A coil carrying a steady current is short-circuited. The current in it decreases `alpha` times in time `t_(0)`. The time constant of the circuit is

Options

A. `tau = t_(0)` In `alpha`
B. `tau = (t_(0))/(In alpha)`
C. `tau = (t_(0))/(alpha)`
D. `tau = (t_(0))/(alpha - 1)`

Explanation

To solve the problem, we need to find the time constant \( \tau \) of a coil that is short-circuited, causing the current to decrease by a factor of \( \alpha \) in a time \( t_0 \).

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a coil with an initial current \( I_0 \).
- When the coil is short-circuited, the current decreases to \( \frac{I_0}{\alpha} \) in time \( t_0 \).

2. **Using the Exponential Decay Formula**:
- The current in an inductor decays according to the formula:
\[ I(t) = I_0 e^{-\frac{t}{\tau}} \]
- At \( t = t_0 \), the current is:
\[ I(t_0) = \frac{I_0}{\alpha} \]

3. **Setting Up the Equation**:
- Substitute \( t = t_0 \) into the exponential decay formula:
\[ \frac{I_0}{\alpha} = I_0 e^{-\frac{t_0}{\tau}} \]

4. **Canceling \( I_0 \)**:
- Since \( I_0 \) is not zero, we can divide both sides by \( I_0 \):
\[ \frac{1}{\alpha} = e^{-\frac{t_0}{\tau}} \]

5. **Taking the Natural Logarithm**:
- To solve for \( \tau \), take the natural logarithm of both sides:
\[ \ln\left(\frac{1}{\alpha}\right) = -\frac{t_0}{\tau} \]
- This can be rewritten using the property of logarithms:
\[ -\ln(\alpha) = -\frac{t_0}{\tau} \]

6. **Rearranging the Equation**:
- Multiply both sides by -1:
\[ \ln(\alpha) = \frac{t_0}{\tau} \]
- Rearranging gives:
\[ \tau = \frac{t_0}{\ln(\alpha)} \]

7. **Final Result**:
- The time constant \( \tau \) of the circuit is:
\[ \tau = \frac{t_0}{\ln(\alpha)} \]

### Conclusion:
- The correct answer is option 2: \( \tau = \frac{t_0}{\ln(\alpha)} \).