Question

An inductor of 3 mH has a current I=5(1-e^-5000t). Find the corresponding maximum energy stored.

Answer 1

Answer:

An inductor of 3mH has a current i = 5(1 – e-5000t). Find the corresponding maximum energy stored. E = \frac{1}{2} LI2 = 0.5 × 3 × 10-3 × 52 = 37.5 mJ.

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