Question 89 of 190

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 $\ln 2=0.693$, ${e}^{\dfrac{-3}{2}}=0.25$)

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

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Answer: (C) $t=7ms;U=1mJ$

Given circuit is $R-L$ growth circuit

The current is the circuit is given by,
$i=\dfrac{E}{R}\left(1-{e}^{-\dfrac{t}{\tau }}\right)$
For current to be half of the peak value,
$i=\dfrac{E}{2R}=\dfrac{E}{R}\left(1-{e}^{-\dfrac{t}{\tau }}\right)$
Solving $t=\tau \ln 2$
$t=\dfrac{L}{R}\ln 2=\dfrac{1}{100}0.693=0.00693$$≃7ms$
At time $t=15ms$,
$i\left(15ms\right)=\dfrac{E}{R}\left(1-{e}^{-\dfrac{15}{10}}\right)$
$\Rightarrow i=\dfrac{6}{100}\left(1-\dfrac{1}{4}\right)=\dfrac{3}{4}\times \dfrac{6}{100}=0.045A$
$\Rightarrow U=\dfrac{1}{2}L{i}^{2}$
by solving we get $U=1mJ$.