A coil of inductance 1H and resistance 100Ω is connected to a...

Given circuit is $R - L$ growth circuit

The current is the circuit is given by,

$i = \dfrac{E}{R}\left(1 - {e}^{- \dfrac{t}{τ}}\right)$

For current to be half of the peak value,

$i = \dfrac{E}{2R} = \dfrac{E}{R}\left(1 - {e}^{- \dfrac{t}{τ}}\right)$

Solving $t = τ\ln 2$

$t = \dfrac{L}{R}\ln 2 = \dfrac{1}{100}0 . 693 = 0 . 00693$$≃ 7ms$

At time $t = 15ms$,

$i\left(15ms\right) = \dfrac{E}{R}\left(1 - {e}^{- \dfrac{15}{10}}\right)$

$⇒ i = \dfrac{6}{100}\left(1 - \dfrac{1}{4}\right) = \dfrac{3}{4} \times \dfrac{6}{100} = 0 . 045A$

$⇒ U = \dfrac{1}{2}L{i}^{2}$

by solving we get $U = 1mJ$.