Consider an RL circuit with R=10 ohms, L=10 microHenry, and V = 30 volts. Suppose that I = 0 at t=0. Find the energy stored in the inductor as t goes to infinity.

Question

anonymous · Answer

It has been a while since I played with induction problems, so I may be a little rusty. The instantaneous power supplied to an inductor is given by $$P=iv=L\frac{di}{dt}$$ and hence the energy $E$ stored on an inductor after time $t$ is given by

$$E=\int\limits_{o}^{t}Pdt=\int\limits_{0}^{I}Li di=\frac{1}{2}LI^2$$, where $I^2$ is given by $$I^2=\frac{V^2}{R^2}$$
Hence $$E=\frac{LV^2}{2R^2}=\frac{(10\times10^{-6})(30)^2}{2(10)^2}=4.5\times10^{-5}\rm{J}$$or 45 $\mu$J.

I think this is correct, but as I said, I'm rusty, so you may want to get a second opinion.

anonymous · Answer

its  1/2 Li^2
i = V/R  there fore energy = 0.5 X 10^-2 X 9
                                     = 4.5 X 10^-2