Find the time after which current in the circuit becomes `80%` of its maximum value

Options

A. `(ln2)/(100)`
B. `(ln3)/(100)`
C. `(ln5)/(100)`
D. `(ln6)/(100)`

Answer

To find the time after which the current in an LR circuit becomes 80% of its maximum value, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the LR Circuit:**
In an LR circuit, the current \( I(t) \) at time \( t \) is given by the formula:
\[ I(t) = I_{\text{max}} \left(1 - e^{-\frac{R}{L} t}\right) \]
where \( I_{\text{max}} \) is the maximum current, \( R \) is the resistance, and \( L \) is the inductance.

2. **Set Up the Equation for 80% of Maximum Current:**
We want to find the time \( t \) when the current \( I(t) \) is 80% of its maximum value:
\[ I(t) = 0.8 I_{\text{max}} \]

3. **Substitute into the Current Equation:**
Substitute \( 0.8 I_{\text{max}} \) into the current equation:
\[ 0.8 I_{\text{max}} = I_{\text{max}} \left(1 - e^{-\frac{R}{L} t}\right) \]

4. **Cancel \( I_{\text{max}} \):**
Since \( I_{\text{max}} \) is common on both sides, we can cancel it out:
\[ 0.8 = 1 - e^{-\frac{R}{L} t} \]

5. **Rearranging the Equation:**
Rearranging gives:
\[ e^{-\frac{R}{L} t} = 1 - 0.8 = 0.2 \]

6. **Taking the Natural Logarithm:**
Taking the natural logarithm on both sides:
\[ -\frac{R}{L} t = \ln(0.2) \]

7. **Solving for Time \( t \):**
Rearranging for \( t \):
\[ t = -\frac{L}{R} \ln(0.2) \]

8. **Substituting Values:**
If we have \( R = 1 \, \Omega \) and \( L = 10 \, \text{mH} = 10 \times 10^{-3} \, \text{H} \):
\[ t = -\frac{10 \times 10^{-3}}{1} \ln(0.2) \]

9. **Calculating \( \ln(0.2) \):**
We know that \( \ln(0.2) \approx -1.6094 \):
\[ t = -10 \times 10^{-3} \times (-1.6094) = 10 \times 10^{-3} \times 1.6094 \approx 0.016094 \, \text{s} \]

10. **Final Answer:**
Thus, the time after which the current becomes 80% of its maximum value is approximately:
\[ t \approx 0.0161 \, \text{s} \]