A coil of inductance `300mh` and resistance `2Omega` is connected to a source of voltage `2V`. The current reaches half of its steady state value in

Options

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

Explanation

To solve the problem step by step, we will follow the same reasoning as outlined in the video transcript.

### Step 1: Identify the given values
- Inductance \( L = 300 \, \text{mH} = 300 \times 10^{-3} \, \text{H} \)
- Resistance \( R = 2 \, \Omega \)
- Voltage \( V = 2 \, \text{V} \)

### Step 2: Calculate the steady state current
The steady state current \( I_s \) can be calculated using Ohm's law:
\[ I_s = \frac{V}{R} \]
Substituting the values:
\[ I_s = \frac{2 \, \text{V}}{2 \, \Omega} = 1 \, \text{A} \]

### Step 3: Determine the current when it reaches half of the steady state value
Half of the steady state current \( I_{1/2} \) is:
\[ I_{1/2} = \frac{I_s}{2} = \frac{1 \, \text{A}}{2} = 0.5 \, \text{A} \]

### Step 4: Use the formula for current in an LR circuit
The current \( I(t) \) in an LR circuit is given by:
\[ I(t) = I_s \left(1 - e^{-\frac{R t}{L}}\right) \]
Setting \( I(t) = I_{1/2} \):
\[ 0.5 = 1 \left(1 - e^{-\frac{R t}{L}}\right) \]

### Step 5: Solve for \( e^{-\frac{R t}{L}} \)
Rearranging the equation:
\[ e^{-\frac{R t}{L}} = 1 - 0.5 = 0.5 \]
This can also be expressed as:
\[ e^{-\frac{R t}{L}} = \frac{1}{2} \]

### Step 6: Take the natural logarithm of both sides
Taking the natural logarithm:
\[ -\frac{R t}{L} = \ln\left(\frac{1}{2}\right) \]
This simplifies to:
\[ -\frac{R t}{L} = -\ln(2) \]
Thus:
\[ \frac{R t}{L} = \ln(2) \]

### Step 7: Solve for time \( t \)
Rearranging gives:
\[ t = \frac{L \ln(2)}{R} \]
Substituting the known values:
\[ t = \frac{300 \times 10^{-3} \, \text{H} \times 0.693}{2 \, \Omega} \]
Calculating:
\[ t = \frac{300 \times 0.693 \times 10^{-3}}{2} = \frac{207.9 \times 10^{-3}}{2} = 0.10395 \, \text{s} \approx 0.1 \, \text{s} \]

### Final Answer
The time when the current reaches half of its steady state value is approximately \( 0.1 \, \text{s} \).

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