The initial rate of increase of current, when a battery of emf 6 V is connected in series with an inductance of 2 H and resistance `12Omega` , is

Options

A. `0.5As^(-1)`
B. `1As^(-1)`
C. `3As^(-1)`
D. `3.5As^(-1)`

Explanation

To find the initial rate of increase of current when a battery of emf 6 V is connected in series with an inductance of 2 H and resistance of 12 Ω, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the RL Circuit**:
- We have a circuit consisting of a resistor (R) and an inductor (L) connected in series with a battery of emf (E).
- Given values:
- \( E = 6 \, \text{V} \)
- \( L = 2 \, \text{H} \)
- \( R = 12 \, \Omega \)

2. **Apply Kirchhoff's Voltage Law (KVL)**:
- According to KVL, the sum of the potential differences around a closed loop is zero.
- The equation for the RL circuit can be written as:
\[ E - L \frac{di}{dt} - Ri = 0 \]
- Rearranging gives:
\[ E = L \frac{di}{dt} + Ri \]

3. **Initial Conditions**:
- At the initial moment (t = 0), the current (i) is zero (i = 0).
- Thus, the equation becomes:
\[ 6 = L \frac{di}{dt} + R \cdot 0 \]
- This simplifies to:
\[ 6 = L \frac{di}{dt} \]

4. **Substituting Values**:
- Substitute \( L = 2 \, \text{H} \) into the equation:
\[ 6 = 2 \frac{di}{dt} \]

5. **Solve for \(\frac{di}{dt}\)**:
- Rearranging gives:
\[ \frac{di}{dt} = \frac{6}{2} = 3 \, \text{A/s} \]

6. **Conclusion**:
- The initial rate of increase of current is \( 3 \, \text{A/s} \).

### Final Answer:
The initial rate of increase of current is \( 3 \, \text{A/s} \).