JEE Main · 2014 · 06 Apr 2014 · Offline · Electromagnetic Induction

While C rests on A the current climbs to its steady value $i_0=\frac{\epsilon }{R}$, at which point the inductor is fully energised and $\frac{di}{dt}=0$. When C is thrown over to B at $t=0$ the cell is cut out and the resistor and inductor are left alone in a single series loop, so the current now decays as $i=i_0{e}^{-Rt/L}$. Kirchhoff's voltage law around that source-free loop reads $iR+L\frac{di}{dt}=0$, which says the drop across the resistor and the drop across the inductor are equal in size but opposite in sign at every instant: $V_R=iR=i_{0}R{e}^{-Rt/L}$ while $V_L=L\frac{di}{dt}=-i_{0}R{e}^{-Rt/L}$. Their ratio is $\frac{V_R}{V_L}=-1$, independent of time and so equal to $-1$ at $t=L/R$ as well. The two elements sit in series in this decay loop, not in parallel, which is exactly why the signed ratio is $-1$ and not $+1$.