1. Given data

   • Inductance: $L = 300\,\text{mH} = 0.3\,\text{H}$
   • Resistance: $R = 2\,\Omega$
   • Voltage: $V = 2\,\text{V}$

2. Current growth in an $RL$ circuit

   When a DC source is connected to an $RL$ circuit, the current grows as

   $$i(t) = I_0\left(1 - e^{-t/\tau}\right)$$

   where

   $$I_0 = \frac{V}{R}, \qquad \tau = \frac{L}{R}$$

3. Find the steady state current

   $$I_0 = \frac{V}{R} = \frac{2}{2} = 1\,\text{A}$$

   Half of steady state current is

   $$\frac{I_0}{2} = 0.5\,\text{A}$$

4. Find the time constant

   $$\tau = \frac{L}{R} = \frac{0.3}{2} = 0.15\,\text{s}$$

5. Use the condition that current is half of steady state value

   $$i(t) = \frac{I_0}{2}$$

   So,

   $$I_0\left(1 - e^{-t/\tau}\right) = \frac{I_0}{2}$$

   Dividing by $I_0$,

   $$1 - e^{-t/\tau} = \frac{1}{2}$$ $$e^{-t/\tau} = \frac{1}{2}$$

   Taking natural log,

   $$-\frac{t}{\tau} = \ln\left(\frac{1}{2}\right) = -\ln 2$$ $$t = \tau \ln 2$$

6. Substitute $\tau = 0.15\,\text{s}$

   $$t = 0.15 \ln 2$$ $$t \approx 0.15 \times 0.693 = 0.104\,\text{s}$$ $$t \approx 0.1\,\text{s}$$

7. Evaluate options

   • A: $0.1\,\text{s}$ ✅
   • B: $0.05\,\text{s}$ ❌
   • C: $0.3\,\text{s}$ ❌
   • D: $0.15\,\text{s}$ ❌

Final Answer: Option A.