The current through a 12-mH inductor is i( t)=30t{e}^{-2t} A, t ≥ 0 .Determine: (a) the voltage across the inductor, (b) the power being delivered to the inductor at t = 1 s, (c) the energy stored in the inductor at t = 1 s.
Question 6.36: The current through a 12-mH inductor is i(t)=30t e^-2t A, t ...
(a) V=L \frac { di }{ dt }12 \mathrm{x}10^{-3}(30e^{-2t}-60te^{-2t})=(0.36-0.72t)e^{-2t}V
(b) P=Vi=(0.36 0.72 1) e^{-2}\mathrm{x}30\mathrm{x}1e^{-2}=0.36\mathrm{x}30e^{-4}=-0.1978 W
(c) W=\frac {1}{2} Li^{2}=0.5\mathrm{x}12\mathrm{x}10^{-3}(30\mathrm{x}1\mathrm{x}e^{-2})^{2}=98.9 mJ
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