The self induced emf of a coil is 25 volts. When the current in it...

Induced emf / back emf in an inductor coil is given by

$\left|ε\right| = \dfrac{Ldi}{dt}$

$⇒ 25 = L \times \dfrac{\left(25 - 10\right)}{1}$

$⇒ L = \dfrac{25}{15} = \dfrac{5}{3}H$

Energy of inductor is given by

$E = \dfrac{1}{2}{Li}^{2}$

Energy change $△ E = \dfrac{1}{2}L\left({{I}_{2}}^{2} - {{I}_{1}}^{2}\right)$

$⇒ ∆ E = \dfrac{1}{2} \times \dfrac{5}{3}\left[625 - 100\right]J$

$∆ E = 437 .5 J$