An inductor of 5 m H is connected in series with a resistor of resistance 4 `Omega` . A battery of 2 V is connected across the circuit through a switch . Calculate the rate of growth of current just after the switch is on .

Answer

To solve the problem of finding the rate of growth of current just after the switch is closed in a circuit with an inductor and a resistor, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Circuit Parameters**:
- Inductance (L) = 5 mH = \(5 \times 10^{-3}\) H
- Resistance (R) = 4 Ω
- Voltage (V) = 2 V

2. **Apply Kirchhoff's Voltage Law**:
When the switch is closed, we can apply Kirchhoff's law to the circuit. The voltage provided by the battery (V) is equal to the sum of the voltage drops across the resistor (IR) and the inductor (L \(\frac{dI}{dt}\)):
\[ V = IR + L \frac{dI}{dt} \]

3. **Rearranging the Equation**:
Rearranging the equation gives us:
\[ L \frac{dI}{dt} = V - IR \]

4. **Finding the Rate of Change of Current**:
We want to find \(\frac{dI}{dt}\) just after the switch is closed (at \(t = 0\)). At this moment, the current (I) is 0 because the inductor initially opposes any change in current. Therefore, substituting \(I = 0\) into the equation:
\[ L \frac{dI}{dt} = V - 0 \cdot R \]
This simplifies to:
\[ L \frac{dI}{dt} = V \]

5. **Solving for \(\frac{dI}{dt}\)**:
Now, we can solve for \(\frac{dI}{dt}\):
\[ \frac{dI}{dt} = \frac{V}{L} \]

6. **Substituting the Values**:
Substitute the values of V and L:
\[ \frac{dI}{dt} = \frac{2 \, \text{V}}{5 \times 10^{-3} \, \text{H}} = \frac{2}{0.005} = 400 \, \text{A/s} \]

### Final Answer:
The rate of growth of current just after the switch is closed is \(400 \, \text{A/s}\).

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