An ideal coil of 10H is connected in series with a resistance of `5(Omega)` and a battery of 5V. 2second after the connections is made, the current flowing in ampere in the circuit is

Options

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

Explanation

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

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

Where:
- \( I(t) \) is the current at time \( t \),
- \( I_0 \) is the maximum current,
- \( R \) is the resistance,
- \( L \) is the inductance,
- \( e \) is the base of the natural logarithm,
- \( t \) is the time.

### Step 1: Determine the maximum current \( I_0 \)

The maximum current \( I_0 \) in the circuit can be calculated using Ohm's law:

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

Where:
- \( E = 5V \) (the voltage of the battery),
- \( R = 5 \Omega \) (the resistance).

Substituting the values:

\[ I_0 = \frac{5V}{5 \Omega} = 1A \]

### Step 2: Substitute values into the current formula

Now we substitute \( I_0 \), \( R \), \( L \), and \( t \) into the current formula:

\[ I(t) = 1 \left(1 - e^{-\frac{5}{10} \cdot 2}\right) \]

### Step 3: Simplify the exponent

Calculate the exponent:

\[ -\frac{R}{L} \cdot t = -\frac{5}{10} \cdot 2 = -1 \]

### Step 4: Calculate the current

Now substitute the exponent back into the equation:

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

### Step 5: Evaluate \( e^{-1} \)

Using the approximate value of \( e \approx 2.718 \):

\[ e^{-1} \approx \frac{1}{2.718} \approx 0.3679 \]

Thus:

\[ I(t) = 1 \left(1 - 0.3679\right) \]
\[ I(t) \approx 1 \times 0.6321 \]
\[ I(t) \approx 0.6321A \]

### Final Answer

The current flowing in the circuit after 2 seconds is approximately **0.6321 A**.

---