Q. 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 1s, the change in the energy of the inductance is :

Solution:

$L \frac{di}{dt} =25$
$ L\times\frac{15}{1} =25 $
$ L= \frac{5}{3}H$
$ \Delta E = \frac{1}{2} \times\frac{5}{3}\times\left(25^{2} - 10^{2}\right) = \frac{5}{6} \times525 = 437.5 J $

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