A coil of `L=5xx10^(-3)` H and R=18 `Omega` is abruptly supplied a potential of 5 volts . What will be the rate of change of current in 0.001 second? `(e^(-3 .6)=0.0273)`

Options

A. 27.3 amp/sec.
B. 27.8 amp/sec.
C. 2.73 amp/sec.
D. 2.78 amp/sec.

Answer

To solve the problem, we need to find the rate of change of current (di/dt) in an LR circuit when a voltage is applied. Here's the step-by-step solution:

### Step 1: Identify the given values
- Inductance, \( L = 5 \times 10^{-3} \, \text{H} \)
- Resistance, \( R = 18 \, \Omega \)
- Voltage, \( e = 5 \, \text{V} \)
- Time, \( t = 0.001 \, \text{s} \)

### Step 2: Calculate the time constant (\( \tau \))
The time constant \( \tau \) for an LR circuit is given by the formula:
\[ \tau = \frac{L}{R} \]
Substituting the values:
\[ \tau = \frac{5 \times 10^{-3}}{18} \approx 2.7778 \times 10^{-4} \, \text{s} \]

### Step 3: Write the expression for current (\( i(t) \))
The current in an LR circuit after applying a voltage is given by:
\[ i(t) = \frac{e}{R} \left(1 - e^{-\frac{t}{\tau}}\right) \]

### Step 4: Differentiate to find \( \frac{di}{dt} \)
To find the rate of change of current, we differentiate \( i(t) \):
\[ \frac{di}{dt} = \frac{e}{R} \cdot \frac{1}{\tau} e^{-\frac{t}{\tau}} \]

### Step 5: Substitute the values into the derivative
Now we substitute the known values into the derivative:
\[ \frac{di}{dt} = \frac{5}{18} \cdot \frac{1}{\tau} e^{-\frac{t}{\tau}} \]
Substituting \( \tau \):
\[ \frac{di}{dt} = \frac{5}{18} \cdot \frac{1}{2.7778 \times 10^{-4}} e^{-\frac{0.001}{2.7778 \times 10^{-4}}} \]

### Step 6: Calculate \( e^{-\frac{t}{\tau}} \)
Calculate \( -\frac{t}{\tau} \):
\[ -\frac{0.001}{2.7778 \times 10^{-4}} \approx -3.6 \]
Thus,
\[ e^{-3.6} \approx 0.0273 \]

### Step 7: Substitute back to find \( \frac{di}{dt} \)
Now substituting back:
\[ \frac{di}{dt} = \frac{5}{18} \cdot \frac{1}{2.7778 \times 10^{-4}} \cdot 0.0273 \]
Calculating \( \frac{5}{18} \cdot \frac{1}{2.7778 \times 10^{-4}} \):
\[ \frac{5}{18} \approx 0.2778 \]
\[ \frac{0.2778}{2.7778 \times 10^{-4}} \approx 1000 \]
Thus,
\[ \frac{di}{dt} \approx 1000 \cdot 0.0273 \approx 27.3 \, \text{A/s} \]

### Final Answer
The rate of change of current \( \frac{di}{dt} \) at \( t = 0.001 \, \text{s} \) is approximately:
\[ \frac{di}{dt} \approx 27.3 \, \text{A/s} \]