In an oscillating LC circuit with L = 50 mH and `C = 4.0 mu F`, the current is initially a maximum.How long will it take before the capaciotr is fully discharged for the first time:

Options

A. `7 xx 10^(-4)s`
B. `14 xx 10^(-4)s`
C. `28 xx 10^(-4)s`
D. none

Explanation

To solve the problem of how long it will take for the capacitor in an oscillating LC circuit to be fully discharged for the first time, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Inductance \( L = 50 \, \text{mH} = 50 \times 10^{-3} \, \text{H} \)
- Capacitance \( C = 4.0 \, \mu\text{F} = 4.0 \times 10^{-6} \, \text{F} \)

2. **Calculate the Angular Frequency \( \omega \):**
The angular frequency \( \omega \) for an LC circuit is given by the formula:
\[ \omega = \frac{1}{\sqrt{LC}} \]
Substitute the values of \( L \) and \( C \):
\[ \omega = \frac{1}{\sqrt{(50 \times 10^{-3}) \times (4.0 \times 10^{-6})}} \]

3. **Calculate the Time Period \( T \):**
The time period \( T \) of the oscillation is related to the angular frequency by:
\[ T = \frac{2\pi}{\omega} \]
We can also express it in terms of \( L \) and \( C \):
\[ T = 2\pi \sqrt{LC} \]
Substitute the values:
\[ T = 2\pi \sqrt{(50 \times 10^{-3}) \times (4.0 \times 10^{-6})} \]

4. **Calculate \( T \):**
First, calculate \( LC \):
\[ LC = 50 \times 10^{-3} \times 4.0 \times 10^{-6} = 200 \times 10^{-9} = 2.0 \times 10^{-7} \]
Now calculate \( \sqrt{LC} \):
\[ \sqrt{LC} = \sqrt{2.0 \times 10^{-7}} \approx 1.414 \times 10^{-4} \]
Now substitute back to find \( T \):
\[ T = 2\pi (1.414 \times 10^{-4}) \approx 8.88 \times 10^{-4} \, \text{s} \]

5. **Determine the Time for Full Discharge:**
The capacitor is fully discharged for the first time at \( \frac{T}{4} \):
\[ \text{Time to fully discharge} = \frac{T}{4} = \frac{8.88 \times 10^{-4}}{4} \approx 2.22 \times 10^{-4} \, \text{s} \]

### Final Answer:
The time it will take before the capacitor is fully discharged for the first time is approximately \( 2.22 \times 10^{-4} \, \text{s} \).

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