The self-inductance of a coil is 2H. The current in the coil changes from 8A to 2.95 A in 0.01 s. The time constant of the coil will be -

Options

A. 2s
B. 0.1 s
C. 10 ms
D. 1 ms

Explanation

To find the time constant of the coil, we can use the formula for the current through an inductor, which is given by:

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

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**:
- Self-inductance \( L = 2 \, H \)
- Initial current \( I_0 = 8 \, A \)
- Final current \( I = 2.95 \, A \)
- Time interval \( t = 0.01 \, s \)

2. **Set up the equation**:
We can rearrange the current equation to solve for \( \tau \):
\[ I = I_0 e^{-t/\tau} \]
Taking the natural logarithm of both sides gives:
\[ \ln(I) = \ln(I_0) - \frac{t}{\tau} \]

3. **Rearranging the equation**:
Rearranging the equation to isolate \( \tau \):
\[ \frac{t}{\tau} = \ln(I_0) - \ln(I) \]
This can be simplified using the properties of logarithms:
\[ \frac{t}{\tau} = \ln\left(\frac{I_0}{I}\right) \]
Therefore, we can express \( \tau \) as:
\[ \tau = \frac{t}{\ln\left(\frac{I_0}{I}\right)} \]

4. **Substituting the values**:
Substitute the known values into the equation:
\[ \tau = \frac{0.01}{\ln\left(\frac{8}{2.95}\right)} \]

5. **Calculating the logarithm**:
First, calculate \( \frac{8}{2.95} \):
\[ \frac{8}{2.95} \approx 2.71 \]
Now, calculate the natural logarithm:
\[ \ln(2.71) \approx 1 \]

6. **Final calculation of \( \tau \)**:
Now substitute back into the equation for \( \tau \):
\[ \tau = \frac{0.01}{1} = 0.01 \, s \]

7. **Convert to milliseconds**:
To convert seconds to milliseconds, multiply by 1000:
\[ \tau = 0.01 \times 1000 = 10 \, ms \]

### Final Answer:
The time constant of the coil is \( 10 \, ms \).