An inductor `(L=100 mH)`, a resistor `(R=100 ohm)` and a battery `(E=100 V)` are. initially connected in series as shown in the figure. Afler a long time, the battery is disconnected after short-circuiting the points `A` and `B`. The current in the circuit, `1 ms` after the short-circuit, is `(k)/(4)`. Find `k`.

Answer

To solve the problem step by step, we will analyze the circuit behavior before and after the battery is disconnected and the points A and B are short-circuited.

### Step 1: Determine the initial current in the circuit

Initially, when the battery is connected for a long time, the inductor behaves like a short circuit (since the current has reached a steady state). The maximum current \( I_0 \) through the circuit can be calculated using Ohm’s law:

\[ I_0 = \frac{E}{R} \]

Where:
- \( E = 100 \, V \) (the voltage of the battery)
- \( R = 100 \, \Omega \) (the resistance)

Substituting the values:

\[ I_0 = \frac{100 \, V}{100 \, \Omega} = 1 \, A \]

### Step 2: Analyze the circuit after the short-circuit

After the battery is disconnected and points A and B are short-circuited, the current will start to decay through the inductor and resistor. The current \( I(t) \) at any time \( t \) after the short-circuit can be described by the formula:

\[ I(t) = I_0 e^{-\frac{R}{L} t} \]

Where:
- \( R = 100 \, \Omega \)
- \( L = 100 \, mH = 0.1 \, H \)
- \( t \) is the time in seconds

### Step 3: Substitute the values into the decay equation

First, we need to calculate \( \frac{R}{L} \):

\[ \frac{R}{L} = \frac{100 \, \Omega}{0.1 \, H} = 1000 \, s^{-1} \]

Now, substituting \( I_0 \) and \( t = 1 \, ms = 1 \times 10^{-3} \, s \) into the equation:

\[ I(1 \, ms) = 1 \, A \cdot e^{-1000 \cdot 1 \times 10^{-3}} \]

Calculating the exponent:

\[ e^{-1000 \cdot 0.001} = e^{-1} \]

Thus, we have:

\[ I(1 \, ms) = 1 \, A \cdot e^{-1} \]

### Step 4: Calculate \( e^{-1} \)

The value of \( e^{-1} \) is approximately \( 0.3679 \). Therefore:

\[ I(1 \, ms) \approx 1 \, A \cdot 0.3679 \approx 0.3679 \, A \]

### Step 5: Relate the current to \( k \)

According to the problem, the current \( I(1 \, ms) \) is given as \( \frac{k}{4} \):

\[ 0.3679 = \frac{k}{4} \]

To find \( k \), we multiply both sides by 4:

\[ k = 4 \cdot 0.3679 \approx 1.4716 \]

### Final Answer

Rounding \( k \) to a suitable number of significant figures, we find:

\[ k \approx 1.47 \]