ln 2$
$ t =\frac{1}{R} \ln 2=\frac{1}{100} 0.693=0.00693$
$ =7 ms$
$ i(15 ms )=\frac{E}{R}\left(1- e ^{-\frac{15}{10}}\right)$
$ i =\frac{6}{100}(1-1 / 4)=\frac{3}{4} \times \frac{6}{100} $
$U=\frac{1}{2} LI ^2$
by solving we get $U =1\, mJ$.