A coil of inductance 300mH and resistance `2 Omega. The current reaches half of its steady state value in

Options

A. 0.1 s
B. 0.05 s
C. 0.3 s
D. 0.15 s

Explanation

To solve the problem, we need to find the time it takes for the current in an RL circuit (a circuit with a resistor and an inductor) to reach half of its steady-state value.

### Step-by-Step Solution:

1. **Identify the given values:**
- Inductance (L) = 300 mH = 0.3 H
- Resistance (R) = 2 Ω
- Voltage (V) = 2 V

2. **Calculate the steady-state current (I₀):**
The steady-state current in an RL circuit can be calculated using Ohm's law:
\[ I_0 = \frac{V}{R} \]
Substituting the values:
\[ I_0 = \frac{2 \, \text{V}}{2 \, \Omega} = 1 \, \text{A} \]

3. **Determine half of the steady-state current:**
Half of the steady-state current (I_half) is:
\[ I_{\text{half}} = \frac{I_0}{2} = \frac{1 \, \text{A}}{2} = 0.5 \, \text{A} \]

4. **Use the time constant (τ) of the RL circuit:**
The time constant (τ) for an RL circuit is given by:
\[ \tau = \frac{L}{R} \]
Substituting the values:
\[ \tau = \frac{0.3 \, \text{H}}{2 \, \Omega} = 0.15 \, \text{s} \]

5. **Use the formula for current in an RL circuit:**
The current (I) at any time (t) in an RL circuit is given by:
\[ I(t) = I_0 \left(1 - e^{-\frac{R}{L}t}\right) \]
To find the time when the current reaches half of its steady-state value:
\[ 0.5 = 1 \left(1 - e^{-\frac{2}{0.3}t}\right) \]
Rearranging gives:
\[ e^{-\frac{2}{0.3}t} = 0.5 \]

6. **Take the natural logarithm of both sides:**
\[ -\frac{2}{0.3}t = \ln(0.5) \]
Therefore:
\[ t = -\frac{0.3}{2} \ln(0.5) \]

7. **Calculate the value of t:**
Using the value of \(\ln(0.5) \approx -0.693\):
\[ t = -\frac{0.3}{2} \times (-0.693) \approx \frac{0.3 \times 0.693}{2} \approx 0.104 \, \text{s} \]

### Final Answer:
The time taken for the current to reach half of its steady-state value is approximately \(0.104 \, \text{s}\).

---