Past exam of the physics 2 paper of the University of Oxford / 2022 / 1

Position A disconnects the battery and closes the resistor-inductor loop. Taking clockwise electric current as positive, Kirchhoff's voltage law for this RL circuit givesThe electric current through an ideal inductor remains continuous across a switch with no impulsive voltage. Thus $I(0^{+})=I_0$, and separation of variables givesIf the switch remains at A, the same result holds for every $t>0$. The time constant is $L/R$; the stored magnetic energy is dissipated by Joule heating.

Put $s=e^{-R\Delta t/L}$ and $J=V/R$, with $0<s<1$. The first interval leaves electric current $I_1=s{I}_0$. Position B reconnects the battery, so the driven RL circuit obeys $L\dot{I}+RI=V$. Multiplication by the integrating factor $e^{Rt/L}$, or subtraction of the constant solution $J$, givesConsequentlyThe same two intervals give the recurrence relation $I_{2n+2}=s^2{I}_{2n}+J(1-s)$. Its fixed point is $I_{\ast }=J/(1+s)$. Subtracting this fixed point makes the recurrence relation homogeneous:This is the upper endpoint of the eventual periodically switched RL circuit response, immediately before switching from B to A. The lower endpoint is $sJ/(1+s)$. Their sum is $J$, consistent with equal charging and discharging intervals.