1. The energy ð‘–, of an inductor with inductance ð¿ is given by
ð‘–= 1/2ð¿âˆ«10 ð‘¡^2ð‘’^−𑡠ð‘‘ð‘¡
For ð¿=(1 ×10−3)ð», Find ð‘–.
Let us find the energy:
"\ud835\udc56= \\frac{1}{2}\ud835\udc3f\\int_0^1 \ud835\udc61^2\ud835\udc52^\u2212\ud835\udc61 \ud835\udc51\ud835\udc61=|u=t^2,\\ dv=e^{-t}dt,\\ du=2tdt,\\ v=-e^{-t}|=\n\\frac{1}{2}\ud835\udc3f(-t^2e^{-t}|_0^1+2\\int_0^1te^{-t}dt)=\\frac{1}{2}\ud835\udc3f(-e^{-1}+2\\int_0^1te^{-t}dt)=|u=t, dv=e^{-t}dt, du=dt, v=-e^{-t}|=\\frac{1}{2}\ud835\udc3f(-e^{-1}+2(-te^{-t}|_0^1+\\int_0^1e^{-t}dt))=\n-e^{-t}|=\\frac{1}{2}\ud835\udc3f(-e^{-1}+2(-e^{-1}-e^{-t}|_0^1))=\n\\frac{1}{2}\ud835\udc3f(-e^{-1}+2(-e^{-1}-e^{-1}+1))=\n\\frac{1}{2}\ud835\udc3f(2-5e^{-1})"
For "\ud835\udc3f=(1 \u00d710^{\u22123})H" we have "i=\\frac{1}{2}(2-5e^{-1})10^{-3}H."
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