= Solution
Opening the switch removes the source and $R_1$ but leaves a closed path through the <inductor> and $R_2$. Set the new time origin at switch-off. <Inductor> current continuity gives $i_L(0)=V_0/R_1=10\,\mathrm A$. Since the two parallel branches are now isolated, <Kirchhoff's current law> gives $I_2=-i_L$, and $L\dot i_L=-R_2i_L$. Thus
$$
i_L(t)=\frac{V_0}{R_1}e^{-R_2t/L},\qquad
\tau_{\rm off}=\frac L{R_2}=7.90\,\mathrm{\mu s}.
$$
All the initially stored <magnetic energy> is ultimately converted to <Joule heating> in $R_2$:
$$
\boxed{Q_{2,\rm off}=\frac12L\left(\frac{V_0}{R_1}\right)^2
=3.95\times10^{-2}\,\mathrm J.}
$$
Direct integration of $R_2i_L^2(t)$ gives the same result. The initial shared <voltage> is $v(0^+)=-R_2i_L(0)=-1000\,\mathrm V$; its reversal drives the return current. The ideal circuit permits this transient <voltage>; real switching would also be limited by stray <capacitance>, breakdown and other parasitic effects.
Back to article page
