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 12 Ω, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values:**
- EMF (E) = 6 V
- Inductance (L) = 2 H
- Resistance (R) = 12 Ω

2. **Calculate the time constant (τ):**
The time constant (τ) for an RL circuit is given by the formula:
\[ \tau = \frac{L}{R} \]
Substituting the values:
\[ \tau = \frac{2 \, \text{H}}{12 \, \Omega} = \frac{1}{6} \, \text{s} \]

3. **Determine the maximum current (I₀):**
The maximum current (I₀) in the circuit can be calculated using Ohm's law:
\[ I_0 = \frac{E}{R} \]
Substituting the values:
\[ I_0 = \frac{6 \, \text{V}}{12 \, \Omega} = 0.5 \, \text{A} \]

4. **Find the expression for the rate of change of current (di/dt):**
The rate of change of current in an RL circuit is given by:
\[ \frac{di}{dt} = \frac{I_0 R}{L} e^{-\frac{R t}{L}} \]
At time \( t = 0 \):
\[ \frac{di}{dt} \bigg|_{t=0} = \frac{I_0 R}{L} e^{0} = \frac{I_0 R}{L} \]

5. **Substitute the values to find di/dt at t = 0:**
\[ \frac{di}{dt} \bigg|_{t=0} = \frac{0.5 \, \text{A} \times 12 \, \Omega}{2 \, \text{H}} = \frac{6}{2} = 3 \, \text{A/s} \]

### Final Answer:
The initial rate of increase of current is **3 A/s**.

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