An inductance `L` and a resistance `R` are first connected to a battery. After some time the battery is disconnected but `L` and `R` remain connected in a closed circuit. Then the current reduces to `37%` of its initial value in

Options

A. RL second
B. `(R)/(L)`second
C. `(L)/(R)`second
D. `(1)/(LR)`second

Explanation

To solve the problem, we need to analyze the behavior of the current in an RL circuit when the battery is disconnected. The current in the circuit decreases over time due to the inductor's property of opposing changes in current.

### Step-by-Step Solution:

1. **Understanding the Circuit**: Initially, we have an inductor (L) and a resistor (R) connected to a battery. When the battery is connected, the current increases until it reaches a maximum value (I₀) determined by the battery voltage and the resistance.

2. **Current Growth Equation**: The current growth in the circuit when the battery is connected can be described by the equation:
\[ I(t) = I_0 (1 - e^{-t/\tau}) \]
where \( \tau = \frac{L}{R} \) is the time constant of the circuit.

3. **Battery Disconnection**: After some time, the battery is disconnected. The inductor will now oppose the change in current, and the current will start to decay.

4. **Current Decay Equation**: The current decay in the circuit after the battery is disconnected is given by:
\[ I(t) = I_0 e^{-t/\tau} \]

5. **Finding Time for 37% Current**: We need to find the time \( t \) when the current reduces to 37% of its initial value \( I_0 \). This means:
\[ I(t) = 0.37 I_0 \]
Substituting this into the decay equation:
\[ 0.37 I_0 = I_0 e^{-t/\tau} \]

6. **Simplifying the Equation**: We can divide both sides by \( I_0 \) (assuming \( I_0 \neq 0 \)):
\[ 0.37 = e^{-t/\tau} \]

7. **Taking Natural Logarithm**: Taking the natural logarithm of both sides gives:
\[ \ln(0.37) = -\frac{t}{\tau} \]

8. **Solving for Time \( t \)**: Rearranging the equation to solve for \( t \):
\[ t = -\tau \ln(0.37) \]

9. **Substituting \( \tau \)**: Recall that \( \tau = \frac{L}{R} \), so we can substitute this into our equation:
\[ t = -\frac{L}{R} \ln(0.37) \]

10. **Final Expression**: Thus, the time taken for the current to reduce to 37% of its initial value is:
\[ t = \frac{L}{R} \ln\left(\frac{1}{0.37}\right) \]