A coil of 10 H inductance has a `5 Omega` resistance and is connected to a 5 volt battery in series . The current in ampere in the circuit 2 seconds after the circuit is switched on is __

Options

A. 1 amp
B. `(1-1//e)` amp
C. `(1-e)` amp
D. `1//e` amp

Answer

To solve the problem, we need to find the current in an RL circuit (a circuit with a resistor and an inductor) after 2 seconds of being connected to a 5V battery. The given values are:

- Inductance (L) = 10 H
- Resistance (R) = 5 Ω
- Voltage (E) = 5 V
- Time (t) = 2 seconds

### Step-by-Step Solution:

1. **Identify the formula for current in an RL circuit**:
The current \( I(t) \) in an RL circuit can be expressed as:
\[ I(t) = \frac{E}{R} \left(1 - e^{-\frac{t}{L/R}}\right) \]
where \( E \) is the voltage, \( R \) is the resistance, \( L \) is the inductance, and \( t \) is the time.

2. **Calculate \( L/R \)**:
We need to calculate the time constant \( \frac{L}{R} \):
\[ \frac{L}{R} = \frac{10 \, \text{H}}{5 \, \Omega} = 2 \, \text{s} \]

3. **Substitute the values into the formula**:
Now we substitute \( E = 5 \, \text{V} \), \( R = 5 \, \Omega \), and \( t = 2 \, \text{s} \) into the current formula:
\[ I(2) = \frac{5}{5} \left(1 - e^{-\frac{2}{2}}\right) \]

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

5. **Calculate \( e^{-1} \)**:
The value of \( e^{-1} \) is approximately \( 0.3679 \):
\[ I(2) = 1 - 0.3679 \approx 0.6321 \, \text{A} \]

6. **Final answer**:
Therefore, the current in the circuit 2 seconds after it is switched on is approximately:
\[ I(2) \approx 0.632 \, \text{A} \]