The current in an `L-R` circuit builds upto `3//4th` of its steady state value in `4sec`. Then the time constant of this circuit is

Options

A. `(1)/("ln 2")s`
B. `(2)/("ln 2")s`
C. `(3)/("ln 2")s`
D. `(4)/("ln 2")s`

Explanation

To find the time constant of the L-R circuit where the current builds up to \( \frac{3}{4} \) of its steady state value in 4 seconds, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula**:
The current \( I(t) \) in an L-R circuit is given by the equation:
\[ I(t) = I_0 \left(1 - e^{-\frac{t}{\tau}}\right) \]
where \( I_0 \) is the steady state current and \( \tau \) is the time constant.

2. **Set Up the Equation**:
We know that at \( t = 4 \) seconds, the current \( I(4) \) is \( \frac{3}{4} I_0 \). Therefore, we can set up the equation:
\[ \frac{3}{4} I_0 = I_0 \left(1 - e^{-\frac{4}{\tau}}\right) \]

3. **Simplify the Equation**:
Dividing both sides by \( I_0 \) (assuming \( I_0 \neq 0 \)):
\[ \frac{3}{4} = 1 - e^{-\frac{4}{\tau}} \]

4. **Rearranging the Equation**:
Rearranging gives:
\[ e^{-\frac{4}{\tau}} = 1 - \frac{3}{4} = \frac{1}{4} \]

5. **Taking the Natural Logarithm**:
Taking the natural logarithm of both sides:
\[ -\frac{4}{\tau} = \ln\left(\frac{1}{4}\right) \]

6. **Simplifying the Logarithm**:
We can express \( \ln\left(\frac{1}{4}\right) \) as:
\[ \ln\left(\frac{1}{4}\right) = \ln(1) - \ln(4) = 0 - \ln(4) = -\ln(4) \]
Thus, we have:
\[ -\frac{4}{\tau} = -\ln(4) \]

7. **Solving for the Time Constant \( \tau \)**:
Rearranging gives:
\[ \frac{4}{\tau} = \ln(4) \]
Therefore, solving for \( \tau \):
\[ \tau = \frac{4}{\ln(4)} \]

8. **Expressing \( \ln(4) \)**:
We can express \( \ln(4) \) as:
\[ \ln(4) = \ln(2^2) = 2\ln(2) \]
Substituting this back into the equation for \( \tau \):
\[ \tau = \frac{4}{2\ln(2)} = \frac{2}{\ln(2)} \]

### Final Answer:
Thus, the time constant \( \tau \) of the circuit is:
\[ \tau = \frac{2}{\ln(2)} \text{ seconds} \]