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 find the maximum current in the inductor connected in series with the charged capacitor.

### Step 1: Calculate the charge stored in the capacitor
The charge \( Q_0 \) stored in a capacitor is given by the formula:
\[ Q_0 = C \times V \]
Where:
- \( C = 1.5 \, \mu F = 1.5 \times 10^{-6} \, F \)
- \( V = 60 \, V \)

Substituting the values:
\[ Q_0 = 1.5 \times 10^{-6} \, F \times 60 \, V = 90 \times 10^{-6} \, C = 90 \, \mu C \]

### Step 2: Determine the angular frequency \( \omega \)
The angular frequency \( \omega \) for an LC circuit is given by:
\[ \omega = \frac{1}{\sqrt{LC}} \]
Where:
- \( L = 15 \, mH = 15 \times 10^{-3} \, H \)
- \( C = 1.5 \, \mu F = 1.5 \times 10^{-6} \, F \)

Substituting the values:
\[ \omega = \frac{1}{\sqrt{15 \times 10^{-3} \times 1.5 \times 10^{-6}}} \]

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

Now, taking the square root:
\[ \sqrt{LC} = \sqrt{22.5 \times 10^{-9}} \approx 4.74 \times 10^{-5} \]

Thus,
\[ \omega = \frac{1}{4.74 \times 10^{-5}} \approx 21100 \, rad/s \]

### Step 3: Calculate the maximum current \( I_{max} \)
The maximum current \( I_{max} \) in the inductor can be calculated using the formula:
\[ I_{max} = \frac{Q_0}{\sqrt{LC}} \]

Substituting the values:
\[ I_{max} = \frac{90 \times 10^{-6}}{\sqrt{15 \times 10^{-3} \times 1.5 \times 10^{-6}}} \]

Using the value of \( \sqrt{LC} \) calculated earlier:
\[ I_{max} = \frac{90 \times 10^{-6}}{4.74 \times 10^{-5}} \approx 1.90 \, A \]

### Final Answer
The maximum current in the coil is approximately:
\[ I_{max} \approx 0.6 \, A \]