In an LR-circuit the current attains one-third of its final steady value in 5s. What is the time-constant of the circuit ? `(log_(2) = 0.405)`

Answer

To solve the problem, we need to find the time constant (τ) of an LR circuit where the current reaches one-third of its final steady value in 5 seconds.

### Step-by-Step Solution:

1. **Understand the Current Growth in an LR Circuit**:
The current \( I(t) \) in an LR circuit can be described by the equation:
\[ I(t) = I_0 \left(1 - e^{-\frac{t}{\tau}}\right) \]
where \( I_0 \) is the final steady value of the current, \( t \) is the time, and \( \tau \) is the time constant.

2. **Set Up the Equation for One-Third of Final Value**:
According to the problem, the current attains one-third of its final steady value in 5 seconds. Therefore, we can write:
\[ I(5) = \frac{I_0}{3} \]
Substituting into the equation gives:
\[ \frac{I_0}{3} = I_0 \left(1 - e^{-\frac{5}{\tau}}\right) \]

3. **Simplify the Equation**:
Dividing both sides by \( I_0 \) (assuming \( I_0 \neq 0 \)):
\[ \frac{1}{3} = 1 - e^{-\frac{5}{\tau}} \]
Rearranging this gives:
\[ e^{-\frac{5}{\tau}} = 1 - \frac{1}{3} = \frac{2}{3} \]

4. **Take the Natural Logarithm**:
Taking the natural logarithm of both sides:
\[ -\frac{5}{\tau} = \ln\left(\frac{2}{3}\right) \]
This can be rewritten as:
\[ \frac{5}{\tau} = -\ln\left(\frac{2}{3}\right) \]

5. **Use the Given Value of \( \ln(2) \)**:
We know from the problem that \( \ln(2) = 0.405 \). We can express \( \ln\left(\frac{2}{3}\right) \) as:
\[ \ln\left(\frac{2}{3}\right) = \ln(2) - \ln(3) \]
Since \( \ln(3) \) can be approximated using \( \ln(3) \approx 1.099 \) (not provided in the question, but can be found), we can calculate:
\[ \ln\left(\frac{2}{3}\right) \approx 0.405 - 1.099 = -0.694 \]

6. **Calculate the Time Constant \( \tau \)**:
Substituting back into the equation gives:
\[ \frac{5}{\tau} = 0.694 \]
Therefore:
\[ \tau = \frac{5}{0.694} \approx 7.21 \text{ seconds} \]

### Final Answer:
The time constant \( \tau \) of the circuit is approximately **7.21 seconds**.