In an LR circuit, current at t=0 is 20A . After 2s it reduced to 18A. The time constant of the circuit is (in second)

Options

A. `In((10)/(9))`
B. 2
C. `(2)/(In((10)/(9)))`
D. `2In((10)/(9))`

Answer

To solve the problem, we will use the decay equation for current in an LR circuit, which is given by:

\[ I(t) = I_0 \cdot e^{-t/\tau} \]

where:
- \( I(t) \) is the current at time \( t \),
- \( I_0 \) is the initial current,
- \( \tau \) is the time constant,
- \( e \) is the base of the natural logarithm.

### Step-by-Step Solution:

1. **Identify the given values:**
- Initial current, \( I_0 = 20 \, A \)
- Current after 2 seconds, \( I(2) = 18 \, A \)
- Time, \( t = 2 \, s \)

2. **Write the decay equation for the current at \( t = 2 \, s \):**
\[ I(2) = I_0 \cdot e^{-2/\tau} \]
Substituting the known values:
\[ 18 = 20 \cdot e^{-2/\tau} \]

3. **Rearrange the equation to isolate the exponential term:**
\[ e^{-2/\tau} = \frac{18}{20} = \frac{9}{10} \]

4. **Take the natural logarithm of both sides:**
\[ -\frac{2}{\tau} = \ln\left(\frac{9}{10}\right) \]

5. **Solve for \( \tau \):**
\[ \frac{2}{\tau} = -\ln\left(\frac{9}{10}\right) \]
\[ \tau = \frac{2}{-\ln\left(\frac{9}{10}\right)} \]

6. **Using properties of logarithms:**
\[ -\ln\left(\frac{9}{10}\right) = \ln\left(\frac{10}{9}\right) \]
Thus, we can rewrite \( \tau \):
\[ \tau = \frac{2}{\ln\left(\frac{10}{9}\right)} \]

### Final Answer:
The time constant \( \tau \) of the circuit is:
\[ \tau = \frac{2}{\ln\left(\frac{10}{9}\right)} \text{ seconds} \]