 Solution

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.