The self induced emf of a coil is 25 volts. When the current in it is changed at uniform rate from 10 A to 25 A in 1 s , the change in the energy of the inductance is:

Step-by-step solution

Induced emf / back emf in an inductor coil is given by ε = Ldi dt ⇒ 25 = L × 25 - 10 1 ⇒ L = 25 15 = 5 3 H Energy of inductor is given by E = 1 2 Li 2 Energy change △ E = 1 2 L I 2 2 - I 1 2 ⇒ ∆ E = 1 2 × 5 3 625 - 100   J ∆ E = 437 .5   J