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 the circuit after 2 seconds when an ideal coil (inductor) of 10 henry is connected in series with a resistance of 5 ohms and a battery of 5 volts.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Inductance (L) = 10 H
- Resistance (R) = 5 Ω
- Voltage (E) = 5 V
- Time (t) = 2 s

2. **Determine the Final Current (I_f):**
The final current (I_f) in the circuit can be calculated using Ohm's law:
\[ I_f = \frac{E}{R} \]
Substituting the values:
\[ I_f = \frac{5 \, \text{V}}{5 \, \Omega} = 1 \, \text{A} \]

3. **Use the Formula for Current in an LR Circuit:**
The current (I) at any time (t) in an LR circuit is given by the formula:
\[ I(t) = I_f \left(1 - e^{-\frac{R}{L}t}\right) \]
Substituting the known values into the formula:
\[ I(t) = 1 \left(1 - e^{-\frac{5}{10} \cdot 2}\right) \]
Simplifying the exponent:
\[ I(t) = 1 \left(1 - e^{-1}\right) \]

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

5. **Final Answer:**
The current flowing in the circuit after 2 seconds is approximately:
\[ I(t) \approx 0.632 \, \text{A} \]