1. Initial steady current before switching

   The circuit has: $L = 100\,\text{mH} = 0.1\,\text{H}, \quad R = 100\,\Omega, \quad E = 100\,\text{V}$

   After a long time with the battery connected, the inductor behaves like a short circuit in DC steady state.

   So the current just before disconnecting the battery is $I_0 = \frac{E}{R} = \frac{100}{100} = 1\,\text{A}$

2. Current after battery is disconnected and points $A$ and $B$ are shorted

   After shorting $A$ and $B$, the battery is removed from the effective loop, and the resistor-inductor loop remains.

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

3. Compute the time constant

   $\tau = \frac{L}{R} = \frac{0.1}{100} = 0.001\,\text{s} = 1\,\text{ms}$

4. Find current at $t = 1\,\text{ms}$

   Since $t = \tau,$ we get $I(t) = I_0 e^{-1} = 1 \cdot e^{-1} = \frac{1}{e}\,\text{A}$

5. Check options

   • A: $\frac{1}{e}\,\text{A}$ ✅
   • B: $e\,\text{A}$ ❌
   • C: $0.1\,\text{A}$ ❌
   • D: $1\,\text{A}$ ❌

Therefore, the correct answer is Option A.