An idal coil of 10 is connected in series with a resitance of `5Omega` and a battery of 5V. After 2s, after the connection is made, the current flowing ( in ampere) in the circuit is

Options

A. (1-e)
B. e
C. `e^(-1)`
D. `(1-e^(-1))`

Explanation

To solve the problem, we need to find the current flowing in the circuit after 2 seconds. We will use the formula for the current in an LR circuit, which is given by:

\[ I(t) = I_0 \left(1 - e^{-\frac{t}{\tau}}\right) \]

where:
- \( I(t) \) is the current at time \( t \),
- \( I_0 \) is the maximum current (steady-state current),
- \( e \) is the base of the natural logarithm,
- \( \tau \) is the time constant of the circuit.

### Step-by-Step Solution:

1. **Identify the given values:**
- Inductance \( L = 10 \, \text{H} \)
- Resistance \( R = 5 \, \Omega \)
- Voltage \( E = 5 \, \text{V} \)
- Time \( t = 2 \, \text{s} \)

2. **Calculate the maximum current \( I_0 \):**
The maximum current \( I_0 \) can be calculated using Ohm's law:
\[ I_0 = \frac{E}{R} = \frac{5 \, \text{V}}{5 \, \Omega} = 1 \, \text{A} \]

3. **Calculate the time constant \( \tau \):**
The time constant \( \tau \) is given by:
\[ \tau = \frac{L}{R} = \frac{10 \, \text{H}}{5 \, \Omega} = 2 \, \text{s} \]

4. **Substitute the values into the current equation:**
Now we can substitute \( I_0 \), \( t \), and \( \tau \) into the current equation:
\[ I(t) = I_0 \left(1 - e^{-\frac{t}{\tau}}\right) = 1 \left(1 - e^{-\frac{2}{2}}\right) = 1 \left(1 - e^{-1}\right) \]

5. **Calculate \( e^{-1} \):**
The value of \( e^{-1} \) is approximately \( 0.3679 \):
\[ I(t) = 1 \left(1 - 0.3679\right) \approx 1 \times 0.6321 \approx 0.6321 \, \text{A} \]

### Final Answer:
The current flowing in the circuit after 2 seconds is approximately \( 0.6321 \, \text{A} \).