A) 740 J
B) 637.5 J
C) 540 J
D) 437.5 J
Correct Answer: D
Solution :
Induced emf \[=\,\,L\frac{di}{dt}\,=\,\frac{L\times (25-10)}{1}\] \[=L\times 15\] \[\,L\times 15=25\] \[L=\frac{25}{15}\] Change in energy of inductor \[\,=\,\,\frac{1}{2}L\left[ {{25}^{2}}-{{10}^{2}} \right]\] \[=\,\,\frac{1}{2}\times \frac{25}{15}\times 15\times 35\] \[=\,\,\frac{25\times 35}{2}\] = 437.5 JouleYou need to login to perform this action.
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