JEE Main 2019 Physics question
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:
Answer: (D)
Solution
L = 25/15 = 5/3 H; ΔU = ½L(25 − 10) = 437.5 J.