A `1.00 muF` capacitor is charged by a `40.0 V` power supply. The fully charged charged capacitor is then discharged through a `10.0 mH` inductor. Find the maximum current in the resultaing oscillations.

Options

A. `400mA`
B. `800 mA`
C. `600 mA`
D. `150 mA`

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the maximum charge on the capacitor
The maximum charge \( Q \) on a capacitor is given by the formula:
\[ Q = C \times V \]
where:
- \( C = 1.00 \, \mu F = 1.00 \times 10^{-6} \, F \)
- \( V = 40.0 \, V \)

Substituting the values:
\[ Q = (1.00 \times 10^{-6} \, F) \times (40.0 \, V) = 40.0 \times 10^{-6} \, C \]
Thus, the maximum charge \( Q \) is:
\[ Q = 40.0 \, \mu C \]

### Step 2: Calculate the angular frequency of the LC circuit
The angular frequency \( \omega \) of an LC circuit is given by:
\[ \omega = \frac{1}{\sqrt{L \times C}} \]
where:
- \( L = 10.0 \, mH = 10.0 \times 10^{-3} \, H \)
- \( C = 1.00 \times 10^{-6} \, F \)

Substituting the values:
\[ \omega = \frac{1}{\sqrt{(10.0 \times 10^{-3} \, H) \times (1.00 \times 10^{-6} \, F)}} \]
Calculating the product inside the square root:
\[ = \frac{1}{\sqrt{10.0 \times 10^{-9}}} = \frac{1}{\sqrt{10^{-8}}} = \frac{1}{10^{-4}} = 10,000 \, rad/s \]

### Step 3: Calculate the maximum current in the oscillations
The maximum current \( I_{\text{max}} \) in the oscillations can be calculated using the formula:
\[ I_{\text{max}} = \omega \times Q \]

Substituting the values:
\[ I_{\text{max}} = (10,000 \, rad/s) \times (40.0 \times 10^{-6} \, C) \]
Calculating this gives:
\[ I_{\text{max}} = 10,000 \times 40.0 \times 10^{-6} = 400 \times 10^{-3} \, A = 0.400 \, A = 400 \, mA \]

### Final Result
The maximum current in the resulting oscillations is:
\[ I_{\text{max}} = 400 \, mA \]