A `1.5 mu F` capacitor is charged of 60 V. The charging battery is then disconnected and a 15 mH coil is connected in series with the capacitor so that LC oscillations occur. Assuming that the circuit contains no resistance, the maximum current in this coil shall be close to

Options

A. `1.4 A`
B. ` 1.2 A`
C. `0.8 A`
D. `0.6 A`

Explanation

To solve the problem step by step, we will follow the process of calculating the maximum current in the LC circuit formed by the charged capacitor and the inductor.

### Step 1: Calculate the Charge on the Capacitor
The charge \( Q \) on a capacitor is given by the formula:
\[ Q = C \times V \]
where:
- \( C = 1.5 \, \mu F = 1.5 \times 10^{-6} \, F \)
- \( V = 60 \, V \)

Substituting the values:
\[ Q = (1.5 \times 10^{-6}) \times 60 = 9 \times 10^{-5} \, C \]

### Step 2: Identify the Inductance
The inductance \( L \) of the coil is given as:
\[ L = 15 \, mH = 15 \times 10^{-3} \, H \]

### Step 3: Calculate the Angular Frequency \( \omega \)
The angular frequency \( \omega \) for an LC circuit is given by:
\[ \omega = \frac{1}{\sqrt{LC}} \]

Substituting the values of \( L \) and \( C \):
\[ \omega = \frac{1}{\sqrt{(15 \times 10^{-3}) \times (1.5 \times 10^{-6})}} \]

Calculating \( LC \):
\[ LC = 15 \times 10^{-3} \times 1.5 \times 10^{-6} = 22.5 \times 10^{-9} \, H \cdot F \]

Now, taking the square root:
\[ \sqrt{LC} = \sqrt{22.5 \times 10^{-9}} = 4.743 \times 10^{-5} \, s \]

Thus,
\[ \omega = \frac{1}{4.743 \times 10^{-5}} \approx 21080 \, rad/s \]

### Step 4: Calculate the Maximum Current \( I_{max} \)
The maximum current \( I_{max} \) in the LC circuit can be calculated using the formula:
\[ I_{max} = \frac{Q_{max}}{\sqrt{LC}} \]

Substituting the values:
\[ I_{max} = \frac{9 \times 10^{-5}}{\sqrt{(15 \times 10^{-3}) \times (1.5 \times 10^{-6})}} \]

Calculating:
\[ I_{max} = \frac{9 \times 10^{-5}}{4.743 \times 10^{-5}} \approx 1.895 \, A \]

### Final Answer
The maximum current in the coil is approximately:
\[ I_{max} \approx 1.895 \, A \]