A coil has an inductance of 2H and resistance of 4Ω . A 10V is...

The data given is

$L = 2HR = 4ΩV = 10V$

Using Ohm's law, $V = IR$,

$I = \dfrac{10V}{4Ω} = \dfrac{5}{2}A$

The energy of an inductor is given by
$E = \dfrac{1}{2}L{I}^{2} = \dfrac{1}{2} \times 2 \times \left(\dfrac{25}{4}\right) = 625 \times {10}^{- 2}J$