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:

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TestHub · Accepted Answer

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

$|ε| = \frac{Ldi}{dt}$

$\Rightarrow 25 = L \times \frac{(25 - 10)}{1}$

$\Rightarrow L = \frac{25}{15} = \frac{5}{3} H$

Energy of inductor is given by

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

Energy change $△ E = \frac{1}{2} L ({I_{2}}^{2} - {I_{1}}^{2})$

$\Rightarrow ∆ E = \frac{1}{2} \times \frac{5}{3} [625 - 100] J$

$∆ E = 437 .5 J$

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