An ideal coil of `10` henry is joined in series with a resistance of `5` ohm and a battery of `5` volt. `2` second after joining, the current flowing in ampere in the circuit will be

Options

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

Explanation

To solve the problem, we need to find the current flowing in an LR circuit after a certain time. The circuit consists of an inductor (L), a resistor (R), and a battery (E). The formula for the current \( I(t) \) in an LR circuit is given by:

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

where:
- \( I_0 = \frac{E}{R} \) is the maximum current,
- \( e \) is the base of the natural logarithm,
- \( R \) is the resistance,
- \( L \) is the inductance,
- \( t \) is the time.

### Step-by-step Solution:

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

2. **Calculate the maximum current \( I_0 \):**
\[ I_0 = \frac{E}{R} = \frac{5 \, \text{V}}{5 \, \Omega} = 1 \, \text{A} \]

3. **Substitute the values into the current formula:**
\[ I(t) = I_0 \left(1 - e^{-\frac{R}{L} t}\right) \]
Substituting \( I_0 = 1 \, \text{A} \), \( R = 5 \, \Omega \), \( L = 10 \, \text{H} \), and \( t = 2 \, \text{s} \):
\[ I(2) = 1 \left(1 - e^{-\frac{5}{10} \cdot 2}\right) \]

4. **Simplify the exponent:**
\[ I(2) = 1 \left(1 - e^{-1}\right) \]

5. **Final expression for the current:**
\[ I(2) = 1 - e^{-1} \, \text{A} \]

### Conclusion:
The current flowing in the circuit after 2 seconds is \( 1 - e^{-1} \, \text{A} \).

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