An inductor of 20 H and a resistance of `10 Omega`, are connected to battery of 5 volt in series, then initial rate of change of current is :-

Options

A. `0.5 amp//s`
B. `2.0 amp//s`
C. `2.5 amp//s`
D. `0.25 amp//s`

Answer

To solve the problem of finding the initial rate of change of current (di/dt) in an LR circuit consisting of an inductor and a resistor connected in series to a battery, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Inductance (L) = 20 H (Henries)
- Resistance (R) = 10 Ω (Ohms)
- Voltage (V) = 5 V (Volts)

2. **Understand the Formula for Current in an LR Circuit:**
The current (I) in an LR circuit as a function of time (t) is given by:
\[ I(t) = I_0 (1 - e^{-\frac{R}{L}t}) \]
where \(I_0\) is the steady-state current.

3. **Determine the Steady-State Current (I0):**
The steady-state current can be calculated using Ohm's law:
\[ I_0 = \frac{V}{R} \]
Substituting the values:
\[ I_0 = \frac{5 \, \text{V}}{10 \, \Omega} = 0.5 \, \text{A} \]

4. **Differentiate the Current Equation:**
To find the rate of change of current (di/dt), we differentiate the current equation with respect to time (t):
\[ \frac{di}{dt} = I_0 \cdot \frac{R}{L} e^{-\frac{R}{L}t} \]

5. **Evaluate at t = 0:**
To find the initial rate of change of current, we substitute \(t = 0\):
\[ \frac{di}{dt}\bigg|_{t=0} = I_0 \cdot \frac{R}{L} e^{0} \]
Since \(e^0 = 1\), we have:
\[ \frac{di}{dt}\bigg|_{t=0} = I_0 \cdot \frac{R}{L} \]

6. **Substitute the Values:**
Now, substituting \(I_0\), \(R\), and \(L\):
\[ \frac{di}{dt}\bigg|_{t=0} = 0.5 \, \text{A} \cdot \frac{10 \, \Omega}{20 \, \text{H}} = 0.5 \cdot 0.5 = 0.25 \, \text{A/s} \]

### Final Answer:
The initial rate of change of current is:
\[ \frac{di}{dt}\bigg|_{t=0} = 0.25 \, \text{A/s} \]

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