In a decaying `L-R` circuit, the time after which energy stores in the inductor reduces to one fourth of its initial values is

Options

A. `(ln 2)L/R`
B. `0.5L/R`
C. `sqrt2L/R`
D. `(sqrt2/(sqrt2-1))L/R`

Explanation

To solve the problem of finding the time after which the energy stored in the inductor of a decaying L-R circuit reduces to one fourth of its initial value, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Initial Energy Stored in the Inductor**:
The energy \( U \) stored in an inductor is given by the formula:
\[ U = \frac{1}{2} L I^2 \]
where \( L \) is the inductance and \( I \) is the current through the inductor. The initial current in the circuit is denoted as \( I_0 \).

2. **Set Up the Equation for One Fourth Energy**:
The initial energy stored in the inductor is:
\[ U_0 = \frac{1}{2} L I_0^2 \]
We need to find the time \( t \) when the energy reduces to one fourth of its initial value:
\[ U = \frac{1}{4} U_0 = \frac{1}{4} \left( \frac{1}{2} L I_0^2 \right) = \frac{1}{8} L I_0^2 \]

3. **Relate Current to Energy**:
The energy at time \( t \) can also be expressed in terms of the current \( I(t) \):
\[ U(t) = \frac{1}{2} L I(t)^2 \]
Setting this equal to one fourth of the initial energy, we have:
\[ \frac{1}{2} L I(t)^2 = \frac{1}{8} L I_0^2 \]
Simplifying, we find:
\[ I(t)^2 = \frac{1}{4} I_0^2 \]
Taking the square root gives:
\[ I(t) = \frac{I_0}{2} \]

4. **Current Decay in an L-R Circuit**:
The current in an L-R circuit decays according to the equation:
\[ I(t) = I_0 e^{-\frac{t}{\tau}} \]
where \( \tau = \frac{L}{R} \) is the time constant of the circuit.

5. **Set Up the Equation**:
Now, substituting \( I(t) = \frac{I_0}{2} \) into the decay equation:
\[ \frac{I_0}{2} = I_0 e^{-\frac{t}{\tau}} \]
Dividing both sides by \( I_0 \) (assuming \( I_0 \neq 0 \)):
\[ \frac{1}{2} = e^{-\frac{t}{\tau}} \]

6. **Take the Natural Logarithm**:
Taking the natural logarithm of both sides:
\[ \ln\left(\frac{1}{2}\right) = -\frac{t}{\tau} \]
This simplifies to:
\[ -\ln(2) = -\frac{t}{\tau} \]
Therefore:
\[ t = \tau \ln(2) \]

7. **Substitute for \( \tau \)**:
Since \( \tau = \frac{L}{R} \):
\[ t = \frac{L}{R} \ln(2) \]

### Final Answer:
The time after which the energy stored in the inductor reduces to one fourth of its initial value is:
\[ t = \frac{L}{R} \ln(2) \]