When a choke coil carrying a steady current is short circuited, the current in it decreases to `beta(lt1)` times its initial value in a time `T`. The time constant of the choke coil is

Options

A. `T/beta`
B. `T/(ln(1/beta)`
C. `T/(lnbeta)`
D. `T ln beta`

Explanation

To find the time constant of the choke coil when it is short-circuited, we can follow these steps:

### Step 1: Understand the Problem
When a choke coil carrying a steady current is short-circuited, the current decreases to a fraction (β) of its initial value (I₀) over a time period (T). We need to find the time constant (τ) of the choke coil.

### Step 2: Write the Current Decay Equation
The current in an inductor decays exponentially when the inductor is short-circuited. The equation for the current (I) at time (t) 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

### Step 3: Set Up the Equation
According to the problem, at time \( T \), the current decreases to \( \beta I_0 \):
\[ \beta I_0 = I_0 e^{-T/\tau} \]

### Step 4: Simplify the Equation
We can cancel \( I_0 \) from both sides (assuming \( I_0 \neq 0 \)):
\[ \beta = e^{-T/\tau} \]

### Step 5: Take the Natural Logarithm
To solve for \( \tau \), we take the natural logarithm of both sides:
\[ \ln(\beta) = -\frac{T}{\tau} \]

### Step 6: Rearrange to Find the Time Constant
Rearranging the equation gives us:
\[ \tau = -\frac{T}{\ln(\beta)} \]

### Final Answer
Thus, the time constant \( \tau \) of the choke coil is:
\[ \tau = -\frac{T}{\ln(\beta)} \]