An inductor of 5 henry and a resistance of 20 ohm are connected in series with a battery of 5 volt. The initial rate of change of current is

Options

A. `1 A s^(-1)`
B. `2.0 A s^(-1)`
C. `2.5 A s^(-1)`
D. `0.25 A s^(-1)`

Explanation

To find the initial rate of change of current in an inductor connected in series with a resistor and a battery, we can use the formula derived from the inductor's behavior in an RL circuit.

### Step-by-Step Solution:

1. **Identify the given values:**
- Inductance (L) = 5 H (henry)
- Resistance (R) = 20 Ω (ohm)
- Voltage (V) = 5 V (volt)

2. **Use the formula for the current in an RL circuit:**
The current (I) in an RL circuit is given by:
\[ I(t) = I_0 \left(1 - e^{-\frac{R}{L}t}\right) \]
where \(I_0\) is the steady-state current when \(t\) approaches infinity.

3. **Calculate the steady-state current \(I_0\):**
The steady-state current can be calculated using Ohm's law:
\[ I_0 = \frac{V}{R} \]
Substituting the values:
\[ I_0 = \frac{5 \, \text{V}}{20 \, \Omega} = 0.25 \, \text{A} \]

4. **Differentiate the current with respect to time to find the rate of change of current:**
To find the initial rate of change of current, we differentiate \(I(t)\) with respect to time \(t\):
\[ \frac{dI}{dt} = \frac{d}{dt} \left(I_0 \left(1 - e^{-\frac{R}{L}t}\right)\right) \]
Using the chain rule:
\[ \frac{dI}{dt} = I_0 \cdot \left(0 - \left(-\frac{R}{L}\right)e^{-\frac{R}{L}t}\right) = I_0 \cdot \frac{R}{L} e^{-\frac{R}{L}t} \]

5. **Evaluate at \(t = 0\):**
At \(t = 0\), \(e^{-\frac{R}{L} \cdot 0} = e^0 = 1\):
\[ \frac{dI}{dt}\bigg|_{t=0} = I_0 \cdot \frac{R}{L} \cdot 1 = I_0 \cdot \frac{R}{L} \]
Substituting \(I_0 = 0.25 \, \text{A}\), \(R = 20 \, \Omega\), and \(L = 5 \, \text{H}\):
\[ \frac{dI}{dt}\bigg|_{t=0} = 0.25 \cdot \frac{20}{5} = 0.25 \cdot 4 = 1 \, \text{A/s} \]

6. **Final Answer:**
The initial rate of change of current is \(1 \, \text{A/s}\).