An Lc circuit contains a 40 mH inductor and a `25 mu F` capacitor. The resistance of the circuit is negligible.The time is measured from the instant the circuit is closed. The energy stored in the circuit is completely magnetic at time (in milliseconds)

Options

A. `0,3.14,6.28`
B. `0,1.57,4.71`
C. `1.57,4.71,7.85`
D. `1.57,3.14,4.71`

Explanation

To solve the problem of finding the time at which the energy stored in an LC circuit is completely magnetic, we can follow these steps:

### Step 1: Understand the LC Circuit
An LC circuit consists of an inductor (L) and a capacitor (C) connected in series. When the circuit is closed, energy oscillates between the capacitor and the inductor. The energy is stored as electric energy in the capacitor and as magnetic energy in the inductor.

### Step 2: Identify the Given Values
- Inductance (L) = 40 mH = \(40 \times 10^{-3}\) H
- Capacitance (C) = 25 µF = \(25 \times 10^{-6}\) F

### Step 3: Calculate the Frequency of Oscillation
The frequency of oscillation (f) in an LC circuit is given by the formula:
\[ f = \frac{1}{2\pi\sqrt{LC}} \]
Substituting the values of L and C:
\[ f = \frac{1}{2\pi\sqrt{(40 \times 10^{-3})(25 \times 10^{-6})}} \]

### Step 4: Calculate the Period of Oscillation
The period (T) is the reciprocal of frequency:
\[ T = \frac{1}{f} \]
Substituting the expression for frequency:
\[ T = 2\pi\sqrt{LC} \]

### Step 5: Substitute the Values to Find T
Calculating \(T\):
\[ T = 2\pi\sqrt{(40 \times 10^{-3})(25 \times 10^{-6})} \]
Calculating the square root:
\[ \sqrt{(40 \times 10^{-3})(25 \times 10^{-6})} = \sqrt{1 \times 10^{-6}} = 10^{-3} \]
Thus,
\[ T = 2\pi \times 10^{-3} \text{ seconds} = 2\pi \text{ milliseconds} \]

### Step 6: Determine the Time for Maximum Magnetic Energy
The energy in the inductor is maximum at specific times during the oscillation:
- This occurs at \(T/4, 3T/4, 5T/4, \ldots\)

Substituting \(T = 2\pi\):
- First maximum: \(T/4 = \frac{2\pi}{4} = \frac{\pi}{2} \text{ milliseconds} \approx 1.57 \text{ milliseconds}\)
- Second maximum: \(3T/4 = \frac{3 \times 2\pi}{4} = \frac{3\pi}{2} \text{ milliseconds} \approx 4.71 \text{ milliseconds}\)
- Third maximum: \(5T/4 = \frac{5 \times 2\pi}{4} = \frac{5\pi}{2} \text{ milliseconds} \approx 7.85 \text{ milliseconds}\)

### Step 7: Conclusion
The times at which the energy is completely magnetic are approximately:
- \(1.57 \text{ ms}\)
- \(4.71 \text{ ms}\)
- \(7.85 \text{ ms}\)

The correct answer is \(1.57 \text{ ms}\), \(4.71 \text{ ms}\), or \(7.85 \text{ ms}\).