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Question:

At a given instant an 62 mH inductor stored an energy of 81 µJ.?

I have no idea what to do with this problem, if you can hel p me out, I'd appreciate it!At a given instant an 62 mH inductor stored an energy of 81 µJ. What is the current in the inductor at this instant?___________ mA If the current is increasing at the rate of 95 mA/s, how long does it take for the energy to increase by a factor of 51 from the initial value? _______________ sPlease show me the work done so I understand what equations you're using and how to utilize them in other problems. Much appreicated!Angel Mina

Answer:

1) for an inductor, the energy stored is U (1/2) L i^2, where 'i' is the current value. then, 81x10^-6 (1/2) x 62x10^-3 i^2 or i^2 2x81x10^-6 / 62x10^-3 26.13 x10^-4 or i 5.11x10^-2 amp 51.1x10^-3 amp 51.1 mA 2) since U (1/2) L i^2 then, U / t^2 (1/2) L (i/t)^2 ie., t^2 2 U / L (1/t)^2 2x65.61x10^-6 / 62x10^-3 x (95x10^-3)^2 131220 / 559550 0.2345 ie., time t 0.484 sec 0.5 sec (nearly)
Dec 7, 2017
Quite simple! Energy stored in an inductor0.5 LI² Given, L62mH62*10^-3 Henry and Energy E stored 81*10^-6 J which gives 81*10^-60.5*62*10^-3*I² or I ≈ 0.05112 A ≈ 51.12 mA
Dec 7, 2017

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