The current through a 10-mH inductor is 6{ e}^{ -t/2}2 A. Find the voltage and the power at t = 3 s.
Question 6.34: The current through a 10-mH inductor is 6 e^-t/2 /2 A. Find ...
i = 6{ e}^{ -t/2}
\begin{array}{l}\begin{aligned}\mathrm{v}&=\mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}=10 \times 10^{-3}(6)\left(\frac{1}{2}\right) \mathrm{e}^{-\mathrm{t} / 2} \\&=-30 \mathrm{e}^{-t/2} \mathrm{mV}\end{aligned} \\\mathrm{v}(3)=-30 \mathrm{e}^{-3 / 2} \mathrm{mV}=-6.694 \mathrm{mV} \\\mathrm{p}=\mathrm{vi}=-180 \mathrm{e}^{-\mathrm{t}} \mathrm{mW}\end{array}p(3) = -180{ e}^{ -3} mW = –8.962 mW
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