Question

A coil of inductance $1H$ and resistance $100\Omega$ is connected to a battery of $6V$. 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 $15ms$ after the circuit is switched on.
(Given $\mathrm{In}2=0.693,e^{-3/2}=0.25$ )

• 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

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