Past exam of the physics 2 paper of the University of Oxford / 2017 / 11 / d / 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 $L{\dot{i}}_L=-R_2{i}_L$. ThusAll the initially stored magnetic energy is ultimately converted to Joule heating in $R_2$:Direct integration of $R_2{i}_L^2(t)$ gives the same result. The initial shared voltage is $v(0^{+})=-R_2{i}_L(0)=-1000\mathrm{V}$; 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.