A simple `LR` circuit is connected to a battery at time `t = 0`. The energy stored in the inductor reaches half its maximum value at time

Options

A. `(R )/(L)` In `[(sqrt(2))/(sqrt(2) - 1)]`
B. `(L )/(R)` In `[(sqrt(2)-1)/(sqrt(2))]`
C. `(L )/(R)` In `[(sqrt(2))/(sqrt(2) - 1)]`
D. `(R)/(L)` In `[(sqrt(2)-1)/(sqrt(2))]`

Explanation

To solve the problem of finding the time at which the energy stored in an inductor reaches half its maximum value in an LR circuit, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Maximum Energy Stored in the Inductor:**
The maximum energy stored in an inductor \( U_{\text{max}} \) is given by the formula:
\[ U_{\text{max}} = \frac{1}{2} L I_0^2 \]
where \( I_0 \) is the steady-state current when \( t \to \infty \).

2. **Current in the LR Circuit:**
The current \( I(t) \) in an LR circuit connected to a battery is given by:
\[ I(t) = \frac{V}{R} \left(1 - e^{-\frac{R}{L}t}\right) \]
where \( V \) is the voltage of the battery and \( R \) is the resistance.

3. **Finding the Current When Energy is Half Maximum:**
We want to find the time \( t \) when the energy stored in the inductor is half of its maximum value:
\[ U(t) = \frac{1}{2} L I(t)^2 = \frac{1}{2} U_{\text{max}} = \frac{1}{4} L I_0^2 \]
Setting \( I(t) = \frac{V}{R} \) when \( t \to \infty \), we have:
\[ I_0 = \frac{V}{R} \]
Thus, we need:
\[ \frac{1}{2} L I(t)^2 = \frac{1}{4} L \left(\frac{V}{R}\right)^2 \]

4. **Substituting for \( I(t) \):**
Substitute \( I(t) \) into the equation:
\[ \frac{1}{2} L \left(\frac{V}{R} \left(1 - e^{-\frac{R}{L}t}\right)\right)^2 = \frac{1}{4} L \left(\frac{V}{R}\right)^2 \]

5. **Simplifying the Equation:**
Cancel \( \frac{1}{2} L \) from both sides:
\[ \left(\frac{V}{R} \left(1 - e^{-\frac{R}{L}t}\right)\right)^2 = \frac{1}{2} \left(\frac{V}{R}\right)^2 \]
Taking square roots:
\[ \frac{V}{R} \left(1 - e^{-\frac{R}{L}t}\right) = \frac{V}{R \sqrt{2}} \]
Cancel \( \frac{V}{R} \) (assuming \( V \neq 0 \)):
\[ 1 - e^{-\frac{R}{L}t} = \frac{1}{\sqrt{2}} \]

6. **Solving for \( e^{-\frac{R}{L}t} \):**
Rearranging gives:
\[ e^{-\frac{R}{L}t} = 1 - \frac{1}{\sqrt{2}} \]

7. **Taking the Natural Logarithm:**
Taking the natural logarithm:
\[ -\frac{R}{L}t = \ln\left(1 - \frac{1}{\sqrt{2}}\right) \]
Thus, solving for \( t \):
\[ t = -\frac{L}{R} \ln\left(1 - \frac{1}{\sqrt{2}}\right) \]

8. **Using the Identity:**
We can simplify \( 1 - \frac{1}{\sqrt{2}} \) as \( \frac{\sqrt{2} - 1}{\sqrt{2}} \):
\[ t = -\frac{L}{R} \ln\left(\frac{\sqrt{2} - 1}{\sqrt{2}}\right) \]
This can be rewritten using logarithmic properties:
\[ t = \frac{L}{R} \left(\ln(\sqrt{2} - 1) - \ln(\sqrt{2})\right) \]

9. **Final Expression:**
This leads us to the final expression:
\[ t = \frac{L}{R} \ln\left(\frac{\sqrt{2}}{\sqrt{2} - 1}\right) \]
Which matches with the options provided.

### Conclusion:
After comparing with the options, we find that the correct answer is:
\[ t = \frac{L}{R} \ln\left(\frac{\sqrt{2}}{\sqrt{2} - 1}\right) \]