A `1.5 mu f` capacitor is charged to 20 V. The charging battery is then disconnected and a 15 mH coil is connected in series with the capacitance so that LC oscillations occur. Write current in the inductor as a function of time

Options

A. `l=(0.2A) sin omega t`
B. `l=(0.3A) sin omega t`
C. `l=(0.5A) sin omega t`
D. None of the Above

Explanation

To find the current in the inductor as a function of time in an LC circuit, we can follow these steps:

### Step 1: Calculate the maximum charge on the capacitor
The maximum charge \( Q_0 \) on the capacitor can be calculated using the formula:
\[ Q_0 = C \times V \]
where:
- \( C = 1.5 \, \mu F = 1.5 \times 10^{-6} \, F \)
- \( V = 20 \, V \)

Substituting the values:
\[ Q_0 = (1.5 \times 10^{-6} \, F) \times (20 \, V) = 30 \times 10^{-6} \, C \]

### 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 \( LC \):
\[ LC = 15 \times 10^{-3} \times 1.5 \times 10^{-6} = 22.5 \times 10^{-9} \, H \cdot F \]
Now, calculating \( \omega \):
\[ \omega = \frac{1}{\sqrt{22.5 \times 10^{-9}}} \approx \frac{1}{4.743 \times 10^{-5}} \approx 21080.5 \, rad/s \]

### Step 3: Write the expression for charge as a function of time
The charge \( Q(t) \) on the capacitor as a function of time is given by:
\[ Q(t) = Q_0 \cos(\omega t) \]
Substituting \( Q_0 \):
\[ Q(t) = 30 \times 10^{-6} \cos(\omega t) \]

### Step 4: Find the current as a function of time
The current \( I(t) \) in the circuit is the rate of change of charge:
\[ I(t) = -\frac{dQ}{dt} \]
Differentiating \( Q(t) \):
\[ I(t) = -\frac{d}{dt}(30 \times 10^{-6} \cos(\omega t)) = 30 \times 10^{-6} \omega \sin(\omega t) \]
Substituting \( \omega \):
\[ I(t) = 30 \times 10^{-6} \times 21080.5 \sin(\omega t) \approx 0.2 \sin(\omega t) \, A \]

### Final Expression
Thus, the current in the inductor as a function of time is:
\[ I(t) = 0.2 \sin(\omega t) \, A \]