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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637.5 J

740 J

540 J

437.5 J

answer is A.

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Detailed Solution

$\begin{array}{l} ΔE = \frac{1}{2} {Li}_{f}^{2} - \frac{1}{2} {Li}_{i}^{2} and L \cdot \frac{d ϕ}{dt} = 25 \\ L \times \frac{25 - 10}{1} = 25 \\ 15 L = 25, L = \frac{25}{15} = \frac{5}{3} H \\ ΔE = \frac{1}{2} L (i_{f}^{2} - i_{i}^{2}) = \frac{1}{2} \times \frac{5}{3} (625 - 100) \\ = \frac{5}{6} \times 525 = 437.5 J \end{array}$

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