1. Initial condition before switching

   Point $C$ is connected to $A$ for a long time, so the circuit reaches steady state with the battery connected across the series combination of $R$ and $L$.

   In steady state for a DC source:

   • Inductor behaves like a short circuit.
   • Current becomes constant.

   Hence the current just before switching is $I_0 = \frac{V}{R}$ where $V$ is the battery emf.

2. After switching at $t=0$

   Now $C$ is disconnected from $A$ and connected to $B$. This removes the battery from the loop, and the $R-L$ circuit undergoes decay of current.

   The current in an $RL$ discharge circuit is $i(t)=I_0 e^{-Rt/L}$

   Substituting $I_0=V/R$, $i(t)=\frac{V}{R}e^{-Rt/L}$

3. Voltage across the resistor

   Voltage across the resistor is $V_R = iR$ So, $V_R = \left(\frac{V}{R}e^{-Rt/L}\right)R = Ve^{-Rt/L}$

4. Voltage across the inductor

   For the inductor, $V_L = L\frac{di}{dt}$

   Since $i(t)=I_0 e^{-Rt/L},$ we get $\frac{di}{dt}=-\frac{R}{L}I_0 e^{-Rt/L}$

   Therefore,

   =-RI_0 e^{-Rt/L}$$ Using $I_0=V/R$, $$V_L=-Ve^{-Rt/L}$$

5. Required ratio at $t=L/R$

   At $t=\frac{L}{R},$ we have $e^{-Rt/L}=e^{-1}=\frac{1}{e}$

   So, $V_R=\frac{V}{e}, \qquad V_L=-\frac{V}{e}$

   Hence the ratio $\frac{V_R}{V_L} = \frac{V/e}{-V/e}=-1$

6. Option check

   • A: $\dfrac{e}{1-e}$ ❌
   • B: $1$ ❌
   • C: $-1$ ✅
   • D: $\dfrac{1-e}{e}$ ❌

Therefore, the correct answer is Option C.